48044934 Life Assurance Mathematics W F Scott

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LIFE ASSURANCE MATHEMATICS

W.F.Scott

c©1999 W.F.ScottDepartment of Mathematical SciencesKing’s CollegeUniversity of AberdeenAberdeenAB24 3UEU.K.

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Preface

This book consists largely of material written for Parts A2 and D1 of the U.K. actuarial exami-nations (old system). It is hoped that the material given here will prove useful for much of Subjects104 and 105 of the new examinations and certain similar examinations at universities, and that itmay also be useful as a general reference work on life assurance mathematics.

William F. Scott. November 10, 1999

Contents

1 NON-SELECT LIFE TABLES 91.1 Survivorship functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2 Probabilities of death and survival . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 The force of mortality, µx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4 The expectation of life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.5 The assumption of a uniform distribution of deaths . . . . . . . . . . . . . . . . . . . 171.6 Central death rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.7 Laws of mortality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2 SELECT LIFE TABLES 292.1 What is selection? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.2 Construction of select tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3 The construction of A1967-70. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.4 Some formulae for the force of mortality. . . . . . . . . . . . . . . . . . . . . . . . . . 322.5 Select tables used in examinations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 ASSURANCES 373.1 A general introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Whole life assurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.3 Commutation functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.4 The variance of the present value of benefits . . . . . . . . . . . . . . . . . . . . . . . 393.5 Assurances payable at the end of the year of death. . . . . . . . . . . . . . . . . . . . 413.6 Assurances payable at the end of the 1/m of a year of death. . . . . . . . . . . . . . 443.7 Temporary and deferred assurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.8 Pure endowments and endowment assurances . . . . . . . . . . . . . . . . . . . . . . 463.9 Varying assurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.10 Valuing the benefits under with profits policies . . . . . . . . . . . . . . . . . . . . . 503.11 Guaranteed bonus policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.13 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4 ANNUITIES 614.1 Annuities payable continuously . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.2 Annuities payable annually . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.3 Temporary annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.4 Deferred annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

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4.5 Annuities payable m times per annum . . . . . . . . . . . . . . . . . . . . . . . . . . 684.6 Complete annuities (or “annuities with final proportion”) . . . . . . . . . . . . . . . 704.7 Varying annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

5 PREMIUMS 815.1 Principles of premium calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.2 Notation for premiums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.3 The variance of the present value of the profit on a policy. . . . . . . . . . . . . . . . 835.4 Premiums allowing for expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 835.5 Premiums for with profits policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.6 Return of premium problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 865.7 Annuities with guarantees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.8 Family income benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

6 RESERVES 956.1 What are reserves? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 956.2 Prospective reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 956.3 Net premium reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 966.4 Retrospective reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 996.5 Gross premium valuations and asset shares. . . . . . . . . . . . . . . . . . . . . . . . 1026.6 The variance of L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046.7 Zillmerised reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1066.8 Full preliminary term reserves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1096.9 Reserves for with-profits policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1106.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

7 APPLICATIONS OF RESERVES 1197.1 Surrender values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.2 Paid-up policy values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1207.3 Alterations and conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.4 The actual and expected death strains . . . . . . . . . . . . . . . . . . . . . . . . . . 1247.5 Mortality profit and loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1257.6 Other sources of profit and loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1297.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1337.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

8 EXTRA RISKS 1378.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1378.2 A constant addition to the force of mortality . . . . . . . . . . . . . . . . . . . . . . 1378.3 A variable addition to the force of mortality . . . . . . . . . . . . . . . . . . . . . . . 1398.4 Rating up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1408.5 Debts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1428.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1458.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

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9 PROFIT-TESTING 1519.1 Principles of profit-testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.2 Cash flow calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1529.3 The profit vector and the profit signature . . . . . . . . . . . . . . . . . . . . . . . . 1559.4 The assessment of profits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1579.5 Some theoretical results about {σt} . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1599.6 Withdrawals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1609.7 The actual emergence of profits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1629.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1659.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

10 STATIONARY POPULATIONS 17110.1 Some Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17110.2 The Central Death Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17510.3 Relationships Between mx and qx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17610.4 Stationary Funds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17610.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17810.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

11 JOINT-LIFE FUNCTIONS 18511.1 Joint-Life Mortality Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18511.2 Select Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18811.3 Extensions to More than 2 Lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18811.4 The Joint Expectation of Life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19011.5 Monetary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19111.6 Last Survivor Probabilities (two lives only) . . . . . . . . . . . . . . . . . . . . . . . 19511.7 Last Survivor Monetary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19611.8 Reserves for Last Survivor Assurances . . . . . . . . . . . . . . . . . . . . . . . . . . 19811.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20011.10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

12 CONTINGENT ASSURANCES 20512.1 Contingent Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20512.2 Contingent Assurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20612.3 Premiums and Reserves for Contingent Assurances . . . . . . . . . . . . . . . . . . . 21012.4 A Practical Application – The Purchase of Reversions . . . . . . . . . . . . . . . . . 21112.5 Extension to Three Lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21212.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21312.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

13 REVERSIONARY ANNUITIES 21913.1 Reversionary Annuities Payable Continuously . . . . . . . . . . . . . . . . . . . . . . 21913.2 Reversionary Annuities Payable Annually or mthly . . . . . . . . . . . . . . . . . . . 22013.3 Widow’s (or Spouse’s) Pension on Death after Retirement . . . . . . . . . . . . . . . 22113.4 Actuarial Reduction Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22413.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22513.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

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14 PROFIT TESTING FOR UNIT-LINKED POLICIES 23114.1 Unit-Linked Policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23114.2 Mechanics of the Unit Fund . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23114.3 The Sterling Fund (or Sterling Reserves) . . . . . . . . . . . . . . . . . . . . . . . . . 23314.4 The Assessment of Profits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23514.5 Zeroisation of the Profit Signature . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23614.6 Withdrawals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23714.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24014.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

15 MULTIPLE-DECREMENT TABLES 24715.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24715.2 The Associated Single-Decrement Tables . . . . . . . . . . . . . . . . . . . . . . . . . 24915.3 The Relationships between the Multiple-Decrement Table and its Associated Single-

Decrement Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24915.4 Dependent Rates of Exit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25015.5 Practical Construction of Multiple-Decrement Tables . . . . . . . . . . . . . . . . . . 25315.6 Further Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25615.7 Generalization to 3 Modes of Decrement . . . . . . . . . . . . . . . . . . . . . . . . . 25715.8 “Abnormal” Incidence of Decrement . . . . . . . . . . . . . . . . . . . . . . . . . . . 25815.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26315.10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

16 FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES 26716.1 Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26716.2 The Use of “Defective” Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26716.3 Evaluation of Mean Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26816.4 Benefits on Death by a Particular Cause . . . . . . . . . . . . . . . . . . . . . . . . . 27216.5 Extra Risks Treated as an Additional Mode of Decrement . . . . . . . . . . . . . . . 27216.6 Calculations Involving a Change of State . . . . . . . . . . . . . . . . . . . . . . . . . 27416.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27516.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

17 MULTIPLE-STATE MODELS 28117.1 Two Points of View . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28117.2 Kolmogorov’s Forward Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28117.3 Life Tables as Stochastic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28317.4 Sickness Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

18 SICKNESS FUNCTIONS 28718.1 Rates of Sickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28718.2 Valuing Sickness Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28918.3 Various Other Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29218.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29518.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298Appendix The Manchester Unity Experience 1893-97 . . . . . . . . . . . . . . . . . . . . . 301

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19 PENSION FUNDS 30319.1 General Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30319.2 Valuation Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30319.3 Service Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30419.4 Salary Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30419.5 The Value of Future Contributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30619.6 The Value of Pension Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30819.7 Fixed Pension Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30919.8 Average Salary Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31119.9 Final Salary Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31319.10Lump Sums on Retirement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31619.11Death and Withdrawal Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31719.12Return of Contributions on Death or Withdrawal . . . . . . . . . . . . . . . . . . . . 31819.13Spouse’s Benefits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32019.14Preserved Pensions on Leaving Service . . . . . . . . . . . . . . . . . . . . . . . . . . 32219.15Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32419.16Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327Appendix Formulae for valuing a return of contributions . . . . . . . . . . . . . . . . . . . 331

APPENDICES 333A Some notes on examination technique . . . . . . . . . . . . . . . . . . . . . . . . . . 333B Some technical points about the tables used in examinations . . . . . . . . . . . . . 334C Some common mistakes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335D Some formulae for numerical integration . . . . . . . . . . . . . . . . . . . . . . . . . 336

SUPPLEMENT 337

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Chapter 1

NON-SELECT LIFE TABLES

1.1 Survivorship functions

We consider a certain population from age α to a “limiting age” w , i.e. the youngest age to whichno-one survives. (In theoretical work we may have w = ∞.) Let (x) be a shorthand notation for “alife aged (exactly) x”; we define the survivorship function

s(x) = Pr{ (α) survives to age x } (1.1.1)

In practice one usually uses the function

lx = lαs(x) (1.1.2)

where lα is called the “radix” of the table, and is usually a large number such as 1, 000, 000. If livesare considered from birth, we naturally have α = 0, and we have lx > 0 for x < w, lw = 0, unlessw is infinite, in which case lx → 0 as x → ∞. The function lx is assumed to be continuous. Byformulae (1.1.1) and (1.1.2) ,

Pr{ (α) survives to age x } =lxlα

(1.1.3)

Example 1.1.1. English Life Table No.12 - Males refers to the male population of England andWales in 1970-72. In this table α = 0 , w may be taken to be about 106, and the radix of the table is100, 000. We observe (for example) that, since l50 = 90, 085 , the probability that a new-born childwill survive to age 50 is s(50) = l50/l0 = 0.90085, or 90.1%.

We now consider survival from age α in terms of a random variable. Let T (or T (x) if the age x

is not clear; Tx is sometimes used, but this may be confused with the symbol Tx = lx◦ex used in

stationary population problems) be the variable representing the future lifetime (in years, includingfractions) of (x). According to our definitions,

Pr{ T (α) ≤ t } = the probability that (α) will die within t years= 1− the probability that (α) will survive for at least t years= 1− s(α + t) (1.1.4)

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10 CHAPTER 1. NON-SELECT LIFE TABLES

1.2 Probabilities of death and survival

We now consider how to deal with lives aged between α and w. To simplify matters, we shall supposehere that w = ∞. The variables {T (x) : x ≥ α} are assumed to be related as follows:for each x1, x2, x3 such that α ≤ x1 ≤ x2 ≤ x3,

Pr{T (x1) > x3 − x1} = Pr{T (x1) > x2 − x1}Pr{T (x2) > x3 − x2} (1.2.1)In particular, when x1 = α, x2 = x and x3 = x + t, we have

Pr{T (x) > t} =Pr{T (α) > x− α + t}

Pr{T (α) > x− α} =s(x + t)

s(x)=

lx+t

lx (1.2.2)

We introduce the important notation:

tpx = Pr{T (x) > t} = Pr{ (x) survives to age x + t} (1.2.3)

tqx = Pr{T (x) ≤ t} = Pr{ (x) dies between ages x and x + t} (1.2.4)

When t = 1 , it may be omitted, so that

px = Pr{ (x) survives for at least a year } (1.2.5)

qx = Pr{ (x) dies within a year } (1.2.6)

qx is called the (q-type) rate of mortality at age x. (The m-type, or central, rate of mortality,mx, is considered in section 1.6.) It is clear that, for all t ≥ 0 ,

tpx + tqx = 1 (1.2.7)

and, by the formula (1.2.2),

tpx =s(x + t)

s(x)=

lx+t

lx(1.2.8)

and

tqx = 1− lx+t

lx(1.2.9)

That is, the (cumulative) distribution function of the variable T is

F (t) = Pr{T ≤ t} = tqx =

{1− lx+t

lx, if t ≥ 0

0 , if t < 0(1.2.10)

We may say that “the expected number of survivors at age x + t among lx lives aged x is lx+t”because the chance that a given life aged x will survive to age x + t is lx+t

lx. In practice, the word

“expected” is sometimes omitted, but it is always to be understood; assuming that the lx lives areindependent, the number of survivors at age x + t will follow a binomial distribution with mean

lx · tpx = lx+t

and variance

lx · tpx(1− tpx) = lx+t

(1− lx+t

lx

)

1.2. PROBABILITIES OF DEATH AND SURVIVAL 11

(Even if the lives are not independent, the formula for the expected number of deaths is correct.)Note that

px =lx+1

lx(1.2.11)

qx = 1− lx+1

lx=

dx

lx(1.2.12)

where

dx = lx − lx+1 (1.2.13)= the (expected) number of deaths at age x last

birthday among lx lives aged x

Example 1.2.1. The following extract from English Life Table No. 12 - Males illustrates thefunctions lx, dx, px and qx .

agex lx dx px qx

0 100,000 2,499 0.97551 0.02449

1 97,551 153 0.99843 0.00157

2 97,398 96 0.99901 0.00099

3 97,302 67 0.99931 0.00069

4 97,235 60 0.99938 0.00062

(The high death rate at age 0, relative to the rates at ages 1 to 4, is quite noticeable.)

We also observe that, for all ages x3 ≥ x2 ≥ x1 ≥ α , we have the “rule of multiplication ofprobabilities of survival”:

x3−x1px1 = x2−x1px1 · x3−x2px2 (1.2.14)

(which is merely a re-statement of equation (1.2.1). )

Note. The definition of tpx is sometimes taken to bePr{(α) survives to age x + t | (α) survives to age x }, which equals

Pr{T (α) > x− α + t and T (α) > x− α}Pr{T (α) > x− α} =

Pr{T (α) > x− α + t}Pr{T (α) > x− α}

= Pr{T (x) > t}

in agreement with our earlier definition. Further, tpx is sometimes written as Sx(t). When α = 0we sometimes write S0(t) = S(t), so S(t) and s(t) are equal. The notation S(t) is much used insurvival analysis, in which t denotes the time lived by a patient since the start of a given treatment.In reliability engineering the notation R(t) may be used instead of S(t).


1.3 The force of mortality, µx

This may be defined as the “instantaneous rate of mortality”, i.e.

µx = limh→0+

hqx

h(1.3.1)

Theorem 1.3.1. Suppose that µx is continuous on [α, w). We have

µx = − l′xlx

(1.3.2)

Proof. we have

µx = limh→0+

−[s(x + h)− s(x)]hs(x)

=−s′+(x)

s(x)

where s′+(x) denotes the R.H. derivative of s(x). But s(x) and s′+(x) = −s(x)µx are continuouson [α,w), so s′(x) exists and equals −s(x)µx (see McCutcheon and Scott, An Introduction to theMathematics of Finance, Appendix 1. )Hence

µx = −s′(x)s(x)

= − l′xlx

We remark thatµx = − d

dxlog lx (1.3.3)

Hence

−∫ x

α

µydy =∫ x

α

d

dy(log ly) dy

= [ log ly ]xα= log lx − log lα

= log(

lxlα

)

from which we obtain the important formulae:

lx = lα exp(−

∫ x

α

µy dy

)(1.3.4)

and

s(x) = exp(−

∫ x

α

µy dy

)(1.3.5)

It follows that

tpx =lx+t

lx= exp

(−

∫ x+t

x

µy dy

)= exp

(−

∫ t

0

µx+r dr

)(1.3.6)

If µx is piecewise continuous, we may apply the above formulae over each age-range and combinethe results; the above formulae remain true, except that l′x does not exist at the points at which

1.3. THE FORCE OF MORTALITY, µx 13

µx is not continuous (the R.H. and L.H. derivatives of lx differ at these points.) Alternatively, wemay regard the actual position as closely approximated by a model in which the force of mortalityis continuous.We may now express the probability density function of the variable T (the future lifetime of (x))in terms of the force of mortality. The p.d.f. f(t) of T is the derivative of the distribution functionF (t), i.e.

f(t) = F ′(t) =

d

dt

(1− lx+t

lx

), if t > 0

0 , if t < 0

By the chain rule

d

dtlx+t = l′x+t

= −lx+tµx+t (by formula (1.3.2))

so we obtain

f(t) =

{lx+tµx+t

lx= tpxµx+t , if t > 0

0 , if t < 0(1.3.7)

In view of the following general formula (connecting the d.f. and p.d.f.):

F (t) =∫ t

−∞f(r) dr

we have

tqx =∫ t

0rpxµx+r dr (1.3.8)

An alternative notationSome statisticians refer to the force of mortality as the “hazard rate” or “transition intensity.”

Let f(t) and F (t) denote the p.d.f. and d.f. respectively of T = T (x) . The hazard rate at age x + t(often considered to be “at time t years after entry to assurance (say) at age x”) is defined in termsof f(t) and F (t) by the formula

h(t) =f(t)

1− F (t)(1.3.9)

which equals µx+t (by formulae (1.2.7) and (1.3.7).) If α = 0 and x = 0, we naturally have T = thefuture lifetime of a new-born child, and

µt =f(t)

1− F (t)= hazard rate at age t years (1.3.10)

Example 1.3.1. Suppose that, for x ≥ α,

µx = cδxδ−1 (δ > 1, c > 0) (1.3.11)

(This is “Weibull’s law of mortality”.) Find a simple expression for Sx(t) = tpx.


Solution.

s(x) = exp(−

∫ x

α

µy dy

)

= exp(−

∫ x

α

cδyδ−1 dy

)

= exp(−c

[yδ

]x

α

)

= exp(−cxδ

). exp

(cαδ

)

hence

tpx = Sx(t) = exp(−c

[(x + t)δ − xδ

])

Numerical estimation of µx from lxIf lx is known only at integer values of x, we may use numerical differentiation to estimate µx. Theformula

f ′(x) + f(x + 1)− f(x− 1)2

(1.3.12)

which is exact if f is a cubic between x− 1 and x + 1 , may be used to show that

µx + − l′xlx

=lx−1 − lx+1

2lx(1.3.13)

Deferred probabilitiesThe symbol m| indicates deferment for m years; for example,

m|nqx = P{(x) will die between ages x + m and x + m + n } (1.3.14)By elementary probability, this equals

Pr{(x) will die before age x + m + n} − Pr{(x) will die before age x + m}= m+nqx − mqx

= mpx − m+npx

=lx+m − lx+m+n

lx

That is,

m|nqx =lx+m − lx+m+n

lx(1.3.15)

This may be remembered by the following rule: of lx lives aged x, lx+m − lx+m+n is the (expected)number of deaths aged between ages x + m and x + m + n. If n = 1 it may be omitted, so we have

m|qx = Pr{(x) will die between the ages x + m and x + m + 1 }

=dx+m

lx(1.3.16)

We remark that there is no such thing as m|npx.

We may also use the result that

m|nqx = Pr{(x) survives to age x + m}.P r{(x + m) dies within n years}= mpx.nqx+m (1.3.17)

1.4. THE EXPECTATION OF LIFE 15

Example 1.3.2. Consider the force of mortality to be as in Example 1.3.1. Find an expression forthe chance that (x) will die between ages x + m and x + m + n.

Solution. We require:

m|nqx =lx+m − lx+m+n

lx

=s(x + m)− s(x + m + n)

s(x)

=exp[−c(x + m)δ]− exp[−c(x + m + n)δ]

exp[−cxδ]

1.4 The expectation of life

Let T be the future lifetime (in years, including fractions) of (x). The mean of T is written◦ex , and

is called the complete expectation of life at age x. That is,

◦ex = E(T ) (1.4.1)

=∫ ∞

0

t · tpxµx+t dt (1.4.2)

by formula 1.3.7. If w < ∞ , the upper limit of the integral should be replaced by w − x.

Theorem 1.4.1.◦ex=

∫ ∞

0tpx dt (1.4.3)

(with ∞ replaced by w − x if w is finite).

Proof. By integration by parts, with u = t , v = −tpx ,

◦ex=

∫ ∞

0

t · tpxµx+t dt = [−t · tpx]∞0 −∫ ∞

0

−tpx dt

=∫ ∞

0tpx dt

since it may be shown that t · tpx → 0 as t →∞ (using the fact that E(T ) < ∞. ).

We may also use the formulaVar(T ) = E(T 2)− [E(T )]2 (1.4.4)

where

E(T 2) =∫ ∞

0

t2 · tpxµx+t dt

=[−t2 · tpx

]∞0

+∫ ∞

0

2t · tpx dt

(on setting u = t2 , v = −tpx)

= 2∫ ∞

0

t · tpx dt


(since t2 · tpx →∞ , because E(T 2) < ∞ ).Hence

Var(T ) = 2∫ ∞

0

t · tpx dt− (◦ex)2 (1.4.5)

The median future lifetime of (x) is the solution, t, of the equation

F (t) = 0.5i.e., tqx = 0.5

i.e., lx+t = 0.5 lx (1.4.6)

x + t may be estimated by linear interpolation, as in the next example.

Example 1.4.1. Find the median future lifetime of (10), according to English Life Table No.12 -Males.

Solution. We must find the value of t such that

l10+t = 0.5 l10 = 48, 469.5

On inspecting the tables, 10 + t lies between 72 and 73. By linear interpolation in y = 10 + t , wehave

ly − l72l73 − l72

= y − 72

i.e. y = 72 +48, 469.5− 48, 62545, 430− 48, 625

= 72.05, so t = 62.05 years.

The modal future lifetime of (x) is the time, t0 years, at which the p.d.f. tpxµx+t (or, equivalently,the function lx+tµx+t) reaches its maximum. The modal age at death of (x) is y = x + t0, whichis the value of y for which lyµy attains its maximum in the range y ≥ x. (The function lyµy issometimes called the “curve of deaths”.) If lyµy has a unique maximum at age 80 (say), this is themodal age at death for all lives aged under 80, and hence the modal future lifetime of a life agedx < 80 is 80− x. (See Exercise 1.7 for an example in which there is a unique modal age at death.)We now consider the discrete random variable K (or K(x) if the age x is not clear) defined by

K = the integer part of T

= the number of complete years to be livedin the future by (x) (1.4.7)

Now it follows by formula 1.3.16 that

Pr{K = k} = k|qx ( k = 0, 1, 2, ...) (1.4.8)

This variable is used in many actuarial calculations. In particular, the curtate expectation oflife at age x, written ex , is the mean of K ; that is

ex =∞∑

k=0

k · k|qx =∞∑

k=1

k.k|qx (1.4.9)

1.5. THE ASSUMPTION OF A UNIFORM DISTRIBUTION OF DEATHS 17

Theorem 1.4.2.

ex =∞∑

k=1

kpx =lx+1 + lx+2 + . . .

lx(1.4.10)

Proof.

ex =dx+1

lx+ 2

dx+2

lx+ 3

dx+3

lx+ . . .

=1lx

[(dx+1 + dx+2 + dx+3 + . . .)

+ (dx+2 + dx+3 + . . .)+ (dx+3 + . . .)+ . . .]

=lx+1 + lx+2 + lx+3 + . . .

lx

[since ly = (ly − ly+1) + (ly+1 − ly+2) + . . .

= dy + dy+1 + . . . ]

We may also evaluate Var(K) by the formula

Var(K) = E(K2)− [E(K)]2

=∞∑

k=1

k2 dx+k

lx− (ex)2 (1.4.11)

As is clear by general reasoning,◦ex+ ex +

12

(1.4.12)

A more precise approximation may be obtained from the Euler-Maclaurin formula, which we shalldiscuss later.

1.5 The assumption of a uniform distribution of deaths

Let x be fixed. We may say there is a uniform distribution of deaths (U.D.D.) between ages xand x + 1 if, for 0 ≤ t ≤ 1,

lx+t = (1− t)lx + t lx+1 (1.5.1)

(i.e. ly is linear for x ≤ y ≤ x + 1)This equation may be written in the form

lx+t = lx − tdx (0 ≤ t ≤ 1) (1.5.2)

Theorem 1.5.1. The following conditions are each equivalent to the assumption of U.D.D. betweenages x and x + 1:

tqx = t · qx (0 ≤ t ≤ 1) (1.5.3)

tpxµx+t = qx (0 ≤ t < 1) (1.5.4)


Proof. Assume 1.5.2 holds. Thentpx = 1− t · qx (1.5.5)

But tpx = 1− tqx , so tqx = t · qx ; therefore 1.5.3 holds.This argument may be reversed, so 1.5.2 and 1.5.3 are equivalent. Now suppose that (1.5.3) holds.By differentiation with respect to t, we obtain (1.5.4), and we may obtain (1.5.3) from (1.5.4) byintegration.

Example 1.5.1. In a certain non-select mortality table, there is a uniform distribution of deathsbetween any two consecutive integer ages. Find formulae in terms of l30, l31 and l32 for

(i) 1.5p30.5

(ii) µ30.5

Solution.

(i)l32l30.5

=l32

12 (l30 + l31)

since l30+t is linear for 0 < t < 1

(ii) tp30µ30+t = q30 for 0 ≤ t ≤ 1, so µ30.5 =q30

12p30

=l30 − l31

l3012

=l30 − l31

12 (l30 + l31)

1.6 Central death rates

For simplicity of notation, we again suppose that the limiting age of our mortality table is infinity.Define

Tx =∫ ∞

0

lx+tdt (1.6.1)

=∫ ∞

x

lydy (on setting y = x + t)

from which we obtain

◦ex =

∫∞x

lydy

lx

=Tx

lx

and hence

Tx = lx◦ex (1.6.2)

We also define

Lx =∫ 1

0

lx+tdt (1.6.3)

1.7. LAWS OF MORTALITY 19

Note that

Lx =∫ x+1

x

lydy

=∫ ∞

x

lydy −∫ ∞

x+1

lydy

= Tx − Tx+1 (1.6.4)

Assume that there is a Uniform Distribution of Deaths (U.D.D) between the ages x and x + 1, i.e.lx+t is linear for 0 ≤ t ≤ 1.We have

Lx =∫ 1

0

lx+tdt = 12 (lx + lx+1) (1.6.5)

(as the trapezoidal rule is exactly correct, not just an approximation).Since lx+1 = lx − dx, we have

Lx = lx − 12dx (1.6.6)

Formulae (1.6.5) and (1.6.6) are sometimes used as approximations when U.D.D. does not hold.The central death rate at age x is defined as

mx =dx

Lx(1.6.7)

Note the following important approximation:

mx =

∫ 1

0lx+tµx+tdt∫ 1

0lx+tdt

+ µx+ 12

(1.6.8)

Relationships between mx and qx

If there is U.D. of D. between ages x and x + 1, formula (1.6.6) shows that

mx =dx

Lx

=dx

lx − 12dx

=qx

1− 12qx

(1.6.9)

We may rearrange this equation to get

qx =mx

1 + 12mx

(1.6.10)

If U.D.D. does not hold, these results may be used as approximations.

1.7 Laws of mortality

The term ‘law of mortality’ is used to describe a mathematical expression for µx (or possibly qx

or mx) which may be explained from biological or other arguments (rather than being just a best-fitting curve.) The most famous law of mortality is that of Gompertz (1825), who postulated that


µx satisfies the following simple differential equation:

dµx

dx= kµx (x ≥ α)

This may be solved to giveµx = Bcx (x ≥ α) (1.7.1)

Gompertz’ Law is often found to be quite accurate (at least as a first approximation) for ages overabout 25 or 30, the value of c usually being between 1.07 and 1.12.

Example 1.7.1. Show that, if Gompertz’ Law holds for all ages greater than or equal to α, thereis a positive constant g such that

tpx = gcx(ct−1) for x ≥ α, t ≥ 0 (1.7.2)

Solution.

tpx = exp[−

∫ t

0

µx+rdr

]

= exp[−Bcx

∫ t

0

crdr

]

= exp[−Bcx(ct − 1)

log c

]

= gcx(ct−1) with g = exp[− B

log c

]

In 1860 Makeham suggested the addition of a constant term to Gompertz’ formula for µx, giv-ing Makeham’s law:

µx = A + Bcx (x ≥ α) (1.7.3)

Example 1.7.2. Show that, if Makeham’s law holds for all ages greater than or equal to α, thereare positive constants s and g such that

tpx = stgcx(ct−1) for x ≥ α, t ≥ 0 (1.7.4)

Solution.

tpx = exp[−

∫ t

0

µx+rdr

]

= exp[−

∫ t

0

(A + Bcxcr)dr

]

= exp[−At]gcx(ct−1) (using Example 1.7.1)

= stgcx(ct−1) with g = exp[− B

log c

], s = e−A

1.7. LAWS OF MORTALITY 21

Weibull’s law of mortality has already been mentioned in Example 1.3.1. In practice it is usu-ally less successful than those of Gompertz and Makeham in representing human mortality. Thefitting of laws of mortality is complicated by the fact that mortality rates may be varying with time.

If a law of mortality holds, certain mortality and monetary functions may be evaluated analyt-ically (rather than numerically), but this point is of little practical importance in the computerage. There are also certain simplifications in the evaluation of joint-life functions (that is, functionsdepending on the survival of more than one life.)


Exercises

1.1 Calculate the following probabilities on the basis of English Life Table No. 12-Males.

(i) 40p25,(ii) 60|10q25,(iii) the probability that (30) survives for at least 10 years,(iv) the probability that (40) survives to age 65,(v) the probability that (50) dies within 10 years,(vi) the probability that (50) fails to reach age 70,(vii) the probability that (60) dies between ages 80 and 85,(viii) the probability that (60) dies within the first five years after retiring at age 65.In addition, express (i) and (ii) in words.

1.2 On the basis of E.L.T. No. 12 - Males, find the probability that a life aged 30 will

(i) survive to age 40,(ii) die before reaching age 50,(iii) die in his 50th year of age, i.e. between ages 49 and 50,(iv) die between his 40th birthday and his 50th birthday,(v) die either between exact ages 35 and 45 or between exact ages 70 and 80.

1.3 A man aged 50 has just retired because of ill health. Up to exact age 58 he will be subject toa constant force of mortality of 0.019803 p.a., after which his mortality will be that of E.L.T.No. 12 - Males. Find the probability that he will

(i) die before age 55,(ii) live to age 65,(iii) die between ages 55 and 60.

1.4 For the first 5 years after arrival in a certain country, lives are subject to a constant force ofmortality of 0.005. Thereafter lives are subject to mortality according to English Life TableNo. 12- Males with an addition of 0.039221 to the force of mortality.

(i) A life aged exactly 30 has just arrived in the country.

(a) Show that the probability that the life will survive to age 35 is 0.97531.(b) Find the probability that the life will survive to age 60.

(ii) What is the probability that a life aged exactly 33 who has been in the country for 3years will die between ages 50 and 51?(Assume that these lives will remain in the given country.)

1.5 For a certain animal population,

lx =l0

(1 + x)2(x ≥ 0)

Calculate

(i) the complete expectation of life at birth,(ii) the force of mortality at age 1 year,(iii) the chance that a newly-born animal will die between ages 1 and 2 years.

1.8. EXERCISES 23

1.6 Suppose that tpxµx+t is decreasing for 0 ≤ t ≤ 1. Show that qx < µx.

1.7 Suppose that Gompertz’ law, µx = Bcx, holds for all x ≥ α, c being greater than 1. Assumethat µα < log c.

(i) Give a formula for s(x).(ii) Show that lxµx attains a maximum when µx = log c, and has no other stationary points.

1.8 (Difficult) Suppose that there is a “uniform distribution of deaths” from age x to age x + 1,i.e. lx+t is linear in t for 0 ≤ t ≤ 1.

Show that, for all 0 ≤ s ≤ t ≤ 1,

t−sqx+s =t− s

1− sqxqx

1.9 (Difficult) Suppose that the “Balducci hypothesis” holds from age x to age x + 1, i.e.1−tqx+t = (1− t)qx for 0 ≤ t ≤ 1.

Show that, for all 0 ≤ s ≤ t ≤ 1,

t−sqx+s =t− s

1− (1− t)qxqx

1.10 Suppose that, for some a > 0,s(x) = (1− x/w)a (0 ≤ x ≤ w)

Give simple formulae for

(i) µx,(ii)

◦ex (Note finite limiting age),

(iii) 10p70,(iv) 40|5q35

1.11 A certain group of lives now aged 60 experience mortality according to a(55) males ultimatewith addition to the force of mortality. The addition is 0.0005 at age 60, increasing linearly to0.0025 at age 80, at which level the addition remains constant.

Find the probability that a life aged exactly 60 dies within 20 years.


Solutions

1.1 (i) 40p25 =l65l25

= 0.71528

(ii) 60|10q25 =l85 − l95

l25= 0.10021

(iii) 10p30 =l40l30

= 0.98452

(iv) 25p40 =l65l40

= 0.73025

(v) 10q50 =l50 − l60

l50= 0.12389

(vi) 20q50 =l50 − l70

l50= 0.39162

(vii) 20|5q60 =l80 − l85

l60= 0.16173

(viii) 5|5q60 =l65 − l70

l60= 0.17338

In words,

(i) is “the probability that a life aged 25 will survive to age 65.”(ii) is “the probability that a life aged 25 will survive to age 85 and die before age 95.”

1.2 (i) 10p30 =l40l30

= 0.98452

(ii) 20q30 =l30 − l50

l30= 0.05437

(iii) 19|q30 =d49

l30= 0.00613

(iv) 10|10q30 =l40 − l50

l30= 0.03889

(v) 5|10q30 + 40|10q30 =

(l35 − l45) + (l70 − l80)l30

= 0.35789

1.3 Let us use an asterisk to denote the mortality table of the life concerned, while lx etc. refersto E.L.T. 12-Males. Let k = 0.019803.

(i) 5q∗50 = 1− 5p

∗50 = 1− exp(−5k) = 0.09427

1.9. SOLUTIONS 25

(ii) 15p∗50 = 8p

∗507p

∗58 = exp(−8k)

l65l58

= 0.71188

(iii) 5|5q∗50 = 5p∗50 − 10p

∗50 = 5p

∗50 − 8p

∗502p

∗58

= exp(−5k)− exp(−8k)l60l58

= 0.08540

1.4 (i) (a) Prob. of survival for 5 years = exp(−

∫ 5

0

0.005dt

)= 0.97531

(b) 0.97531 exp (−0.039221× 25)lELT

60

lELT

35

= 0.3051

(ii) Prob. = prob. of survival for 17 years - prob. of survival for 18 years

= exp(−2× 0.005) exp (−0.039221× 15)lELT

50

lELT

35

− exp(−2× 0.005) exp (−0.039221× 16)lELT

51

lELT

35

= 0.02377

1.5 (i)◦e0=

∫ ∞

0

lxl0

dx =∫ ∞

0

(1 + x)−2 dx =[ −11 + x

]∞

0

= 1

(ii) µx = − l′xlx

=2

1 + x, so µ1 = 1

(iii) 1|q0 =l1 − l2

l0=

14− 1

9= 0.13889

1.6 Let f(t) = tqx − µx.t (0 ≤ t ≤ 1)

f ′(t) =d

dt

∫ t

0rpxµx+r dr − µx = tpxµx+t − µx < 0 for 0 < t < 1

Now f(0) = 0, and f(t) is decreasing for 0 ≤ t ≤ 1, so f(1) < 0; therefore qx < µx.

1.7 (i) For x ≥ α,

tpx = exp[−

∫ t

0

Bcx+s ds

]

= exp[−Bcx(ct − 1)/ log c

]

Therefore s(x) = x−αpα = exp [−B(cx − cα)/ log c].


(ii) It follows that lx = k. exp [−Bcx/ log c] for some k. Therefored

dx(lxµx) =lxBcx log c

+ Bcxk exp [−Bcx/ log c][−Bcx log c]

log c

=lxµx(log c−Bcx) (∗)=0 when µx = Bcx = log c

Let x0 be the unique point at which this occurs. Notice that (by equation (∗) above)

d

dx(lxµx) > 0 for x < x0, and is negative for x > x0.

Therefore x0 is a maximum point of lxµx.

1.8 Under U.D.D.,lx+t = tlx+1 + (1− t)lx (0 ≤ t ≤ 1)

and similarly for lx+s.

Thereforelx+s − lx+t

lx=

(t− s)lx − (t− s)lx+1

lx= (t− s)qx

Therefore t−sqx+s =(t− s)qx

1− sqx, since

lxlx+s

=1

1− sqx.

1.9

t−sqx+s = 1− lx+t

lx+s= 1− lx+1

lx+s− lx+t

lx+s

(1− lx+1

lx+t

)

= 1−sqx+s − (1− t−sqx+s)1−tqx+t

Therefore t−sqx+s = 1−sqx+s − 1−tqx+t

1− 1−tqx+t(on rearranging).

Now apply “Balducci” to the R.H.S. of the equation. This gives the desired result.

1.10 (i) µx = − d

dx[log s(x)] =

a

w − x(ii)◦ex =

∫ w−x

0tpx dt =

∫ w

x

ly dy/lx

=

∫ w

x(1− y/w)a dy

(1− x/w)a

Substitute z = 1− y/w to obtain (calculus exercise!) the result that◦ex=

w − x

a + 1

(iii) 10p70 =s(80)s(70)

=(

w − 80w − 70

)a

(iv) 40|5q35 =s(75)− s(80)

s(35)=

(1− 75

w

)a − (1− 80

w

)a

(1− 35

w

)a

1.9. SOLUTIONS 27

1.11 Addition to force of mortality = 0.0005 + 0.0001t (0 ≤ t ≤ 20)

Prob. of survival for 20 years = exp{−

∫ 20

0

(µ60+t + 0.0005 + 0.0001t)dt

}

= 20p60 exp(−0.003) on integration

= 0.4108

∴ Ans. = 0.5892


Chapter 2

SELECT LIFE TABLES

2.1 What is selection?

In the previous chapter we considered mortality rates to depend only on age: life tables of this formare sometimes called aggregate life tables. We now consider the situation when mortality rates (orthe force of mortality ) depend on two factors: (i) age, and (ii) the time (duration) since a certainevent, known as “selection”.

One important example of “selection” is the acceptance of a proposal for life assurance at normalrates of premium: the mortality rates of lives who have been recently accepted for life assuranceat normal rates may be expected to be lower than those of the general population. After a certainperiod the difference in mortality between those who have been accepted for life assurance and thegeneral population of the same age decreases, but it is not correct to say that the effect of selection“wears off” entirely (since the general population contains some lives who would never have beenaccepted for life assurance at normal premium rates). This point may be confirmed by a comparisonbetween the mortality rates of English Life Table No.12 - Males and A1967-70 ultimate, which wewill discuss below. (The process by which life offices decide whether to accept lives for assurance,and on what terms, is called “underwriting”.)

Another form of selection is the “self-selection” exercised by those who buy annuities: it may nor-mally be assumed that such lives are in good health (for their age), for the purchase of an annuityis otherwise likely to be a poor investment.

In these examples, the mortality of those selected is lower than that of the general population,particularly in the period soon after selection. In “reverse selection”, mortality rates are higher thanthose of the general population (or some other reference group). An example of reverse selectionis early retirement due to ill-health. After a certain period from the date of ill-health retirement,the mortality of these lives may be expected to become closer to that of lives who retired in normalhealth, or who are not yet retired.

29

30 CHAPTER 2. SELECT LIFE TABLES

2.2 Construction of select tables

Let us, for definiteness, consider “selection” to be the acceptance of a proposer for life assurance atnormal rates by a life office.Let

x = age at entry to assurance, i.e. at the date of selectiont = duration (in years) since the date of selection

The current (or attained) age of such a life is

y = x + t

We write “[x] + t” as a shorthand notation for a “life aged x at selection and duration t years sinceselection”. Thus for example,

hq[x]+t = Pr{“[x] + t” will die within h years}= Pr{a life aged x + t, who has selected t years ago,

will die within h years} (2.2.1)µ[x]+t = the force of mortality of a life aged x + t who has selected t years ago

= limh→0+

hq[x]+t

h(2.2.2)

The select period. We assume that there is a period, s years, such that the mortality of thoseselected at least s years ago depends only on the attained age, x + t. That is,

q[x]+t = qx+t (t ≥ s) (2.2.3)µ[x]+t = µx+t (t ≥ s) (2.2.4)

where qx+t and µx+t refer to those who were selected at least s years ago: such people are calledultimate lives.

For each fixed entry age x, we may define the family of random variables.

T ([x] + t) = the future lifetime of “[x] + t” (t ≥ 0)

Regarding x as fixed, we may construct a “life table” for those selected at age x by methods similarto those of chapter 1 (with t in place of x, and α = 0). We have, for example,

hq[x]+t = Pr{T ([x] + t) ≤ h}

= 1− l[x]+t+h

l[x]+t

where l[x]+t is the expected number number of survivors at age x + t of l[x] lives selected at age x.But instead of fixing each radix l[x] arbitrarily, we choose them to be such that l[x]+t depends onlyon x + t when t ≥ s. More precisely, we first construct a life table for the ultimate lives by themethods given in chapter 1; the function l[x]+t is then constructed to be such that

l[x]+t = lx+t (t ≥ s) (2.2.5)

This ensures that relationships such as

hp[x]+t =l[x]+t+h

l[x]+t(2.2.6)

2.3. THE CONSTRUCTION OF A1967-70. 31

(which are found by replacing “x” by “[x] + t” in the formulae of chapter 1) are true. We may omitthe square brackets enclosing x in expressions such as hp[x]+t and l[x]+t+h if t ≥ s or t + h ≥ srespectively.

We now give formulae for l[x]+t for 0 ≤ t ≤ s. If q[x]+t is given for t = 0, 1, 2, . . . , s−1 , we proceedrecursively, using the formulae

l[x]+t+1

l[x]+t= 1− q[x]+t (t = s− 1, , s− 2, . . . , 0) (2.2.7)

That isl[x]+s−1 = l[x]+s/(1− q[x]+s−1)l[x]+s−2 = l[x]+s−1/(1− q[x]+s−2)

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

l[x] = l[x]+1/(1− q[x])(2.2.8)

If µ[x]+t is given for 0 ≤ t ≤ s , we use the formula

l[x]+t =lx+s

exp(− ∫ s

tµ[x]+r dr

) (t < s) (2.2.9)

2.3 The construction of A1967-70.

We illustrate the construction of select tables by reference to A1967-70, which has a select period of2 years. This table refers to the mortality of male assured lives in the U.K. during the period 1967 to1970, and is based on data collected by the Continuous Mortality Investigation Bureau (C.M.I.B.).Mortality rates were extended to young ages (including “ultimate” lives aged under 2) for which nodata existed.

The “building blocks” of the table were the mortality rates:

qy (y = 0, 1, 2, . . . )q[x]+t (x = 0, 1, 2, . . . ; t = 0, 1)

The stages of construction of ly and l[x]+t were as follows:

1. l0 was fixed arbitrarily at 34,489.

2. l1, l2, l3, . . . , were computed as shown (working downwards in the third column of Table2.3.1).

3. By working to the left from l2 , l[0]+1 and l[0] were found; similarly for l[1]+1 and l[1], etc.

select selectduration 0 duration 1 ultimate

34, 489 = l034, 489(1− q0) = 34, 463.8 = l1

l[0] = l[0]+1

1−q[0]← l[0]+1 = l2

1−q[0]+1← 34, 463.8(1− q1) = 34, 440.4 = l2

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ← 34, 440.4(1− q2) = 34, 418.7 = l3...


Table 2.3.1

2.4 Some formulae for the force of mortality.

We assume that l[x]+t and µ[x]+t are continuous in t (t ≥ 0).

µ[x]+t = limh→0+

hq[x]+t

h

= limh→0+

l[x]+t − l[x]+t+h

hl[x]+t

=− d

dtl[x]+t

l[x]+t(t ≥ 0) (2.4.1)

(the derivative is, strictly speaking, a L.H. derivative, but arguments similar to those given in chapter1 show that the l[x]+t is differentiable in t ).We may thus write

µ[x]+t = f ′(t) (t ≥ 0)where f(t) = − log(l[x]+t).We now consider how best to estimate µ[x]+t for t = 0, 1, 2, . . . , s− 1 from a table of l[x]+t (t =0, , 1, 2, . . . , s). To estimate µ[x] = µ[x]+0, we may use the differentiated form of Newton’s forwarddifference formula, which gives

f ′(0) + ∆f(0)− 12∆2f(0) (2.4.2)

where

∆f(0) = f(1)− f(0)

∆2f(0) = f(2)− 2f(1) + f(0)

To estimate µ[x]+t , where t = 1, 2, . . . , s− 1 , we may use the formula

f ′(t) + f(t + 1)− f(t− 1)2

to obtain

µ[x]+t + −12

log(l[x]+t+1/l[x]+t−1) (2.4.3)

2.5 Select tables used in examinations.

Formulae and Tables for Actuarial Examinations give

1. A1967-70 (select period 2 years) and

2. a(55) males and females.

Note that a(55) consists of 2 tables, a(55) males and a(55) females. In each case the select period is1 year, and functions are given only for ages 60 and over. These tables were constructed on the basisof C.M.I.B. data and were intended to be suitable for those purchasing annuities in about 1955, anallowance being made for improvements in mortality .One should be careful not to look up “select” values (such as q[x] )when ultimate values (such as qx)are required, or vice versa. In the case of a(55), one should be careful not to look up the “males”table when the “females” table is required, and vice versa.

2.6. EXERCISES 33

Exercises

2.1 A mortality table has a select period of three years.

(i) Find expressions in terms of the life table functions l[x]+t and ly for

(a) q[50]

(b) 2p[50]

(c) 2|q[50]

(d) 2|3q[50]+1

(ii) Calculate 3p53 given that:

q[50] = 0.01601, 2q[50] = 0.96411,

2|q[50] = 0.02410, 2|3q[50]+1 = 0.09272

2.2 In its premium rate basis, an office assumes a 3-year select period. Functions on this table areindicated by an asterisk. The table is such that:

q∗[x+7] = q∗[x+3]+1 = q∗[x]+2 = q∗x+1

and ultimate mortality follows A1967-70 ultimate. Assuming further that l∗y = ly on A1967-70ultimate, calculate l∗[45]+t for t = 0, 1 and 2.

2.3 A certain life table has a select period of 1 year. At each integer age x , the select rate mortalityis 50% of the ultimate rate. Calculate e[60] , given that e60 = 17.5424 and q60 = 0.0142.

2.4 Explain briefly the concept of selection in relation to mortality tables. Define the term “selectperiod” and, using the A1967-70 table as an example, explain how a select table differs inconstruction from an aggregate table such as English Life Table No. 12 - Males.


Solutions

2.1 (i) (a) q[50] =l[50] − l[50]+1

l[50]= .01601 (A)

(b) 2p[50] =l[50]+2

l[50]= .96411 (B)

(c) 2|q[50] =l[50]+2 − l53

l[50]= .02410 (C)

(d) 2|3q[50]+1 =l53 − l56l[50]+1

= .09272 (D)

(ii) Put l50 = 100, 000 (say); thenl[50] = 98, 399 from (A)

l[50]+2 = 96, 411 from (B)l53 = 94, 001 from (C)l56 = 84, 877 from (D)

Then 3p53 = l56l53

= .90294.

2.2l∗[45]+2 =

l481− q∗[45]+2

=l48

1− q46= 33, 033

l∗[45]+1 =l[45]+2

1− q∗[45]+1

=33, 0331− q43

= 33, 101

l∗[45] =l∗[45]+1

1− q∗[45]=

33, 1011− q39

= 33, 144

2.3e[60] =

l61 + l62 + . . .

l[60]

=l60l[60]

· l61 + l62 + . . .

l60

=l60l[60]

e60

But q[60] = 0.5q60 , which implies that

l[60] =l61

1− 12q60

Therefore l60l[60]

= l60(1− 12 q60)

l61= (1− 1

2 q60)

1−q60= 1.007202 so

e[60] = 1.007202× 17.542 = 17.668

2.4 Mortality rates depend on the age x at “selection” [e.g. entry to assurance, purchase of anannuity (self selection)] and duration t years since selection. Thus q[x]+t = Pr{ a life aged x+ t

2.7. SOLUTIONS 35

who was selected t years ago will die within a year }, etc. The select period, s years, is suchthat

q[x]+t = qx+t for t ≥ s

i.e. for lives who were selected at least s years ago (these being called “ultimate” lives) mortalitydepends only on the attained aged. The l[x]+t function is constructed to be such that

l[x]+t = lx+t (t ≥ s)

In A1967-70, s = 2. In a non-select or aggregate life table, such as E.L.T. No. 12 - Males,mortality depends on one variable only (age) and there is no “selection”.Note. It is not correct to say that the effects of selection “wear off” entirely (and that themortality of ultimate lives is the same as that of the general population). If this were true,A1967-70 ultimate would resemble ELT 12 - Males, and it does not. The general populationincludes some lives who would never have been acceptable for assurance cover (at normalpremium rates).


Chapter 3

ASSURANCES

3.1 A general introduction

A life assurance (or life insurance) policy is a contract which promises to pay a specified sum,S say, on the death of a given life (the life assured) at any future time, or between certain specifieddates. The death benefit S is called the sum assured; it may be level (constant) or it may increaseor decrease in a manner specified in the contract. Policies under which payments depend on thedeath or survival of more than one life may also be issued.

In the case of with profits policies, the sum assured may be increased by additions, called bonuses,which depend on the experience of the office and its bonus distribution policy. With profits policiesmay be found in traditional or unitised form (see later discussion). One also encounters unit-linked policies, in which the benefits are directly linked to the performance of certain assets, e.g. acertain portfolio of equities. If the benefits (and premiums) are completely specified in money termsin the contract, the policy is said to be without profits (or non-profit). Policies may also provideendowment benefits, i.e. sums payable on the survival of the life (or lives) assured until a certaindate (or dates).

In this chapter we shall consider contracts issued on the life of one person. The benefits and expensesare paid for by premiums, which we shall discuss later.

3.2 Whole life assurances

Suppose that, when calculating the value of benefits, the life office assumes that its funds will earninterest at a constant rate, i per annum. The corresponding force of interest per annum is

δ = log(1 + i)

and the present value of 1 due at time t years is

vt = (1 + i)−t = e−δt (3.2.1)

Now consider a life aged x, who is subject to a certain non-select mortality table, and let T be thefuture lifetime of (x). A whole life assurance is a policy providing a certain sum assured, S, onthe death of (x) at any future date. The present value of this benefit (assumed for the moment tobe payable immediately on death) is

Z = g(T ) = SvT (3.2.2)

37

38 CHAPTER 3. ASSURANCES

The mean of this variable, E(Z), is called the mean (or expected) present value (M.P.V.) ofthe whole life assurance benefit, and is

E(Z) =∫ ∞

0

g(t)f(t) dt (3.2.3)

where f(t) is the p.d.f. of T (for t > 0); hence

E(Z) = S

∫ ∞

0

vttpxµx+t dt (3.2.4)

When S = 1 , we write

E(Z) =∫ ∞

0

vttpxµx+t dt = Ax (3.2.5)

where A stands for “assurance” and the bar indicates that the sum assured is payable immediatelyon the death of (x). The M.P.V. of the sum of £S payable immediately on the death of (x) istherefore

SAx

Example 3.2.1. Find the M.P.V. of a whole life assurance of £10, 000 , payable immediately onthe death of (40), according to E.L.T. No. 12-Males with interest at 4% p.a.

Solution.

M.P.V. = 10, 000 A40 on E.L.T. No. 12-Males, 4% interest= 10, 000× 0.31981 from Tables= £3, 198.10

Select tablesFor a select life (i.e. a person who has just been selected) aged x, we replace x by [x], and for an“[x] + t” life (i.e. a life now aged x + t who was selected t years ago) we replace x by [x] + t. Thus,letting T = T ([x]) denote the future lifetime of “[x]”, we write

A[x] = M.P.V. of vT

=∫ ∞

0

vttp[x]µ[x]+t dt (3.2.6)

If T = T ([x] + t) = future lifetime of “[x] + t” , we have

A[x]+t = M.P.V. of vT

=∫ ∞

0

vrrp[x]+tµ[x]+t+r dr (3.2.7)

3.3. COMMUTATION FUNCTIONS 39

3.3 Commutation functions

Functions such as Ax may easily be evaluated by numerical integration, and are often tabulateddirectly at various rates of interest. (In fact, a modern computer can easily compute Ax directly atany rate of interest required, e.g. 4.032% ).

Commutation functions are numerical devices (developed by Griffith Davies and others) which allowthe calculation of certain common assurance (and annuity) values at a specified rate of interest froma small number of columns, labelled Dx , Mx etc. In view of their importance in the Tables forActuarial Examinations we shall look at them in detail.

At this point we consider only whole life assurances payable immediately on death. Define

Dx = vxlx (3.3.1)

Cx =∫ 1

0

vx+tlx+tµx+t dt (3.3.2)

Mx =∞∑

t=0

Cx+t (3.3.3)

Now

Ax =

∫∞0

vx+tlx+tµx+t dt

vxlx

=Cx + Cx+1 + Cx+2 + . . .

Dx

(on writing∫ r+1

r

vx+tlx+tµx+t dt =∫ 1

0

vx+r+ulx+r+uµx+r+u du

by the change of variable u = t− r)

so that

Ax =∑∞

t=0 Cx+t

Dx=

Mx

Dx(3.3.4)

We shall consider commutation functions again in relation to temporary and deferred assurances,etc.

3.4 The variance of the present value of benefits

We recall that Z = g(T ) = SvT is the present value of S due immediately on the death of (x), andthat the mean of Z is SAx. What is the variance of Z?Answer. We have

var(Z) = E(Z2)− [E(Z)]2

=∫ ∞

0

S2v2ttpxµx+t dt− (

SAx

)2

= S2

[∫ ∞

0

(v∗)ttpxµx+t dt− (

Ax

)2]

(where v∗ = v2)

= S2[A∗x −

(Ax

)2]

(3.4.1)


where the rate of interest i∗ is such that

v∗ =1

1 + i∗= v2 =

1(1 + i)2

i.e. i∗ = 2i + i2 (3.4.2)

Note that the force of interest, δ∗, corresponding to i∗ is such that

v∗ = e−δ∗ = v2 = e−2δ

i.e. δ∗ = 2δ (3.4.3)

The standard deviation of Z is, of course, found by taking the square root of var(Z). If we considera block of n identical whole life policies on independent lives aged x, the total present value of theassurance benefits is

Z = Z1 + Z2 + . . . + Zn

where, for i = 1, 2, . . . , n,

Zi = present value of benefit under ith policy

= SvT (i)

with T (i) = future lifetime of ith life. Since the variables {T (i)} are assumed to be independent, thevariance of Z is

Var(Z) =n∑

i=1

Var(Zi)

=n∑

i=1

S2[A∗x −

(Ax

)2]

= nS2[A∗x −

(Ax

)2]

(3.4.4)

and so the standard deviation of the total present value is

S√

n

√[A∗

x −(Ax

)2]

(3.4.5)

We remark that the mean present value of a group of policies is the sum of their separate M.P.V’seven if the lives are not independent. Formulae 3.4.4 and 3.4.5 may be generalised to cover the casewhen the sums assured and/or ages of the lives are different (see exercise 3.3).

Jensen’s inequalityLet g(t) be convex, which is the case of g′′(t) ≥ 0 for all t > 0. Jensen’s inequality shows that

E[g(T )] ≥ g[E(T )] (3.4.6)

An application. Let g(t) = e−δt , where δ > 0. We have

g′′(t) = δ2e−δt ≥ 0 for all t > 0

3.5. ASSURANCES PAYABLE AT THE END OF THE YEAR OF DEATH. 41

By Jensen’s inequality,

E(vT ) ≥ vE(T )

but

E(T ) =◦ex

and

E(vT ) = Ax

so we have shown that

Ax ≥ v◦ex (3.4.7)

That is, at any positive rate of interest, v◦ex (which is the present value of £1 on the assumption

that (x) dies when he reaches his expectation of life) is less than or equal to the mean present value,Ax.

Example 3.4.1. Verify formula 3.4.7 for x = 50 on the basis of E.L.T. No.12-Males, 4% p.a. interest.

Solution.◦e50 = 22.68

so v◦e50 = 0.41085

which is less than A50 = 0.44429

3.5 Assurances payable at the end of the year of death.

Suppose that the death benefit S is payable at the end of the year of death (years being measuredfrom the date of issue of the policy). It is now convenient to use the random variable

K = [T ] = the integer part of T

which has discrete probabilities

Pr{K = k} = k|qx (k = 0, 1, 2, . . . )

The present value of the assurance is now

Z = g(K) = SvK+1

which has mean

E[g(K)] =∞∑

k=0

Svk+1k|qx (3.5.1)

= SAx


where

Ax =∞∑

k=0

vk+1k|qx (3.5.2)

We define the commutation functions

Cy = vy+1dy (3.5.3)

My =∞∑

k=0

Cy+k (3.5.4)

and find that

Ax =∑∞

k=0 vx+k+1dx+k

Dx

=Mx

Dx(3.5.5)

Example 3.5.1. Show that the variance of Z (= SvK+1) is

S2[A∗x − (Ax)2

](3.5.6)

where ∗ indicates at the rate of interest 2i + i2.

Solution.

Var(Z) =E(Z2)− [E(Z)]2

=S2

[ ∞∑

k=0

(v∗)k+1k|qx − (Ax)2

]

=S2[A∗x − (Ax)2

]

The relationship between Ax and Ax.Assuming that deaths occur on average mid-way through each policy year (i.e. the year between twoconsecutive policy anniversaries), benefits payable at the end of the year of death will be received,on average, 6 months later than those immediately payable on death. We thus have the approximaterelationship

Ax + (1 + i)12 Ax (3.5.7)

This may be confirmed by establishing the following approximate relationship

Cx + (1 + i)12 Cx (3.5.8)

which follows from the mean value theorem for integrals. Formulae (3.3.3) and (3.5.4) now showthat

Mx + (1 + i)12 Mx (3.5.9)

3.5. ASSURANCES PAYABLE AT THE END OF THE YEAR OF DEATH. 43

and hence

Ax =Mx

Dx+ (1 + i)

12 Mx

Dx= (1 + i)

12 Ax (3.5.10)

Theorem 3.5.1. If there is a uniform distribution of deaths between the ages x + k and x + k + 1(for k = 0, 1, 2, . . . ), we have

Ax =i

δAx (3.5.11)

Proof. It is sufficient (since Ax =[∑∞

k=0 Cx+k

]/Dx and Ax = [

∑∞k=0 Cx+k]/Dx ) to show that

Cy =i

δCy for y = x + k (k = 0, 1, 2, . . . )

Now, since we have U.D. of D. between ages y and y + 1,

ly+tµy+t = dy (0 ≤ t < 1)

Therefore

Cy =∫ 1

0

vy+tly+tµy+t dt

= dyvy

∫ 1

0

vt dt

= dyvy

(1− v

δ

)

= dyvy+1

(i

δ

)

=i

δCy

Note. By the mathematics of finance, if i is small we have

i

δ+ 1 +

12i

and

(1 + i)12 + 1 +

12i

so we sometimes find the following approximations (which are suitable only when i is small):

Ax = (1 +12i)Ax

Cx = (1 +12i)Cx

Mx = (1 +12i)Mx

(3.5.12)


3.6 Assurances payable at the end of the 1/m of a year ofdeath.

Suppose that £1 is payable at the end of the 1/m year (measured from the issue date) following thedeath of (x); for example, if m = 12 , the sum assured is payable at the end of the month of death.The position is illustrated in Figure 3.6.1 below, in which ∗ indicates the time when (x) dies.

-

0

|1m

|2m

|. . . . . . . . . r

m

|(r+1)

m

|death occurs here

JJ

payment is made here

À time(years)

Figure 3.6.1:

The mean present value of this benefit is

A(m)x =

∞∑r=0

vr+1m r

m | 1m

qx (3.6.1)

As m →∞ , we find that (as one would expect from general reasoning)

A(m)x −→ Ax (3.6.2)

Corresponding to formula 3.5.7 , we have the useful approximation

Ax + v1/(2m)Ax (3.6.3)

where the factor of v1/(2m) allows for interest for 1/(2m) year, which is the average “delay” inreceiving death benefit. The function A

(m)x is not often used in practice.

3.7 Temporary and deferred assurances

A term (or temporary) assurance contract pays a sum assured of £S, say, on the death of (x)within the term of the policy, which we usually write as n years. If the benefit is payable immediatelyon death, the present value (as a random variable) is

Z = g(T ) ={

SvT if T ≤ n0 if T > n

The mean of this variable is written as S A1x:n , the “1” indicating that the status (x) must fail

before the status n (a fixed period of n years) for a payment to be due (and that the payment isdue when this event occurs). Thus

A1x:n =

∫ n

0

vttpx µx+t dt (3.7.1)

This may be evaluated by numerical integration or by commutation functions, i.e.

A1x:n =

n−1∑t=0

Cx+t

Dx=

Mx − Mx+n

Dx(3.7.2)

3.7. TEMPORARY AND DEFERRED ASSURANCES 45

(provided that n is an integer)If the sum assured is payable at the end of year of death (if this occurs within the n year term), wehave the present value

Z = g(K) ={

SvK+1 if K < n0 if K ≥ n

The M.P.V. of this benefit is written as S A1x:n , and we have

A1x:n =

n−1∑t=0

vt+1t|qx (3.7.3)

(It is assumed that n is an integer.) In terms of commutation functions,

A1x:n =

n−1∑t=0

Cx+t

Dx=

Mx −Mx+n

Dx(3.7.4)

In view of the relationship between Cx and Cx, we have the approximations

A1x:n ; (1 + i)

12 A1

x:n ; i

δA1

x:n (3.7.5)

A deferred assurance provides a sum of £S (say) on the death of (x) if this occurs after a certainperiod, called the deferred period (or, more correctly, the period of deferment), which we usuallywrite as n years. If the benefit is payable immediately on death of (x) after the deferred period haselapsed, the present value is

Z = g(T ) ={

SvT if T > n0 if T < n

if the sum assured is payable immediately on death, or

Z = g(K) ={

SvK+1 if K ≥ n0 if K < n

if the sum assured is payable at the end of year of death.The corresponding M.P.V.’s are written as

n|Ax = vnnpx Ax+n (3.7.6)

andn|Ax = vn

npx Ax+n (3.7.7)

We have the formulae

n|Ax =∫ ∞

n

vttpxµx+t dt (3.7.8)

=

∞∑t=n

Cx+t

Dx

=Mx+n

Dx(3.7.9)


and

n|Ax =∞∑

t=n

vt+1t|qx (3.7.10)

=

∞∑t=n

Cx+t

Dx

=Mx+n

Dx(3.7.11)

In view of the relationship between Cx and Cx , we have the approximations:

n|Ax ; (1 + i)12 n|Ax ; i

δn|Ax (3.7.12)

The evaluation of term and deferred assurance functions is often simplified by the observation that awhole life policy may be thought of as a combination of a term assurance and a deferred assurance,so that

Ax = A1x:n + n|Ax (3.7.13)

and Ax = A1x:n + n|Ax (3.7.14)

3.8 Pure endowments and endowment assurances

We now consider a pure endowment of £1 of term n years, i.e. a policy providing the sum of £1at time n years if (x) is then alive.The present value of this benefit is

Z ={

vn if T ≥ n0 if T < n

and hence the M.P.V. is

vnnpx

(since Pr{T ≥ n} = npx ). This M.P.V. is written as

nEx or A 1x:n or A 1

x:n

i.e.nEx = A 1

x:n = A 1x:n = vn

npx (3.8.1)

In terms of commutation functions we have the useful result that

nEx =Dx+n

Dx(3.8.2)

Returning to the formulae (3.7.6) and (3.7.7) , we see that the M.P.V. of a deferred assurance may bewritten as the product of nEx and Ax+n or Ax+n. That is, the value is obtained by multiplying theM.P.V. of the assurance at the “vesting date” (when the life attains age x+n) by a pure endowmentfactor.

We finally consider an endowment assurance of term n years with sum assured £S. In thispolicy, the sum assured is payable if (x) dies within n years or on maturity of the contract at time

3.8. PURE ENDOWMENTS AND ENDOWMENT ASSURANCES 47

n years, whichever occurs first. It follows that an endowment assurance is a combination of a termassurance and a pure endowment (of the same term). If we assume that the death benefit is payableimmediately on death, if death occurs within n years, the present value of the benefits is

Z = g(T ) ={

SvT if T < nSvn if T ≥ n

The mean of Z is written S Ax:n .Thus

Ax:n = A1x:n + A 1

x:n (3.8.3)= A1

x:n + nEx

=Mx − Mx+n + Dx+n

Dx(3.8.4)

Similarly, if the death benefit is payable at the end of the year of death, the present value is

Z = g(K) ={

SvK+1 if K < nSvn if K ≥ n

and the M.P.V. is written as S Ax:n . Thus

Ax:n = A1x:n + A 1

x:n (3.8.5)= A1

x:n + nEx

=Mx −Mx+n + Dx+n

Dx(3.8.6)

In view of the fact that the term Dx+n occurs in formulae (3.8.4) and (3.8.6), it is not correct towrite

Ax:n ; (1 + i)12 Ax:n A COMMON

MISTAKE

Correct relationships are (for example):

Ax:n ; (1 + i)12 A1

x:n + A 1x:n (3.8.7)

and

Ax:n ; (1 + i)12 (Mx −Mx+n) + Dx+n

Dx(3.8.8)

Select tables The adjustments to the formulae when a select table is used are straightforward: oneneed only replace ‘x + t’ by ‘[x] + t’ (or ‘[x]’ if the life assured has just been selected) and dispensewith [ ] if the duration since selection is at least equal to the select period, s years, and in compoundinterest functions.For example,

C[x]+t = vx+t+1d[x]+t

Example 3.8.1. On the basis of A1967-70 select mortality and 4% p.a. interest, calculate the meanpresent value of each of the following assurance benefits for a life aged 30:


(i) A whole life assurance for £10, 000, payable immediately on death;

(ii) A 20-year term assurance for £50, 000, payable at the end of the year of death;

(iii) A 20-year endowment assurance for £50, 000, with the death benefit payable immediately ondeath;

(iv) A deferred temporary assurance for £100, 000, payable at the end of the year of death, if deathoccurs between ages 40 and 50 exactly.

[Use the factor (1+ i)12 for accelerating payments from end of year of death to the moment of death.]

Solution.

(i) 10, 000A[30] = 10, 000(1.04)12 A[30]

= £1, 934.87

(ii) 50, 000(M[30] −M50)

D[30]= £1, 012.90

(iii) (1.04)12 × 1, 012.90 + 50, 000

D50

D[30]= £23, 070.56

(iv) 100, 000(M40 −M50)

D[30]= £1, 360.91

3.9 Varying assurances

Suppose that a contract provides the sum of β(t) immediately on the death of (x) at time t years.The present value of this benefit is

Z = g(T ) = β(T ) vT (T > 0)

and hence the M.P.V. is∫ ∞

0

vtβ(t)tpxµx+t dt (3.9.1)

If β(t) = t for all t > 0 , the M.P.V. is written as (IA)x , i.e.

(IA)x =∫ ∞

0

tvttpxµx+t dt (3.9.2)

3.9. VARYING ASSURANCES 49

If β(t) = [t]+1 , where [t] indicates the integer part of t, the M.P.V. is written as (IA)x, and it mayeasily be shown that

(IA)x =

∞∑t=0

(t + 1)Cx+t

Dx

={[

Cx + Cx+1 + · · · ] +[Cx+1 + Cx+2 + · · · ] + · · ·

}/Dx

=Mx + Mx+1 + · · ·

Dx

=Rx

Dx(3.9.3)

where

Rx =∞∑

k=0

Mx+k (3.9.4)

In view of the fact that the average benefit is 12 less when β(t) = t than in the present case, we have

the approximation

(IA)x ; (IA)x − 12Ax (3.9.5)

We now consider the case when a death benefit of β(k) in year k is payable at the end of the yearof death. The present value is

Z = g(K) = β(K + 1) vK+1 (K = 0, 1, 2, . . .)

and hence the M.P.V. is ∞∑

k=0

β(k + 1)vk+1k|qx (3.9.6)

When β(k) = k for k = 1, 2, . . . , the M.P.V. is written (IA)x, so we have

(IA)x =∞∑

k=0

(k + 1)vk+1k|qx (3.9.7)

In terms of commutation functions we have

(IA)x =

∞∑

k=0

(k + 1)Cx+k

Dx

=Mx + Mx+1 + · · ·

Dx

=Rx

Dx(3.9.8)

where

Rx =∞∑

k=0

Mx+k (3.9.9)


In view of the approximations Cx ; (1 + i)12 Cx, etc., we find that

Rx ; i

δRx ; (1 + i)

12 Rx (3.9.10)

and

(IA)x ; i

δ(IA)x ; (1 + i)

12 (IA)x (3.9.11)

We may also define similar symbols for increasing temporary and endowment policies, e.g.

(IA)1x:n =n−1∑

k=0

(k + 1)vk+1k|qx

=

n−1∑

k=0

(k + 1)Cx+k

Dx

={

[Cx + 2Cx+1 + · · · ]− [Cx+n + 2Cx+n+1 + · · · ]

− n [Cx+n + Cx+n+1 + · · · ]}/

Dx

=Rx −Rx+n − nMx+n

Dx(3.9.12)

and

(IA)x:n = (IA)1x:n + n · nEx (3.9.13)

3.10 Valuing the benefits under with profits policies

So far we have studied without profits contracts, for which the premiums and benefits are com-pletely fixed (in money terms). In with profits contracts, the premiums are higher but the policy-holder has the right to a share of the profits of the office. In the U.K., the usual system for sharingprofits is by means of additions to the basic benefit (sum assured, pension, etc.) in the form ofbonuses. These may be either reversionary bonuses, which are usually added annually and are notsubsequently reduced, or terminal bonuses, which are paid only on death or maturity (and possiblyon surrender) and are not guaranteed to continue.

Nowadays one may also encounter unitised with profit policies. These are not the same asunit-linked policies: the maturity proceeds of the latter are linked to the performance of a unittrust or internal fund of the life office, while for U.W.P. policies the office credits each policy withinternal units which are subject to certain guarantees.

In this book we shall study reversionary bonuses only. There are 3 systems of adding bonusesin common use:

(1) the simple bonus system;

(2) the compound bonus system, and

(3) the two-tier (or supercompound) bonus system.

3.10. VALUING THE BENEFITS UNDER WITH PROFITS POLICIES 51

(1) Simple bonuses. In this system, bonuses are calculated only on the basic sum assured (B.S.A.),which we shall denote by S. Consider a whole life with profits policy issued t years ago to a lifethen aged x. Suppose that bonuses vest annually in advance, let the bonuses already vested be B,and suppose that a bonus is about to be added. the formula used to calculate new bonus additionis

new bonus = S × rate of bonus p.a. (3.10.1)

To value the benefits we must estimate the rate of simple bonus p.a. (applying now and for theforseeable future) as b (say). The total sum assured is thus

S + B + bS in year 1 from the present time,S + B + 2bS in year 2 from the present time,

and so on, giving a total benefit in year k ofS + B + kbS

(3.10.2)

If the benefits are payable at the end of the year of death, the M.P.V. of the benefits is∞∑

k=0

[S + B + (k + 1)bS] vk+1k|qx+t

=(S + B)Ax+t + bS(IA)x+t (3.10.3)

Notes. (1) At the inception of the policy, we have t = 0 and B = 0.(2) If the benefits are payable immediately on death, the M.P.V. is

(S + B)Ax+t + bS(IA)x+t (3.10.4)

(3) In the case of an n-year endowment assurance with profits, the benefits on maturityare normally equal to those on death in the final year. This leads to the followingformulae for the value of benefits:

at end of yearof death

: (S + B)Ax+t:n−t + bS(IA)x+t:n−t (3.10.5)

immediatelyon death

: (S + B)Ax+t:n−t + bS(IA)x+t:n−t (3.10.6)

(2) Compound bonuses In this system, the formula for bonus additions is

new bonus =

basic sumassured

plus bonusesalreadyvesting

× rate of bonus p.a. (3.10.7)

Assuming a bonus rate of b p.a. applies now and for the forseeable future, the benefits will be

(S + B)(1 + b) in year 1

(S + B)(1 + b)2 in year 2and so on, giving

(S + B)(1 + b)k in year k(3.10.8)


If the benefits are payable at the end of the year of death, their M.P.V. is

∞∑

k=0

(S + B)(1 + b)k+1vk+1k|qx+t

=(S + B)∞∑

k=0

(1 + b

1 + i

)k+1

k|qx+t

=(S + B)A∗x+t (3.10.9)

where A∗ is at rate of interesti− b

1 + b(3.10.10)

If the benefits are payable immediately on death, their M.P.V. is approximately

(S + B)(1 + i)12 A∗x+t (3.10.11)

(NOT (S + B)A∗x+t

)

For endowment assurances, the corresponding formulae are

(S + B)A∗x+t:n−t

(3.10.12)

and(S + B)

[(1 + i)

12 A∗ 1

x+t:n−t + A∗ 1x+t:n−t

](3.10.13)

Notes. (1) In formula (3.10.13), we observe that the pure endowment part is as in formula(3.10.12): only the term assurance part is “accelerated”

(2) At inception, we have t = 0 and B = 0.

(3) Supercompound (two-tier) bonuses The formula is

new bonus =(B.S.A.)

(bonus rate p.a.in respect ofB.S.A.

)

+(

bonuses alreadyvesting

)

bonus ratep.a. in respectof bonusesalready vesting

(3.10.14)

Let us assume that the bonus rate p.a. in respect of the B.S.A. and previous bonus additions area and b respectively (now and for the forseeable future).

Let B(k) = total bonus in year k (from the present time), and let us define B(0) = B. We have

B(1) = B(1 + b) + aS

B(2) = B(1 + b)2 + aS[1 + (1 + b)]

and so on, giving the conjecture

B(k + 1) = B(1 + b)k+1 + aS × sk+1 b (3.10.15)

3.10. VALUING THE BENEFITS UNDER WITH PROFITS POLICIES 53

Proof by induction. Since

B(1) = (aS + bB) + B = B(1 + b) + aS

the result is true for k = 0. Now assume formula (3.10.15) is true for B(k +1), k being some value.We have

B(k + 2) = [aS + bB(k + 1)] + B(k + 1)

= (1 + b)[(1 + b)k+1B

]+ aS

[(1 + b)sk+1 b + 1

]

= (1 + b)k+2B + aS(sk+2 b

)

so the result is also true for B(k + 2), as required.

It follows that, in this bonus system, the M.P.V. of a whole life assurance is

∞∑

k=0

[S + B(k + 1)]vk+1k|qx+t

(assuming death benefits are payable at end of the year of death)

=∞∑

k=0

{S + aS

[(1 + b)k+1 − 1

b

]+ B(1 + b)k+1

}k|qx+t

=∞∑

k=0

{S

(1− a

b

)+

(a

bS + B

)(1 + b)k+1

}k|qx+t

=S(1− a

b

)Ax+t +

(a

bS + B

)A∗x+t (3.10.16)

where A∗x+t is at ratei− b

1 + b

Notes. (1) If the sum assured is payable immediately on death, the M.P.V. is approximately

S(1− a

b

)Ax+t +

(a

bS + B

)(1 + i)

12 A∗x+t (3.10.17)

(2) At inception of the policy, we have t = 0 and B = 0.(3) The “ordinary” compound bonus system is a special case of the two-tier system with

a = b.(4) For an endowment assurance the formulae are:

at end of yearof death

: S(1− a

b

)Ax+t:n−t +

(a

bS + B

)A∗

x+t:n−t (3.10.18)

immediatelyon death

: S(1− a

b

)Ax+t:n−t

+(a

bS + B

) [(1 + i)

12 A∗ 1

x+t:n−t + A∗ 1x+t:n−t

](3.10.19)

(Note that in formula (3.10.19) the term assurance benefits are “accelerated”but the pure endowment parts are the same as in formula (3.10.18).)


3.11 Guaranteed bonus policies

As their name suggests, these are not with profits contracts, as they provide additions to the basicbenefit at rates which are fixed at the outset and do not depend on future the experience of the lifeoffice. But many of the formulae used to value these benefits are the same as those we have derivedin the previous section.

Example 3.11.1. A company issues 20-year endowment assurances each with a basic sum assuredof £1, 000 to male lives aged 45.

Guaranteed simple reversionary bonuses at the rate of 2.25% of the current sum assured vest on pay-ment of each annual premium. Alternatively, at the outset of a policy, a life assured may elect thatcompound reversionary bonuses should be added to the policy instead of simple reversionary bonuses.

The sum assured and added bonuses will be payable at maturity, or at the end of the year ofdeath, if earlier.

Derive an expression from which can be calculated the guaranteed rate of compound bonus thecompany should offer so that the value of the benefits at the outset is the same.

Solution. Let b be the annual rate of guaranteed compound bonus. The equation of value for b is:

1000 [A45:20 + 0.0225(IA)45:20 ]

= 1000A∗45:20

at rate i∗ = i−b1+b

This may be solved for i∗, and hence b may be found.

3.12. EXERCISES 55

Exercises

3.1 Evaluate (a) A60 and (b) A60 on the bases:

(i) A1967-70 ultimate, 4% p.a. interest;(ii) E.L.T. No.12 - Males, 4% p.a. interest.

3.2 (i) Show that Ax = vqx + vpxAx+1

(ii) Given that p60 = 0.985, p61 = 0.98, i = 0.05 and A62 = 0.6 , evaluate A61 and A60

3.3 Consider n lives now aged x1, x2, . . . , xn respectively. Let Z be the total present value of acontract providing the sum of £Si immediately on death of (xi), i = 1, 2, . . . , n. The n livesare subject to the same non-select mortality table, Table B, and the interest is taken to befixed at rate i p.a.

(i) Show that E(Z) = S1Ax1 + · · ·+ SnAxnon Table B with interest at rate i p.a.

(ii) Assuming further that the future lifetimes T (xi) of the lives are independent variables,show that

Var(Z) =n∑

j=1

S2j [A∗xj

− (Axj )2]

where ∗ refers to an interest rate of 2i + i2 p.a.

3.4 Using commutation functions or otherwise calculate the values of the following:

(i) A[40]:10 on A1967-70, 4% p.a. interest;(ii) A 1

30:20 on A1967-70 Ultimate, 4%;(iii) A 1

30:20 on A1967-70 Ultimate, 4%;(iv) A30:20 on A1967-70 Ultimate, 4%;(v) A30:20 on English Life Table No.12 Males, 4%.

3.5 A life aged 50 who is subject to the mortality of the A1967-70 Select table, effects a pureendowment policy with a term of 20 years for a sum assured of £10,000.

(i) Write down the present value of the benefits under this contract, regarded as a randomvariable.

(ii) Assuming an effective rate of interest of 5% per annum, calculate the mean and thevariance of the present value of the benefits available under this contract.

3.6 What are the random variables (in terms of K = curtate future lifetime of (x)) whose meansare represented by the following symbols?

(i) nEx

(ii) A1x:n

(iii) A 1x:n

3.7 A life aged exactly 60 wishes to arrange for a payment to be made to a charity in 10 years’time. If he is still alive at that date the payment will be £1000. If he dies before the paymentdate, the amount given will be £500. Assuming an effective interest rate of 6% per annum and


mortality according to ELT No.12-Males, calculate the standard deviation of the present valueof the liability.

3.8 (Difficult)You are given that

(i) 1000 (IA)50 =4,996.75 ,(ii) 1000 A 1

50:1 =5.58,(iii) 1000 A51 = 249.05 and(iv) i =0.06.

Calculate 1000(IA)51.

3.13. SOLUTIONS 57

Solutions

3.1 (i) (a) A60 = (1 + 0.04)12 A60

= (1.04)12 0.51726

= 0.52750(b) A60 = 0.51726 from tables

(ii) (a) A60 = 0.58317 from tables(b) A60 = (1.04)−

12 A60

= 0.57185

3.2 (i) Ax =∞∑

k=0

k|qx vk+1

= qxv +∞∑

k=1

px k−1|qx+1 vk+1

= qxv + pxv

∞∑

k=0

k|qx+1 vk+1

= qxv + pxvAx+1

(ii) A61 =1

1.05(0.02 + 0.98 · 0.6) = 0.57905

A62 =1

1.05(0.015 + 0.985 · 0.57905) = 0.55749

3.3 (i)

We have

Z =n∑

j=1

SjvTj

so

E(Z) =n∑

j=1

SjE(vTj ) =n∑

j=1

SjAxj

(ii) By independence of Txj , we have

Var(Z) =n∑

j=1

S2j var(vTj )

=n∑

j=1

S2j [A∗xj

− (Axj )2]

3.4 (i)M[40] −M50 + D50

D[40]=

1904.8595− 1767.5555 + 4597.06076981.5977

= 0.67812

(ii)M30 −M50

D30=

1981.9552− 1767.555510433.310

= 0.020550


(iii) A 130:20 ; (1 + i)

12 A 1

30:20 = 0.020956

(iv) A30:20 = A 130:20 +

D50

D30= 0.46157

(v) A 130:20 ;

A 130:20

(1 + i)12

= (1 + i)−12M30 − M50

D30= 0.03286

3.5 (i)

P.V. ={

10, 000v20 if T ≥ 200 if T < 20

where T = future lifetime of (50).(ii)

E(P.V.) = (20p[50] × v20 + (1− 20p[50])× 0)10, 000

= 10, 000× 0.725539× 1.05−20 = 2734.48

E[(P.V.)]2 = 20p[50](10, 000v20)2 = 0.725539× 10, 0002 × 1.05−40 = 10, 305, 968

⇒ V ar(P.V.) = E[(P.V.)2]− [E(P.V.)]2

= 2, 828, 587 = 1, 681.842

3.6 (i)

Z ={

vn if K ≥ n0 if K < n

(ii)

Z ={

0 if K ≥ nvK+1 if K < n

(iii) Same as 3.6(i)

3.7 Let X payment to be made at time 10 years.

X ={

1000 with prob. 10p60

500 with prob. (1− 10p60)

∴ E(X) = 100010p60 + 500(1− 10p60)= 500(1 + 10p60) = £847.207

Var(X) = 1000210p60 + 5002(1− 10p60)− (mean)2

= 1, 000, 000[10p60 +14(1− 10p60)]− (mean)2

= 1, 000, 000[0.25 + 0.7510p60]− (mean)2

= 770, 811.1− (847.207)2

= 53, 051.39∴ s.d. of X = 230.33

Present value is v10(X) and so has s.d. 230.33v10 at 6% = £128.615

3.13. SOLUTIONS 59

3.8

(IA)50 =C50 + 2C51 + · · ·

D50= A 1

50:1 +(C51 + C52 + · · · ) + (C51 + 2C52 + · · · )

D50

= A 150:1 +

D51

D50(A51 + (IA)51)

Now vq50 = A 150:1 = 0.00558

∴ D51

D50= v − vq50 = 0.93782

∴ 4.99675 = 0.00558 + 0.93782[0.24905 + (IA)51]∴ (IA)51 = 5.07305

∴ 1, 000(IA)51 = 5, 073


Chapter 4

ANNUITIES

4.1 Annuities payable continuously

Let (x) be entitled to £1 p.a. for life, payable “continuously”, i.e. the rate of payment at time tyears is £1 p.a. if (x) is then alive. The present value of this benefit, expressed as a random variable,is

g(T ) = aT for T > 0 (4.1.1)

This variable has mean ax, i.e.

ax = E(aT ) =∫ ∞

0

at tpxµx+t dt (4.1.2)

An alternative definition is:

ax =∫ ∞

0

vt

︸︷︷︸interest (or

discounting) factor

tpx

︸︷︷︸probabilityof survival

of (x) for t years

dt (4.1.3)

To show that these definitions are equivalent, we may use integration by parts, as follows:

∫ ∞

0

at

︸︷︷︸

at =∫ t

0

vr dr

Sod

dt(at ) = vt

tpxµx+t

︸︷︷︸

d

dt(−tpx) =

d

dt(tqx)

= tpxµx+t

dt

61

62 CHAPTER 4. ANNUITIES

∫ ∞

0

at tpxµx+tdt =[at (−tpx)

]∞0

+∫ ∞

0

vttpx dt

=∫ ∞

0

vttpx dt

Note the following conversion relationship (i.e. a formula connecting assurance and annuityvalues):

Ax = 1− δax (4.1.4)

which follows from:

ax =∫ ∞

0

at tpxµx+t dt

=∫ ∞

0

(1− vt

δ

)tpxµx+t dt

=1δ(1− Ax).

We may also argue as follows. By the Mathematics of Finance,

1︸︷︷︸Capitalinvested

= δaT

︸︷︷︸p.v.

of interest

+ vT

︸︷︷︸p.v. of

return of capital

(for all T > 0)

Take expected values of each side, which gives

1 = δax + Ax

The variance of the present value of an annuity. Since

aT =1− vT

δ

we have

Var(aT ) =1δ2

Var(vT )

=A∗x −

(A2

x

)

δ2(4.1.5)

where ∗ indicates “at the rate of interest 2i + i2 p.a.”; the corresponding force of interest is δ∗ = 2δ,so some writers use 2Ax rather than A∗x.

Example 4.1.1. Express Var(aT ) in terms of ax and a∗x.

Solution.A∗x = 1− (2δ)a∗x since force of interest is 2δ

so, by formula (4.1.5),

Var(aT ) =1− 2δ(a∗x)− (1− δax)2

δ2

=1− 2δ(a∗x)− 1− δ2(ax)2 + 2δax

δ2

=1δ

[2(ax − a∗x)− δ(ax)2

]

4.2. ANNUITIES PAYABLE ANNUALLY 63

Commutation Functions. Define

Dx = vxlx (as stated in Chapter 3) (4.1.6)

Dx =∫ 1

0

vx+tlx+t dt (4.1.7)

Nx =∞∑

t=0

Dx+t (4.1.8)

Example 4.1.2. Show that

ax =Nx

Dx(4.1.9)

Solution.

Nx =∞∑

t=0

Dx+t

but (on change of variable) Dx+t =∫ t+1

t

vx+rlx+r dr for t = 0, 1, 2, . . ., so

Nx =∫ ∞

0

vx+rlx+r dr

from which we obtainNx

Dx=

∫ ∞

0

vttpx dt = ax.

4.2 Annuities payable annually

Now consider the cases when an annuity of £1 p.a. is payable (i) annually in advance, (ii) annuallyin arrear. (The former is called an annuity-due.)In case (i), the p.v. of the benefit is

g(K) = aK+1 (4.2.1)

In case (ii), it isg(K) = aK = aK+1 − 1 (4.2.2)

where we take a0 = 0. The means are denoted by ax and ax respectively. Note that

ax = 1 + ax VITAL! (4.2.3)

(since the annuitant gets extra immediate payment of 1 in case (i)).Now

ax =E(aK+1

)

=∞∑

k=0

ak+1 k|qx (4.2.4)

But

aK+1 =1− vK+1

d(K = 0, 1, 2 . . . )


so

ax =E

[1− vK+1

d

]=

1− E(vK+1)d

=1−Ax

d(4.2.5)

We thus have the conversion relationship

Ax = 1− dax (4.2.6)

Important formulae for ax and ax. We note the following results:

ax =∞∑

t=1

vttpx (4.2.7)

ax =∞∑

t=0

vttpx (4.2.8)

It is sufficient to prove (4.2.8) since ax = ax − 1 then gives (4.2.7).

Two Proofs(1) Regard the annuity as sum of pure endowments due at times 0,1,2,. . .. Thus

ax = 0Ex + 1Ex + 2Ex . . .

=∞∑

t=0

vttpx

(2) By the definition,

ax =∞∑

k=0

ak+1 k|qx

= a1

(dx

lx

)+ a2

(dx+1

lx

)+ a3

(dx+2

lx

)+ . . .

=1lx

{dx + (1 + v)dx+1 + (1 + v + v2)dx+2 + . . .

}

=

{(dx + dx+1 + dx+2 + . . . ) + v(dx+1 + dx+2 + . . . ) + v2(dx+2 + . . . ) + . . .

}

lx

=lx + vlx+1 + v2lx+2 + . . .

lx

=∞∑

t=0

vttpx,

as required.

Evaluation of ax by Commutation Functions Define

Nx =∞∑

t=0

Dx+t

4.3. TEMPORARY ANNUITIES 65

We obtain

ax =vxlx + vx+1lx+1 + . . .

vxlx

=Dx + Dx+1 + . . .

Dx

=Nx

Dx

Remark If we require ax, use ax = ax − 1 or

ax =Nx+1

Dx(since Nx+1 = Dx+1 + Dx+2 + . . . )

The variance of aK+1 .We use the formula

Var(aK+1

)= Var

(1− vK+1

d

)

=1d2

Var(vK+1

)

=A∗x − (Ax)2

d2(4.2.9)

where ∗ indicates the rate of interest i∗ = 2i + i2 p.a.Note For an annuity payable in arrear, we use the result that

Var (aK ) = Var(aK+1 − 1

)

= Var(aK+1

)(4.2.10)

4.3 Temporary annuities

Let us first consider the “continuous payments” case, and suppose that (x) is entitled to £1 p.a.payable continuously for at most n years. (that is, payments cease when x dies or after n years,whichever is earlier.) The p.v. is

g(T ) =

{aT if T < n

an if T ≥ n= a

min{T,n} (4.3.1)

The mean of g(T ) is

ax:n =∫ n

0

at tpxµx+t dt + an

∫ ∞

ntpxµx+t dt

=∫ n

0

vttpx dt (4.3.2)

by integration by parts. (Check this!)

Note. THERE ARE NO SUCH THINGS AS a1x:n

and ax:

1n

!!

Commutation Functions.If n is an integer,

ax:n =

∫ n

0vx+tlx+t dt

vxlx=

Nx − Nx+n

Dx(4.3.3)


If payments are made annually in advance, and are limited to at most n payments, we obtain

ax:n = mean of

{aK+1 if K < n

an if K ≥ n= E

[amin{K+1,n}

](4.3.4)

=n−1∑

k=0

ak+1 k|qx + an npx

which may be shown to be equal to

1 + vpx + v22px + · · ·+ vn−1

n−1px

(i.e., the sum of the values of n pure endowments each for £1)

=Dx + Dx+1 + · · ·+ Dx+n−1

Dx

Hence

ax:n =Nx −Nx+n

Dx(4.3.5)

If the payments are made annually in arrear, we obtain

ax:n = mean of

{aK if K < n

an if K ≥ n= E

[amin{K,n}

](4.3.6)

leading to

ax:n =Dx+1 + Dx+2 + · · ·+ Dx+n

Dx=

Nx+1 −Nx+n+1

Dx(4.3.7)

Note that

(a) ax:n = 1 + ax:n−1 (4.3.8)(b) ax:n and a[x]:n are tabulated for certain values of x+n in the A1967-70 section of “Formulae

and Tables.”

Example 4.3.1. By expressing Ax:n and ax:n as expectations of appropriate random variables, orotherwise, prove the conversion relationships

(i) Ax:n = 1− dax:n , and

(ii) Ax:n = 1− δax:n .

4.4. DEFERRED ANNUITIES 67

Proof.

(i) By Maths. in Finance, amin{K+1,n} =

1− vmin{K+1,n}

d.

Take expected values to obtain

ax:n =1−Ax:n

d

which gives the required result.

(ii) Take expected values in the equation

amin{T,n} =

1− vmin{T,n}

δ.

4.4 Deferred annuities

These are annuities which commence in m (say) years’ time, provided that the annuitant is thenactive. Thus

m|ax = M.P.V. of annuity of 1 p.a. to (x), payable con-tinuously, beginning in m years’ time

(4.4.1)

It is often best to evaluate m|ax by the formula

m|ax =(

Dx+m

Dx

)

︸︷︷︸pure

endowmentfactor

× ax+m

︸︷︷︸annuityfactor

at age x + m

(4.4.2)

Similarly,

m|ax =(

Dx+m

Dx

)ax+m (4.4.3)

and so on. Note thatax = ax:n + n|ax (4.4.4)

and

m|ax =Dx+m

Dx

(Nx+m

Dx+m

)=

Nx+m

Dx(4.4.5)

N.B. Pensions are (essentially) deferred annuities.

Select Tables The modifications needed for select tables are straightforward, and are illustrated inTable 4.4.1 below.


Type of Annuity Symbol In terms of tp[x]

In terms of CommutationFunctions

Immediate a[x]

∞∑t=1

vttp[x]

N[x]+1

D[x]

Annuity-due a[x]

∞∑t=0

vttp[x]

N[x]

D[x]

Temporary (n years) a[x]:n

n∑t=1

vttp[x]

N[x]+1 −Nx+n+1

D[x](n ≥ 1)

Temporary annuity-due(n years’ payments) a[x]:n

n−1∑t=0

vttp[x]

N[x] −Nx+n

D[x](n ≥ 2)

Deferred (m years) m|a[x]

∞∑t=m+1

vttp[x]

Nx+m+1

D[x](m ≥ 1)

Deferred annuity-due(m years) m|a[x]

∞∑t=m

vttp[x]

Nx+m

D[x](m ≥ 2)

Table 4.4.1: Expression for Values of Single-Life Curtate Annuities. Life aged x.Select Mortality Table (select period 2 years)

4.5 Annuities payable m times per annum

We require the following formula from numerical analysis:The Euler-Maclaurin formula:

∫ ∞

0

f(t) dt +∞∑

t=0

f(t)− 12f(0) +

112

f ′(0) (4.5.1)

Woolhouse’s formula may be then deduced:

1m

∞∑t=0

f

(t

m

)+

∞∑t=0

f(t)−(

m− 12m

)f(0) +

(m2 − 112m2

)f ′(0) (4.5.2)

(We have assumed that f(t) → 0 and f ′(t) → 0 as t →∞.)

Example 4.5.1. Use the Euler-Maclaurin formula to deduce an approximate formula for ax in termsof ax

4.5. ANNUITIES PAYABLE M TIMES PER ANNUM 69

Solution. Let f(t) = vttpx = exp

(− ∫ t

0(δ + µx+r) dr

). Hence

f ′(t) = −(δ + µx+t) exp(− ∫ t

0(δ + µx+r) dr

). ∴ f(0) = 1 and f ′(0) = −(δ + µx). By E.-M.

ax + ax − 12− 1

12(µx + δ) or ax +

12− 1

12(µx + δ) (4.5.3)

In practice the final term is usually ignored, giving

ax + ax − 12

= ax +12

(4.5.4)

Example 4.5.2. Find an approximate formula for◦ex in terms of ex.

Solution. Set i = 0 in example 4.5.1. This gives

◦ex + ex +

12− 1

12µx (4.5.5)

The final term is usually omitted, giving

◦ex + ex +

12

(4.5.6)

The symbols a(m)x and a

(m)x refer to the expected present values of an annuity of 1 per annum payable

monthly in advance and monthly in arrear respectively. Thus

a(m)x =

∞∑t=0

1m

vt

mtm px

(4.5.7)

and

a(m)x =

∞∑t=0

1m

vt

mtm px

(4.5.8)

Example 4.5.3. Find an approximate formula for a(m)x in terms of ax.

Solution. Apply Woolhouse’s formula to f(t) = vttpx. This gives

a(m)x + ax −

(m− 12m

)−

(m2 − 112m2

)(µx + δ) (4.5.9)

In practice one usually uses

a(m)x + ax −

(m− 12m

)(4.5.10)

Note.

a(m)x = a(m)

x − 1m

= ax −(

m− 12m

)− 1

m(ignoring the final term)

= ax +(

m− 12m

)(4.5.11)


Temporary mthly annuities

Define

a(m)x:n = M.P.V. of £1 p.a. payable mthly in advance to (x)

for at most n years

= a(m)x − n|a(m)

x (4.5.12)

= a(m)x − Dx+n

Dxa(m)x+n

+[ax −

(m− 12m

)]−

(Dx+n

Dx

)[ax+n −

(m− 12m

)]

= ax:n −(

m− 12m

)×

(1− Dx+n

Dx

)

︸︷︷︸CARE!

DO NOT OMITTHIS TERM

(4.5.13)

We also have

a(m)x:n = ax:n +

(m− 12m

) (1− Dx+n

Dx

)(4.5.14)

4.6 Complete annuities (or “annuities with final proportion”)

Suppose that (x) is to receive 1 p.a. monthly in arrear with a final payment immediately on death at

time r since the last full payment, i.e. on death at timet

m+ r, where t = 0, 1, 2, . . . and 0 ≤ r < 1

m

This final payment varies from 0 to 1m and hence is on average about 1

2m . This annuity is called“complete” or “with final proportion”, and we define

◦a(m)

x = M.P.V. of complete annuity

+ a(m)x +

12m

Ax (4.6.1)

When m = 1 we may write◦a(m)

x =◦ax, giving

◦ax + ax +

12Ax (4.6.2)

More accurate formulae for◦ax. If we have U.D. of D in each of the age-ranges x to x+1, x+1

to x + 2, etc., we have the exact formulae

◦ax= ax +

(i− δ

δ2

)Ax = ax +

(i− δ

iδ

)Ax (4.6.3)

Proof. The final payment is shown below: Hence (using the formula for a varying assurance)

4.7. VARYING ANNUITIES 71final

payment(on death of (x) )

·········

·········

·········

·········

6

-···

......

1 2 3 40

1

time (years)

◦ax= ax +

∞∑t=0

vttpx

︸︷︷︸pure endowment

factor to age x + t

∫ 1

0

rvrrpx+tµx+t+r

︸︷︷︸value at time tof death benefit

in year t + 1

dr

By U.D. of D.,

◦ax = ax +

∞∑t=0

vttpxqx+t

∫ 1

0

rvr dr

︸︷︷︸

= (I a)1 = a1 −vδ

= ax +∞∑

t=1

vt+1t|qx(1 + i)

[a1 − v

δ

]

= ax + Ax

(s1 − 1

δ

)= ax + Ax

((1 + i)− 1

δ− 1

)/δ

= ax + Ax

(i− δ

δ2

)

The second part of (4.6.3) follows from Ax = iδ Ax (which holds by U.D.D.)

4.7 Varying annuities

Suppose, firstly, that payments are made continuously so long as (x) survives at the rate b(t) p.a. attime t. The present value of these payments (as a random variable depending on the future lifetimeT of (x)) is (by the Mathematics of Finance)

g(T ) =∫ T

0

vtb(t) dt (4.7.1)

The mean present value is thus

E[g(T )] =∫ ∞

0

g(t)tpxµx+t dt (4.7.2)


Using the following important formula (obtained from integration by parts):

∫ ∞

0

u(t)(∫ t

0

v(r) dr

)dt =

∫ ∞

0

v(t)(∫ ∞

t

u(r) dr

)dt (∗)

we obtain the alternative expression

E[g(T )] =∫ ∞

0

vt

︸︷︷︸interestfactor

b(t)︸︷︷︸rate of

paymentat time t

tpx

︸︷︷︸probabilityof survival

dt (4.7.3)

To prove this, letv(t) = vtb(t)

andu(t) = tpxµx+t

so that ∫ ∞

t

u(r) dr = tpx then use formula (∗)

Example 4.7.1. Define (I a)x = M.P.V. of an increasing annuity to (x) in whichthe rate of payment at time t is t.

Find an expression for (I a)x in terms of an integral.

Solution. Let b(t) = t, t ≥ 0. This gives g(T ) = (I a)T , and we have

(I a)x = M.P.V. of annuity to (x) with payment at rate t p.a. at time t

=∫ ∞

0

tvttpx dt (4.7.4)

An approximation. Let f(t) = tvttpx, (t ≥ 0). By Euler-Maclaurin,

(I a)x +∞∑

t=0

tvttpx − 1

2f(0) +

112

f ′(0)

But f(0) = 0 and f ′(0) = [vttpx + t d

dt (vttpx)]t=0 = 1, so

(I a)x +∞∑

t=1

tvttpx +

112

= (Ia)x +112

(see definition of (Ia)x below) (4.7.5)

Now suppose that

b(t) = [t] + 1 =

1 in year 12 in year 2. . .

4.7. VARYING ANNUITIES 73

The M.P.V. of the corresponding annuity is written (Ia)x. Thus

(Ia)x =∫ ∞

0

([t] + 1)vttpx dt

=(∫ ∞

0

+∫ ∞

1

+∫ ∞

2

)vt

tpx dt

= ax + 1|ax + 2|ax + . . .

=Nx + Nx+1 + . . .

Dx

=Sx

Dx(4.7.6)

where

Sx =∞∑

t=0

Nx+t (4.7.7)

Now suppose that benefits are paid annually so long as (x) survives, the benefit at time t years beingb(t) (t = 0, 1, 2, . . . ). The present value of the benefits (regarded as a random variable) is

g(K) =K∑

t=0

vtb(t) (4.7.8)

since the last payment is made at time K. (If b(0) > 0, this is a variable annuity-due). The M.P.V.of the varying annuity is thus

E[g(K)] =∞∑

k=0

g(k)k|qx

=1lx{g(0)dx + g(1)dx+1 + . . . }

=1lx{g(0)[dx + dx+1 + . . . ]

+ [g(1)− g(0)][dx+1 + dx+2 + . . . ]+[g(2)− g(1)][dx+2 + dx+3 + . . . ] + . . . }

=1lx

{b(0)lx + vb(1)lx+1 + v2b(2)lx+2 + . . .

}

E[g(K)] =∞∑

t=0

b(t)︸︷︷︸benefit

at time t

vt

︸︷︷︸interestfactor

tpx

︸︷︷︸survivalfactor

(4.7.9)

This formula forms the basis of “spreadsheet”calculations for pension schemes, etc.

[This may also be found by summing the pure endowments]


ApplicationsLet b(t) = t + 1 (t = 0, 1, 2, . . . ). The M.P.V. is defined as (Ia)x, so we have

(Ia)x =∞∑

t=0

(t + 1)vttpx

=

[ ∞∑t=0

(t + 1)Dx+t

]/Dx

=Dx + 2Dx+1 + . . .

Dx

=(Dx + Dx+1 + . . . ) + (Dx+1 + Dx+2 + . . . ) + . . .

Dx

=Nx + Nx+1 + . . .

Dx

=Sx

Dx(4.7.10)

where

Sx =∞∑

t=0

Nx+t (4.7.11)

Similarly, when b(t) = t (t = 1, 2, . . . ) we get the M.P.V.

(Ia)x =∞∑

t=1

tvttpx =

∞∑t=1

tDx+t

Dx

=Sx+1

Dx(4.7.12)


(Ia)x = (Ia)x − ax

Solution.

(Ia)x =∞∑

t=0

tvttpx

=∞∑

t=0

(t + 1− 1)vttpx

=∞∑

t=0

(t + 1)vttpx −

∞∑t=0

vttpx

=(Ia)x − ax

Temporary increasing annuities

4.7. VARYING ANNUITIES 75

Define

(Ia)x:n = M.P.V. of payments of t + 1 at time t(t = 0, 1, 2, . . . , n − 1), provided that (x) isthen alive

=n−1∑t=0

(t + 1)vttpx =

[n−1∑t=0

(t + 1)Dx+t

]/Dx

=Dx + 2Dx+1 + · · ·+ (n− 1)Dx+n−2 + nDx+n−1

Dx

=(Dx + 2Dx+1 + . . . )

Dx− (Dx+n + 2Dx+n+1 + . . . )

Dx

− (nDx+n + nDx+n+1 + . . . )Dx

=Sx − Sx+n − nNx+n

Dx(4.7.13)

Similarly,

(Ia)x:n =Dx+1 + 2Dx+2 + · · ·+ nDx+n

Dx

=(Dx+1 + 2Dx+2 + . . . )

Dx− (Dx+n+1 + 2Dx+n+2 + . . . )

Dx

− (nDx+n+1 + nDx+n+2 + . . . )Dx

=Sx+1 − Sx+n+1 − nNx+n+1

Dx

Also,

(I a

)x:n

=∫ n

0

tvttpx dt

(Ia)x:n =Sx − Sx+n − nNx+n

Dx(4.7.14)

Note that (for example)

m| (Ia)x:n =(

Dx+m

Dx

)(Ia)x+m:n

which enables us to express m| (Ia)x:n in terms of commutation functions, viz

m| (Ia)x:n =Sx+m − Sx+m+n − nNx+m+n

Dx

Conversion relationships for increasing annuities and assurances

ax = δ(I a

)x

+(IA

)x

(1)

ax = d (Ia)x + (IA)x (2)

ax = δ (Ia)x +(IA

)x

(3)

ProofTo show (1), take expected values on each side of the maths of finance formula

aT = δ(I a

)T

+ TvT


To show (2), take expected values on each side of

aK+1 = d (Ia)K+1 + (K + 1)vK+1

To show (3), we need to use the formula

(Ia)t =a[t]+1

− ([t] + 1)vt

δfor all t ≥ 0 (∗∗)

where

(Ia)t = p.v. of an annuity-certain payable continuously at rate 1 p.a. in year1, 2 p.a. in year 2, . . . , ceasing at time t exactly.

=∫ t

0

vr ([r] + 1) dr

(∗∗) holds for t = 0, 1, 2, 3, . . . by McCutcheon and Scott, formula 3.6.6. It may be proved for generalt ≥ 0 by letting t = n + r ( n integer, 0 ≤ r < 1) and observing that

(Ia)t = (Ia)n +∫ n+r

n

vs(n + 1) ds

=an − nvn

δ+ (n + 1)vn

(1− vr

δ

)

=an+1 − (n + 1)vt

δ

Note. There are also conversion relationships for temporary increasing annuities, e.g.

ax:n = δ(I a)x:n + (IA)x:n

4.8. EXERCISES 77

Exercises

4.1 The following is an extract from a life table with a select period of 1 year.

Age x l[x] lx+1 age x + 155 90,636 90,032 5656 89,739 89,132 57

88,151 5887,094 5985,874 6084,586 61

Evaluate a56:5 and a[56]:5 at 5 % per annum interest.

4.2 Given that ax = 20, ax:n = 18 and ax+n = 8, find the values of nEx and ax:n .

4.3 (i) Write down an expression for ax in terms of v, px, and ax+1.(ii) On a certain select mortality table the select period is one year. Express a[x] in terms of

v, p[x] and ax+1. Given that q[x] = .6qx and that at 4 14% p.a. interest a45 = 15.719 and

a46 = 15.509, find the value of a[45] (at the same rate of interest).

4.4 Using the A1967-70 table with 4% p.a. interest find the values ofa[40]:30 , a[39]+1:30 , a40:30 , (Ia)40, 5|a[40]:25 , 5|(Ia)40:25

4.5 (i) Find the present value of a deferred annuity of £1000 p.a. to a man aged 40. Paymentscommence on his 60th birthday, if he is then alive, and continue thereafter annually forlife.Basis: A1967-70 ultimate mortality, 4 % p.a. interest.

(ii) As above, but select mortality (at entry.)(iii) As above, but A1967-70 ultimate mortality to exact age 60 and a(55) males ultimate

mortality above exact age 60.

4.6 According to a certain mortality table, which has a select period of 1 year and is such thatq[x] = 0.6qx for each x,

a70 at 10% interest = 5.641, and,a71 at 10% interest = 5.449.

Find a[70] at 10% interest.


Solutions

4.1

a56:5 =l56 + vl57 + · · ·+ v4l60

l56= 4.451

a[56]:5 =l[56] + vl57 + · · ·+ v4l60

l[56]= 4.463

4.2ax = ax:n + nExax+n =⇒ nEx =

14

ax:n = ax:n − nEx + 1 = 18.75

4.3 (i) ax = vpx(1 + ax+1)(ii) Note that with a one-year select period

a[45] = vp[45]a46

= (1.0425)−116.509p[45]. (a)

We must determine p[45] from the given data. Nowa45 = vp45a46

i.e.

p45 =(1 + i)a45

a46= 1.0425× 15.719

16.509= .9926136

Henceq45 = 0.0073864

Thenq[45] = 0.6q45 = 0.0044318

Hencep[45] = 1− q[45] = 0.9955682

From (a) we then geta[45] = 15.7658

4.4a[40]:30 = 16.960 directly tabulated

a[39]+1:30 = (N[39]+1 −N70)/D[39]+1 = 16.953

a40:30 = 16.949 directly tabulated

(Ia)40 = S40/D40 = 276.468

5|a[40]:25 = (N45 −N70)/D[40] = 12.342

5|(Ia)40:25 = (S46 − S71 − 25N71)/D40 = 122.337

4.9. SOLUTIONS 79

4.5 (i) 1000.N60/D40 = 5130.08, say £5130(ii) 1000.N60/D[40] = 5133.68, say £5134

(iii) 1000.(

D60D40

).a∗60, where the factor D60

D40is evaluated on A1967-70 mortality and the factor

a∗60 is evaluated on a(55) males mortality. The value is thus 1000 × .40873 × 12.625 =5160.22, say £5160.

4.6 a70 = v(1 − q70)a71, so 1 − q70 = 0.96218. Hence q[70] = 0.6q70 = 0.022918, and q[70] =v(1− q[70])a71 = 5.730


Chapter 5

PREMIUMS

5.1 Principles of premium calculations

Premiums calculated without an allowance for expenses are called net premiums, whilst premiumswhich are actually charged are called office or gross premiums. Office premiums are usually cal-culated with an explicit allowance for expenses, and in some cases for a profit to the office. A lifeassurance policy may be issued with (a) a single premium payable at the date of issue, or (b) reg-ular premiums payable in advance and usually (but not always) of level amount. The frequency ofregular payments may be, for example, yearly or monthly, and the maximum number of premiumspayable may be limited to (for example) 20. Premiums should, of course, cease on the death of theassured life, when the policy matures, or when there is no longer any possibility of future benefits:for example, on the expiry of the term of a temporary (term) assurance policy.

Premiums are usually calculated by the equivalence principle, which may be stated as follows:

E(Z) = 0 (5.1.1)

whereZ = present value of profit to the life office on the contract

In some cases an explicit loading for profit is included in the calculation; that is, we replace equation(5.1.1) by

E(Z) = B (5.1.2)

where B = M.P.V. of profit on the contract. In the absence of information to the contrary, however,we shall assume that the equivalence principle (5.1.1) is to be used, although in some cases the useof “conservative” assumptions regarding mortality, interest and expenses means that the office hasan implicit margin of expected profit. The rate of interest is usually taken to be fixed (not random).

Example 5.1.1. Consider a whole life assurance of 1 payable immediately on the death of (x),and suppose that premiums are payable continuously at rate P p.a. until the death of (x). Give aformula for P in terms of annuity functions, assuming that the equivalence principle applies.

Solution. Let

Z = g(T ) = P.V. of profit to office on this contract

= P aT − vT

81

82 CHAPTER 5. PREMIUMS

Since we require that E{g(T )} = 0, we have

E{P aT − vT } = 0

∴ P.E (aT ) = E(vT

)

∴ P ax = Ax

∴ P =Ax

ax=

1ax− δ using the conversion relationship Ax = 1− δax

The equations (5.1.1) and (5.1.2) may be expressed as equations of payments, or equations ofvalue, i.e.

M.P.V. of premiums =M.P.V. of benefits+ M.P.V. of expenses (if any)+ M.P.V. of profit to the office (if any)

(5.1.3)

In example 5.1.1, the equation of value is

P ax = Ax

5.2 Notation for premiums

The International Actuarial Notation (see Formulae and Tables for Actuarial Examinations) shouldbe used, at least for straightforward policies. If the policy is complicated it is best to just use thesymbol P (or P ′, P ′′) for the premium (single or regular). The general symbols P, P ′, P ′′ may beused for any sum assured, but the standard symbols Px, P (Ax) , etc. , refer to a sum assured of£1.

Some of the more important rules of the International Notation are the following:

1. The symbols P ( ), or P ( ), or P (m)( ) indicate the level net annual premium for the benefitindicated in the brackets. Premiums are assumed to continue for as long as the contract canprovide benefits, i.e. n years for a n-year contract, provided of course that the life assured isstill alive. P, P , P (m) indicate that payments are made annually in advance, continuouslyand mthly in advance respectively.

2. If premiums are limited to, at most t years’ payments (where t < the term of contract, i.e. theterm of the benefits), we write tP ( ), tP ( ), tP

(m)( ).e.g.

10P (Ax:20 ) = net annual premium for 20-year EA with S.A. =1(payable at end of year of death or on survivalfor 20 years.), limited to at most 10 payments(i.e. payments cease on death or after 10payments are made, whichever occurs first)

=Ax:20

ax:10

3. When benefits are payable at end of year of death we may (optionally) shorten the notationas follows:

(e.g.) P (Ax) = Px

5.3. THE VARIANCE OF THE PRESENT VALUE OF THE PROFIT ON A POLICY. 83

Table 5.2.1 relates to a mortality table with select period 2 years, for example A1967-70, and refersto policies in which premiums are payable annually in advance.

5.3 The variance of the present value of the profit on a policy.

Consider again the policy of Example 5.1.1. We have

g(T ) = p.v. of profit to office on contract

= P aT − vT

(We need not assume the equivalence principle holds: if it does, E[g(T )] = 0.) Note that

g(T ) = P

(1− vT

δ

)− vT

=P

δ−

(P + δ

δ

)vT

∴ Var[g(T )] =(

P + δ

δ

)2

Var(vT )

=(

P + δ

δ

)2 [A∗x

︸︷︷︸at rate2i + i2

− (Ax

)2] (5.3.1)

Note the idea of “collecting” all the terms involving vT before taking the variance: a similar techniquemay be used for endowment assurance policies (but not generally).

5.4 Premiums allowing for expenses

Most expenses may be classified as either:

(a) initial (incurred at the outset only) , or

(b) renewal (incurred on the payment of later premiums).

There may also be expenses of payment of benefits (especially for pensions and annuities) and theexpenses of maintaining records of policies with continuing benefits after premiums have ceased.Expenses may also be divided into:

(i) commission payments, and

(ii) other costs.

Some offices employ a policy fee system, whereby a fixed addition of (say) £15 is added to theannual premium; for example the office annual premium for a policy with a sum assured for £20,000may be quoted as “£10 per £1,000 plus a policy fee of £15”, giving £20× 10 + £15 = £215. Thissystem reflects the fact that certain administrative costs do not depend on the size of the benefit.

Example 5.4.1. Consider n-year endowment assurance without profits with sum assured £1 issuedto a select life aged x. Expenses are e per premium payment (including the first) plus I at issuedate (so the total initial expense is I + e ). Find formulae for the level office annual premium, P ′′.


Net

Annual

Prem

iumpayable

throughoutduration

ofcontract

Net

Annual

Prem

iumLim

itedto

tpaym

entsT

ypeof

Assurance

Symbol

Interm

sof

Aand

a

Interm

sof

comm

utationFunctions

(n≥

2)Sym

bolIn

terms

ofA

anda

Interm

sof

comm

utationFunctions

(n≥

2and

t≥2)

Tem

poraryP

1[x]:n

A1[x]:n

a[x

]:n

M[x

] −M

x+

n

N[x

] −N

x+

nt P

1[x]:n

A1[x]:n

a[x

]:t

M[x

] −M

x+

n

N[x

] −N

x+

t

Deferred

Tem

poraryP

(m |A

1[x]:n

)m |A

1[x]:n

a[x

]:m+

n

Mx+

m−

Mx+

m+

n

N[x

] −N

x+

m+

nt P

(m |A

1[x]:n

)m |A

1[x]:n

a[x

]:t

Mx+

m−

Mx+

m+

n

N[x

] −N

x+

t

Whole-life

P[x

]

A[x

]

a[x

]

M[x

]

N[x

]t P

[x]

A[x

]

a[x

]:t

M[x

]

N[x

] −N

x+

t

Deferred

Whole-life

P(m |A

[x] )

m |A[x

]

a[x

]

Mx+

m

N[x

]t P

(m |A

[x] )

m |A[x

]

a[x

]: t

Mx+

m

N[x

] −N

x+

t

Endow

ment

Assurance

P[x

]:n

A[x

]: n

a[x

]:n

M[x

] −M

x+

n+

Dx+

n

N[x

] −N

x+

nt P

[x]:n

A[x

]: n

a[x

]:t

M[x

] −M

x+

n+

Dx+

n

N[x

] −N

x+

t

Table

5.2.1:E

xpressionsfor

Annual

Prem

iums.

SelectM

ortalityTable

(selectperiod

2years)

Note

.W

hen

n<

2or

t<

2,th

esefo

rmula

esh

ould

be

suita

bly

modifi

ed.

5.5. PREMIUMS FOR WITH PROFITS POLICIES 85

Solution. The equation of value is

P ′′a[x]:n = A[x]:n + ea[x]:n + I (5.4.1)

where P ′′ = level annual premium. Suppose further that e = kP ′′ (i.e. renewal expenses are aproportion k of the office premium): we have

(1− k)P ′′a[x]:n = A[x]:n + I

∴ P ′′ =A[x]:n + I

(1− k)a[x]:n

=1

(1− k)

[P[x]:n +

I

a[x]:n

](5.4.2)

Sometimes I is a proportion of the first premium, cP ′′, say. This leads to

(1− k)P ′′a[x]:n = A[x]:n + cP ′′

∴ P ′′ =A[x]:n

(1− k)a[x]:n − c

(For example, k = 2 12% and c = 47 1

2% if expenses are 50% of the first premium and 2 12% of all

subsequent premiums .)

5.5 Premiums for with profits policies

In early days of life assurance, premiums for policies with the right to participate in the profits of thelife office were calculated on conservative assumptions but without an explicit allowance for bonusdeclarations. In modern conditions there is often an explicit allowance for possible future bonusrates for with profits (or “participating”) policies, as illustrated in the next example.

Example 5.5.1. A with-profits whole-life assurance is about to be issued to a man aged 45. Thebasic sum assured is £12, 000. The office assumes that at the start of each policy year there will bebonus additions at the rate of 1% per annum compound. The basic sum assured and bonuses arepayable immediately on the man’s death. The policy has half-yearly premiums, payable for at most20 years. Calculate the half-yearly premium on the following basis: (Note: at 3.75% per annuminterest, the value of A[45] is 0.34587).

Solution. Let annual premium be P .

Value of benefits + 12, 000(1 + i)12 [1.01vq[45] + (1.01)2v2

1|q[45] + · · · ]= 12, 000(1 + i)

12 A∗[45] (see formula (3.10.11))

where ∗ is at ratei− 0.011 + 0.01

+ 3.96%

By linear interp., A∗[45] ; 0.32804.

∴ Value of benefits = £4033.69


Equation of value is

0.96P a (2)[45]:20

︸︷︷︸12.362

= 0.16P a (2)[45]:1

︸︷︷︸0.98753

+ 200 + 4033.69

∴ P = £361.58

Hence the half-yearly premium is £180.79

5.6 Return of premium problems

Consider, as an example, a policy issued to (x) with level annual premiums, P , providing:

(i) £1, 000 on survival for n years, and

(ii) a return of all premiums paid, at the end of the year of death, if (x) dies within n years.

Assume firstly that the premiums are returned without interest (sometimes described as “ReturnNo Interest”, abbreviated to R.N.I.). If there are no expenses, the equation of value for P is:

P ax:n = 1000Dx+n

Dx

︸︷︷︸survivalbenefit

+ P (IA)1x:n

︸︷︷︸return ofpremiumson death

(5.6.1)

This may be solved for P .Note. If the premiums are returned immediately on death, replace (IA)1x:n by (IA)1x:n .

Now suppose that the premiums are returned with compound interest at rate j p.a. In practice,such policies are often described as “ Return With Interest” (R.W.I.). If the death benefit is paidat the end of the year of death, the term P (IA)1x:n in equation (5.6.1) must be replaced by

n−1∑

k=0

vk+1k|qx

{P sk+1 j

}

︸︷︷︸death benefitin year k + 1

= P

n−1∑

k=0

[(1 + j)k+1 − 1

dj

]vk+1

k|qx

=P

[j/(1 + j)]. {A∗1

x:n −A1x:n } (5.6.2)

where ∗ indicates the rate of interest i−j1+j

Notes. 1. If the death benefit is payable immediately on death, formula (5.6.2) should be multi-plied by (1 + i)

12 /(1 + j)

12 = (1 + i∗)

12 .

2. If i (the rate of interest used to discount premiums and benefits in the equation ofvalue) equals j, some of these problems may be simplified: if there are no expenses,mortality (within n years) may be ignored giving the compound interest equation

P an = 1000vn

5.7. ANNUITIES WITH GUARANTEES 87

This result may best be proved by reference to reserves, which we shall discuss later.

Example 5.6.1. A life office sells policies each providing a cash sum at age 65. Premiums of £1, 000are payable annually in advance during the deferred period. On the death of the policyholder duringthe period of deferment, the premiums paid are returned immediately without interest. In respectof a life now aged 45, find the cash sum at age 65, given that the office uses the following basis:

A1967-70 select

4% p.a. interest

expenses are ignored

Solution. Let C be available at age 65. The equation of value is

1000a[45]:20 = CD65

D[45]+ 1000(IA) 1

[45]:20

∴ C =1000

[(N[45] −N65)− (1.02)(R[45] −R65 − 20M65)

]

D65

=1000[76, 722.7− 1.02× 7, 407.24]

2144.1713= £32, 258

5.7 Annuities with guarantees

Annuities are sometimes sold with the provision that payments will certainly continue until theirtotal equals the purchase price, B say (or possibly some proportion of this purchase price.)

Let us consider an annuity of £1 p.a., payable continuously with this guarantee to a life aged xat the issue date. Ignoring expenses, the equation of value for B is

B = aB + B |ax (5.7.1)

Theorem 5.7.1. Equation (5.7.1) has a unique solution

Proof. Let

f(B) = aB + B |ax −B (B ≥ 0)

=∫ B

0

vtdt +∫ ∞

B

vttpxdt−B

Hence

f ′(B) = vB − vBtpx − 1

= vBBqx − 1

< 0 for all B > 0

Now f(0) = ax > 0 and, as B → ∞, f(B) → −∞ (as ax → 1δ and B |ax → 0). It follows that

f(B) = 0 has a unique solution


Note. The purchase price of an annuity with this guarantee may be considerably larger than foran ordinary annuity.

If the annuity instalments are paid annually in arrear we must solve the equation

B = an + n|ax (5.7.2)

subject to the condition n−1 < B ≤ n. (Here n is the guarantee period, which must be an integer.)This is solved by trial and error.

Another type of guaranteed annuity is that in which the balance (if any) of the purchase price(or a proportion of it) over the total annuity payments received is paid immediately on the deathof the annuitant. For example, if the guarantee consists of a payment of the balance of 85% of thepurchase price over the total annuity instalments received, equation (5.7.1) would be replaced by

B = aB +∫ n

0

[0.85B − t]vttpxµx+tdt (5.7.3)

where n = 0.85B (the time when the annuity payments equal 85% of the purchase price).

Example 5.7.1. A man aged 65 buys a guaranteed annuity payable continuously for a purchaseprice of £32, 258. The annuity payments are guaranteed to continue until the total payments reach£20, 000. The office issuing the contract uses the following basis:

A1967-70 ultimate;

4% p.a. interest;

expenses are ignored.

Let n be the guarantee period. Give an equation for n, and prove that this equation has a uniquesolution, which lies between 6 and 7 years.

Solution. Let Z be the annual rate of payment of the annuity. We have the equation

32, 258 = Z(an + n|a65) (1)

subject to nZ = 20, 000 (2)Replace Z by 20, 000/n in (1) to obtain

an + n|a65 =32, 25820, 000

n = 1.6129n

i.e. solve f(n) = an + n|a65 − 1.6129n = 0

Now f(0) = a65 > 0, and, as n →∞, f(n) → −∞.

f ′(n) =d

dn

(∫ n

0

vtdt +∫ ∞

n

vttp65dt− 1.6129n

)

= vn − vnnp65 − 1.6129

= vnnp65 − 1.6129 < 0, since, for n > 0, vn

np65 < 1

∴ f(n) decreases, so f(n) = 0 has a unique solution.

5.8. FAMILY INCOME BENEFITS 89

Try n = 7

f(n) = 6.12136 +1, 289.75672, 144.171

× 7.772− 11.2903

< 0

Try n = 6

f(n) = 5.34626 +1, 401.60932, 144.171

× 8.112− 9.6774

> 0

Hence 6 < n < 7

5.8 Family income benefits

A family income benefit of term n years is a series of instalments payable from the date of death ofthe assured life, if he dies within n years, for the balance of the term. It may be considered as adecreasing term assurance in which the benefit on death is an annuity-certain for the balance of theterm.

Case 1.The benefits are payable continuouslySuppose that F.I.B. payments are at rate £B per annum, and the life assured is aged x at the issuedate. The present value of the benefits is

B(an − aT ) if T < n0 if T ≥ n

(5.8.1)

The P.V. of the benefits may thus be written in the form

B[an − a

min{T,n}]

(5.8.2)

and hence their M.P.V. isB (an − ax:n ) (5.8.3)

Case 2.The benefits made mthly in arrear, beginning at the end of the 1/m year of death (mea-sured from the issue date.)Consider the combination of a F.I.B. and an mthly temporary annuity of term n years, payable inarrear. It is clear that their total present value is Ba

(m)n no matter when (x) dies. Hence the M.P.V.

of the family income benefit isB

(a(m)

n − a(m)x:n

)(5.8.4)

Case 3. As in case 2, but with payments beginning immediately on death.As payments are received on average 1

2m year earlier than in case 2, the M.P.V. is approximately

B(1 + i)1

2m

(a(m)

n − a(m)x:n

)(5.8.5)

Premiums. These are found by an equation of value in the usual way. There is, however, adanger that, if premiums are payable for the full n-year term, the policy may have a negative reserveat the later durations (see later discussion of reserves.) This means that the policyholder may beable to lapse the contract leaving the office with a loss, and possibly obtain the same benefits morecheaply by effecting a new policy. In practice, the effect of expenses and other factors may make thispossibility hardly profitable, especially if the F.I.B. is part of a more general assurance contract, asin Example 5.8.1 below. The possibility of lapse option may be avoided by making the premium -paying term shorter than the benefit term.


Example 5.8.1. Ten years ago a woman aged exactly 35 effected an assurance policy by level annualpremiums payable for a maximum of 25 years. The policy provided the following benefits:

(i) a whole life assurance benefit of £2, 000 payable at the end of the year of death,

(ii) a family income benefit of term 25 years, with payments of £300 per month, beginning im-mediately on death, if this occurs within 25 years. The final payment is made in the monthending 25 years after the issue date.

Calculate the annual premium on the basis given below:

A1967-70

6% interest

expenses are 3% of all premiums.

Solution. Let P be the annual premium.

0.97P a35:25 = 2000A25 + 3600(1 + i)124

(a(12)

25− a(12)

35:25

)

P =2000× 0.12187 + 3600(1.06)

124 (13.1312− 12.8527)

0.97× 13.282= £96.93

Mortgage protection policies are similar to those for family income benefits, although thedeath benefits are payable in one sum rather than in instalments over balance of the term. Thesimilarity arises from the fact that the loan outstanding after the tth payment has been made isequal to the value of the future loan instalments (see McCutcheon and Scott, “An Introduction tothe Mathematics of Finance”, Table 3.8.1).

5.9. EXERCISES 91

Exercises

5.1 A life aged 40 effects a 25-year without profits endowment assurance policy with a sum assuredof £50,000 (payable at the end of the year of death or on survival to the end of the term). Levelpremiums are payable annually in advance throughout the term of the policy or until earlierdeath of the life assured. Calculate the level premium, P , using the following premium basis

Mortality: A1967-70 Ultimate;Interest: 6% p.a.Expenses: none

5.2 An office issues a large number of 25-year without-profit endowment assurances on lives agedexactly 40. Level annual premiums are payable throughout the term, and the sum assured ofeach policy is £10,000, payable at the end of the year of death or on survival to end of theterm. The office’s premium basis is:

A1967-1970 ultimate;4% p.a. interest;expenses are 5% of each annual premium including the first, with additional initial ex-penses of 1% of the sum assured.

Calculate the annual premium for each policy.

5.3 A 5-year temporary assurance, issued to a woman aged 55, has a sum assured of £50,000in the first year, reducing by £10,000 each year. The sum assured is payable at the end of theyear of death. Level premiums, limited to at most 3 years’ payments, are payable annually inadvance. Calculate the annual premium.

Basis:A1967-1970 select mortality4% p.a. interestexpenses are 10% of all office premiums

5.4 A life office sells immediate annuities, using English Life Table No. 12 - Males, 4% p.a. interestwith no expenses as the premium basis. Assuming that the mortality of annuitants does followthis table, that investments will earn 4% per annum, and that expenses are negligible, find theprobability that the office will make a profit on the sale of an annuity payable continuously toa life aged 55.

5.5 (i) Let g(T ) be the present value of the profit to the life office, at the issue date, in respectof an n-year without profits endowment assurance to (x) with sum assured (payableimmediately on death if this occurs within n years) and premium P per annum, payablecontinuously for the term of the policy. Expenses are ignored in all calculations.(a) Write down an expression for g(T ).(b) Derive expressions for

(1) the mean, and(2) the variance of g(T ).

(c) For what value of P is the mean of g(T ) equal to zero?(ii) An office issues a block of 400 without profits endowment assurances, each for a term of

25 years, to lives aged exactly 35. The sum assured under each policy is £10, 000 and thepremium is £260 per annum, payable continuously during the term. The sum assured ispayable immediately on death, if death occurs within the term of the policy.


Assuming that the office will earn 4% interest per annum, that the future lifetime of thelives may be described statistically in terms of the A1967-70 ultimate table, and thatexpenses may be ignored, find(a) the mean present value of the profit to the office on the block of policies, and(b) the standard deviation of the present value of this profit.

[A35:25 = 0.15646 on A1967-70 ultimate 8.16%

]

5.6 A life office issued a certain policy to a life aged 40. The benefits under this contract are asfollows:

On death before age 60: an immediate lump sum of £1, 000

On survival to age 60: an annuity of £500 p.a., payable continuously for theremaining lifetime of the policyholder.

Level annual premiums are payable continuously until age 60 or earlier death.Premiums are calculated according to the following basis:

Mortality: English Life Table No.12-Males

Interest: 4% p.a.

Expenses: Nil

Calculate the annual premium.

5.10. SOLUTIONS 93

Solutions

5.1 Equation of value is:M.P.V. of premiums = M.P.V. of benefits

P a40:25 = 50000A40:25 at A67-70 ultimate, 6% p.a.interest=⇒ P × 13.081 = 50000× 0.25955 (see page 67 of tables)

=⇒ P = £992.09

5.2 Let the annual premium be P ′. We have100 + 10000A40:25 = 0.95P ′a40:25

∴ P =10000A40:25 + 100

0.95a40:25

= £276.70

5.3 Let the annual premium be P . M.P.V. of assurance benefits is (from first principles)10000D[55]

{5C[55] + 4C[55]+1 + 3C57 + 2C58 + C59

}[= 60000A 1

[55]:5 − 10000(IA) 1[55]:5 ]

M.P.V. of premiums less expenses is

0.9P a[55]:3 = 0.9P[N[55] −N58]

D[55]

∴ P =10000{5C[55] + 4C[55]+1 + 3C57 + 2C58 + C59}

0.9[N[55] −N58]

=10000× 371.511

0.9× 10449= £395.06

5.4 Consider £1 p.a. of annuity. The purchase price is a55 and the office will make a profit if deathoccurs before time t, where

a55 = at at 4%

That is,1− vt

δ= a55, so t =

log [1− δa55]log v

i.e.

t =log A55

log v=

log A55

−δ= 16.99

The probability of making a profit is thus

tq55 = 1− l55+t

l55= 1− l71.99

l55= 0.434 (or 43.4%)

(by interpolation)

5.5 (i) (a) g(T ) =

{P aT − SvT if T < n

P an − Svn if T ≥ n


(b) (1) E[g(T )] = P ax:n − SAx:n

(2) Write g(T ) = P

[1− h(T )

δ

]− Sh(T )

where h(T ) =

{vT if T < n

vn if T ≥ n

=P

δ−

(P

δ+ S

)h(T )

∴ Var[g(T )] =(

P

δ+ S

)2

Var[h(T )]

=(

P

δ+ S

)2 [A∗x:n − (

Ax:n

)2]

where ∗ indicates the rate of interest 2i + i2 p.a.

(c) By (b)(1), E[g(T )] = 0 when P =SAx:n

ax:n

(ii) (a) First consider 1 policy.E[g(T )] = 260a35:25 − 10, 000A35:25

= 260×15.542︸︷︷︸using

a35:25 + a35 − 12

− D60D35

(a60 − 12 )

− 10, 000[ 1− δa35:25

︸︷︷︸0.39043

]

= 136.60

∴ For 400 policies, M.P.V. of profit = £54, 640(b) For 1 policy, variance of p.v. of profit is(

260δ

+ 10, 000)2 [

A 8.16%

35:25

︸︷︷︸0.15646

− ( A35:25

︸︷︷︸0.39043

)2]

= (1.662915× 104)2 × 0.004024415 = 1, 112, 866

∴ s.d. for 400 policies = 20× 1.662915× 104 ×√

0.004024415= £21, 098

5.6 Let P = annual premium.

P a40:20 = 1, 000A 140:20 + 500

N60

D40

∴ P =95.84 + 2, 040.81

13.261= £161.12

Chapter 6

RESERVES

6.1 What are reserves?

Reserves, or policy values, are sums of money held by financial institutions such as life officesand pension funds to cover the difference between the present value of future liabilities (includingexpenses) and the present value of future premiums or contributions. Alternatively, the reserve ofa contract may be considered as an accumulation of past premiums less expenses and the cost ofdeath claims (and other benefits). Reserves are required for various purposes, e.g.

(1) to pay surrender values (or transfer values in a pension fund);

(2) to work out the revised premium or sum assured if a policy is altered or converted to anothertype;

(3) for inclusion in statutory returns to the Department of Trade and Industry (or other supervisorybodies) for the purpose of demonstrating the office’s solvency;

(4) for internal office calculations to decide the bonus rates of with-profits policies, the distributionof profits to shareholders, etc.

The bases used to calculate “reserves” for each of these purposes may be different, and somepractical considerations are beyond our present scope.

6.2 Prospective reserves

Consider a life assurance policy issued t years ago to a life then aged x. Define the NET LIABILITYor PROSPECTIVE LOSS of the office in respect of this policy to be the random variable

L = present value of future benefits+ present value of future expenses− present value of future premiums (6.2.1)

The reserve or policy value of the contract (calculated prospectively, i.e. by reference to futurecashflows) is defined as

tV = E(L) (6.2.2)= M.P.V. of future benefits and expenses−M.P.V. of future premiums (6.2.3)

95

96 CHAPTER 6. RESERVES

If expenses may be ignored, we have

tV = M.P.V of future benefits−M.P.V of future premiums (6.2.4)

The mortality, interest and expense assumptions used to evaluate tV are known as the reservingbasis. This may or may not agree with the premium basis. If these bases agree (or are assumedto agree) and there are no expenses, we obtain net premium reserves, which we shall considerin the next section. By convention, the reserve tV is calculated just before receipt of any premiumthen due: the reserve just after payment of this premium is

tV + P ′′ − e (6.2.5)

where P ′′ is the premium paid at duration t years and e is the expense then incurred.

Are negative reserves allowable?

Certain formulae may give negative values of tV, at least for some policies and at early durations t.A negative reserve should not normally be held because the life office is thereby treating the contractas an asset: if the policy is discontinued there is no way in which the policyholder can be made topay money to the office! Similarly, the reserves stated in Statutory Returns should not be negative,although negative reserves may be permissible in certain internal office calculations.The general rule is therefore that, if a formula gives a negative value of tV, this should be replaced byzero. Policies should in general be designed so that negative reserves do not arise (cf. the discussionof family income benefit policies in Section 5.8).

6.3 Net premium reserves

These are reserves calculated without allowance for expenses, and on the assumption that the pre-mium and the reserve bases agree. (In some cases, the actual premium basis is different, and thepremiums valued are calculated on the reserve basis). A net premium reserve basis is unambiguouslyspecified by

(i) a mortality table, and

(ii) a rate of interest.

By formula (6.2.4),

tV = net premium reserve

= M.P.V. of future benefits - M.P.V. of future premiums (6.3.1)where the valuation assurance and annuity factors are calculated on the specified mortality andinterest basis. The premiums valued are also calculated on this basis.

Example 6.3.1. Consider a whole life policy with sum assured £1 without profits, payable imme-diately on the death of (x). The policy was issued t years ago by level annual premiums payablecontinuously throughout life. Find a formula for the net premium reserve tV (on a given mortalityand interest basis).

6.3. NET PREMIUM RESERVES 97

Solution. Let L be the net liability (a random variable). We have

L = vU − P aU

where U = future lifetime of (x + t), and P = P (Ax).∴ tV = E(L) = Ax+t − P (Ax)ax+t.

Notation. The International Notation for net premium reserves is similar to that for premiums,with the addition of “t” to indicate the duration. If premiums are limited to (say) h years of payment,the symbol h is placed above the duration t, as shown in Table 6.3.1. The general symbol tV maybe used for any sum assured and any reserve, whether net premium or not, but tVx, tV (Ax), etc.,refer to net premium reserves for a policy with a sum assured of £1. If select mortality tables areused, we write tV[x], etc.CONVENTIONS

(1) The symbols tVx, etc., refer to the reserves just before payment of the premium due at timet (if a premium is then payable.) The net premium reserve just after receipt of this premiumis of course tVx + Px, etc. Similar considerations apply if premiums are payable half-yearly,monthly, etc.

(2) By formula (6.3.1) with t = 0, we have 0V = 0 for all net premium reserves. In the case ofn-year endowment assurances or pure endowments, it is not always clear whether to take nVas zero or the sum assured (S, say), i.e. whether to assume that the sum assured has or hasnot already been paid. It appears that the convention nV = 0 is used in profit testing but notelsewhere.

Type ofPolicy Notation Prospective Formula for Reserve

Whole LifeAssurance tV(Ax) Ax+t − P (Ax)ax+t

n-year TermAssurance tV(A1

x:n ) A 1x+t:n−t − P (A1

x:n )ax+t:n−t

n-year EndowmentAssurance tV(Ax:n ) Ax+t:n−t − P (Ax:n )ax+t:n−t

h-Payment YearsWhole LifeAssurance

ht V (Ax)

Ax+t − hP(Ax)ax+t:h−t t < hAx+t t ≥ h

h-Payment Yearsn-year

EndowmentAssurance

ht V (Ax:n )

Ax+t:n−t − hP(Ax:n )ax+t:h−t t < hAx+t:n−t t ≥ h

n-year PureEndowment tV(A 1

x:n ) A 1x+t:n−t − P (A 1

x:n )ax+t:n−t

Table 6.3.1: Net Premium Reserves (premiums payable continuously)


Type ofPolicy Symbol Prospective Policy Value Retrospective Policy Value

Whole LifeAssurance tVx Ax+t − Pxax+t

Dx

Dx+t

[Pxax:t −A1

x:t]

EndowmentAssurance tVx:n Ax+t:n−t − Px:n ax+t:n−t

Dx

Dx+t

[Px:n ax:t −A1

x:t]

TemporaryAssurance tV1

x:n A 1x+t:n−t − P 1

x:n ax+t:n−t

Dx

Dx+t

[P 1

x:n ax:t −A1x:t

]

PureEndowment tV 1

x:n A 1x+t:n−t − P 1

x:n ax+t:n−t

Dx

Dx+t[P 1

x:n ax:t ]

Table 6.3.2: Net Premium Reserves (premiums payable annually)

Example 6.3.2.

(i) What is the International Notation for the reserve of Example 6.3.1 ?

(ii) Express this reserve in terms of annuity functions.

Solution.

(i) tV (Ax)

(ii)

tV (Ax) = Ax+t − P (Ax)ax+t

= 1− δax+t − (1ax− δ)ax+t using conversion relationships

= 1− ax+t

ax(6.3.2)

Reserves at non-integer durationsLet us consider (for example) a whole life non-profit policy with sum assured £1 payable at the endof the year of death, effected by (x) by level annual premiums, Px. The net premium reserve atinteger duration t is

tVx just before premium is paidtVx + Px just after premium is paid

Now consider duration r + k, integer r, 0 < k < 1. The reserve r+kVx is estimated by linearinterpolation between rVx + Px and r+1Vx, i.e.

r+kVx ; (1− k)(rVx + Px) + kr+1Vx (0 < k < 1) (6.3.3)

[The actual value is

v1−k[1−kqx+r+k + (1− 1−kqx+r+k)Ax+r+1]− Pxv1−k(1− 1−kqx+r+k)ax+r+1

6.4. RETROSPECTIVE RESERVES 99

which is complicated to evaluate.] Similar formulae apply for other classes of business with premi-ums payable annually in advance.

Reserves when premiums are payable mthlySuppose that premiums are payable mthly in advance (e.g. m=12, which corresponds to monthlypremiums, which are very common in practice.) To simplify matters let us suppose that the sumassured is payable immediately on death: this avoids the calculation of awkward assurance factorsat non-integer ages. Consider a policy providing £1,000 immediately on the death of (x), effectedby mthly premiums during the lifetime of (x). The annual premium, payable mthly in advance, isP (m)(Ax) = P , say, and the reserve at duration t (where t is an integer multiple of 1

m ) is

tV = 1, 000Ax+t − P a(m)x+t just before payment of the premium then due,

or

tV +P

mjust after payment of this premium.

The reserve at duration t = r + km (where r is an integer multiple of 1

m and 0 < k < 1) is usually

estimated by linear interpolation between rV + Pm and r+ 1

mV .

NOTE One may use standard symbols, e.g.

tV(12)(Ax) = Ax+t − P (12)(Ax)a(12)

x+t

tV(4)(A1

x:10 ) = A 1x+t:10−t − P (4)(A1

x:10 )a (4)x+t:10−t

if t is an integer multiple of 112 and 1

4 respectively, but these are seldom employed.

6.4 Retrospective reserves

So far we have calculated the reserve of a policy by referring to future cashflows, assuming thatthe contract will not be surrendered. In practice, the policyholder may wish to surrender thecontract, and in that event will, at least in the early years of the policy, probably expect a surrendervalue related to the accumulation of his premiums less expenses and the cost of life assurancecover. Such a surrender value is related to the retrospective reserve of the contract, which isobtained by accumulating the premiums less expenses and the cost of benefits, of a hypothetical largegroup of identical policies whose mortality follows the office’s tables exactly (i.e. without randomfluctuations), and then dividing the hypothetical funds among the survivors.Let us again assume that there are no expenses and that the premium and reserve bases agree.Consider, for example, a whole-life policy, with sum assured £1 payable at the end of the yearof death, issued t years ago by level annual premiums to a life then aged x. The (prospective)net premium reserve (on a specified mortality and interest basis, e.g. A1967-70 ultimate, 6% p.a.interest) just before payment of the premium now due is

tVx = Ax+t − Pxax+t

The corresponding retrospective reserve is found by accumulating the funds of (say) lx identicalpolicies until time t, and sharing the money out among the survivors. Since mortality is assumed tofollow the table exactly, there are dx deaths in the first policy year, dx+1 in the second policy year,and so on, with lx+t survivors at time t. Interest is assumed to be earned at the rate i p.a. used in


the premium and the reserving bases, so the accumulated funds at time t are

[lxPx(1 + i)t + lx+1Px(1 + i)t−1 + · · ·+ lx+t−1Px(1 + i)]

− [dx(1 + i)t−1 + dx+1(1 + i)t−2 + · · ·+ dx+t−1]

=lx(1 + i)t[Pxax:t −A1x:t ]

On division by lx+t (the assumed number of survivors at policy duration t), we find that

retrospectivereserve at

duration t years=

Dx

Dx+t[Pxax:t −A1

x:t ]

More generally, we have


duration t years=

Dx

Dx+t

M.P.V. (at issue date) ofpremiums, minus M.P.V.of benefits and expenses, inthe first t years

(6.4.1)

If there are no expenses, we have


duration t years=

Dx

Dx+t

M.P.V. (at issue date) ofpremiums, less benefits infirst t years

(6.4.2)

Some examples of formulae for retrospective reserves are given in Table 6.3.2 above.

Theorem 6.4.1. If the premium and the reserving bases agree, the prospective and the retrospectivereserves of a given policy are equal.

Proof. We illustrate the argument by means of the whole life policy discussed earlier in this section.The retrospective and prospective reserves at policy duration t are

VR =Dx

Dx+t[Pxax:t −A1

x:t ]

and

VP = Ax+t − Pxax+t

respectively. Hence

Dx+t

Dx(VR − VP ) = Pxax:t −A1

x:t

− Dx+t

DxAx+t +

Dx+t

DxPxax+t

= Pxax −Ax

(using the facts that ax = ax:t + t|ax and Ax = A1x:t + t|Ax.)

But Pxax − Ax = 0, by the equation of value for the level annual premium, so we have proved thedesired result.This argument may easily be extended to cover other policies, including situations in which thereare expenses.

6.4. RETROSPECTIVE RESERVES 101

It follows that, if the premium and reserving bases agree (or are assumed to agree in reserve calcu-lations), one need not specify whether a prospective or retrospective reserve is required. In practice,however, the prospective method is more often used.

Example 6.4.1. Give formulae for the following net premium reserves in terms of other monetaryfunctions:

(i) tVx:n ,

(ii) tV (Ax:n ),

(a)by the prospective method, and

(b) the retrospective method.

Solution.

(i) (a) Ax+t:n−t − Px:n ax+t:n−t

(b)Dx

Dx+t{Px:n ax:t −A1

x:t }

(ii) (a) Ax+t:n−t − P (Ax:n )ax+t:n−t

(b)Dx

Dx+t{P (Ax:n )ax:t − A1

x:t }


(i) tVx:n = 1− ax+t:n−t

ax:n(6.4.3)

(ii) tV (Ax:n ) = 1− ax+t:n−t

ax:n(6.4.4)

Solution.

(i)

tVx:n = Ax+t:n−t − Px:n ax+t:n−t

= 1− dax+t:n−t − (1

ax:n− d)ax+t:n−t

= 1− ax+t:n−t

ax:n


(ii)

tV (Ax:n ) = Ax+t:n−t − P (Ax:n )ax+t:n−t

= 1− δax+t:n−t − (1

ax:n− δ)ax+t:n−t

= 1− ax+t:n−t

ax:n

6.5 Gross premium valuations and asset shares.

We now allow for the possibility of expenses in the premium and reserving bases, which are nolonger assumed to be the same. The reserves obtained by the formula (6.2.3) are called prospectivegross premium reserves, whereas those obtained by a retrospective accumulation of premiums lessexpenses and the cost of life assurance benefits are called retrospective gross premium reserves. (Inpractice, however, the phrase “gross premium reserve” usually refers to prospective reserves only.) Ifthe premium and reserve bases do not agree, the two methods will normally give different results. Animportant feature of gross premium reserves are the facts that the prospective reserve at inceptionis not normally equal to zero, and the retrospective gross premium reserve of an n-year policy justafter it matures is not generally equal to zero.

We begin by considering prospective gross premium reserves. The premium basis (or the premi-ums themselves) must be specified, together with the reserving basis, which gives

(i) a mortality table,

(ii) an interest rate, and

(iii) the allowance for future expenses.

If the allowance for future expenses is a proportion k of future premiums (including any duenow), the formula for the prospective reserve is

tV = M.P.V. of future benefits −M.P.V. of future premiumsmultiplied by thefactor (1− k)

(6.5.1)

Example 6.5.1. Consider the whole life policy discussed in section 6.4 above, and suppose that

(i) the premium basis is A1967-70 ultimate, 6% p.a. interest, with no allowance for expenses; and

(ii) reserves are to be calculated prospectively by the gross premium method on the followingbasis:

mortality: A1967-70 ultimate;

interest: 4% p.a.;

expenses: 5% of all future premiums.

Find a formula for the reserve at duration t years (just before receipt of the premium then due.)

6.5. GROSS PREMIUM VALUATIONS AND ASSET SHARES. 103

SolutionThe annual premium is

P ′ = Px on A1967-70 ultimate, 6% p.a.

so the formula for the reserve istV = Ax+t − 0.95P ′ax+t

where the factors Ax+t and ax+t are on A1967-70 ultimate, 4% interest.If a retrospective gross premium reserve is required, one must specify the allowance for past expenses.If, in the policy of example 6.5.1, one allowed for expenses of 10% of all past premiums, the formulafor the retrospective gross premium reserve would be

tV =Dx

Dx+t

[0.9P ′ax:t −A1

x:t]

where the annuity and assurance factors, and Dx/Dx+t, are on A1967-70 ultimate at 4% p.a. inter-est (or whatever mortality and interest basis is to be used.)

Asset sharesThe asset share of a life assurance policy is a retrospective gross premium reserve calculated on thebasis of the mortality, interest and expenses actually experienced by the office. In order to illustratethe calculations, let us consider an n-year policy issued t years ago to (x), and define

St = sum assured in year t (payable at end of year of death);Pt = premium payable at the start of policy year t;et = actual expenses incurred at the start of policy year t;it = rate of interest earned by the office in policy year t;

q[x]+t = observed mortality rate among lives aged x + t whoentered assurance at age x.

Let tV denote the asset share at duration t of the policy, before payment of the premium then due.We clearly have 0V = 0, and the following relation holds:

t+1V = (tV + Pt+1 − et+1)(1 + it+1)− death cost, Dt+1 0 ≤ t ≤ n− 1 (6.5.2)

To calculate the “death cost”, we must consider the given policy’s share of the cost of making theasset share (at the end of the policy year) up to the sum assured, St+1, for those policies whichbecame claims in the year. Since the proportion of policies in force at the start of the year whichbecame claims was q[x]+t, the death cost is

Dt+1 =q[x]+t(St+1 − t+1V) (6.5.3)

(which is negative if St+1 < t+1V).

Notes

1. As will be stated in the next chapter, the quantity St+1 − t+1V is called the death strain at riskof the policy in year t + 1.

2. By substituting formula (6.5.3) in formula (6.5.2), we may obtain an expression for t+1V in termsof tV and other known quantities. In practice, however, one may perhaps replace Dt+1 by asuitable approximate value and use formula (6.5.2) directly.


3. The asset share at policy duration n years may of course differ from the maturity value, leadingto a profit or loss to the office at that time.

4. In the unlikely event that mortality , interest and expenses follow a certain valuation basisexactly, the asset share of a policy will equal the retrospective gross premium reserve on thatbasis.

5. Asset shares are particularly important for with profit policies, as their surrender and maturityvalues are not fixed in advance and the asset share gives one method of deciding “fair” surrenderand maturity values. A number of other considerations are, however, also involved, and we shallnot pursue the discussion of asset shares.

Net premium versus gross premium reservesA full description of the advantages and disadvantages of the various approaches is outside the scopeof this course, but we make some remarks.

It would at first seem obvious that a prospective gross premium valuation should be used, asthis represents the mean of the present value of the net liability. But this assumes that the policywill not be surrendered: the policyholder may do this at any time, and (at least in the early yearsof the contract) will expect a surrender value related to the accumulation of his past premiumsless expenses, i.e. a retrospective gross premium reserve. This suggests that the office should holdreserves equal to the greater of

(i) a prospective premium reserve, and

(ii) a retrospective gross premium reserve (or asset share),

the bases being perhaps different (reflecting expected future conditions and actual past conditionsrespectively.) The position is further complicated by the possible existence of guaranteed surrendervalues and the difficulty of estimating future conditions. In practice, a net premium valuation ata low rate of interest may be used, either by choice or because this is required by the supervisoryauthorities: such a basis (though apparently artificial, since the premium value is unrelated tothe actual premium being charged) may have the property that the corresponding reserves arecomparable with, or generally greater than, the more “scientific” reserves given above. Similararguments lead to the “net premium method” for with-profit policies (provided that terminal bonusesare allowed for separately): cf. Section 6.9 below.

The reason for the use of a low rate of interest in net premium reserves is that, by Lidstone’stheorem, net premium reserves normally increase as the rate of interest falls.

The supervisory authorities may be prepared to allow Zillmerisation (see Section 6.7 below) ofthe reserves (subject to limits on I), but many offices prefer to publish unmodified net premiumreserves, in case Zillmerisation is seen as a sign of financial weakness.

The above discussion refers mainly to the calculation of reserves for solvency testing, but reservesare also used for internal purposes: deciding upon bonus levels, alterations to policies, etc. For thesepurposes the office may use other bases.

The office must also allow for random fluctuations from the mean numbers of deaths (especiallywhen there is a small number of policies with very large sums assured or death strains at risk) andconsider the exchange or reinsurance of large risks.

6.6 The variance of L

We return to the definition of the random variable L given by formula (6.2.1). Its mean is definedas the reserve tV, but in certain problems we also require its variance, Var(L). How may this bestbe calculated?

6.6. THE VARIANCE OF L 105

Let us consider a policy providing £S immediately on the death of a life aged x at issue, withpremiums of P p.a. payable continuously throughout life. The random variable L is defined as

L = prospective loss on contract (at duration t)

= SvU − P aU (6.6.1)

where U is the future lifetime of the policyholder, who is now aged x + t, and expenses are ignored.Now we cannot use the formula

Var(L) = Var(SvU ) + Var(P aU ) wrong!

(because SvU and P aU are not independent), so we express L in terms of vU . That is,

L = SvU − P

(1− vU

δ

)

=(

S +P

δ

)vU − P

δ


Var(L) =(

S +P

δ

)2

Var(vU )

=(

S +P

δ

)2 [A∗x+t −

(Ax+t

)2]

(6.6.2)

where ∗ indicates a rate of interest of 2i + i2 p.a. This technique may also be used for certain othertypes of contracts.

We now consider a number of life assurance policies, labelled from j = 1 to j = N . The totalreserves for the group of policies is

N∑

j=1

E[Lj ] = V1 + V2 + · · ·+ VN (6.6.3)

where Lj and Vj are, respectively, the net liability and the reserve of the jth policy. If we furtherassume that the lives assured are independent,

Var

N∑

j=1

Lj

=

N∑

j=1

Var(Lj) (6.6.4)

where Var(Lj) may be calculated by (say) formula (6.6.2). If we now suppose that all the policiesare identical (i.e. they are of the same sum assured, type and duration, and were issued at the samedate to N different lives all of the same age), the average of the net liabilities is

L =

N∑

j=1

Lj

/N (6.6.5)

which has mean E(L) and variance Var(L)/N , where L is the net liability of a given policy. Theseresults are sometimes used in solvency calculations, etc. It is sometimes also assumed that N is so

large that the distribution ofN∑

j=1

Lj or L is approximately normal (by the central limit theorem

and related results), although this may only be accurate for very large values of N .


6.7 Zillmerised reserves

Let us now suppose that the premium basis is the same as the reserving basis (or is assumed to bethe same for the purpose of calculating reserves.) As stated earlier, the retrospective and prospectivemethods give the same results, but reserves may not be “net premium reserves” because of the effectof expenses. In certain important classes of policy, however, the reserve is obtainable by means ofa simple adjustment to the net premium reserve: the corresponding reserve is called a “Zillmerisedreserve”, “Zillmerised net premium reserve” or sometimes “modified net premium reserve”.

Let us consider an n-year endowment assurance for a sum assured of £1 without profits, payableat the end of the year of death (if this occurs within the term of the policy.) There are level premiumsof P ′′, payable annually in advance for n years or until earlier death of the policyholder, who wasaged x at the issue date. The reserve is required at duration t, just before payment of the premiumthen due.

The premium and reserving basis includes the following allowances for expenses: expenses of eon the payment of each premium, with additional initial expenses of I (so the total initial expenseis I + e).

Theorem 6.7.1. The reserve for the above policy is

tVZ = (1 + I)tV − I (6.7.1)

where tV denotes the corresponding net premium reserve, i.e.

tV = tVx:n =Ax+t:n−t − Px:n ax+t:n−t

=1− ax+t:n−t

ax:n

Proof. We first note that the annual premium P ′′ is such that

P ′′ax:n = Ax:n + eax:n + I

∴ P ′′ − e =I + Ax:n

ax:n

Now consider the reserve at duration t by the prospective method.

tVZ = M.P.V. of future benefits and expenses less premiums

= Ax+t:n−t − (P ′′ − e)ax+t:n−t

= Ax+t:n−t −[I + Ax:n

ax:n

]ax+t:n−t

=[Ax+t:n−t − Px:n ax+t:n−t

]−(

I

ax:n

)ax+t:n−t

(where Px:n is the net annual premium)

= tVx:n − I

[ax+t:n−t

ax:n

]

Now we use the fact that (by formula 6.4.3)

tVx:n = 1− ax+t:n−t

ax:n

6.7. ZILLMERISED RESERVES 107

to give the formula

tVZ = tVx:n − I [1− tVx:n ]

= (1 + I)tVx:n − I

as required.

Notes

1. The case of a whole life policy is similarly dealt with (put n = ∞ in the above formulae.)

2. A similar argument may be used to establish formula (6.7.1) if the sum assured is payableimmediately on death and premiums are payable continuously. In this case, tV = tV(Ax:n ).

3. Zillmer’s formula does not in general hold for other classes of policy, nor when the premiumdue at time t has been paid. (It is, however, sometimes used in practice for all policies and alldurations.)

4. When the duration t is short (e.g. t = 1) Zillmer’s formula may give a negative reserve, whichshould, as a rule, be replaced by zero, i.e. a policy should not be treated as an asset to theoffice. Note that formula (6.7.1) gives

0VZ = −I

which is correct if the additional initial expense I is thought of as being disbursed before thefirst premium is received: if it is not, one should write

0VZ = 0

(as is always assumed in profit-testing: see later.)

5. If the sum assured is not £1, we naturally multiply formula (6.7.1) by the sum assured: forexample, the reserve per £1000 sum assured of a whole life without profits policy might bequoted as

1020tVx − 20

(corresponding to I= 2% ).

Example 6.7.1. Ten years ago life office issued a 20-year endowment assurance without profits to(35). The sum assured is £10,000, payable at the end of the year of death (or on survival for 20years), and premiums are payable annually in advance. The basis for premiums and reserves is:

A1967-70 ultimate;

6% p.a. interest;

expenses are 3% of all office premiums (including the first) with additional initial expenses of1.5% of the sum assured.

Calculate

(i) the annual premium, and

(ii) the reserve,

(a) just before receipt of the premium now due, and


(b) just after receipt of the premium now due.

Solution.

(i) The office premium is

P ′′ =100000.97

[A35:20 + 0.015

a35:20

]

= £288.67

Note that tables at 6% interest are limited, so we use these formulae:

a35:20 = a35 − v20 l55l35

a55

= 11.9969

and

A35:20 = 1− da35:20

(ii) One may use Zillmer’s formula in this case (or other methods). The reserve is

(a) 10000[1.01510V35:20 − 0.015] = £3497

(b) £3497 + 0.97×£288.76 = £3777 (see formula (6.2.5))

Example 6.7.2. Suppose that the policyholder of example 6.7.1 were to surrender his policy (beforepaying the premium now due) and the office grants a surrender value equal to the reserve, as definedin that question. What yield per annum would the policyholder have obtained?

Solution. The equation of value is

288.67s10 = 3497 at rate i=yield p.a.∴ s10 = 12.11

We use the compound interest tables, i.e.

i s10

0.03 11.810.04 12.49

Interpolate:

i− 0.030.04− 0.03

; 12.11− 11.8112.49− 11.81

= 0.44

∴ i ; 3.44%

6.8. FULL PRELIMINARY TERM RESERVES. 109

6.8 Full preliminary term reserves.

We again assume that the premium and reserving bases agree, and suppose that the reserve for awhole life or endowment assurance policy is such that

1V = 0 (6.8.1)

That is, the reserve just before payment of the second annual premium is zero. Let us again supposethat the sum assured is £1, payable at the end of the year of death, and that expenses are as in theprevious section. We must have

1V = Ax+1:n−1 − (P ′′ − e)ax+1:n−1 = 0.

Hence

P ′′ − e =Ax+1:n−1

ax+1:n−1

= Px+1:n−1

Hence

tV = Ax+t:n−t − Px+1:n−1 ax+t:n−t

= A(x+1)+(t−1):(n−1)−(t−1)

− Px+1:n−1 a(x+1)+(t−1):(n−1)−(t−1)

= t−1Vx+1:n−1

= the net premium reserve for the correspondingpolicy on a life aged x + 1 at entry with termn− 1 years at duration t− 1

This reserve is called the Full Preliminary Term reserve, which we shall write as tVFPT . We have

shown thattV

FPT = t−1Vx+1:n−1 (6.8.2)

In the case of a whole life policy, we may set n = ∞ to obtain

tVFPT = t−1Vx+1 (6.8.3)

If the sum assured is not £1, we naturally multiply these expressions by the sum assured. Similarformulae hold if the sum assured is payable immediately on death, when premiums are payablecontinuously or monthly, etc., and for certain other types of policy.

What level of initial expenses leads to a F.P.T. reserve? We observe that, by the retrospectivemethod,

1VFPT =

Dx

Dx+1

[(P ′′ − e)ax:1 − I −A1

x:1]

= 0

from which we obtainP ′′ = e + I + A1

x:1 (6.8.4)

This states that the first premium is exactly sufficient to pay the initial expenses, I + e, and thecost of the first year’s life assurance cover, A1

x:1 .

The use of full preliminary term reserves was suggested by T. B. Sprague in 1870. Other actu-aries (particularly in North America) have devised formulae for “modified full preliminary termreserves”, in which the allowance for total initial expenses may be smaller than that of formula(6.8.4), but we do not pursue this topic.


6.9 Reserves for with-profits policies

We shall deal with “traditional” with profits policies only, and ignore terminal bonuses (althoughthese may in practice be very important.) Let us consider an n-year endowment assurance issued tyears ago to a life then aged x with basic sum assured S (payable, with attaching bonuses, at theend of the year of death or on maturity.) We suppose that the policy was issued with level annualpremiums, and that the premium now due has not yet been paid. Reversionary bonuses are addedon the payment of each premium, and we denote the total bonus added to date by B (hence thetotal death benefit in the year just ending is S + B.)The principal methods used for calculating reserves for with profits policies are:

(1) the net premium method;

(2) the bonus reserve (or gross premium) method; and

(3) the asset share method.

(1) The net premium method. The reserve is taken to be the net premium reserve for corre-sponding non profit policy, plus the mean present value of the bonuses already declared. In thecase of the endowment assurance policy discussed above, the reserve is given by the formulae:

tVWP = StVx:n + BAx+t:n−t (6.9.1)

= (S + B)Ax+t:n−t − SPx:n ax+t:n−t (6.9.2)

where all the actuarial functions are calculated on the given mortality and interest basis.

The rationale of this method is that the additional premiums for a with profits policy (relativeto the corresponding without profits policy) are considered to have “earned” the bonus B so fardeclared, so an additional reserve is required to cover the value of these bonuses.

If the sum assured is payable immediately on death, we have

tVWP = (S + B)Ax+t:n−t − P ax+t:n−t (6.9.3)

where P is the annual premium for the corresponding non-profit contract, i.e.

P =SAx:n

ax:n= SP (Ax:n ) (6.9.4)

Similar formulae may be given for whole life policies and for policies with monthly premiums,etc.

Example 6.9.1. On 1st January 1993 a life office issued a with-profit assurance endowmentpolicy with a term of 10 years to each of 100 male lives then aged exactly 50. The basic sum as-sured under each contract was £20, 000, and the basic sum assured, plus any bonuses, is payableon maturity or immediately on earlier death. Level annual premiums are payable in advancethroughout the term. Among the group of policies there were 2 deaths during 1993 and 4 deathsduring 1994. There were no lapses or surrenders during 1993 or 1994. For this group of policies,the office has declared a compound reversionary bonus of 3 3

4% per annum vesting in advance oneach 1st January from outset. The office values the policies by a net premium method, usingA1967-70 ultimate mortality and interest at 3%p.a.

Calculate the total reserves which the office held for the group of policies on 31st December 1994.

6.9. RESERVES FOR WITH-PROFITS POLICIES 111

Solution. We require 94×reserve of one policy. Using the net premium method for with profitspolicies,

reserveper policy

= S[2V(A50:10 )] + BA52:8

where S = 20, 000, B = 20, 000[(1.0375)2 − 1] (i.e. bonuses already vesting)i.e.

reserveper policy

= 20, 000(1.0375)2A52:8 − P a52:8

where P = net premium for corresponding non-profit policy

= 20, 000A50:10

a50:10

= 1, 760.94

[where we have used

A50:10 + (A 150:10 )(1.03)

12 + A 1

50:10= (1.03)

12 (A50:10 − v10

10p50) + v1010p50

= 0.068 + 0.68419 = 0.75219 ]

∴ Reserves needed = 94× [21, 528.125 A52:8

︸︷︷︸

= (1.03)12 (A52:8 − v8

8p52) + v88p52

= (1.03)12 (0.06142) + 0.73329

= 0.79562

− 1, 760.94a52:8 ]

= £443, 417

(2) The bonus reserve method. As the name suggests, reserves are calculated on the assumptionthat reversionary bonuses will be declared at certain annual rates. The value of the benefitsis therefore calculated by the formulae given in Section 3.10 above, according to the system ofbonus declarations used by the office, and the reserve is then found by subtracting the value ofthe office premiums less projected future expenses. If, in the case of the endowment assurancepolicy discussed above, it is assumed that future bonuses will be at the rate b per annum on thecompound system and a proportion k of future premiums will be absorbed in expenses, a formulafor the reserve is

tVWP = (S + B)A∗

x+t:n−t− (1− k)P ′′ax+t:n−t (6.9.5)

where ∗ indicates the rate of interest (i − b)/(1 + b). Similar formulae may be derived for otherpolicies.

Since the premium valued is the actual (or office) premium for the policy, this method is some-times referred to as the gross premium method for with profits policies.

As might be expected, the reserves calculated by the bonus reserve method depend greatly on theassured future levels of bonus, and in the early years of a contract the method may give reserveswhich are quite unrelated to the accumulation of premiums.


(3) The asset share method. This has been described (for non-profit policies) in Section 6.5above; the formulae are almost unchanged for with profit policies, the only difference being thefact that the sum assured in each policy year includes the vested bonuses. Since these only affectthe death cost, the reserves given by the asset share method do not depend greatly on past bonusdeclarations.

A complete description of the advantages and disadvantages of these methods of calculatingreserves is beyond the scope of this book.

6.10. EXERCISES 113

Exercises

6.1 (i) Express tVx in terms of Ax and Ax+t. Hence, or otherwise, find the values of nVx andnVx+n, given that

1−Ax+2n = Ax+2n −Ax+n = Ax+n −Ax

(ii) Calculate 1V40 given that P40 = .01536, p40 = .99647 and i = .05.

6.2 Consider an n-year pure endowment policy, issued to (x), with sum assured 1 and with annualpremiums payable throughout the duration of the policy. In the event of death within the nyears, all premiums paid will be returned without interest at the end of the year of death.

Obtain expressions for the reserve at duration t

(i) prospectively and

(ii) retrospectively.

Using commutation functions, show that your expressions are equal. (Assume that the premiumdue at time t has not yet been paid.)

6.3 (Difficult)

Given that Px = 0.02, nVx = 0.06, and P 1x:n = 0.25, find P 1

x:n .

6.4 Ten years ago a life office issued a 20-year endowment assurance without profits to (35). Thesum assured is £10000, payable at the end of the year of death (or on survival for 20 years),and premiums are payable annually in advance. Using A1967-70 ultimate 6%, and ignoringexpenses, calculate

(i) the annual premium;

(ii) the reserve, assuming that the premium now due has been paid.

6.5 (Difficult) You are given:

(i) Px = 0.01212

(ii) 20Px = 0.01508

(iii) P 1x:10 = 0.06942

(iv) 10Vx = 0.11430

Calculate 2010Vx .

6.6 A whole life assurance with sum assured £100,000 payable at the end of the year of death waspurchased by a life aged 30. The policy has annual premiums payable throughout life.The basis for calculating reserves for this policy is as follows:

net premium method: A1967-70 ultimate, 5% p.a. interest.Estimate the policy value at duration 28 1

4 by interpolation.

6.7 Describe the following terms briefly:

(i) net premium reserves;

(ii) Zillmerised reserves;


(iii) Full Preliminary Term reserves; and

(iv) gross premium reserves,

in each case giving suitable formulae in respect of a whole life assurance issued t years ago toa life then aged x, with level premiums payable annually in advance throughout the term andwith a sum assured of £1 payable at the end of the year of death. Assume that the premiumnow due has not yet been paid.

6.11. SOLUTIONS 115

Solutions

6.1 (i)

tVx = 1− ax+t

ax=

dax − dax+t

dax=

Ax+t −Ax

1−Ax

The given information implies thatAx+n = 2Ax+2n − 1

andAx = 2Ax+n −Ax+2n = 3Ax+2n − 2

Hence

nVx =Ax+n −Ax

1−Ax=

1−Ax+2n

3− 3Ax+2n=

13

and

nVx+n =Ax+2n −Ax+n

1−Ax+n=

1−Ax+2n

2− 2Ax+2n=

12

(ii) Retrospective method gives

1V40 = [P40 − vq40](1 + i)l40l41

=P40(1 + i)− q40

p40= 0.01264

6.2 The equation of value for the annual premium P is:

P ax:n = nEx + P (Rx −Rx+n − nMx+n)/Dx

orP (Nx −Nx+n) = Dx+n + P (Rx −Rx+n − nMx+n) (∗)

The expressions for the reserves are as follows:

Retrospective:tV = [P (Nx −Nx+t)− P (Rx −Rx+t − tMx+t)] /Dx+t (∗∗)

Prospective:

tV = n−tEx+t + tPA 1x+t:n−t + P (IA) 1

x+t:n−t − P ax+t:n−t

= [Dx+n + tP (Mx+t −Mx+n)+ P (Rx+t −Rx+n − [n− t]Mx+n)

− P (Nx+t −Nx+n)] /Dx+t (∗ ∗ ∗)

The expressions (∗∗) and (∗ ∗ ∗) are equal if and only if

PNx − PRx = Dx+n − tPMx+n − PRx+n − P [n− t]Mx+n + PNx+n

i.e. if and only ifDx+n + P (Rx −Rx+n − nMx+n) = P (Nx −Nx+n)

which is true by (∗).6.3 We know that

nVx = 1− ax+n

ax= 0.06, so

ax+n

ax= 0.94


1ax:n

− 1ax

=ax − ax:n

axax:n

= nExax+n

axax:n

= 0.94P 1x:n

= 0.235

Px:n − Px =(

1ax:n

− d

)−

(1ax− d

)=

1ax:n

− 1ax

= 0.235

so

Px:n = 0.255P 1

x:n = Px:n − P 1x:n = 0.005

6.4 (i) The annual premium is

10000P35:20 = 10000(

1a35:20

− d

)= £267.51

sincea35:20 = a35 − v20 l55

l35a55 = 11.997

(ii) The reserve is

1000010V35:20 = 10000(

1− a45:10

a35:20

)= £3593

sincea45:10 = a45 − v10 l55

l45a55 = 7.687

∴ Reserve after premium is paid = £3593 + £267.51= £3861 (to nearest £1)

6.5 The prospective method does not work. By the retrospective method,

(reminder: 2010Vx refers to limited payments policy)

2010Vx − 10Vx =

Dx

Dx+10[20Pxax:10 − Pxax:10 ] (since benefits are the same

under both policies)

= (20Px − Px)(

Dxax:10

Dx+10

)

= 20Px − Px

P 1x:10

= 0.04264

∴ 2010Vx = 0.15694

6.628V = 100, 000 28V30 = 33, 10129V = 100, 000 29V30 = 34, 754

Interpolate between 28V + P and 29V, where P is the annual premium, i.e. P = 100, 000P30 = 724.00. This gives

28 14V =

34(33, 101 + 724) +

14(34, 754)

= £34, 057

6.11. SOLUTIONS 117

6.7 (i) Net premium reserve = reserve calculated ignoring expenses andassuming premium basis agrees with re-serving basis

= tVx = Ax+t − Pxax+t

in example given

(ii) Zillmerised reserve = (1 + I)tV − I per unit sum assured

(where tV = net prem. reserve, I = additional initial expenses per £1 sum assured)

= (1 + I)tVx − I in example given.

(iii) F.P.T. reserve = net premium reserve for corresponding policy ofterm n− 1, entry age x + 1, duration t− 1

= t−1Vx+1:n−1 per unit sum assured for E.A.

= t−1Vx+1 in example given

(iv) Gross prem. reserve =( value of future

benefits

)−

( value of futuregross premiumsless expenses.

)

= Ax+t − (1− k)P ′ax+t

in example, whereP ′ = office A.P. andk = propn. of premiums allowed

for future expenses.


Chapter 7

APPLICATIONS OF RESERVES

7.1 Surrender values

Most life assurance policies (except term assurances) provide for a sum of money known as a sur-render value to be paid to the holder if the payment of premiums is discontinued and the contractterminated. If the scale of surrender values is fixed in advance, the policy is said to have guaranteedsurrender values. For example, a whole life contract (with level annual premiums payable through-out life) may be issued with guaranteed surrender values equal to the full preliminary term reserve,i.e.

(SV )t = S.t−1Vx+1 (7.1.1)

where x is the age at inception of the policy, t is the duration, (SV)t is the surrender value, S is thesum assured, and tVx is calculated on a specified mortality and interest basis. (It is assumed thatthe premium due at time t is unpaid.) More usually, however, surrender values are not guaranteed,although there may be guaranteed minimum surrender values, as is the case for Industrial Assurancepolicies in the U.K. (i.e. those in which the premiums are collected regularly from the home of thepolicyholder.)

Surrender values are normally related in a simple way to the reserve of the contract, but themethod of calculating the reserve for this purpose may be different from those used for other purposes,e.g. statutory returns to the Department of Trade and Industry. Surrender value scales must takeseveral factors into account:

(i) the experience of the office since the policy was issued,

(ii) the expected future experience,

(iii) consistency between different classes of business,

(iv) competition between life offices (and possibly with a market for “second hand” life policies),and

(v) guaranteed minimum surrender values (if applicable.)

A full discussion of all points involved is beyond our present scope. In some cases surrender valuesare calculated from paid up policy values, which we discuss in the next section.

Transfer values in pension schemes correspond to surrender values of life assurance policies: atransfer value is, however, payable only to another pension scheme and not to the scheme memberdirectly.

119

120 CHAPTER 7. APPLICATIONS OF RESERVES

7.2 Paid-up policy values

Instead of surrendering his policy, a policyholder who wishes to stop paying the premiums may askfor the contract to continue in force with reduced benefits (payable when the original benefits wereto be paid.) The reduced benefit is called the paid-up sum assured (P.U.S.A.).If the original sum assured on death differs from that on maturity, the paid-up benefits will normallybe in the same ratio.The general symbol tW denotes the paid-up sum assured at duration t years, assuming that the

premium then due is unpaid. If expenses are ignored and all calculations are conducted on the samemortality and interest basis, net premium paid-up policy values may be defined, using the notationtWx, tWx:n , etc. For example,

tWx = the paid-up sum assured at duration t in respectof a whole-life assurance issued to a life then agedx by level annual premiums, with sum assured £1payable at the end of the year of death

(7.2.1)

Suppose that, at the date of discontinuance of premiums of this whole life policy, the life office wishesto make the mean net liability for the altered contract the same as that of the original contract.This leads to the equation

tVx = tWxAx+t (7.2.2)


tWx = tVx

Ax+t(7.2.3)

Similar calculations hold for other policies, e.g.

tWx:n = tVx:n

Ax+t:n−t

(7.2.4)

More generally, we obtain the following general formula for the P.U.S.A.:

tW = tV

A(7.2.5)

where tV is the reserve or surrender value and A is an assurance factor, such as Ax+t or Ax+t:n−t .

(We note that no expenses have been allowed for at or after the date of alteration: if such an allowanceis desired, the formulae should be modified.)

In the important case of endowment assurances, the proportionate rule is sometimes used inpractice, i.e.

tW =t

nS (7.2.6)

where S is the original sum assured, t is the total number of premiums actually paid and n is thenumber originally payable.Surrender values of endowment assurances (or other policies) are sometimes found from the P.U.S.A.by the formula

(SV )t = (P.U.S.A.)A (7.2.7)

where A is the appropriate assurance factor at age x + t. In particular, if the proportionate rule forendowment assurances is used, the surrender value at duration t years is sometimes taken as

t

nSAx+t:n−t (7.2.8)

7.3. ALTERATIONS AND CONVERSIONS 121

if the sum assured is payable at the end of the year of death, if this occurs within the balance of theterm, n− t years, or

t

nSAx+t:n−t (7.2.9)

if the sum assured is payable immediately on death within this period.

Example 7.2.1.

(i) Four years ago a life then aged 35 effected a 25-year without profits endowment assurance byannual premiums for a sum assured of £50,000 payable at the end of the year of death or onsurvival for 25 years. He surrendered the policy just before paying the premium now due.

The life office calculated the premiums for this policy on the basis of A1967-70 select at 4%interest with the following expense loadings:

2.5% of each premium, including the first, plus a furtherinitial expense of 1% of the sum assured.

The office also calculates reserves on this basis, and surrender values are equal to 95% ofreserves.

(a) Calculate the office annual premium.

(b) Calculate the surrender value of the policy.

(ii) Suppose that the policyholder of (i) had made his policy paid-up instead of surrendering it,and that the office calculates paid-up values by the proportionate rule. Calculate the yield perannum (to the nearest 0.1%) that he would have obtained on his entire transaction, assumingthat he survives to maturity date.

Solution.

(i) (a) Let A.P. be P ′. We have0.975P ′a[35]:25 = 50000(A[35]:25 + 0.01)

∴ P ′ = £1289.10

(b)Reserve = 50000A39:21 − 0.975P ′a39:21

= £4790.00∴ S.V. = 0.95× reserve = £4550.50

(ii) PUSA=50000× 4/25 = 8000. SolveP ′s4 = 8000v21

i.e.(1 + i)21s4 = 6.20588

By trials and interpolation, i ; 1.9%.

7.3 Alterations and conversions

Life offices are frequently asked to alter a policy; for example, a policyholder with an endowmentassurance maturing at age 65 may request that the policy should be changed to an endowment


assurance maturing at age 60. The policyholder might also request his policy to be converted toanother class of business: for example, from a whole life assurance to an endowment assurance.

The usual rule when carrying out these calculations is to equate the reserves before and afterconversion. If V1 and V2 denote the reserves before and after the conversion respectively, the rulemay be stated as follows:

V1 = V2 (7.3.1)

If there are expenses of C for the conversion itself, the formula becomes

V1 − C = V2 (7.3.2)

The bases used to calculate V1 and V2 may in practice be different, and V2 must be calculatedprospectively by the gross premium method.

To show that formulae (7.3.1) and (7.3.2) hold, we argue as follows. At the date of conversion,the policyholder’s original policy is worth V1 on immediate surrender. (In practice, however, thereserve V1 used in conversion calculations may exceed the surrender value, as the office is not losingbusiness on a conversion or alteration.) Imagine that V1 is applied as a special single premiumtowards the new contract. The equation of value of the new contract is thus

V1 + M.P.V. of future premiums (including any due now)= M.P.V. of future benefits + future expenses + C

Rearranging this equation gives:

V1 − C = M.P.V. of future benefits + expenses − premiums= V2

The office then solves this equation for the unknown quantity (usually the new sum assured orpremium.) An example of the use of formula (7.3.1) has already been encountered in equation(7.2.2) above.

Example 7.3.1. Ten years ago, a man now aged 40 effected a whole of life assurance for a sumassured of £10000, payable at the end of the year of death, by level annual premiums. The premiumswere calculated using A1967-70 ultimate 4% and an allowance for expenses of 50% of the first year’spremium and 5% of each subsequent premium.

Immediately before payment of the eleventh annual premium the man requests that the policy beconverted into an endowment assurance maturing on his sixtieth birthday, with the annual premiumremaining unaltered. Calculate his revised sum assured using a full preliminary term reserve (onA1967-70 ultimate, 4% interest) for finding the reserve of the original policy, and

A1967-70 ultimate, 4% interest, expenses of 5% of premiumsfor calculating the reserve of the new policy.

Solution. Annual premium is P ′′, where 0.95P ′′a30 = 10000A30 + 0.45P . ∴ P ′′ = £97.13.

V1 = Reserve of original policy = 10000 9V31

︸︷︷︸

1− a40

a31

= £949.85

7.3. ALTERATIONS AND CONVERSIONS 123

Equation of value at date of conversion is

reserve before conversion = (prosp.) reserve after conversion∴ 949.85 = SA40:20 − 0.95× 97.13

︸︷︷︸allowing 5%of premiumsfor expenses

a40:20

∴ S = £4717

Example 7.3.2. An office issued a ten-year endowment assurance to a man aged exactly 45. Themonthly premium was £25 during the first five years increasing to £50 thereafter. The sum assured,payable immediately on death, was calculated using A1967-70 select 4%, allowing for expenses of£1 per month with additional initial expenses of 2 1

2% the sum assured.After five years the man requested that the premium remain at £25 per month for the next five

years, with the death benefit staying unaltered. Calculate the reduced sum payable on survival usingthe premium basis and allowing for a £30 alteration charge.

Solution. Let S =original sum assured.

300a(12)[45] :10 + 300

D50

D[45]a(12)

50 :5 = S[0.025 + A[45]:10

]+ 12a(12)

[45] :10

∴ S = £4871

(using a(12)[45] :10 = 8.154, a(12)

50 :5 = 4.489,D50

D[45]= 0.8093, A[45]:10 = 0.6808 )

Let S.A. on survival be reduced to S′. Equate reserves before and after conversion [allowing forcost of conversion]:

4871A50:5 − (600− 12)a(12)50 :5 − 30

= 4871A 150:5 + S′

D55

D50− (300− 12)a(12)

50 :5

∴ (4871− S′)D55

D50= 300a(12)

50 :5 + 30

∴ S′ = £3144

Example 7.3.3. In a certain country, the Universities operate the following superannuation schemefor academic staff, of whom there are two grades, Lecturer and Professor. Lecturers retire at age 65with a lump sum of £30, 000, while Professors retire at age 70 with a pension of £12, 000 per annumpayable monthly in advance. Staff pay for their benefits by means of level annual premiums, payableannually in advance until retirement or earlier death. Premiums are returned without interest ondeath before retirement, and withdrawals may be ignored. The employers do not contribute to thescheme. Death benefits are paid at end of year of death. The benefits are provided by policies issuedby a life office which uses the following basis for all calculations:

A1967-70 ultimate


4% per annum interest


A Professor cannot be demoted to Lecturer.

(i) Calculate the annual premium payable by a Lecturer on entry to the scheme at 25.

(ii) Calculate the reserve, just before payment of the premium now due, for a Lecturer aged 45with 20 years’ service.

(iii) The Lecturer of (ii) is now promoted to Professor. Calculate his or her new annual premium,the first due at once.

Solution.

(i) Let the annual premium be P . The equation of value is

P a25:40 = 30000A 125:40 + P (IA) 1

25:40=⇒ P = £274.44

(ii)

V1 = 30000D65

D45+ 20PA 1

45:20 + P (IA) 145:20 − P a45:20

= £8536

(iii) Let the new premium be P ′. We have,

V1 = V2

That is,

8536 = value of future benefits− future premiums for new contract

= 12000D70

D45a(12)70 + 20PA 1

45:25 + P ′(IA) 145:25 − P ′a45:25

=⇒ P ′ = £1525.87

7.4 The actual and expected death strains

Consider N identical policies on lives aged x at entry and now aged x + t, and assume that thepremiums now due are unpaid. The death strain at risk for a given contract in the coming policyyear is

S − t+1V (7.4.1)

where S denotes the sum payable on death (assumed payable at the end of the policy year) andt+1V denotes the reserve at the end of the policy year. Since the life office may be considered to beholding the reserve t+1V for each policy in force at time t + 1 in an imaginary “bank account” forthe policyholder, the death strain at risk measures the extra cash required if he dies in policy yeart+1. Let θx+t denote the number of deaths in the coming year. Note that (in the notation used formortality investigations)

N = Ex+t = the q-type exposed to risk at age x + t(since each life is exposed to the risk ofdeath for the whole policy year)

7.5. MORTALITY PROFIT AND LOSS 125

Assuming independence of the lives we have

θx+t ∼ binomial(N, qx+t)

which has mean Nqx+t.Now consider the random variable

A.D.S. = actualdeathstrain

= θx+t(S − t+1V) (7.4.2)

The mean of this variable (even without the assumption of independence of the lives) is

EDS = expecteddeathstrain

= Nqx+t(S − t+1V) (7.4.3)

Note. Once the number of deaths θx+t, is known (at the end of the policy year), ADS ceases to bea random variable: it is then a realisation of the variable.

The position may be generalised to cover cases in which policies considered are not identical,provided that their premiums are due on the same date. The actual death strain is

ADS =∑

death claims

in policy

year t + 1

(S − t+1V) (7.4.4)

and the expected death strain is

EDS =∑

all policies in

force at the

start of policy

year t + 1

qx+t(S − t+1V) (7.4.5)

Note. The sum assured S may depend on the policy year, t + 1.

7.5 Mortality profit and loss

Consider a block of business in force at the start of policy year t + 1, the premiums now due beingunpaid, and suppose that the life office holds funds equal to the total reserves for this block of busi-ness. Let us (i) ignore expenses, (ii) suppose that the premium and reserving bases agree, and (iii)assume that interest will be earned in policy year t + 1 at the valuation rate, i p.a. The mortalityprofit in policy year t+1 is defined as the funds at the end of the year less the money needed to paydeath claims and to set up reserves for the survivors. Hence

mortality profit =∑

all in force

at the

start of

year t + 1

(tV + P )(1 + i)−∑

death

claims

S −∑

survivors

t+1V (7.5.1)

Hence


mortality profit =∑

all in force

at the

start of

year

(tV + P )(1 + i)−∑

deaths

(S − t+1V)−∑

all in force

at the

start of

year

t+1V (7.5.2)

Theorem 7.5.1. Under the present assumptions, the mortality profit to the life office in respect ofthis group of policies, i.e. the difference between the actual funds at the end of the policy year andthe funds required to pay claims and set up reserves, is equal to E.D.S. minus A.D.S.; that is,

mortality profit in policy year t + 1 = E.D.S.−A.D.S. (7.5.3)

Proof. Consider a particular policy in the group. We have equation

(tV + P )(1 + i) = Sqx+t + (1− qx+t)t+1V (7.5.4)

which may be thought of as saying that the accumulated funds at the end of the year, (tV + P )(1 + i) ,

must provide:

(i) the expected cost of death claims, Sqx+t, plus

(ii) the expected cost of setting up reserves for the survivors, (1− qx+t)t+1V.

Equation (7.5.4) may be rearranged to give

(tV + P )(1 + i) = t+1V + qx+t(S − t+1V) (7.5.5)

On summing over all policies in force at the start of the year, we find that∑

all in force

at the

start of

year

(tV + P )(1 + i) =∑

all in force

at the

start of

year

t+1V + EDS (7.5.6)

Hence, by equation (7.5.2),mortality profit = EDS−ADS

as required.

Notes. 1. The sum assured S and the premium P may depend on the policy year, t + 1.2. Equation (7.5.4) may be verified directly for particular policy types: e.g., for a whole

life policy with sum assured £1,

(tVx + Px)(1 + i) = qx+t + (1− qx+t)t+1Vx (7.5.7)

Example 7.5.1. On 1st January 1988, a certain life office sold 100 10-year without profits endow-ment assurance policies to lives then aged 50. Premiums are payable annually in advance, and thesum assured under each contract is £30000, payable on maturity or at the end of the year of previousdeath. The office calculates premiums and reserves on the following basis:

7.5. MORTALITY PROFIT AND LOSS 127

mortality: A1967-70 ultimate

interest: 5% per annum

expenses: nil

All 100 policies were still in force on 1st January 1994, and 2 policyholders died during 1994.Calculate the mortality profit or loss to the office during 1994 in respect of this business, at 31December 1994.

Solution. Death strain at risk for each policy at 31.12.94

= 30000− 300007V50:10

= 30000(

a57:3

a50:10

)= 10753

Hence

mortality profit = E.D.S.−A.D.S.= (100q56 − 2)× 10753= −11378 (Loss of £11378)

Example 7.5.2. On the 1st January 1979 a life office issued a number of 20-year non-profit endow-ment assurance policies, with annual premiums payable in advance throughout the term and sumsassured payable on maturity or at the end of the year of death, to lives then aged exactly 45. At31st December 1993, total sums assured of £4,000,000 remained in force. During 1993 sums assuredof £50,000 became death claims, and there were no other exits in 1993.

The office calculates premiums, and maintains net premium reserves, on the basis given below.Calculate the profit or loss from mortality for this group of policies for the year ending 31st

December 1993.

Basis: mortality: A1967-70 ultimate


expenses: nil.

Solution.Sums assured in force on 1.1.1993 = £4050000

∴ Expected death strain = q59(4050000)[1− 15V45:20 ]

= q59(4050000)(

a60:5

a45:20

)

Actual death strain = 50000(1− 15V45:20 )

= 50000(

a60:5

a45:20

)


∴ mortality profit = E.D.S.−A.D.S.

= (4050000× 0.01299373− 50000)(

4.48913.488

)

= £874

Some generalisations

1. If expenses are included in the calculations, we need only alter tV + P to tV + P − e, where eis the expense incurred at the beginning of policy year t + 1. The equation corresponding to(7.5.4) is

(tV + P − e)(1 + i) = Sqx+t + (1− qx+t)t+1V (7.5.8)

and formula (7.5.3) remains correct.

(It is assumed that the actual expenses are as allowed for in the basis used for premiums andreserves. If not, one obtains expense profits or losses, which we do not discuss here.)

2. If the premium basis differs from the reserving basis, formula (7.5.3) remains true provided thatreserves are calculated by a gross premium method (either retrospective or prospective.) Thisfollows from the fact that equation (7.5.8) remains true (with mortality, interest and expensesaccording to the reserving basis and P on the premium basis.)

Example 7.5.3. For several years an insurance company has issued a large number of special en-dowment assurances. Each policy matures at exact age 65 and is effected by annual premiumspayable on each January 1st throughout the term. The sum assured, payable at the end of the yearof death during the term, is one half of the sum assured that will be paid if the policyholder survivesto age 65. Details of the policies in force on 31 December 1988 are as follows:

Exact age Total sums assuredpayable on death (£)

Total annual premiums(£)

55 3500000 250000

The claims in 1989 were on policies with the following total sums assured and premiums:

Total sums assuredpayable on death (£)

Total annualpremiums (£)

50000 4000

Assuming that at 31.12.88 the office’s funds were equal to its reserves, calculate the mortality profitor loss in 1989, given that the following reserving basis is used:prospective method, gross premium reserve using A1967-70 ult., 4% p.a. interest, no expenses.

Solution. Let EDS and ADS denote the expected death strain and the actual death strain in 1989.Then

EDS =∑

all j

q55 [Sj − {Sj(A56:9 + 9E56)− Pj a56:9 }]

7.6. OTHER SOURCES OF PROFIT AND LOSS 129

where the summation is over all policies in force at the start of the year, i.e.

EDS = q55

[(∑

Sj)− (∑

Sj)(A56:9 + 9E56) + (∑

Pj)a56:9

]

= q55(694048) (since∑

all j

Sj = 3500000 and∑

all j

Pj = 250000)

= 5859

The actual death strain is obtained by a summation of the death strains at risk over those policieswhich become claims. Thus

ADS =∑

Claims

[Sj − {Sj(A56:9 + 9E56)− Pj a56:9 }]

= (∑

Claims

Sj)− (∑

Claims

Sj)(A56:9 + 9E56) + (∑

Claims

Pj)a56:9

= 13082 (since∑

Claims

Sj = 50000 and∑

Claims

Pj = 40000)

The mortality profit is thus = 5859−13082=−7223, i.e. a loss of £7,223.

7.6 Other sources of profit and loss

Although some of these topics may also be considered under the heading of “profit-testing”, we shallgive certain straightforward results on interest and surrender profits here.

Interest profits. Suppose that the office earns interest at a rate i′ p.a. in policy year t + 1, theinterest rate in the reserving basis being i p.a. Assuming as before that the funds at the end ofpolicy year t equal the reserves for the business under consideration, the interest profit is

[∑

all in force

at the

start of policy

year t + 1

(tV + P − e)](i′ − i) (7.6.1)

Example 7.6.1. On 1st January 1978 a special twenty-year endowment assurance policy for aninitial sum assured of £20,000 was issued to a life then aged 40 exactly. The sum assured, which ispayable at the end of the year of death or on survival to maturity date, increased to £25,000 after 10years and to £30,000 on maturity. The annual premium was P for the first ten years, and thereafterincreased to 1.25P . The office’s basis for premiums and reserves is

A1967-70 ultimate

4% per annum interest


(i) Calculate the initial annual premium P , and the reserve for the contract at (a) 31st December1989, and (b) 31st December 1990.


(ii) Find the profit or loss from mortality for the year ending 31st December 1990 for a group ofsuch policies, all effected on 1st January 1978 by lives then aged 40 exactly, given that onepolicyholder dies in 1990 and 120 policies remained in force at 31st December 1990. Therewere no withdrawals in 1990.

(iii) The office found that it made neither profit nor loss in 1990 in respect of the policies of (ii)above. What rate of interest did it earn in 1990?(Expenses are negligible.)

Solution.

(i) Let initial premium be P .

P [a40:20 + 0.2510|a40:10 ] = 20000A40:20 + 500010|A40:10 + 5000D60

D40

∴ P =20000M40 + 5000M50 − 25000M60 + 30000D60

N40 + 0.25N50 − 1.25N60

= £907.00

(a)

12V = 25000A52:8 + 5000D60

D52− 1.25P a52:8

= £14085

(b)

13V =[(12V + 1.25P )(1.04)− 25000q52]

1− q52

= £15772 (this may also be calculated directly)

(ii)

E.D.S. = 121q52(25000− 13V)= £6734

A.D.S. = 25000− 13V = £9228∴ mortality profit = 6374− 9228 = −£2494

(iii) Let interest rate earned be i′. The interest profit is

(i′ − 0.04)[121](12V + 1.25P ) = 2494

∴ i′ = 0.0414, or 4.14%

Surrender profits. Assume that surrenders may take place only at the end of a policy year (justbefore payment of any premium then due). The surrender profit is

∑

surrenders

(t+1V − S.V.) (7.6.2)

7.6. OTHER SOURCES OF PROFIT AND LOSS 131

where S.V. is the surrender value paid.

Summary of profit and loss

Profit or loss for year = accumulation of funds at end of year− cost of death claims− cost of surrenders− cost of setting

up reserves forremaining policies

= Mortality profit+ interest profit+ surrender profit (7.6.3)

Example 7.6.2. On 1st January 1977 an office issued a block of without profits endowment assur-ance policies and a block of without profits pure endowment policies to men aged exactly 40. Eachpolicy was for a term of 20 years and a sum assured of £1000. Premiums are payable annually inadvance during the term of the policies and death claims are paid on 31st December each year.

On 1st January 1987 there were 1000 endowment assurance policies and 5000 pure endowmentpolicies in force. During the year to 31st December 1987 sixty endowment assurance policies becameclaims by death and twenty pure endowment policies were terminated, without payment of any sumassured, by the deaths of the assured lives.

On 31st December 1987 ten endowment assurance policies and five pure endowment policies weresurrendered.

Calculate the profit or loss arising during the year 1987, showing separately the contributionsfrom mortality and from surrender if the office uses the following basis:

Premiums and reserves: A1967-70 ultimate at 4% p.a. interest

Surrender values: Net premium reserve on A1967-70 ultimate at6% p.a. interest

Ignore expenses.

Solution. Let us consider each type of policy separately.

Endowment Assurances.Reserve per policy at end of 1987 = 100011V40:20

= 1000(

1− a51:9

a40:20

)

= 1000(

1− 753113.764

)

= 452.85

∴ Death strain at risk per policy in force at start of 1987 is 1000− 452.85 = 547.15

∴ E.D.S. = 1000q50 × 547.15 = 26202

A.D.S. = 60× 547.15 = 32829


∴ mortality profit = E.D.S.−A.D.S. = −6627

Surrender profit=10(Reserve-S.V.), where

S.V. = 100011V40:20 at 6 %

= 1000(

1− 7.02911.871

)= 407.88

∴ Surrender profit = 449.70

Pure Endowments. Annual premium is

P =1000D60

D40

a40:20

= 29.70

Reserve per policy at end of 1987 is

P

(N40 −N51

D51

)= 29.70

(132002− 68970

4399.08

)= 425.55

Death strain at risk is −425.55 per policy.

E.D.S. = 5000q50(−425.55) = −10189

A.D.S. = −20× 425.55 = −8511

∴ mortality profit = E.D.S.−A.D.S.= −1678

S.V.= net premium reserve at 6%. It is easier to find the reserve prospectively.

Define P ′ = net premium at 6% p.a. interest =1000v20 l60

l40

a40:20

=279.2511.871

= 23.52

∴ Reserve = 1000v9 l60l51

− P ′a51:9 at 6% p.a. interest

= 546.87− 165.32 = 381.55

∴ Surrender profit is 5(425.55-381.55)=220.00

Summary

Mortality profit = −£8305Surrender profit = £670

Total profit = −£7635 (a loss of £7,635)

7.7. EXERCISES 133

Exercises

7.1 Explain briefly how a life office calculates the new sum assured or annual premium on thealteration or conversion of an existing policy.

7.2 A life office issued a whole life without profits policy to a woman aged 30, with sum assured of£110,000. Premiums were payable annually in advance throughout life, and the sum assuredwas payable at the end of the year of death. Immediately before payment of the 15th premium,the policyholder requests that the policy be converted to a without profits endowment assurancefor the same sum assured, payable on maturity at age 60 or at the end of the year of death, ifbefore age 60. The office calculates premiums and maintains reserves on the basis of A1967-70ultimate mortality and 4% per annum interest, with expenses of 4% of each premium. Expensesof alteration may be ignored.

Find the revised annual premium.

7.3 On 1 January 1972 an office issued a large number of 40-year endowment assurance policies,each with sum assured £1000 (payable at the end of the year of death, if this occurs within theterm) to a group of lives all then aged 25. On 31 December 1988 there were 8567 policies stillin force. During 1989 there were 13 deaths among the policyholders.

Find the actual death strain, the expected death strain and the mortality profit or loss for thebusiness in 1989, using the following basis for all calculations:

mortality: A1967-70 select

interest: 4% p.a.

expenses: none

7.4 On 1 January 1991, a life office issued a number of identical special endowment assurances tolives then aged 45. Each policy had a term of 15 years and level annual premiums were payablethroughout the term. On survival to the end of the term, a sum assured of £5000 is payable.On death before the end of the term, a sum assured of £1000 is payable at the end of the yearof death and, in addition, all premiums paid are returned without interest.

The premium basis for these policies was as follows:


interest: 4% p.a.

expenses: Nil

(i) Show that the annual premium for each of these policies is £242.27.

(ii) Calculate the reserve, on the premium basis, on 1 January 1996 for one of these policies,assuming that the premium then due is unpaid.

(iii) On 1 January 1995, a total of 10,000 of these policies were still in force. In 1995, 47of these lives died, ten policies were surrendered, and the life office earned 4% intereston its investments. Expenses were negligible in 1995. The surrenders all took place atthe end of 1995, and the office gave surrender values equal to 95% of the reserve on thepremium basis.Calculate the profit or loss to the office in 1995 from this block of business in respectof

(a) mortality , and


(b) surrenders.

7.5 A life office has issued a 5-year decreasing temporary assurance to a life aged 60. The sumassured, payable at the end of the year of death, is £100,000 if the life dies in the first year,£90,000 if the life dies in the second year, and so on, decreasing by £10,000 each year. Levelannual premiums are payable throughout the term of the policy.

The premium and reserve basis for this policy is as follows:

mortality: A1967-70 Ultimate

interest: 4% p.a.

expenses: Nil

(i) Calculate the annual premium for the policy.

(ii) Write down a relationship between the reserve at integer duration t and the reserve atduration t + 1.

7.8. SOLUTIONS 135

Solutions

7.1 The office calculates V1, the reserve for the original policy at the date of conversion. The officethen sets up the equation

V1 − C = V2

whereC = cost of alteration (if any)

andV2 = (prospective) reserve of new policy.

Solve for unknown quantity (usually a sum assured or premium.)

7.2 Let

P = original A.P. =1100001− 0.04

P30

Reserve before conversion = V1 = 110, 000A44 − 0.96P a44

= 110, 00014V30

= 110, 000[1− a44

a30

]

= 16, 923

Let new A.P. be P ′. We solve for P ′ in the equation V1 = V2, i.e.

V1 = 110, 000A44:16 − 0.96P ′a44:16

∴ P ′ =110, 000× .54703− 16, 923

0.96× 11.777= £3, 825

7.3 On 31/12/88 the duration is 17 years. The reserve at the end of 1989 (per unit sum assured)therefore equals

18V[25]:40 = 1− a43:22 /a[25]:40 = .28397

Hence in 1989 we have, per policy in force at the start of the year, a death strain at risk of(1000-283.97)=£716.03.

The expected death strain per policy is 716.03× q42 = 1.31105.

For the block of business, therefore, we have:

total expected death strain = 8567× 1.31105 = 11, 234.52total actual death strain = 13× 716.03 = 9, 308.4

The 1989 mortality profit is therefore (11,234.52−9,308.4)=£1,926.14.


7.4 (i) Let A.P. be P .P a[45]:15 (= 11.250) = 1, 000A[45]:15 (= 0.56732)

+ 4000(

D60

D[45](= 0.50271)

)

+ P (IA) 1[45]:15 (= 0.60855)

∴ P = £242.27

(ii)

5V = 1000A50:10 + 4000D60

D50+ 5PA 1

50:10 + P (IA) 150:10 − P a50:10

= £1, 349

(iii) (a)D.S.A.R. in 1995 in respect of 1 policy = (1000 + 5P )− 5V

= £861.97EDS = 10, 000q49 ×DSAR = £36, 714

ADS = 47×DSAR = £40, 513∴ mortality profit = EDS−ADS = −£3, 798 (loss of £3,798)

(b)Surrender profit per policy = 0.05× reserve

∴ Total surrender profit = 10× 0.05× 5V

= £675.

7.5 (i) Let the Annual Premium be P .P a60:5 = 110, 000A 1

60:5 − 10, 000(IA) 160:5

which givesP = £1, 347.03

(ii)(tV + P )(1.04) = q60+t(100, 000− 10, 000t) + (1− q60+t)t+1V

(since the sum assured on death in year t + 1 is 100, 000− 10, 000t).

Chapter 8

EXTRA RISKS

8.1 Introduction

It is usual in practice for a life office to accept the majority of proposers for life assurance policies on“normal” (or “standard”) terms, the remainder being either declined or accepted on special terms,i.e. they are accepted but with higher premiums or reduced benefits. Those accepted on specialterms are considered to be subject to an extra mortality risk due to a health impairment, a dangerousoccupation or a dangerous spare-time activity (such as motor racing.) The assessment of extra risksis usually carried out by experienced underwriters using a “numerical rating system”; we shall notgo into this, but shall assume that the addition to the normal rates or force of mortality is known.

Some offices make extensive use of “rating up”: that is impaired lives are treated as normallives a number of years older than actual age. (Females are sometimes treated as males 5, say, yearsyounger, although this rule is very inaccurate at certain young ages.) Smokers are sometimes treatedas non-smokers 6 years older.

8.2 A constant addition to the force of mortality

Let µ[x]+t refer to the force of mortality at duration t years for a life who was selected for assuranceto age x on the office’s normal terms. (If a non-select table is used, omit [ ].) Let n years be the

duration of the policy, and let µ∗[x]+t be the corresponding force of mortality for a life accepted on

special terms. We now assume that there is k > 0 such that:

µ∗[x]+t = µ[x]+t + k (0 ≤ t ≤ n) (8.2.1)

Hence

tp∗[x] = exp

[−

∫ t

0

µ∗[x]+r dr

]

= exp[−

∫ t

0

(µ[x]+r + k) dr

]

= e−kttp[x] (8.2.2)

137

138 CHAPTER 8. EXTRA RISKS

Also,

tE∗[x] = vt

tp∗[x]

= e−(δ+k)ttp[x]

= e−δ′ttp[x] (8.2.3)

where

δ′ = δ + k (8.2.4)

and hence i′ = eδ′ − 1 = eδ+k − 1.Since an annuity factor is a sum or integral of pure endowment factors, we have (for example)

a∗[x] = a[x] on normal mortality but with force of interest δ′ = δ + k (8.2.5) This rule does not

apply to assurances and premiums, but one may sometimes use conversion relationships to expressthese in terms of annuity factors, as in the following example.

Example 8.2.1. A certain life office uses the following basis for calculating premiums for assurancepolicies on lives accepted at normal rates:

A1967-70 ultimate

4% interest


A certain proposer, aged 50, for a 15-year endowment assurance without profits by annual premiumsis considered by the office to be subject to the mortality of the office’s normal table with an additionof 0.019048 to the force of mortality for the next 15 years. The sum assured is £10,000, payableat the end of the year of death, or on survival for 15 years. Calculate the addition to the normalannual premium.

Solution. Let ∗ indicate special mortality basis.Extra annual premium is

10, 000[P ∗50:15

− P50:15 ]

Note that

P ∗50:15

=1

a∗50:15

− d by a conversion relationship

=1

a50:15 0.06

− d since δ.04 + 0.019048 = δ.06

But d is still at 4% interest.Hence the extra premium is

10, 000[

1a50:15 .06

− 1a50:15 .04

]= £108.51

8.3. A VARIABLE ADDITION TO THE FORCE OF MORTALITY 139

In some problems we must evaluate assurance factors directly, without the use of conversion rela-tionships; e.g.

(IA)∗1x:n = M.P.V. of an increasing temporary assurance

for an impaired life aged x at entry

=∫ n

0

tvttp∗xµ∗x+t dt

=∫ n

0

tvt(tpxe−kt)(µx+t + k) dt

=∫ n

0

t (v′)t

︸︷︷︸v′ is at rate i′

(with δ′ = δ + k)

tpx(µx+t + k) dt

= (IA)1x:n at rate of interest i′

+ k(I a)x:n at rate of interest i′ (8.2.6)

A formula for the expectation of life of an impaired lifeWe have

oe∗x = E(T ∗) =

∫ ∞

0tp∗x dt (by formula (1.4.3))

=∫ ∞

0tpxe−kt dt

= ax at force of interest k p.a. (8.2.7)

Note also that

E[(T ∗)2] = 2∫ ∞

0

t.tp∗x dt (by integration by parts)

= 2∫ ∞

0

t.tpxe−kt dt

= 2(I a)x at force of interest k p.a. (8.2.8)

Hence

Var(T ∗) = 2(I a)x − (ax)2 at force of interest k p.a. (8.2.9)

Remark. To find the median future lifetime of the impaired life, we solve the equation

tq∗x = 0.5

i.e. tp∗x =

lx+te−kt

lx= 0.5

8.3 A variable addition to the force of mortality

Let us now suppose that there is ξ(t) ≥ 0 such that

µ∗[x]+t = µ[x]+t + ξ(t) (0 ≤ t ≤ n) (8.3.1)


We obtain the formula

tp∗[x] = tp[x] exp

(−

∫ t

0

ξ(r) dr

)(8.3.2)

In some cases∫ t

0

ξ(r) dr may be calculated analytically, as in the next example.

Example 8.3.1. A certain impaired life aged 75 experiences mortality according to a(55) malesultimate with an addition to the force of mortality. The addition is 0.005 at age 75, increasinglinearly to 0.020 at age 90.

(i) Find a formula for the probability that this life will be alive at age 75 + t years (0 ≤ t ≤ 15).

(ii) Estimate, by approximate integration using the trapezoidal rule (not repeated), the singlepremium for a temporary assurance of £50,000 payable immediately on death of (75), if thisoccurs within 15 years, on the following basis:

mortality: as described above

interest: 10% p.a.

expenses: 2% of the single premium.

Solution.

(i)µ∗75+t = µ75+t + 0.005 + 0.001t (0 ≤ t ≤ 15)

∴ tp∗75 = tp75 exp

[−

∫ t

0

(0.005 + 0.001r) dr

]

= tp75 exp[−(0.005t + 0.0005t2)

]

(ii) Let single premium be P .

0.98P = 50, 000∫ 15

0

vttp∗75µ

∗75+t dt

; 50, 000× 152

[µ∗75 + v15

15p∗75µ

∗90

]

= 50, 000× 7.5 [0.06550 + 0.1104× 0.23939× 0.2636]∴ P = £27, 730

8.4 Rating up

In practice, impaired lives are often considered to experience the mortality of a “normal” life r yearsolder (where r depends on the severity of the impairment.) This allows the life office to base all itspremiums on one set of tables.

8.4. RATING UP 141

A practical method of finding “r” is as follows. Suppose that the normal mortality table followsGompertz’ law, so there are constants B, c such that

µx+t = Bcx+t (0 ≤ t ≤ n)

(This is usually a fairly accurate representation of mortality for all ages over about 30.) Supposethat the impairment is such that

µ∗x+t = (1 + θ)µx+t (0 ≤ t ≤ n) (8.4.1)

for some constant, θ. (This formula is usually suitable for “medical” impairments; for accident risksit is generally better to add a constant to the normal force of mortality.) Hence

µ∗x+t = (1 + θ)Bcx+t

= Bc(x+r)+t (0 ≤ t ≤ n)

where cr = 1 + θ, i.e.

r =log(1 + θ)

log c(8.4.2)

In practice c is usually about 1.08, so we have the approximate rule

r =log(1 + θ)log 1.08

(8.4.3)

For example, if θ = 0.4 (corresponding to a 40% increase in the force of mortality),

r ; log 1.4log 1.08

= 4.37

For practical purposes this value of r would be rounded to 4 or 5 years. This rule is not precise,but it may be sufficiently accurate in practice since the value of θ may in general be estimated onlyrather approximately.

Example 8.4.1. A certain life office’s premium basis for lives accepted at normal rates is as follows:

Mortality: A1967-70 ultimate

Interest: 4% per annum

Expenses: none

A certain impaired life, aged 50, is considered to be subject to the office’s normal mortality ratesfor a life 10 years older. This life requires a contract providing the following benefits:

(i) £30, 000 at the end of the year of death, if within 10 years;

(ii) on survival for 10 years, the sum of £(30, 000 + S).

The impaired life finds that the level annual premium for this policy is the same for a policy providingidentical benefits for a life his age accepted at normal rates. Find S.


Solution.

normal premium =30, 000A50:10 + S D60

D50

a50:10

= 30, 000P50:10 + S

(D60

D50

/a50:10

)

“impaired life” = 30, 000P60:10 + S

(D70

D60

/a60:10

)

Equate these to give

S =30, 000(P60:10 − P50:10 )(

D60D50

/a50:10

)−

(D70D60

/a60:10

)

=30, 000(0.08986− 0.08339)

0.075689− 0.068168= £25, 809

8.5 Debts

Instead of paying higher premiums, a proposer accepted on special terms may elect to pay the normalpremiums with reduced death benefits. Such reductions in the death benefits are called “debts” or“liens”, and may be either

(a) a level amount; or

(b) diminishing in arithmetic progression to zero in the final policy year.

We first give an example involving a level debt. (Note that a debt is a reduction in the death benefitonly: the maturity benefit is not reduced.)

Example 8.5.1. A life office calculates annual premiums for without profits endowment assurancesusing the following basis:

Mortality: A1967-70 select

Interest: 4%

Initial expenses: 50% of the first premium

Renewal expenses: 5% of each premium after the first

(i) Calculate the annual premium payable for a 25 year endowment assurance taken out by a malelife aged exactly 45 for a sum assured of £50,000. The death benefit is payable at the end ofthe year of death.

(ii) A man aged exactly 45 effects a policy identical to that in (i) above but rated up 7 years.

(a) Calculate the level extra premium, payable throughout the term of the policy.

(b) Alternatively, the life office offers to charge the standard premium but to impose a leveldebt. Calculate the amount of the debt.

8.5. DEBTS 143

Solution.

(i) Let A.P. be P .0.95P a[45]:25 = 50, 000A[45]:25 + 0.45P

∴ P = £1, 492.41

(ii) (a) Consider a “normal” (select) life aged 52. Let P ′ be premium.0.95P ′a[52]:25 = 50, 000A[52]:25 + 0.45P ′

∴ P ′ = 1, 752.32

∴ Extra premium = £259.91

(b) Let death benefit be reduced by D. The equation of value is0.95P a[52]:25 = 50, 000A[52]:25 −DA 1

[52]:25 + 0.45P

∴ D = £12, 649

Now suppose that we are dealing with an n-year endowment assurance, and that the debt diminishesin arithmetical progression to zero in the final policy year; that is, the debts run as follows:

(n− 1)D in year 1

(n− 2)D in year 2

..................

D in year n− 1

0 in year n

Consequently the M.P.V. of the debts, at the issue date, is

D{nA∗ 1[x]:n − (IA)∗ 1

[x]:n } (8.5.1)

where ∗ indicates impaired lives’ mortality; the death benefits are taken to be payable at the end ofthe year of death (if these benefits are payable immediately on death, a bar should be placed overthe A symbols.) The calculations are illustrated in the next example.

Example 8.5.2. A life office issues 20-year without-profit endowment assurance policies providinga sum assured of £8,000 on maturity or at the end of the year of earlier death. Level monthlypremiums are payable in advance throughout the term or until earlier death.

A certain impaired life aged 45 is considered by the office to have the mortality rates of a “nor-mal” (select) life 5 years older.

Assuming that the ordinary premium appropriate to this life’s actual age is payable, calculate theinitial amount of the debt (i.e. the deduction from the sum assured) such that the debt reducesuniformly each year to nil in the last policy year. Basis:

Mortality of “normal” lives: A1967-70 select

Interest: 4%


expenses: 30% of premiums in first year

5% of later premiums

Solution. LetP

12be “normal” monthly premium. The equation of value for P is

0.95P a(12)

[45]:20− 0.25P a

(12)

[45]:1= 8000A[45]:20

P =8000× [0.48051]{

.95[a[45]:20 − 11

24

(1− D65

D[45]

)]− 0.25

[1− 11

24

(1− D[45]+1

D[45]

)]}

=3844.08

{0.95[13.507− 0.45833× 0.62253]− 0.25[1− 0.45833× 0.04014]}

=3, 844.08

12.5606− 0.24541

= £312.14 (£26.01 per month)

Let the initial debt be 19D. The equation of value for D is

0.95× 312.14a(12)

[50]:20− 0.25× 312.14a

(12)

[50]:1= 8000A[50]:20 − [

20DA 1[50]:20 −D(IA) 1

[50]:20]

D ={

8000× 0.49664− 312.14× 0.95×[a[50]:20 − 0.45833

(1− D70

D[50]

)]

+0.25× 312.14[1− 0.45833

(1− D[50]+1

D[50]

)]}

/{20(M[50] −M70)− (R[50] −R70 − 20M70)

D[50]

}

=3973.12− 296.53[13.087− 0.45833× 0.66887] + 78.035[0.98111]

14581.322 [20× 758.289− 9360.22]

=259.8981.2672

= 205.09

∴ Initial Debt is 19D = £3, 896.71

8.6. EXERCISES 145

Exercises

8.1 An explorer aged exactly 57 has just made a proposal to a life office for a whole life assurancewith a sum assured of £10,000 payable at the end of the year of death. For lives accepted atnormal rate, level annual premiums are payable until death under this policy.

The explorer is about to undertake a hazardous expedition which will last three years. The lifeoffice estimates that during these three years the explorer will experience a constant addition of0.02871 to the normal force of mortality, but after three years will experience normal mortality.The life office quotes a level extra premium payable for the first three years.

Calculate this level extra premium on the following basis:

normal mortality: A1967-70 ultimate


expenses: none

8.2 An impaired life aged exactly 55 wishes to effect a without profit endowment assurance for asum assured of £1,000 payable at the end of 10 years or at the end of the year of earlier death.Level annual premiums are payable throughout the term of the policy.

Special terms are offered on the assumption that the life will experience mortality which canbe represented by:

(a) for the first five years, a constant addition of 0.009569 to the normal force of mortality,and

(b) for the remaining five years, the mortality of a life 8 years older.

The life office quotes a level extra premium payable throughout the term. Calculate this levelextra premium. Basis:

normal mortality: A1967-70 ultimate


expenses: none

8.3 A group of impaired lives now aged 40 experience mortality according to A1967-70 ultimatewith an addition to the force of mortality. The addition is 0.0005 at age 40, increasing linearlyto 0.0025 at age 60, at which level the addition remains constant.

Find the probability that an impaired life aged exactly 40

(i) will die within 20 years;

(ii) will die within 30 years;

(iii) will die between 20 and 30 years from the present time.

8.4 (i) Can you envisage circumstances under which an office could offer an impaired life whowishes to pay the “normal” premiums a level debt but not a diminishing debt?Hint Consider a life who is very severely impaired, and think of an approximate rela-tionship between

(a) the level debt and(b) the initial debt when debts decrease linearly to zero.


(ii) A certain proposer, aged 40, for a 20-year endowment assurance by annual premiumswith sum assured £40,000 without profits (payable at the end of the year of death) isconsidered by your office to be subject to the mortality of the normal table with anaddition of 10 years to the age. Your office’s basis for calculating premiums for “normal”lives is

A1967-70 ultimate4% p.a. interestexpenses of 6% of all premiums.

(a) The proposer asks to pay the annual premium for a “normal” life aged 40, andsuggests that the office should make the normal death benefits subject to a debtwhich decreases in arithmetical progression to zero in the final year. Calculate theinitial debt.

(b) Suppose that the proposer also asks your office to quote a level debt. Calculate thelevel debt.

8.5 The mortality of a certain impaired life aged exactly 60 may be represented as follows:

(1) at ages up to exact age 65, there is a constant addition of 0.009569 to the “normal” forceof mortality, which is that of A1967-70 select; and

(2) at ages between exact age 65 and exact age 70, mortality follows that of a “normal”(ultimate) life 4 years older.

Suppose that this life requires a 10-year endowment assurance without profits with sum assured£10, 000 payable on maturity or at the end of the year of earlier death. The life wishes, however,to pay the “normal” level annual premium for his age. Calculate the level debt which wouldbe deducted from the normal death benefit, using the basis given below:

mortality: as given above

interest: 4% p.a.

expenses: 2 12% of all premiums, with an additional initial expense of1% of the sum assured (before deducting any debt.)

8.7. SOLUTIONS 147

Solutions

8.1 Let P be the normal annual premium. Then

a57 × P = 10, 000A57 at 3%

P ; 377.41

Let E be the extra premium charged for 3 years. Then, the equation of value will be (* denotesimpaired mortality):

(P + E)a∗57:3

+ P 3|a∗57 = 10, 000A∗57

wherea∗57:3 3%

= a57:3 6% = 2.804

3|a∗57 = 1.06−3 l60l57

a60 at 3% = 11.076

A∗57 = A∗ 157:3 + 3E

∗57A60 at 3%

= A∗57:3

− 3E∗57 + 3E

∗57A60 at 3%

= 1− d3%a57:3 6% − 1.06−3 l60l57

(1−A60 at 3%)

= 0.59572

Hence

E =10, 000A∗57 − P (a∗

57:3+ 3|a∗57)

a57:3

= 256.3207

8.2 PNORMAL = 1000P55:10 = £85.84

P ∗ = 1000

[1

a∗55:10

︸︷︷︸

a∗55:10

= a∗55:5

+ [5E∗55]a

∗60:5

= a55:5 at 5% + v5 l60l55

at 5% [a68:5 at 4%]

= 7.677

− d

︸︷︷︸at 4%

]

∴ P ∗ = £91.79 so extra annual premium = £5.95

8.3 Let µx refer to A67-70 ult., µ∗x to impaired lives.

µ∗40+t =

{µ40+t + 0.0005 + 0.0001t (0 ≤ t ≤ 20)µ40+t + 0.0025 (t > 20)


(i)

20p∗40 = exp

{−

∫ 20

0

(µ40+t + 0.0005 + 0.0001t) dt

}

= 20p40 exp

{−

[0.0005t +

12(0.0001t2)

]20

0

}

=l60l40

× exp {−(0.01 + 0.02)}= 0.895579× 0.9704455 = 0.86911

∴ ans = 0.1309

(ii)

30p∗40 = 0.86911× 10p

∗60

= 0.86911× l70l60

× exp[−10× 0.0025]

= 0.66656∴ Ans = 0.3334

(iii) 0.3334−0.1309=0.2025

8.4 (i) The debt cannot, of course, exceed the “normal” sum assured on death. The level debtmust be (very roughly) about one-half of the initial debt (since the average dim. debt isone-half of the initial debt), so we may have cases in which:

initial debt(when debts diminish)

> S.A. > level debt(in cases of serious impairment)

(ii) (a)

Normal A.P. =40, 000P40:20

0.94= 1454.89

If initial debt is 19D, we have the equation of value0.94(1454.89)a50:20 = 40, 000A50:20 −D{20A 1

50:20 − (IA) 150:20 }

i.e.D =

40, 000× 0.49818− 0.94× 1454.89× 13.04720× 0.16819− 2.04074

= 1, 575.26

∴ Initial debt = £29, 929.94 (£29, 930)

(b) Replace D{20A 150:20 − (IA) 1

50:20 } in above by the term DA 150:20 , to give

D =2084.170.16819

= £12, 392

8.5 Let P be the normal A.P.

0.975P a[60]:10

︸︷︷︸7.884

= 10, 000 A[60]:10

︸︷︷︸0.69677

+ 100

∴ P = £919.45

8.7. SOLUTIONS 149

Let ∗ indicate special mortality.

a∗[60]:10

= a[60]:5 0.05

︸︷︷︸4.452

+ v50.05

l65l[60]

︸︷︷︸0.72603

a69:5 0.04

︸︷︷︸4.29496

= 7.570

Let level debt be D.

0.975P a∗[60]:10

︸︷︷︸7.570

= 10, 000 A∗[60]:10

︸︷︷︸1− da∗

[60]:10= 0.708835

− D A∗ 1[60]:10

︸︷︷︸

0.708835− [v5.05v

5.04]

l65l[60]

l74l69

= 0.708835− 0.478114= 0.230719

+ 100

∴ D =402.1190.230719

= £1, 743


Chapter 9

PROFIT-TESTING

9.1 Principles of profit-testing

In this book we shall consider “conventional” life assurance business only, not unit-linked policies(which are considered in part D1.) The main new ideas in profit-testing are as follows:

(1) The sale of a without-profits policy is considered by the life office as an “investment” onthe part of the shareholders and/or with-profits policyholders. (In a mutual life office thereare no shareholders.) Similarly, the sale of a with-profits policy may be considered to be an“investment” on the part of the office’s shareholders, though the position is complicated by thefact that future profits are shared between the with-profits policyholders and the shareholders.

(2) The office calculates the expected net cash flow, and the expected profit, in each policy year intwo stages.

Stage A. The net cash flow in each policy year, and the profit after allowing for appropriatereserves, of a given contract is calculated on the assumption that the policy is still in force atthe start of that year. The resulting net cash flow is called the in force net cash flow, and thevector of profits (indexed by the policy year, t) is called the profit vector, which is written as{(PROt)}.Stage B. The estimated net cash flow per policy sold in policy year t is called the initialnet cash flow for that year, and is found by multiplying the “in force” net cash flow by theprobability that the policy is still in force at the start of the tth policy year, t−1px. (We assumehere that the age at entry to assurance is x, and that a non-select mortality table is used.) Thecorresponding vector of profits (after allowing for reserves) is called the profit signature, whichis written as {σt}.

(3) All cash flows are considered at the end of the policy year, so premiums less expenses areaccumulated to the end of the appropriate policy year at the rate of interest which the officeassumes it will earn on the life funds.

(4) Random fluctuations in mortality are ignored; that is, each policy is considered as if it were oneof a large number of identical contracts whose experience exactly follows the mortality tableused by the office in its projections.

151

152 CHAPTER 9. PROFIT-TESTING

9.2 Cash flow calculations

We consider an n-year “generalised endowment assurance”, issued to (x), and use the followingnotation:

Pt = premium payable in policy year t (at the start of that year, i.e. payable at time t− 1 yearsfrom the issue date);

et = expenses assumed to be incurred in policy year t (at the start of that year);

Dt = death benefit in policy year t (paid at the end of the policy year);

St = survival benefit at the end of policy year t (St is usually zero for t < n);

i = rate of interest p.a. which the life office assumes it will earn on its life funds;

The symbols qx+t, t−1px, etc, refer to the mortality table which the life office assumes will apply tothe policyholders under consideration.

Note. In general, three distinct bases are used in profit-testing:

Basis 1: the premium basis, which is used only to calculate the premiums;

Basis 2: the reserving basis, which is used only to calculate the reserves;

Basis 3: the experience basis, which is that which, the office assumes, will be experienced.

In some cases all three of these bases agree, or two of them agree. Note that the symbols et, i andqx+t given above all refer to Basis 3, while Pt is calculated using Basis 1.We now define, for 1 ≤ t ≤ n,

(CF)t = the “in force” expected net cash flow in policy year t (9.2.1)We clearly have

(CF)t = (Pt − et)(1 + i)︸︷︷︸accumulationof premium

less expenses toend of year

− qx+t−1

︸︷︷︸expectedcost ofdeath

benefits

Dt − px+t−1

︸︷︷︸expectedcost ofsurvivalbenefits

St (9.2.2)

The “initial” expected net cash flow in year t, i.e. the net cash flow in year t per policy sold, isfound from the important equation:

“initial” net cash flow in year t = t−1px

︸︷︷︸prob. that policy will

still be in force attime t− 1, i.e. at

start of policy year t

(CF)t (9.2.3)

Example 9.2.1. An endowment assurance policy, with sum assured £5000, term five years and levelannual premiums, is issued to a life aged 55.

The annual premium is calculated on the following basis:

9.2. CASH FLOW CALCULATIONS 153


Interest: 6% per annum

Initial expenses: £250

Renewal expenses (associated with the payment of the second and each subsequent premium):£42 at the time of payment of the second premium, increasing thereafter by 5% per annum(compound).

The death benefit is payable at the end of the year of death.

(i) Show that the annual premium is £948.74

(ii) Assume that in calculating cash flows and profit signature for the policy the office uses thepremium basis.

On this basis determine for each year of the policy’s duration

(a) the ‘in force’ expected cash flow, and

(b) the ‘initial’ expected cash flow.

Solution.

(i) Let the annual premium be P . Then

P a55:5 = 5000A55:5 + 250 +4∑

t=1

42(1.05)t−1tp55v

t

(where all functions are on A1967-70 ultimate, 6%) , from which it follows that

P =5000× 0.75171 + 250 + 152.62

4.386= £948.74


(ii) We draw up the following table:

(1) (2) (3) (4) (5) (6)

Year Premium Expenses Interest DeathCost

SurvivalCost

t Pt et .06(Pt − et) 5000q54+t Stp54+t

1 948.74 250.00 41.92 42.20 0.00

2 948.74 42.00 54.40 47.10 0.00

3 948.74 44.10 54.28 52.50 0.00

4 948.74 46.30 54.14 58.45 0.00

5 948.74 48.62 54.01 64.95 4935.05

(7) (8) (9)

In forceexpectedcash flow

Prob. inforce

Initial expectedcash flow

(CF)t t−1p55

698.46 1.00000 698.46

914.04 .99156 906.33

906.42 .98222 890.30

898.13 .97191 872.90

−4045.87 .96054 −3886.22

Column(7) = (2)− (3) + (4)− (5)− (6)Column(9) = (7)× (8),

where column (8) gives the probability that the policy will be in force at the start of year t.

9.3. THE PROFIT VECTOR AND THE PROFIT SIGNATURE 155

9.3 The profit vector and the profit signature

We now suppose that the office maintains suitable reserves tV at the end of each policy year. Weassume that 0V = 0 and nV = 0 (i.e., any survival benefit payable when the contract matures attime n years has been paid.) Corresponding to the “in force” and “initial” net cash flows, we havethe profit vector and profit signature; that is,

(PRO)t = cash which will, according to office’sprojections, be transferred from thelife fund to the with-profits policyhold-ers/shareholders at the end of policy yeart, per policy in force at the start of thatyear

(9.3.1)

σt = as for (PRO)t, but per policy sold (9.3.2)

The relationship between these quantities is therefore

σt = t−1px(PRO)t (9.3.3)

Calculation of (PRO)t

We observe that(PRO)t = t−1V(1 + i)

︸︷︷︸accum. ofreservesuntil endof year

+ (CF)t

︸︷︷︸“in force”net cashflow inyear t

− px+t−1.tV

︸︷︷︸money neededfor reservesat end ofthe year

(9.3.4)

(since the chance that a given policy in force at the start of the policy year remains in force at theend is px+t−1.) This formula is sometimes written in the form

(PRO)t = (CF)t + it−1V

︸︷︷︸interest

onreserve

− (IR)t

︸︷︷︸increase

in reservesrequired

(9.3.5)

where

(IR)t = px+t−1.tV − t−1V (9.3.6)= money needed at end of the year

for reserves, less money held atthe start

We may then calculate {σt} by formula (9.3.3).

Example 9.3.1. Consider the policy described in example 9.2.1. Assuming that the premium andexperience basis are as described in that question and that the reserves held are as follows:

0V=0


1V=919

2V=1,876

3V=2,873

4V=3,914

5V=0

Calculate the profit vector and profit signature.

Solution. We draw up the following table:

(1) (2) (3) (4) (5) (6)

YearReserveat startof year

Survivalprob.

Increasein

reserve

Intereston

reserve

In forceexpectedcash flow

t t−1V p54+t (IR)t .06t−1V (CF)t

1 0.00 .99156 911.24 0.00 698.462 919.00 .99058 939.33 55.14 914.043 1876.00 .98950 966.83 112.56 906.424 2873.00 .98831 995.25 172.38 898.135 3914.00 .98701 −3914.00 234.84 −4045.87

(7) (8) (9)In forceprofit

for year

Prob.in force

Profitsignature

(PRO)t t−1p55 σt

−212.78 1.00000 −212.7829.85 .99156 29.6052.15 .98222 51.2275.26 .97191 73.15

102.97 .96054 98.91

Column(4) = p54+t.tV − t−1V

Column(7) = (6) + (5)− (4)Column(9) = (7)× (8),

where column (8) is the probability that the policy will be in force at the start of year t.

9.4. THE ASSESSMENT OF PROFITS 157

Example 9.3.2. A life office issues a 3-year without profits endowment assurance policy to a lifeaged 62. The sum assured of £1, 500 is payable on maturity or at end of the year of death, if within3 years, and there are level annual premiums of £472.50 payable in advance.The office uses the following “experience” basis:



initial expenses: £20

renewal expenses: £5 at the beginning of the second and third policy years.

The office’s reserve basis is as follows:

net premium method, using A1967-70 ultimate mortalityand 3% p.a. interest.

Determine the profit signature of this policy.

Solution. We first work out the “in force” net cash flows, (CF)t.

(1) (2) (3) (4) (5) (6)t Prem. Exp. Interest Death cost Maturity (CF)t = (1) + (2)

= 0.06((1)− (2)) = 15, 000q61+t Cost +(3)− (4)− (5)

1 472.50 20 27.15 26.62 0 453.032 472.50 5 28.05 29.48 0 466.073 472.50 5 28.05 ←−−−−− 1, 500 −−−−−→ −1,004.45

Now we find the profit vector, (PRO)t, and the profit signature, σt.

(1) (2) (3) (4) (5) (6) (7)Reserve at

startof year

Intereston

Reservep61+t (IR)t (PRO)t = t−1p62 σt

t = 0.06(1) (CF)t + (2)− (4) = (5)× (6)

1 0 0 0.98225 468.12 −15.09 1 −15.092 476.58 28.60 0.98035 466.07 14.71 0.98225 14.453 975.71

︸︷︷︸by

net premium methoddo not use office

premium in valuation

58.54 0.97826 −975.71 29.80 0.96295 28.70

9.4 The assessment of profits

The profit signature {σt} can be assessed in one or more of the following ways.


(1) One could work out the Internal Rate of Return (or yield) by solving the equation

n∑t=1

vtσt = 0 (9.4.1)

(the internal rate of return, i0, being the solution of this equation.)

(2) The shareholders may value the net profits at a certain rate of interest, j per annum. Thisrate is called the Risk Discount Rate, and may reflect uncertainties in {σt}, with j normallyhigher than i.

The net present value of the profits is thus

NPV(j) =n∑

t=1

vtσt at rate j (9.4.2)

(3) The Profit Margin is defined as

n.p.v. of profitsn.p.v. of premiums

, both at some rate of interest, im, say

=∑n

t=1 vtσt∑n−1t=0 Pt+1tpxvt

, at rate im (9.4.3)

(im is often equal to the risk discount rate, j.)

Example 9.4.1. A life office issues a three-year non-profit endowment assurance policy to a managed 30. The sum assured is £60,000 on maturity or at the end of the year of earlier death. Levelpremiums of £19,000 are payable annually in advance.The office maintains reserves as follows:

policy year reserve at endof policy year

1 £19,0002 £38,000

The office expects that its life funds will earn interest at 7% p.a. over the next 3 years.The office expects expenses to be as follows:

initial expenses: 10% of the first year’s premium,

renewal expenses: 2% of later premiums.

Mortality is expected to follow A1967-70 ultimate.Calculate

(i) the profit signature;

(ii) the net present value at the issue date of the profit to the office, using a risk discount rate of10% p.a.

9.5. SOME THEORETICAL RESULTS ABOUT {σT } 159

Solution.

(i) We work out the net cash flows, (CF)t, per policy in force at the start of the policy year.

(1) (2) (CF)t

t (Pt − et) (Pt − et)(1 + i) death + mat. costs = (1)− (2)1 17,100 18,297 39.22 18,2582 18,620 19,923 40.26 19,8333 18,620 19,923 60,000 −40,077

[ death cost = 60, 000q30+t−1

mat.cost (in final year only) = 60, 000p30+t−1 ]

We now work out the profit vector.

(1) (2) (3) (4)t tV (CF)t i.t−1V p30+t−1.tV − t−1V (PRO)t = (1) + (2)− (3)1 19,000 18,258 0 18,987 −7292 38,000 19,883 1330 18,975 2,2383 0 −40,077 2660 −38,000 583

Hence profit signature is:

−7292, 238× 0.999346 = 2, 237

583× 0.998676 = 582

(ii)−729v + 2, 237v2 + 582v3 at 10% = £1, 623

9.5 Some theoretical results about {σt}Theorem 9.5.1. If all three bases defined in section 9.2 agree,

σt = (PRO)t = 0 for all t

Proof. This follows from the formula

(PRO)t = t−1V(1 + i) + (CF)t − px+t−1.tV = 0

which is merely a restatement of the equation

(t−1V + Pt − et)(1 + i) = qx+t−1Dt + px+t−1St + px+t−1.tV

which holds (cf chapter 7) since all three bases agree.

Theorem 9.5.2. If the premium basis and experience basis agree, the internal rate of return earnedby the office on a sale of a policy is equal to the rate i p.a. earned by the office’s life funds.


Proof. We must show thatn∑

t=1

vtσt = 0 at rate i p.a.

Now we haven∑

t=1

vtσt =n∑

t=1

vt[t−1px(CF)t + t−1px.t−1V(1 + i)− tpx.tV]

Butn∑

t=1

vtt−1px(CF)t = 0

as this is merely a restatement of the equation of value for calculating premiums. Hence

n∑t=1

vtσt =n∑

t=1

vtt−1px(1 + i)t−1V −

n∑t=1

vttpx.tV

=n−1∑r=1

vrrpx.rV −

n∑t=1

vttpx.tV

= 0, using the fact that 0V = nV = 0

Corollary. If the life office wishes to obtain an I.R.R. on the sale of a policy in excess of the rateof interest it believes it will earn on the life funds, premiums must be higher than those calculatedon the “experience” basis.

Example 9.5.1. What is the I.R.R. associated with the profit signature of example 9.3.1?

Solution. Since the premium and reserving bases agree, the I.R.R. must be i = 6% p.a.

9.6 Withdrawals

So far we have ignored the possibility of surrender. We shall now assume that surrenders may occur,but only at the end of a policy year (just before payment of the premium then due). Let wt denotethe chance that a policy will be surrendered at the end of year t (t = 1, 2, . . . , n − 1). We mayassume that wn = 0, since the policyholder will receive the maturity benefits (if any) at that time.

Let (SV)t denote surrender value at time t. We observe that (PRO)t is as before, except thatthere are additional profits due to the surrender of

(px+t−1)wt

︸︷︷︸prob. that a

policy in forceat time t− 1

will be surrenderedat time t

[tV − (SV)t]︸︷︷︸profit at time tper surrenderat that time

(9.6.1)

9.6. WITHDRAWALS 161

Hence

(PRO)′t = profit vector at time t, allowing for surrenders= (PRO)t + px+t−1wt[tV − (SV)t] (9.6.2)

Note that (PRO)′t = (PRO) if tV = (SV)t for t = 1, 2, . . . , n− 1.We must also adjust σt, to allow for

(i) the change from (PRO) to (PRO)′, and

(ii) the changed probability of the policy being in force at time t− 1.

We now have

t−1p′x = probability that policy is in force at time t− 1

= t−1px(1− w1)(1− w2) . . . (1− wt−1)(9.6.3)

Henceσ′t = revised profit signature = t−1p

′x(PRO)′t (9.6.4)

Example 9.6.1. Consider the policy of Example 9.3.1, but now assume that at each of the durations1,2,3,4 years, 5% of the surviving policyholders will surrender, and that the S.V. is 98% of the office’sreserve. Find the new profit signature.

Solution. We work out the revised profit vector, (PRO)′t:

t (PRO)t + p55+t−1(0.05)(0.02× tV)∗1 −212.78 + 0.99156× .05× .02× 919 = −211.872 29.85 + 0.99058× .05× .02× 1, 876 = 31.713 52.15 + 0.98950× .05× .02× 2, 873 = 54.994 72.26 + 0.98831× .05× .02× 3, 914 = 79.135 102.97 + 0 = 102.97

* when t = 5, (PRO)′t = (PRO)t.

Hence we have:

t (PRO)′t t−1p′x σ′t

1 −211.87 1 −211.872 31.71 .99156× .95 29.873 54.99 .98222× .952 48.754 79.13 .97191× .953 65.935 102.97 .96054× .954 80.56

Notes

1. The I.R.R. allowing for withdrawals may be found by solving the equationn∑

t=1

vtσ′t = 0 (9.6.5)

2. Even if (PRO)′t = (PRO)t for t = 1, 2, . . . , n, {σ′t} is not the same as {σt} because the factors{t−1p

′x} are not the same as {t−1px}.

3. Unlike mortality rates, which are fairly predictable by actuaries, withdrawal rates depend to agreat extent on economic and commercial factors which are less easy to forecast accurately.


9.7 The actual emergence of profits

The actual profits emerging from the sale of a given policy depend, of course, on whether thepolicyholder dies (or surrenders the policy) during the term, and the year in which such an eventoccurs.Let us suppose here that there are no withdrawals, and use the symbols it and et to denote the actualinterest rate earned by the life funds in year t, and the actual expenses incurred at the beginning ofthat policy year. The actual year-end profits emerging in the n years (maximum) of the policy areas follows:

(1) If the policy is no longer in force at the beginning of policy year t, the profit emerging in thatyear is zero;

(2) If the policy is still in force at the beginning of the year, the profit is

(t−1V + Pt − et)(1 + it)−Dt (9.7.1)

if the life dies during the year, and

(t−1V + Pt − et)(1 + it)− St − tV (9.7.2)

if the life survives the year.

These formulae may be easily modified if the policy is altered, as in the following example.

Example 9.7.1. A life office issues a 5-year with-profits endowment assurance policy to a life agedexactly 60. The policy has a basic sum assured of £10,000 payable at the end of the year of deathor at the maturity date. Level premiums are payable annually is advance throughout the term ofthe policy. Simple reversionary bonuses vest at the start of each year, including the first.The premium is calculated according to the following basis:



simple reversionary bonuses at the rate of 4% per annum are assumed

initial expenses: 60% of the first premium

renewal expenses: 5% of each premium after the first

(i) Show that the premium is equal to £2,627.

(ii) The office holds net premium reserves using a rate of interest of 3% per annum and A1967-70ultimate mortality .

Calculate the profit signature for this policy, assuming that the office will earn interest at7% per annum on its assets, mortality follows the A1967-70 ultimate table, and expenses andbonuses will follow the premium basis.

9.7. THE ACTUAL EMERGENCE OF PROFITS 163

(iii) Immediately before the fourth premium was due (before the fourth bonus declaration) thepolicy was made paid-up, with no entitlement to further bonuses. The paid up sum assuredwas 60% of the benefits immediately before the alteration, including declared bonuses.

The policyholder survived to the maturity date. Interest was earned on the life funds was at6% per annum over the period of the contract, and bonuses in the first three years followedthe premium assumptions. Expenses followed the premium assumptions up to the alterationdate, and no expenses were incurred after the policy was made paid-up.

For each of the five years of the policy term, calculate the actual year-end profit earned onthe policy.

Solution.

(i) Let P be the annual premium. M.P.V. of premiums less expenses is

0.95P a[60]:5 − 0.55P = 3.7563P

M.P.V. of benefits is

10, 000A[60]:5 + 400(IA)[60]:5 = 10, 000× 0.82565 + 400[R[60] −R65 − 5M65 + 5D65

D[60]

]

= 9867.88

HenceP = £2, 627.02

(ii) The reserves held are:

1V = 10, 400A61:4 − 10, 000P60:5 a61:4 at 3% interest

= 10, 000(

1− a61:4

a60:5

)+ 400A61:4

= 2191.60

2V = 10, 000(

1− a62:3

a60:5

)+ 800A62:3 = 4, 475.67

3V = 10, 000(

1− a62:3

a60:5

)+ 1200A63:2 = 6, 862.31

4V = 10, 000(

1− a64:1

a60:5

)+ 1600A64:1 = 9, 366.17


This gives the following table:

YearPremiums Expenses Reserve

t−1Vat start

Interest Cost ofClaims

Cost ofReservesat End

Profit perpolicy in force

at start ofyear

t (1) (2) (3) (4) (5) (6) (7)1 2627.02 1576.21 0 73.56 150.10 2159.97 −1185.712 2627.02 131.35 2191.60 328.11 172.95 4404.00 438.433 2627.02 131.35 4475.67 487.99 198.80 6740.51 520.024 2627.02 131.35 6862.31 655.06 227.99 9182.08 602.975 2627.02 131.35 9366.17 830.33 12,000 0 692.17

Col. (4) = 0.07[(1)− (2) + (3)]Col. (5) = cost of death and maturity claims

(including bonuses)Col. (6) = p60+t−1.tV

Col. (7) = (1)− (2) + (3) + (4)− (5)− (6)

Hence the profit signature for the contract is

(−1185.71, 483.43p60 . . . 692.174p60) = (−1185.71, 432.10, 504.31, 574.37, 646.38)

(iii) We draw up the following table:

Year

Pt − et

Reserveat start

t−1V

Interest Reserveat end

tV

Survivalbenefits

Profit

t (1) (2) (3) (4) (5) (6)1050.81 0 63.05 2191.60 0 −1077.74

2 2495.67 2191.60 281.24 4475.67 0 492.843 2495.67 4475.67 418.28 6862.31 0 527.314 0 6862.31 411.74 6524.25∗ 0 749.805 0 6524.25 391.46 0 6720 195.71

∗6720A64:1 at 3% = 6524.25; the sum assured after alteration is 0.6[10, 000+3×400] = 6720,and the office calculates reserves by the net premium method.

Col. (3) = 0.06[(1) + (2)]Col. (6) = (1) + (2) + (3)− (4)− (5)

9.8. EXERCISES 165

Exercises

9.1 (i) In the context of profit-testing, explain the difference between the “profit vector” andthe “profit signature”.

(ii) A certain life office sells assurance policies with term 3 years to lives aged 70. For eachpolicy, the profit vector is estimated to be (−50, 30, 30). Given that the mortality of thepolicyholder is expected to follow A1967-70 ultimate, calculate(a) the profit signature per policy sold;(b) the net present value of the profit to the office on the basis of a risk discount rate of

8% per annum.

9.2 A life office issues a 5-year guaranteed bonus endowment assurance policy to a life aged 60,with basic sum assured £30,000. The sum assured, with attaching bonuses, is payable at theend of the year of death or at maturity. Level premiums are payable annually in advance.The office holds net premium reserves, using the basis A1967-70 ult at 3% p.a. Interest onpremiums and reserves is expected to be earned at an effective rate of 8% p.a. Bonuses will bedeclared annually at a rate of 3% of the basic sum assured.

Bonuses vest at the start of each policy year. Expenses of 40% of the first year’s premiums and5% of subsequent years’ premiums will be incurred. Mortality is expected to follow A1967-70ultimate. Withdrawals may be ignored.

(i) For each policy year, calculate the total sum assured.

(ii) For each duration t = 1, 2, 3, 4 years, calculate the reserve (tV) immediately before pay-ment of the premium then due, given the following values on A1967-70 ultimate, 3% p.a.interest:

(IA)60:5 = 4.1853(IA)61:4 = 3.4681(IA)62:3 = 2.6973(IA)63:2 = 1.8672(IA)64:1 = v = 0.97087

(iii) (Difficult.) Calculate the annual premium (P ) required for the shareholders/with profitspolicyholders to achieve an internal rate of return of 12% p.a. on the sale of this contract.

9.3 Ten years ago a life office issued a large block of 10-year without profits endowment assurancesto lives then aged 30. Each policy was effected by annual premiums, and had a sum assuredof £40,000, payable on survival or at the end of the year of death. The office’s premium basiswas

A1967-70 ultimate

5% interest

expenses of 2% of all premiums with additional initial expenses of 0.5% of the sum assured.

It was found that mortality , interest and expenses followed these assumptions, but there weresurrenders just before payment of the premiums due at durations 1,2 and 3 years. At each ofthese times, 3% of the surviving policyholders surrendered their contracts, and were given asurrender value equal to the office’s reserve, which was calculated on the premium basis, minusa surrender penalty of £40, which the office transferred to the surplus account. By using aprofit-testing approach, or otherwise, calculate the surplus accruing to the office at the end ofeach of the first three policy years, per policy sold.Hint. Note that (PRO)t = 0, so profits arise only from surrenders.


9.4 Your office is considering the issue of 3-year annual-premium endowment assurance policieswithout profits to lives aged 62. In respect of a policy with sum assured £10, 000, payable atthe end of the year of death (if within 3 years) or on maturity, calculate the net present valueof the profit signature on the following assumptions:

premium basis: mortality: A1967-70 ultimateinterest: 6% p.a.expenses: 3% of all premiums.

reserve basis: net premium method using A1967-70 ultimate,4% p.a. interest

rate of interest tobe earned in life fund: 8% p.a.

expenses: 3% of office premiums


risk discount rate: 10% p.a.

9.5 A life office issues 3-year term assurance policy to a man aged exactly 59. The sum assuredis £15, 000, payable at the end of the year of death. Level premiums are payable annually inadvance. Expenses are expected to be as follows:

initial expenses: £10

renewal expenses: £2 incurred at the beginning of the 2nd and each subsequent policy year.

It is assumed that interest of 7% per annum will be earned on the life funds, and that mortalityfollows the A1967-70 ultimate table. The risk discount rate used by the office is 15% per annum.The office calculates the annual premium by requiring that the net present value of the expectedprofit on each policy is equal to 20% of one office premium. Calculate the office premium oneach of the following reserving bases:(i) The office holds zero reserves at each year-end.(ii) The office sets up a reserve at each year-end (except the last) equal to 80% of one office

premium.

9.9. SOLUTIONS 167

Solutions

9.1 (i) The profit vector refers to expected profits per policy in force at start of year, whilst theprofit signature refers to profits per policy sold. Relationship is

σt = (PRO)t × prob. that policy is in force at time t− 1

(ii) (a) σt = t−1p70(PRO)t, so

σ1 = −50

σ2 = 30(

l71l70

)= 28.83

σ3 = 30(

l72l70

)= 27.59

(b) N.P.V. = σ1v + σ2v2 + σ3v

3 at 8%

= −50v + 28.83v2 + 27.59v3

= −46.30 + 24.72 + 21.90 = £0.32

9.2 Note: In (ii), we need the increasing assurance factors (A65+t:5−t and a65+t:5−t are directlytabulated)

(i) policy year t total S.A.1 30,9002 31,8003 32,7004 33,6005 34,500

(ii) The formula istV = (30, 000 + 900t)A60+t:5−t + 900(IA)60+t:5−t − P1a60+t:5−t on A67-70 ult. 3%

whereP1 = net premium on A67-70 ult. 3%

=30, 000A60:5 + 900(IA)60:5

a60:5

=30, 000× 0.86682 + 900× 4.1853

4.572= £6511.67

This leads to the table (note that 0V = 5V = 0):

t tV1 30, 900A61:4 + 900(IA)61:4 − P1a61:4 = 6, 3532 31, 80062:3 + 900(IA)62:3 − P1a62:3 = 12, 9473 32, 70063:2 + 900(IA)63:2 − P1a63:2 = 19, 8114 33, 60064:1 + 900(IA)64:1 − P1a64:1 = 26, 983

(iii) We first work out (CF)t (in terms of P ):


(1)prs. less exp., with interest (2)death costs (3)survival costs (1)−(2)−(3)t (P − et)(1.08) q59+tDt p59+tSt (CF)t

1 0.648P 445.96 0 0.648P − 445.962 1.026P 509.23 0 1.026P − 509.233 1.026P 580.42 0 1.026P − 580.424 1.026P 660.39 0 1.026P − 660.395 1.026P 34500

︸︷︷︸since sum paid ondeath or survival

1.026P − 34, 500

We now work out (PRO)t:

(1)interest on reserve (2)increase in reserve (3)cash flow (1)-(2)+(3)t 0.08t−1V (IR)t = p59+t.tV − t−1V (CF)t (PRO)t

1 0 6261.31 0.648P − 445.96 0.648P − 67072 508.24 6386.67 1.026P − 509.23 1.026P − 63883 1035.76 6512.36 1.026P − 580.42 1.026P − 60574 1584.88 6641.66 1.026P − 660.39 1.026P − 57175 2158.64 −26983.00 1.026P − 34, 500 1.026P − 5358

We have the equation5∑

t=1

vtt−1p60(PRO)t = 0 at 12% interest

i.e.(0.648P − 6707)v

+(1.026P − 6388)(0.985568)v2

+(1.026P − 6057)(0.969785)v3

+(1.026P − 5717)(0.952572)v4

+(1.026P − 5358)(0.93385)v5

= 0

∴ 3.2577P = 21, 488∴ P = £6, 596

(Rough check: right order of magnitude:- P =' P1, and 5P ' 34, 500 (maturity benefit)).

9.3 Profits arise only in respect of surrenders at times 1,2,3. The profit at time t years (t = 1, 2, 3)per policy in force at start of year is

(PRO)′t = (1− q30+t−1)(0.03) (40)︸︷︷︸

surrenderpenalty

= 1.2p30+t−1

The probability of being in force at the start of year t ist−1p

′30 = t−1p30(1− 0.03)t−1 t=1,2,3

∴ Expected surplus arising at time t, per policy sold, isσ′t = (1.2)0.97t−1

tp30

9.9. SOLUTIONS 169

t 1.2(0.97)t−1tp30 = σ′t

1 1.2× 0.99935 = £1.202 1.2× 0.97× 0.99868 = £1.163 1.2××0.972 × 0.99798 = £1.13

9.4 Let P = annual premium

P =10, 000P62:3 .06

0.97=

10, 000× 0.302630.97

= £3, 119.90

Reserves: 0V = 0, 1V = 10, 0001V62:3 .04 = 10, 000(

1− a63:2

a62:3

)at 4%

= 3, 146.38

2V = 10, 000(

1− 5.1386.519

)

= 6, 472.66, 3V = 0

Net cash flows (ignoring reserves)

t (CF)t

1 0.97P (1.08)− 10, 000q62 = 3, 090.912 0.97P (1.08)− 10, 000q63 = 3, 071.863 0.97P (1.08)− 10, 000 = −6, 731.59

Profit vector and signaturet tV (PRO)t = (CF)t + i.t−1V − [p62+t−1.tV − t−1V] t−1p62 σt

1 3,146.38 3, 090.91 + 0− 3, 090.53 = 0.38 1 0.382 6,472.66 3, 071.86 + 251.71− 3, 199.06 = 124.51 0.98225 122.303 0 −6, 731.59 + 517.81 + 6, 472.66 = 258.88 0.96294 249.29

∴ N.P.V. = 0.38v + 122.30v2 + 249.29v3 at 10%= £288.71


9.5 (i) Let A.P. be P(1) (2) (3) (4) (5) (6)

Year Premiums Interest Expected (CF)t = (PRO)t t−1p59 σt = (4)× (5)less expenses = 0.07× (1) death costs = (1) + (2)− (3)

1 P−LO 0.07P − 0.7 194.91 1.07P−205.61 1 1.07 P−205.612 P−2 0.07P − 0.14 216.49 1.07P−218.63 0.987006 1.0561P−215.793 P−2 0.07P − 0.14 240.20 1.07P−242.34 0.972761 1.0409P−235.74

We solve the equation:0.2P = vσ1 + v2σ2 + v3σ3 at 15%

∴ P =205.61× 1.15−1 + 215.79× 1.15−2 + 235.74× 1.15−3

−0.2 + 1.07× 1.15−1 + 1.0561× 1.15−2 + 1.0409× 1.15−3

=496.962.2133

= £224.53

(ii)(1) (2) (3) (4) (5) (6)

Year Reserve Interest (CF)t Reserve Increase in (PRO)t =at start on reserve at end reserve

of year = 0.07× (1) of year = p59+t−1.tV − t−1V = (2) + (3)− (5)

1 0 0 1.07P−205.61 0.8P 0.7896 0.28039P−205.612 0.8P 0.056P 1.07P−218.63 0.8P −0.01155 1.13755P−218.633 0.8P 0.056P 1.07P−242.34 0 −0.8 1.926 P−242.34

(7)σt =

t−1p59.(PRO)t

0.28039P−205.611.12277P−215.761.87354P−235.74

We solve the equation:0.2P = vσ1 + v2σ2 + v3σ3 at 15%∴ P = £233.90

Chapter 10

STATIONARY POPULATIONS

10.1 Some Definitions

We recall that◦ex = complete expectation of life at age x

= E(T )

where T refers to the future lifetime of (x) in years, including fractions.Note that

◦ex=

∫ ∞

0

t.tpxµx+t dt (10.1.1)

Integrating by parts gives

◦ex = [t(−tpx)]∞0 +

∫ ∞

0tpx dt

=∫ ∞

0tpx dt

=∫ ∞

0

lx+t

lxdt (10.1.2)

(as it may be shown that limt→∞(t.tpx) = 0 when◦ex is finite.)

We recall also that

ex = curtate expectation of life at age x

= E(K)

where K is the integer part of T .We find that

ex =∞∑

t=1

tpx (10.1.3)

and, using the Euler–Maclaurin formula,

ex '◦ex −12

171

172 CHAPTER 10. STATIONARY POPULATIONS

Define

Tx =∫ ∞

0

lx+t dt

=∫ ∞

x

ly dy (on setting y = x + t)


◦ex =

∫∞x

ly dy

lx

=Tx

lx

and hence

Tx = lx◦ex (10.1.4)

A stationary population is an idealised large population such that (in an old-fashioned mathe-matical notation) the number of lives in the population between ages x and x + dx is always equalto

klx dx

where k is called the scaling factor and lx is according to some life table.By integrating over all ages greater than or equal to x, we can see that

Total population aged ≥ x =∫ ∞

x

kly dy

= k.Tx (10.1.5)

The position is illustrated in Figure (10.1.1).

Figure 10.1.1: a stationary population

(It is nearly always advisable to draw a diagram in stationary population questions.) In any time-interval of length t years, the number of lives attaining a specified exact age x is klx.t, so the rate atwhich lives attain age x is klx per annum, i.e. klx lives ‘flow’ continuously over exact age x each year.

Note that (if lives are considered at all ages) there will be kl0 births each year. Also, the num-ber of deaths each year between ages x1 and x2 (x1 < x2) is

k(lx1 − lx2) (10.1.6)

The number of lives in the population between any two ages x1 and x2 (x1 < x2) is

k(Tx1 − Tx2) (10.1.7)

10.1. SOME DEFINITIONS 173

Example 10.1.1. The membership of a certain learned society is stationary at 11,500. Membersenter only at exact age 50. They are subject to the mortality of English Life Table No.12 - Males,and there are no withdrawals.

What is the annual number of entrants ?

SolutionOne knows that

k.T50 = 11, 500

Sokl50

◦e50= 11, 500

The annual number of entrants is

kl50 =11, 500◦e50

= 507.05 or 507.

Example 10.1.2. A large company has for many years maintained a staff in a stationary conditionby recruiting 500 annual entrants at exact age 20, uniformly over the year. If the staff retire at age60, there are no withdrawals, and English Life No. 12 – Males mortality is experienced, find:

(a) the size of the staff,(b) the number of staff who retire each year,(c) the number of pensioners.

Solution

(a) We havekl20 = 500.

The total number of staff is

k(T20 − T60) =500l20

(l20◦e20 −l60

◦e60)

= 19, 115.

(b) Number who retire each year

= kl60

=(

500l20

)l60

= 410

(c) Number of pensioners

= kT60

=(

500l20

l60

)◦e60

= 6, 172.


Some more symbolsWe let

Lx = the number of lives between ages x

and x + 1 in a stationary population with scaling factor k = 1.= Tx − Tx+1

=∫ 1

0

lx+t dt (10.1.8)

We may use the approximations

Lx ' lx+ 12' 1

2(lx + lx+1)

Another definition is

◦ex:n = average lifetime lived by (x) between ages x and x + n

= E(T ∗)

where the random variable T ∗ is defined as follows:

T ∗ =

{T if (x) dies within n yearsn if (x) lives for n years.

The variable T ∗ has p.d.f. tpxµx+t for 0 < t < n, and there is discrete probability npx that T ∗ = n.Hence

E(T ∗) =∫ n

0

t.tpxµx+t dt + n.npx

= [−t.tpx]n0 +∫ n

0tpx dt + n.npx (by integration by parts)

=∫ n

0tpx dt

Hence

◦ex:n =

Tx − Tx+n

lx

=◦ex − lx+n

lx

◦ex+n (10.1.9)

(Notice that ax:n =◦ex:n if i = 0 .)

Similarly,

ex:n = average number of complete years lived by(x) between ages x and x + n

=n∑

t=1

tpx =lx+1 + lx+2 + ... + lx+n

lx

=ex − lx+n

lxex+n (10.1.10)

10.2. THE CENTRAL DEATH RATE 175

Example 10.1.3. You are given that

lx = l1

[1− log x

9 log 10

](1 ≤ x ≤ 11)

Find◦e1:10 .

Solution

◦e1:10 =

∫ 10

0

lt+1

l1dt =

∫ 11

1

lxl1

dx

=∫ 11

1

[1− log x

9 log 10

]dx

= 10− 19 log 10

[x log x− x]x=11x=1

= 10− 11 log 11− 109 log 10

= 9.2097.

The average age at death of those who die between ages x and x + n may be worked out asfollows. We have

E(T ∗) = E(T ∗|T ∗ < n)Pr{T ∗ < n}+ E(T ∗|T ∗ ≥ n)Pr{T ∗ ≥ n}so that ◦

ex:n = E(T ∗|T ∗ < n)nqx + n.npx

Hence the average age at death of those who die between ages x and x + n is

x + E(T ∗|T ∗ < n)

=x +◦ex:n −n.npx

nqx

=x +Tx − Tx+n − nlx+n

lx − lx+n(10.1.11)

Note that this result refers to a randomly-chosen life aged x: we need not assume that there is astationary population.

10.2 The Central Death Rate

Define

hmx = the central death rate between ages x and x + h

=lx − lx+h∫ x+h

xly dy

=number of deaths each year between ages x and x + h

population aged between x and x + h in a stationary population

If h = 1, it is omitted, giving

mx =dx

Lx(10.2.1)


Note that

mx =

∫ 1

0lx+tµx+t dt∫ 1

0lx+t dt

' µx+ 12

(10.2.2)

10.3 Relationships Between mx and qx

Assume that there is a Uniform Distribution of Deaths (U.D.D.) between ages x and x+1, i.e. lx+t

is linear for 0 ≤ t ≤ 1.We have

Lx =∫ 1

0

lx+t dt =12(lx + lx+1) (10.3.1)

(as the trapezoidal rule is exactly correct, not just an approximation).Since lx+1 = lx − dx, we have

Lx = lx − 12dx (10.3.2)

Hencemx =

dx

Lx

=dx

lx − 12dx

=qx

1− 12qx

(10.3.3)

We may rearrange this equation to getqx =

mx

1 + 12mx

(10.3.4)

If U.D.D. does not hold, these results may be used as approximations.

10.4 Stationary Funds

Consider a pension scheme, etc., such thatB = annual benefit outgo for scheme (plus expenses, if any)

andC = annual contribution income

are constant.If the funds, F , are such that the interest on them pays B − C, then the funds will also remain

constant.There are two cases:(a) If income and outgo are received and paid continuously, then

δ.F = B − C (10.4.1)

(B, C are rates of payment per annum).(b) If income and outgo are received and paid only at the end of each year, then

iF = B − C (10.4.2)

Note: In case (b), F denotes the fund at the start of the year.

10.4. STATIONARY FUNDS 177

Example 10.4.1. Each year for many years a life office has issued 10,000 temporary assurancepolicies each with a term of ten years and a sum assured of £5,000 to lives aged 25 exactly.

One third of those who survive to age 35 then effect a without profits whole life policy for thesame sum assured as the term policy, and one quarter effect a 25-year without profits endowmentassurance for twice that sum assured. All premiums are payable annually in advance and deathclaims are paid at the end of the year of death. Policies are issued uniformly throughout the year.

The office calculates premiums on A1967-70 ultimate 4%, ignoring expenses. If the office’s ex-perience follows this basis, calculate the size of the fund held for these contracts.

SolutionThe term assurance premium = 5, 000M25−M35

N25−N35= £3.29.

Whole life premium = 5, 000P35 = £56.85.Endowment assurance premium = 10, 000P35:25 = £245.30.

Annual premium income is

10, 000[1 + e25 − l35

l25(1 + e35)

]× 3.29

+ 10, 000l35l25

× 13(1 + e35)× 56.85

+ 10, 000l35l25

× 14

[1 + e35 − l60

l35(1 + e60)

]× 245.30

= 328, 003 + 7, 634, 065 + 14, 769, 057= £22, 731, 125.

Annual rate of payment of claims is

10000[5000

(l25 − l35

l25

)+

l35l25

(13× 5000 +

14× 10, 000

)]

= £41, 723, 716.

Hence δ.F = 41, 723, 716− 22, 731, 125= 18, 992, 591

where i = 0.04.Therefore Fund = 18992591

δ = £484, 249, 000.


Exercises

10.1 The staff of a large company is maintained as a stationary population by 500 new entrantseach year at exact age 20. One third of those reaching age 30 leave immediately. Of theremainder, 1

4 of those attaining age 60 retire immediately and the survivors retire at age 65.The only other decrement is death.Calculate

(i) the number of staff,

(ii) the number of deaths in service each year.

Basis: English Life Table No. 12 - Males

10.2 A certain country’s school system provides education for all children between the ages of 5and 16 exactly. The country’s population is stationary, there being 100,000 births per year,uniformly distributed over the year. The population of the country is subject to the mortalityof English Life Table No. 12 - Males.

(i) Find the number of pupils at any given time.

(ii) The country’s teacher training colleges are such that a constant flow of new entrants tothe profession is maintained. Teachers are recruited uniformly over the year, and the ratio ofpupils to teachers is 20 to 1. All teachers enter the profession at age 21 and retire at age 60,there being no withdrawals. Find the annual number of new teachers recruited.

10.3 For many years a company has recruited, uniformly over each year, 200 employees on their20th birthdays and a fixed number of additional employees on their 25th birthdays. Mor-tality has followed English Life Table No. 12 - Males. Employees may retire on their 60thor 65th birthdays, and one third of employees reaching their 60th birthdays retire on thatdate. Employees leave the company only through death or retirement, and the total numberof employees is 10,000.

Find the total number of new recruits each year.

10.4 The male population of a certain country has been stationary for many years, there being100,000 male births per annum, spread uniformly over the year; the mortality of males followsEnglish Life Table No.12 - Males, and migration may be ignored.

(a) Calculate the size of the country’s male population at any time.

(b) The government of the country has decided to introduce a social security plan, underwhich all employed men between ages 15 and 65 must contribute a fixed sum every week, thesame sum being also payable by their employers. Men over age 65 will receive a pension of 100units of currency per week, and those sick or unemployed between ages 15 and 65 also receivethis amount. If it may be assumed that at any time 95% of men between ages 15 and 65 areemployed, while the remaining 5% are sick or unemployed, calculate the weekly contributionpayable by each employed worker. The scheme is to be financed on a pay-as-you-go basis (nofund is built up) and administrative costs are to be ignored.

10.5 A large manufacturing company has for many years staffed one of its divisions by the recruit-ment, uniformly over each year, of 1,000 staff at exact age 20. At the end of one year in thejob, new staff are examined for suitability, and 20% are dismissed. All employees are assessedat age 35, and 50% are immediately moved out of the division. At age 40, all remaining em-ployees are moved out of the division. Death is the only other reason for leaving the division.Staff experience mortality according to English Life Table No. 12 - Males.Calculate the number of staff in the division.

10.5. EXERCISES 179

10.6 In a certain country, all civil servants are recruited on their 25th birthdays. (Birthdays areassumed to occur uniformly over the calendar year). Of those who reach exact age 40, 10%obtain employment in private companies and leave the civil service immediately. All remainingcivil servants retire when they reach age 60 or die in service before this age. Civil servantshave for many years experienced the mortality of English Life Table No. 12 - Males, and willdo so for the indefinite future. The population of civil servants has been stationary for manyyears, and civil servants are recruited at the rate of 3,000 each year.

(i) How many civil servants are in service at any given date?

(ii) The government of the country has just decided to reduce the size of the civil service by10%. This is to be done by immediately lowering the retirement age of civil servants. Findthe new retirement age (to the nearest month), assuming that there are no other changes.

10.7 In a mortality table with a one-year select period, the following relationships apply at age x:

l[x] =12(lx + lx+1)

tq[x] = t.q[x] (0 ≤ t ≤ 1)

(i) Express p[x] in terms of px.

(ii) Express e[x] in terms of ex and px.

(iii) Express◦e[x] in terms of

◦ex+1 and px.

(iv) Express m[x] in terms of px.

10.8 For many years a life office has issued a steady flow of 10-year endowment assurances withoutprofits to lives aged 55. Premiums are payable continuously and the sum assured is payableimmediately on death, if death occurs before age 65. Premiums are calculated on the followingbasis:mortality: English Life Table No. 12 – Malesinterest: 4%expenses: 5% of all office premiums, plus 25% of office premiums in the first year.

Mortality, interest and expenses have always been in accordance with this basis. The sumsassured issued each year have always been £5,000,000, spread uniformly over the year. Nopolicies terminate except by death or maturity.In respect of this business, calculate

(i) the annual rate of gross premium income,

(ii) the annual rate of payment of maturity claims,

(iii) the annual rate of payment of death claims, and

(iv) the size of the fund which has been built up.


Solutions

10.1 (i) kl20 = 500.

Number of Staff = k

(T20 − 1

3T30 − 1

6T60 − 1

2T65

)

=500l20

(l20

◦e20 −1

3l30

◦e30 −1

6l60

◦e60 −1

2l65

◦e65

)

= 15, 361

(ii) Number of deaths is

500− 500l20

(13l30 +

16l60 +

12l65

)

= 89.

10.2 (i) kl0 = number of births each year = 100,000As l0 = 100, 000, k = 1.

Number of pupils = k(T5 − T16)

= l5◦e5 −l16

◦e16

= 1, 066, 409

(ii) Number of teachers = 120 × 1, 066, 409 = 53, 320.

Let annual number of recruits be c.Then the equation is

c

l21(T21 − T60) = 53, 320

Hence c = 1, 431 teachers per annum.

10.3 kl20 = 200 employees enter at age 20 per annum. Let cl25 be the number who enter at age 25per annum.Then the number of employees is

k

(T20 − 1

3T60 − 2

3T65

)+ c

(T25 − 1

3T60 − 2

3T65

)= 10, 000

where k = 200l20

.Solving this for c gives

c =10, 000− 8157.81

3, 447, 483= 0.00053436

Hence number recruited each year = 200 + cl25

= 251

10.4 (a) kl0 = 100, 000Hence k = 1.

10.6. SOLUTIONS 181

Total male population = kT0

= 100, 000◦e0

= 6, 809, 000

(b) Number of pensioners = kT65

Number aged between 15 and 65 who receive benefit = 0.05k(T15 − T65).Number of contributors = 0.95k(T15 − T65).

Let the employee s contribution be c per week.Hence 2c× (number of contributors) = 100× (no. of beneficiaries)Therefore

c =100[T65 + 0.05(T15 − T65)]

2× 0.95(T15 − T65)T15 = 5, 352, 735 and T65 = 818, 455 givingc = 12.13 units.

10.5 kl20 = number of staff recruited per annum

= 1, 000

Number of staff = k(T20 − T21) + 0.8k(T21 − T35) + 0.4k(T35 − T40)

=1000l20

(T20 − 0.2T21 − 0.4T35 − 0.4T40)

= 14, 060

10.6 (i) kl25 = 3, 000Number of civil servants is

k(T25 − 0.1T40 − 0.9T60)

=3000l25

(l25◦e25 −0.1l40

◦e40 −0.9l60

◦e60)

= 94713

(ii) Number of civil servants required is0.9× 94713 = 85, 242

Let x be the new age of retirement.Then

3000l25

(T25 − 0.1T40 − 0.9Tx) = 85, 242

Hence Tx = 1, 524, 475

T55 = 1, 602, 333T56 = 1, 516, 571

So using linear interpolation givesx ' 55.91' 55 years and 11 months


10.7 (i)

p[x] =lx+1

l[x]=

lx+112 (lx + lx+1)

=2px

1 + px

(ii)

e[x] =lx+1 + lx+2 + ...

l[x]

=lxl[x]

· ex =p[x]

px· ex =

(2

1 + px

)ex

(iii)◦e[x] =

1l[x]

[∫ ∞

0

l[x]+t dt]

=1

l[x][∫ 1

0

l[x]+t dt +∫ ∞

1

lx+t dt]

=1

l[x][12(l[x] + lx+1) + lx+1.

◦ex+1]

(as tq[x] = t.q[x], l[x]+t is linear for 0 ≤ t ≤ 1)

=12(1 + p[x]) + p[x].

◦ex+1

=12

(1 + 3px

1 + px

)+

(2px

1 + px

)◦ex+1

(iv)

m[x] =

∫ 1

0l[x]+tµ[x]+t dt∫ 1

0l[x]+t dt

=l[x] − lx+1

12 (l[x] + lx+1)

=1− p[x]

12 (1 + p[x])

=2(1− px)1 + 3px

(using part (i))

10.8 (i) Let P ′ be the office annual premium per £1 sum assured.We have

0.95P ′a55:10 = A55:10 + 0.25P ′a55:1

Hence P ′ = 0.100663.

The total sums assured in force at any given time are5, 000, 000

l55(T55 − T65)

(£5,000,000 “flows across” age 55, so the scaling factor is5, 000, 000

l55.)

Total sums assured = 5, 000, 000(◦e55 − l65

l55

◦e65

)

= £45, 618, 854.

10.6. SOLUTIONS 183

Hence rate of gross premium income = £4, 592, 130 p.a.

(ii) 5, 000, 000l65l55

= £3, 985, 870

(iii) 5, 000, 000− 3, 985, 870 = £1, 014, 130

(iv) Annual rate of payment of benefits = £5, 000, 000

Annual rate of payment of expenses = 5% of gross premium income+ 25% of gross premium income in first policy year= 0.05× 4, 592, 130

+ 0.25[5, 000, 000

(◦e55 − l56

l55

◦e56

)× 0.100663

]

= £355, 209

Let funds be F . We haveδF = 5, 000, 000 + 355, 209− 4, 592, 130

Hence F = £19, 456, 000 (final figures are unreliable.)


Chapter 11

JOINT-LIFE FUNCTIONS

11.1 Joint-Life Mortality Tables

Consider 2 independent lives aged x and y respectively, subject to the mortality of Tables I and IIrespectively.

Let T = min{T1, T2},where T1 = future lifetime of (x)

and T2 = future lifetime of (y).

That is, T is the “joint future lifetime” of (x), (y), i.e. the time until the first of them dies.

RemarkIf (x) and (y) are a married couple, or business partners, the assumption of independence of T1, T2

is questionable: consider, for example, a car accident in which both lives are killed. But in practiceindependence is normally assumed.

Define

tpI IIxy = Pr{both (x), (y) will survive for t years at least}

= tpIx · tp

IIy (by independence of T1, T2) (11.1.1)

and

tqI IIxy = Pr{T ≤ t}

= the distribution function of T (11.1.2)

= 1− tpI IIxy (11.1.3)

Note A common error is to writetq

I IIxy = tq

Ix · tq

IIy

If Table I = Table II, we may omit the superscripts, so (for example)

tpxy = tpx · tpy

on the table under consideration.Note In the a(55) tables, the notation tpxy, etc. refers to a male life aged x and a female life

aged y (the male life comes first.) That is, Table I is a(55) males ultimate, Table II is a(55) femalesultimate.

185

186 CHAPTER 11. JOINT-LIFE FUNCTIONS

From now on we will omit “I, II” (unless needed.) Define

lxy = lxly

(x ≥ α1, y ≥ α2, where α1, α2 are the youngest ages in tables I, II respectively). By equation 11.1.1,

tpxy =lx+t:y+t

lxy

Note that

tqxy = 1− tpxy

= 1− tpx.tpy

= 1− (1− tqx)(1− tqy)

= tqx + tqy − tqx.tqy (11.1.4)

Using equation 11.1.3,

tqxy =lxy − lx+t:y+t

lxy

anddxy = lxy − lx+1:y+1.

The force of mortality is defined as

µxy = limh→0+

hqxy

h

= limh→0+

[hqx

h+ hqy

h−

(hqx

h

)·(

hqy

h

)h]

= µx + µy (11.1.5)

(since hqx

h → µx, hqy

h → µy as h → 0+).

Example 11.1.1.

Given lxy = 1, 000lx+10:y = 960lx:y+10 = 920

calculate the probability that, of two lives aged x and y, only one will survive for 10 years.

Solution

10px =lx+10:y

lx:y= 0.96

10py =lx:y+10

lxy= 0.92

Required probability =10 px(1−10 py) +10 py(1−10 px)= 0.1136.

11.1. JOINT-LIFE MORTALITY TABLES 187

Theorem

tpxy = exp(−∫ t

0

µx+r:y+r dr) (11.1.6)

Proof. From (11.1.1),

tpxtpy = exp[−∫ t

0

µx+r dr] · exp[−∫ t

0

µy+r dr]

= exp[−∫ t

0

(µx+r + µy+r) dr]

= exp[−∫ t

0

µx+r:y+r dr] by (11.1.5)

The probability density function (pdf) of T.Note that

F (t) = Pr{T ≤ t} = tqxy

is the distribution function of T .(When t < 0, the distribution function of T is zero).Now

f(t) = p.d.f. of T

= F ′(t)

=d

dt[1− tpxy] (for t > 0)

= − d

dt

[exp

(−

∫ t

0

µx+r:y+r dr

)](from (11.1.6)).

= exp[−

∫ t

0

µx+r:y+r dr

]µx+t:y+t (using Chain Rule)

Thus f(t) =

{tpxyµx+t:y+t (t > 0)0 (t < 0)

(11.1.7)

In particular, we must have∫ ∞

0tpxyµx+r:y+r dr =

∫ ∞

0tpxy(µx+t + µy+t) dt

= 1

since the integral of a p.d.f. is always equal to 1.

The discrete variable K = integer part of TThis is the number of complete years to be lived in the future by the “joint life status” of (x, y).Note that

K = min{K1, K2}.


where K1 = integer part of T1, K2 = integer part of T2. This variable is discrete, with probabilities

Pr{K = k} = Pr{1st death occurs between times k and k + 1}= k|qxy

=k pxy −k+1 pxy (by an argument similar to that for a single life)

=lx+k:y+k − lx+k+1:y+k+1

lxy

=dx+k:y+k

lxy(k = 0, 1, 2 . . .) (11.1.8)

11.2 Select Tables

We replace (x), (y) by [x] + r, [y] + u (referring to lives aged x + r, y + u respectively who wereselected r, u years ago respectively). Define

l[x]+r:[y]+u = l[x]+r · l[y]+u

which leads to

tp[x]+r:[y]+u =l[x]+r+t:[y]+u+t

l[x]+r:[y]+u

Example 11.2.1. On A67− 70 calculate 3p[59]+1:[69]+1.

Solution

3p[59]+1:[69]+1 =l[59]+4:[69]+4

l[59]+1:[69]+1

=l63.l73

l[59]+1:[69]+1

= 0.856.

We also defineµ[x]+r:[y]+u = lim

h→0+

(hq[x]+r:[y]+u

h

)= µ[x]+r + µ[y]+u

The p.d.f. of T = future lifetime of ([x] + r, [y] + u) is

f(t) =

{tp[x]+r:[y]+u.µ[x]+r+t:[y]+u+t (t ≥ 0)0 (t < 0).

11.3 Extensions to More than 2 Lives

We extend the definitions of T , tpxy, etc., as follows: consider 3 lives aged x, y, z respectively, whosefuture lifetimes are T1, T2, T3 respectively (assumed to be independent).

tpxyz = Pr{(x), (y) and (z) all survive for at least t years}= tpx.tpy.tpz

=lx+t:y+t:z+t

lxyz

11.3. EXTENSIONS TO MORE THAN 2 LIVES 189

where lxyz = lxlylz for all x, y, z ≥ α1, α2, α3 (α1, α2, α3 being the youngest ages in tables I, II, III).Other functions follow similarly; in particular

µxyz = limh→0+

hqxyz

h= µx + µy + µz

and T = min{T1, T2, T3} has p.d.f.

f(t) =

{tpxyzµx+t:y+t:z+t = tpxyz(µx+t + µy+t + µz+t) (t ≥ 0)0 (t < 0)

Example 11.3.1. Evaluate ∫ ∞

0tpxxxxµx+t:x+t dt

Solution ∫ ∞

0tpxxxxµx+t:x+t dt = 0.5

∫ ∞

0tpxxxx(4µx+t) dt.

= 0.5.

To find the probability that exactly one life out of three survives for t years, use the formula

Pr{(x) survives but (y), (z) die}+Pr{(y) survives but (x), (z) die}+Pr{(z) survives but (x), (y) die}

(note that these events are mutually exclusive)

= tpx(1− tpy)(1− tpz) + tpy(1− tpx)(1− tpz) + tpz(1− tpx)(1− tpy)

Similar calculations apply toPr{exactly one life dies}

and other possibilities.Also

Pr{at least 1 dies within t years} = tqxyz

= 1− Pr{all 3 survive}= 1− tpxyz

= 1− tpxtpytpz

= 1− (1− tqx)(1− tqy)(1− tqz).

Example 11.3.2. The probability that exactly one of three lives aged x will survive for n years is27 times the probability that all three will die within n years. Find the probabilities that

(a) at least two will survive n years(b) at least one will die within n years

Solution

(a) 3npx(1−n px)2 = 27(1−n px)3

thus npx = 9(1−n px)hence npx = 0.9, so nqx = 0.1.


By the binomial distribution, the probability of at least two survivors

=(

32

)(npx)2(1−n px) + (npx)3

= 3(npx)2 − 2(npx)3 = 0.972

(b) Pr{at least one will die} = 1− (npx)3

= 1− (0.9)3

= 0.271

11.4 The Joint Expectation of Life

Define

◦exy = the complete joint expectation of life for the pair aged (x, y)

= E(T ),where T = min{T1, T2} as before.

Hence◦exy=

∫ ∞

0

t.tpxy.µx+t:y+t dt

Proceeding as in single-life case, using integration by parts, we get

◦exy=

∫ ∞

0tpxy dt (11.4.1)

We also have

exy = curtate joint expectation of life of (x, y)= E(K) where K = integer part of T

=∞∑

t=1

tpxy (11.4.2)

The Euler–Maclaurin formula gives the approximation

exy '◦exy −12

The variance of T is calculated as follows:

Var(T ) = E(T 2)− [E(T )]2

=∫ ∞

0

t2.tpxyµx+t:y+t dt− (◦exy)2

=∫ ∞

0

2t.tpxy dt−[∫ ∞

0tpxy dt

]2

(Use integration by parts; similar to the single-life case).

11.5. MONETARY FUNCTIONS 191

11.5 Monetary Functions

Notation is very similar to single-life case (changing ‘x’ to ‘xy’). See “International Actuarial Nota-tion” in “Formulae and Tables for Actuarial Examinations”.(a) AnnuitiesWe define

axy = mean present value (m.p.v.) of an annuityof £1 per annum (p.a.) payable continuouslyso long as both (x), (y) are alive= E[aT ] where T = future lifetime of joint-life status

=∫ ∞

0

at|.tpxyµx+t:y+t dt

=∫ ∞

0

vttpxy dt (11.5.1)

(the proof is similar to single-life case; note that v = 11+i ). Note also that

axy = m.p.v. of an annuity of £1 p.a. payable at thestart of each year so long as both (x), (y) are alive= E[aK+1 ] where K = integer part of T

=∞∑

t=0

vttpxy. (11.5.2)

Other joint-life annuity functions are

axy =∞∑

t=1

vt.tpxy = axy − 1

axy:n = axy −n pxyvnax+n:y+n

Woolhouse’s approximation for annuities payable mthly hold for joint lives, i.e.

a(m)xy ' axy − m− 1

2m

a(m)xy:n ' axy:n − m− 1

2m(1− npxyvn)

The Euler–MacLaurin formula gives

axy ' axy − 12' axy +

12.

Evaluation of annuities using commutation functions.Commutation functions are not often used, and the only functions available in the examination


tables are

Dxy = v12 (x+y)lxy (11.5.3)

Nxy =∞∑

t=0

Dx+t:y+t (11.5.4)

From these we get

nExy = m.p.v. of pure endowment of £1 payableat time n if (x), (y) are both alive= vn.npxy

=Dx+n:y+n

Dxy

and

axy =∞∑

t=0

vt lx+t:y+t

lxy

=Nxy

Dxy.

Note: In A67− 70 we only find Dxx and Nxx (i.e. equal ages only.)

(b) Assurances

Axy = m.p.v. of £1 payable immediately on the failure of the joint-life status

= E(vT ) where T = future lifetime of joint-life status

=∫ ∞

0

vt.tpxyµx+t:y+t dt (11.5.5)

This can be evaluated numerically by approximate integration. Note that

axy = E

(1− vT

δ

)=

1− Axy

δ

so we have the “conversion” relationship

Axy = 1− δaxy.

We also have

Axy = E[vK+1] =∞∑

t=0

vt+1t|qxy (11.5.6)

and it will now be shown thatAxy = 1− daxy

Proof

axy = E[aK+1 ] = E

(1− vK+1

d

)

=1d(1−Axy)

11.5. MONETARY FUNCTIONS 193

So Axy = 1− d.axy

An approximation which can often be used is

Axy ' (1 + i)−12 Axy

Write A 1︷︸︸︷xy :n

to indicate that a sum assumed is payable on the first death of (x), (y) if within n

years.

Example 11.5.1. Show how one can evaluate A 1︷︸︸︷xy :n

on the a(55) tables (male/female.)

Solution. Method 1:

A 1︷︸︸︷xy :n

' (1 + i)12 A 1︷︸︸︷

xy :n

= (1 + i)12 [Axy:n − vn ·n pxy] (as in single-life case).

= (1 + i)12

{1− daxy:n − vn lx+n

lx· ly+n

ly

}

Now use axy:n = axy − vn lx+n

lx· ly+n

lyax+n:y+n

and finally use axy = axy + 1.

Method 2:

A 1︷︸︸︷xy :n

= Axy:n − vn ·n pxy

= (1− δaxy:n )− vn · lx+n

lx· ly+n

ly

' 1− δ(axy +12)− vn · lx+n

lx· ly+n

ly

[1− δ(ax+n:y+n +

12)]

.

(c) PremiumsWe employ equations of value in the same way as for single-life policies. If there are no expenses weobtain net premiums, e.g., in the International Actuarial Notation,

P (Axy) =Axy

axy=

1axy

− δ

If there are expenses, calculate office premiums as for single-life policies by setting up an equationof value. The general symbols P and P ′ may be used for any net or gross premium.


(d) ReservesCalculated as for single-life policies, e.g.,

tV (Axy) = net premium reserve at duration t years forthe joint-life assurance of £1, payable immediatelyon the first death of two lives aged x and y

at the issue date, affected by annualpremiums payable continuously during the joint-lifetime.= Ax+t:y+t − P (Axy)ax+t:y+t

(using prospective method).Note: The formula for retrospective reserve in this example is

Dxy

Dx+t:y+t[P axy:t −A 1︷︸︸︷

xy :t

].

Also, one may show that

tV (Axy) = 1− ax+t:y+t

axy

using conversion relationships. We use the general symbol tV for the reserve at duration t, justbefore payment of any premium then due.

Exam tables(i) A67-70 gives:

axx, Nxx, Dxx, a[xx](= a[x]:[x]), axxx, axxxx

all at 4% p.a. interest.(ii) a(55) gives axy at 2-year intervals for male aged x and female aged y (x, y ≥ 60 only), at 4%,

6% and 8% interest.Interpolation is needed to get (for example) a61:69, as follows:

a61:69 ' 14[a60:68 + a62:68 + a60:70 + a62:70].

Example 11.5.2. Two lives, aged 50 and 60, effect a 10-year temporary joint life assurance withsum assured £60,000. The sum assured is payable immediately on the first death, provided thatthis occurs within 10 years of the issue date. The policy has annual premiums, payable during thejoint lifetime of both lives for at most 10 years.

(i) Express A 1z }| {50 : 60:10

as an integral. Estimate the value of the integral by the following form of

Simpson’s rule: ∫ 10

0

f(t) dt =106· [f(0) + 4f(5) + f(10)]

(ii) Hence, or otherwise, estimate the values of A50:60:10 and a50:60:10 , and calculate the annualpremium for the policy.

Basis: A1967− 70 ultimate mortality4% p.a. interestNo expenses.

11.6. LAST SURVIVOR PROBABILITIES (TWO LIVES ONLY) 195

Solution(i) A 1z }| {

50 : 60:10=

∫ 10

0vt.tp50:60(µ50+t + µ60+t) dt.

Simpson’s rule gives an approximate answer of

A 1z }| {50 : 60:10

= 0.2238.

(ii)

A50:60:10 = A 1z }| {50 : 60:10

+ v10 l60l50

· l70l60

' (1.04)−12 .A 1z }| {

50 : 60:10+ v10 l70

l50= 0.70792.

a50:60:10 =1−A50:60:10

d= 7.594.

The premium P p.a. is

P = 60000A 1z }| {

50 : 60:10

a50:60:10

= £1, 768.24

11.6 Last Survivor Probabilities (two lives only)

Define,

tpxy = Pr{at least 1 of (x), (y) will be alive in t years’ time}= 1− tqxy.

where tqxy = Pr{last survivor of (x), (y) dies within t years}.By elementary probability theory,

tpxy = tpx + tpy − tpxy

and hencetqxy = tqx + tqy − tqxy

These formulae are true even if (x), (y) are not independent. If they are independent, we may write

tpxy = tpx · tpy

givingtpxy = tpx + tpy − tpx · tpy (11.6.1)

andtqxy = tqx · tqy. (11.6.2)

Note that tqxy is the distribution function of the variable

Tmax = max{T1, T2}.

and the p.d.f. of Tmax = time to death of last survivor is

f(t) =d

dt(tqxy) =

d

dt(tqx) +

d

dt(tqy)− d

dt(tqxy)


Assume independence of T1, T2. We get:

f(t) = tpxµx+t + tpyµy+t − tpxtpy(µx+t + µy+t) (t > 0)

Now the “force of mortality” has to be defined as follows:

µxy(t) = hazard rate at time t

=f(t)

1− F (t)

where

f(t) = the p.d.f. of Tmax,

F (t) = the distribution function of Tmax.

This gives

µxy(t) =tpxµx+t + tpyµy+t − tpx.tpy(µx+t + µy+t)

tpxy

Note This is not often used in practice.

11.7 Last Survivor Monetary Functions

(a) AnnuitiesDefine, for example,

axy = m.p.v. of an annuity of £1 p.a. payable annuallyin advance so long as at least one life is alive.

On multiplying the equationtpxy = tpx + tpy − tpxy

by vt, and summing or integrating over t, one obtains the following relationships for annuity functions

axy = ax + ay − axy

axy = ax + ay − axy (11.7.1)

a(m)xy = a(m)

x + a(m)y − a(m)

xy

We may also define temporary last survivor annuity functions, e.g.

axy:n =n−1∑t=0

vt.tpxy

=n−1∑t=0

vt[tpx + tpy − tpxy]

= ax:n + ay:n − axy:n

Note: these formulae hold even without the assumption of independence of the lives.(b) AssurancesTaking expected values on each side of the relationship

vT1 + vT2 = vmin{T1,T2} + vmax{T1,T2}

11.7. LAST SURVIVOR MONETARY FUNCTIONS 197

one obtains

Axy = m.p.v. of a last survivor assurance of £1payable immediately on the death of (x), (y),= Ax + Ay − Axy (11.7.2)

Similarly, when the sum assured is payable at the end of the year of death, we have

vK1+1 + vK2+1 = vmin{K1+1,K2+1} + vmax{K1+1,K2+1}

and henceAxy = Ax + Ay −Axy (11.7.3)

(which is still true even when the lives are not independent.)Note also the conversion relationships

Axy = Ax + Ay −Axy

= (1− dax) + (1− day)− (1− daxy)= 1− daxy

and, similarly, Axy = 1− δaxy, etc.

Example 11.7.1. Evaluate A40:40 on A1967− 70 ultimate, 4% interest.

Solution

A40:40 = 1− da40:40

= 1− d[2a40 − a40:40]= 1− d[2× 18.894− 17.052]= 0.20246

(c) Premiums.We may use an equation of value to calculate net or gross premiums. For example, using the

International Actuarial Notation, we have

Pxy = net annual premium payable annually in advance for a policy providing£1 at the end of the year of the death of the second to die of (x), (y)

=Axy

axy

Using the conversion relationships given above, the following equations can be derived:

P (Axy) =Axy

axy=

1axy

− δ

and

Pxy =1

axy− d.


11.8 Reserves for Last Survivor Assurances

Consider a whole-life assurance providing £S at the end of the year of death of the last survivor of(x), (y). We ignore expenses and assume that the premium and reserve bases are the same. Then

Annual premium, P = SPxy = S

[1

axy− d

]

The value of the prospective reserve at duration t, just before payment of the premium then due,depends on whether or not both lives are still alive at time t.Let

tV1 = reserve assuming both lives are still alive

tV2 = reserve assuming (x) now alive but (y) has died

tV3 = reserve assuming (y) now alive but (x) has died

We find that

tV1 = S[Ax+t:y+t − Pxyax+t:y+t

]

tV2 = S [Ax+t − Pxyax+t]

tV3 = S [Ay+t − Pxyay+t]

Theorem When the first life dies, the prospective reserve will increase (from tV1 to tV2 or tV3).Proof

tV1 = S[(1− dax+t:y+t)− Pxyax+t:y+t

]

= S[1− (Pxy + d)ax+t:y+t

]

< S [1− (Pxy + d)ax+t] (as ax+t < ax+t:y+t)

= S [(1− dax+t)− Pxyax+t]= S [Ax+t − Pxyax+t]= tV 2( and similarly for tV3)

If it is not known whether one or both of (x), (y) survive at time t (and in practice this may bethe case since one death does not result in a claim or a change in the premium), one may value thepolicy as a ‘weighted average’ of tV 1, tV 2 and tV 3, with weights proportional to the probabilitiestpxy, tpx(1− tpy) and tpy(1− tpx) respectively. Hence the weights are

tpxy

tpxy

,tpx(1− tpy)

tpxy

,tpy(1− tpx)

tpxy

respectively

in order for them to add up to 1. Therefore the weighted average reserve is

V =tpxy

tpxy

· tV 1 +tpx(1− tpy)

tpxy

· tV 2 +tpy(1− tpx)

tpxy

· tV 3

Finally, it can be shown that the retrospective reserve is equal to the weighted average reserve,i.e.

V =1

vt.tpxy

S

Pxyaxy:t −A 1︷︸︸︷

xy :t

.

11.8. RESERVES FOR LAST SURVIVOR ASSURANCES 199

Example 11.8.1. A last-survivor policy provided £10, 000 immediately on the death of the sec-ond to die of a man aged 65 and his wife aged 60. Premiums were payable monthly in advance solong as at least one of the couple survived. The office which issued the policy used the following basis:

mortality : a(55)ultimate, males/females as appropriateinterest : 8% per annum ,

expenses : 2.5% of all office premiums .

(i) Find the monthly premium.(ii) Just after the payment of the first premium, the man died. Find the office’s prospective

reserve just after this event, on the premium basis.

Solution(i) Let annual premium, payable monthly, be P ′.

0.975P ′a(12)

m65:

f

60

= 10, 000(1 + i)12 A

m65:

f

60

a(12)m65:

f

60

' am65:

f

60− 11

24= 7.228

a(12)m65

' am65− 11

24= 7.942

a(12)f

60

' a f

60− 11

24= 9.836

Hence a(12)

m65:

f

60

' am65:

f

60

− 1124

= 11.0075− 1124

= 10.549

Am65:

f

60

= 1− dam65:

f

60

= 0.18463

Hence P ′ =10, 000× (1.08)

12 × 0.18463

0.975× 10.549= £186.55

Therefore monthly premium = £15.55 .

(ii) Reserve = 10000(1 + i)12 Af

60 − 0.975P ′a(12)f

60

= 2, 476.96− 1, 773.82

(using Af60 = 1− d.af

60, and a(12)f

60

= a f

60+

1124

)

Hence Reserve = £694.15.


Exercises

11.1 Given that npx = 0.3, npy = 0.4, npz = 0.6, find the probability that, of the lives (x), (y) and(z),(a) none will survive n years(b) exactly one will survive n years(c) at least one will survive n years.

11.2 Prove that(IA)xy = axy − d.(Ia)xy

11.3 Express in terms of px, py, and pz the probabilities that, of three lives (x), (y) and (z),(a) all three will survive one year(b) at least one will survive one year(c) exactly two will survive one year(d) at least two will survive one year

11.4 Derive the formula

exy = E(K) =∞∑

t=1

tpxy where K = integer part of T (T = min{T1, T2})

using Pr{K = k} = k|qxy.11.5 Evaluate A75:75 on the basis of A1967− 70 ultimate at 4% interest.11.6 The probability that at least one of three lives aged 60 will survive to age 65 is eight times the

probability that exactly one will survive to age 65. Assuming that the 3 lives are independentand subject to the same table of mortality, find the probability that exactly one life willsurvive to age 65.

11.7 (i) Define tpxy and show that tpxy = tpx + tpy − tpxy

(ii) Hence, or otherwise, show that axy = ax + ay − axy

11.8 12 years ago a man then aged 48 effected a without profits whole life assurance for £10,000(payable at the end of the year of death) by annual premiums. The premium now due isunpaid. He now wishes to alter the policy so that the same sum assured will be payable atthe end of the year of the first death of himself and his wife, who is 4 years older than himself.Calculate the revised office annual premium, ceasing on the first death, if the office uses thefollowing basis for premiums and reserves.

mortality: A1967-70 ultimate, rated down 4 years for female lives,interest: 4% per annum,expenses: 3% of all office premiums including the first, with additional initial expenses of 1 1

2%of the sum assured. (This additional initial expense is not charged again on the conversion ofan existing policy, providing that the sum assured does not increase.)

11.9 Consider the random variable L equal to the present value of £1 payable immediately on(i) the first death of (x) and (y), and(ii) the second death, in each case at a given rate of interest i p.a.

Show that, in case (i),var(L) = A∗xy − (Axy)2

where ∗ indicates a rate of interest of 2i+ i2 p.a., and give a corresponding result for case (ii).

11.10. SOLUTIONS 201

Solutions

11.1 (a) (1− npx)(1− npy)(1− npz) = 0.7× 0.6× 0.4= 0.168

(b) npx(1− npy)(1− npz) + npy(1− npx)(1− npz) + npz(1− npx)(1− npy)

= 0.3(0.6)(0.4) + 0.4(0.7)(0.4) + 0.6(0.7)(0.6) = 0.436

(c) 1− (1− npx)(1− npy)(1− npz) = 1− 0.168= 0.832.

11.2 Let K = curtate future lifetime of (x, y)(Ia)xy = E

[(Ia)K+1

]

= E

[aK+1 − (K + 1)vK+1

d

]

=1d

[axy − (IA)xy] (as (IA)xy = E[(K + 1)vK+1])

Hence (IA)xy = axy − d(Ia)xy.

11.3 (a) px . py . pz

(b) 1− (1− px)(1− py)(1− pz)(c) pxpy(1− pz) + pxpz(1− py) + pypz(1− px)(d) pxpy(1− pz) + pxpz(1− py) + pypz(1− px) + pxpypz

11.4exy = curtate joint expectation of life

=∞∑

k=1

k.k|qxy

=1

lxy

∞∑

k=1

k.dx+k:y+k

=1

lxy[dx+1:y+1 + 2dx+2:y+2 + 3dx+3:y+3 + · · · ]

=1

lxy[(dx+1:y+1 + dx+2:y+2 + · · · ) + (dx+2:y+2 + dx+3:y+3 + · · · ) + · · · ]

=1

lxy[lx+1:y+1 + lx+2:y+2 + · · · ]

=1

lxy

∞∑t=1

lx+t:y+t

=∞∑

t=1

tpxy

11.5A75:75 = 2A75 − A75:75

' 2(1.04)12 A75 − (1.04)

12 [1− d.a75:75]

= 0.64676.


11.6 Let 5p60 = p. We have (using binomial distribution)

1− (1− p)3 = 8× 3p(1− p)2

Hence 3p− 3p2 + p3 = 24p(1− p)2

3− 3p + p2 = 24− 48p + 24p2

So 23p2 − 45p + 21 = 0

p =45±

√(−45)2 − 4× 23× 21

46

=45±√93

46= 0.76862 (ignoring root > 1)

Hence probability of exactly one survivor = 3p(1− p)2 = 0.1234511.7 (i)

tpxy = Pr{ the last survivor of (x) and (y) survives for t years}= Pr{(x) survives for t years but (y) dies}+ Pr{(y) survives for t years but (x) dies}+ Pr{(x) and (y) both survive for t years}= tpx(1− tpy) + tpy(1− tpx) + tpxtpy

= tpx + tpy − tpxtpy

(ii)axy =

∞∑t=0

tpxyvt

=∞∑

t=0

vt(tpx + tpy − tpxtpy)

=∞∑

t=0

vttpx +

∞∑t=0

vttpy −

∞∑t=0

vttpxy

= ax + ay − axy11.8

V1 = reserve of original policy= 10, 000[1.01512V48 − 0.015] (using Zillmer’s formula).

= 10, 000[1.015(1− a60

a48)− 0.015]

= £2, 341.89

(there is no need to find the original premium)

Let P be the revised annual premium.Equation: V1 = 10, 000A60:60 − 0.97P a60:60

= 10, 000(1− d.a60:60)− 0.97P a60:60

= 6175.77− 9.6447P

Hence P = £397.5111.9 In case (i), L = vT where T = min{T1, T2}. We have

var (L) = E(L2)− [E(L)]2

= E[(v∗)T ]− (Axy)2

= A∗xy − (Axy)2 .


The corresponding result for case (ii) isvar (L) = A∗xy − (Axy)2 .


Chapter 12

CONTINGENT ASSURANCES

12.1 Contingent Probabilities

Suppose we have 2 independent lives (x), (y), subject to the same mortality table. (If these tablesare different, it is easy to make the necessary adjustments in the formulae.) Define

tq1xy

= Pr{(x) dies within t years, and before (y)}i.e. (x) dies within t years and (y) dies after this event (not necessarily within t years).

LetT1 = future lifetime of (x) with pdf f1(t1) = t1pxµx+t1 (t1 > 0)T2 = future lifetime of (y) with pdf f2(t2) = t2pyµy+t2 (t2 > 0)Then

tq1xy

= Pr{T1 ≤ t and T1 ≤ T2}

=∫∫

0<t1≤tt1≤t2

f1(t1)f2(t2) dt1 dt2

=∫ t

0

[∫ ∞

t1t1pxµx+t1 · t2pyµy+t2 dt2

]dt1

=∫ t

0t1pxµx+t1 · t1py dt1

=∫ t

0rpxyµx+r dr (12.1.1)

If t = 1 we may omit it, giving

q1xy

=∫ 1

0tpxyµx+t dt

Note also that ∞q1xy

=∫ ∞

0tpxyµx+t dt = Pr{(x) dies before (y)}

If x = y, we clearly havetq1

x:x=

∫ t

0rpxxµx+r dr

=12

∫ t

0rpxxµx+r:x+r dr

=12 tqxx (12.1.2)

205

206 CHAPTER 12. CONTINGENT ASSURANCES

By general reasoning, the following result holds:

tq1xy

+ tqx

1y

= tqxy (12.1.3)

Note also that ∞q1xy

+∞qx

1y

= ∞qxy = 1

One may also define deferred contingent probabilities, and we find that

t|q1x:y

= tpxy.q 1x+t:y+t

and t|∞q1x:y

= tpxy.∞q 1x+t:y+t

Consideration of the second deathWe also have the definition,

tq2xy

= Pr{(x) dies after (y) and within t years}

By calculations similar to those for tq1xy

, we have

tq2xy

=∫ t

0

(1− rpy)rpxµx+rdr

= tqx − tq1xy

Thus, as expected by general reasoning,

tqx = tq1xy

+ tq2xy

(12.1.4)

which follows from

Pr{T1 ≤ t} = Pr{T1 ≤ t and T1 ≤ T2}+ Pr{T1 ≤ t and T1 > T2}

(the two terms on the R.H.S. are the probabilities of mutually exclusive events.)

12.2 Contingent Assurances

Consider a benefit of £1 payable immediately on the death of (x) if this occurs before the death of(y). The M.P.V. (at a specified rate of interest) is:

A1xy

= E

[vT1 if T1 ≤ T2

0 if T1 > T2

]

=∫∫

t1≤t2

vt1(t1pxµx+t1)(t2pyµy+t2) dt2 dt1

=∫ ∞

0

vt1t1pxµx+t1

(∫ ∞

t1t2pyµy+t2 dt2

)dt1

=∫ ∞

0

vt1t1pxµx+t1(t1py) dt1

=∫ ∞

0

vttpxyµx+t dt (12.2.1)

This is often evaluated by approximate integration.

12.2. CONTINGENT ASSURANCES 207

Example 12.2.1. Using A1967-70 mortality and 4% interest estimate, by approximate integration,the value of

A 150:60

[Assume l108 = 0, so that the integral is over a range of 48 years. Break this into 4 sub-intervals anduse Simpson’s rule over each.]

Solution

A 150:60

=∫ ∞

0

vttp50µ50+t tp60 dt (at 4% interest)

=∫ 48

0

f(t) dt

where f(t) = vttp50:60µ50+t.

Using Simpson’s rule

A 150:60

= 2 [f(0) + 4f(6) + 2f(12) + 4f(18) + 2f(24)

4f(30) + 2f(36) + 4f(42) + f(48)]= 0.1546.

Temporary contingent assurance functions

A1x:y:n

= m.p.v. of £1 payable immediately on death of (x)

if this occurs before the death of (y) and within n years

=∫ n

0

vttpxyµx+t dt (12.2.2)

Note also thatA1

xx=

12Axx (12.2.3)

(since there is a 50% chance that the first “x” of (x, x) will be the first to die).We may therefore evaluate contingent assurances on 2 equal ages from tables of Axx. As expected

by general reasoning, we also haveAxy = A1

xy+ A

x1y

(12.2.4)

Example 12.2.2. (a) Estimate, by approximate integration, the value of a contingent assurance of£100,000 payable immediately on the death of a female aged 60, provided a male aged 50 is still aliveand provided her death occurs within 15 years. The mortality of the female follows a(55) ultimate(females), and that of the male follows A1967-70 ultimate. The interest rate is 7 1

2% per annum andexpenses are ignored.Note. A very accurate approximation to the value is not expected.

(b) It has been suggested that a policy providing the above benefit should be issued by annualpremiums ceasing on the death of the female life or after 15 years, whichever is earlier. State withreasons whether you agree with this suggestion. If you do not, suggest a more suitable premium-paying term, other than issuing the contract by a single premium.


Solution

(a) Value = 100, 000Am50:

1f

60:15

= 100, 000∫ 15

0

vttp50:60µ60+t dt

Evaluate integral by (for example) the three-eighths rule:∫ 15

0

f(t) dt ' 158

[f(0) + 3f(5) + 3f(10) + f(15)]

= 0.12657.

Hence value = £12, 657

(b) No: premiums must also cease on death of (50), otherwise they may still be payable with noprospect of claim: negative prospective reserve. Restrict premium-paying period so that premiumscease on first death or after 15 years, if earlier.

We may also encounter contingent assurances payable on the second death.Define

A2xy

=m.p.v. of £1 payable immediately on death

of (x) if this occurs after the death of (y).

By a similar argument to that for A1xy,

A2xy =

∫ ∞

0

vttpxµx+t(1− tpy) dt

Hence

Ax = A1xy + A2

xy (12.2.5)

as expected.

We may also encounter temporary contingent assurances payable on the second death, e.g. them.p.v. of £1 payable immediately on death of (x) within n years, provided that (x) dies after (y),equals

∫ n

0


= A1x:n

−A1xy:n

Note also thatA2

xx =12Axx (12.2.6)

since there is a fifty per cent chance that the first “x” will be the last survivor.

Example 12.2.3. Using the fact that A1xy

+ Ax

1y

= Axy, find on A1967− 70 ultimate at 4% interest

the values of (i) A60:

160

, (ii) A60:

260

and (iii) A60:

160:10

.

12.2. CONTINGENT ASSURANCES 209

Solution.

(i) A60:

160

= 0.5A60:60

' 0.5(1.04)12 [1− da60:60]

' 0.31491

(ii) Use A60:

260

= A60 −A60:

160

to obtain

A60:

260

= 0.21259

(iii) A60:6

10:10

=12A 1z }| {

60 : 60:10

' 12(1.04)

12

[A60:60 − D70:70

D60:60A70:70

]

= 0.15513

The symbols A1xy

, etc.

These are the same as A1xy

, etc., but with the benefit payable at the end of the year of death.

An exact expression is

A1xy

=∞∑

t=0

vt+1tpxyq 1

x+t:y+t(12.2.7)

In practice, we may use the approximations

A1xy' (1 + i)−

12 A1

xy, etc. (12.2.8)

Note the following results, similar to those given above for A1xy

, etc.:

Axy = A1xy

+ Ax

1y

Ax = A1xy

+ A2xy

A1xx

=12Axx

A2xx

=12Axx

The variance of the present value of a contingent assuranceLet

Z = present value of £1 payable on death of (x), if this occurs before the death of (y)

=

{vT1 if T1 ≤ T2 where v = 1

1+i

0 if T1 > T2

As we have shown above, this has mean

E(Z) = A1xy


The variance of Z isE(Z2)− [E(Z)]2

where

E(Z2) = E

[(v2)T1 if T1 ≤ T2

0 if T1 > T2

]

= E

[(v∗)T1 if T1 ≤ T2

0 if T1 > T2

]where v∗ =

11 + i∗

with i∗ = i2 + 2i

= A∗1xy

at rate of interest i2 + 2i (or force of interest 2δ)

HenceVar(Z) = A∗1

xy− (A1

xy)2 (12.2.9)

12.3 Premiums and Reserves for Contingent Assurances

Premiums are calculated by the usual equation of value (including expenses if necessary). Reservesare usually calculated prospectively, making allowance for any deaths which have already occurred.

Example 12.3.1. Two lives (A and B) are both aged 30. Calculate, on the basis of A1967 − 70ultimate mortality and 4% p.a. interest, the annual premium, payable during the lifetime of A, toprovide an insurance of £1000 payable at the end of the year of death of A, provided A dies afterB.

Find the policy value (on the premium basis) after 10 years (before the premium then due ispaid) if

(i) only A is then alive;

(ii) both lives are then alive.

Solution

A 230:30

= A30 −A 130:30

= A30 − 12A30:30 = (1− da30)− 1

2(1− da30:30)

=12− d(a30 − 1

2a30:30)

Hence

premium =A 2

30:30

a30· 1000 = 1000

[1

2a30− d

(1− 1

2a30:30

a30

)]

= 1000.(0.00327) = 3.27 = P, say.

Hence the reserves are as follows:

(i) 10V = 1000A40 − P a40 if only A is alive = 211.56(ii) 10V = 1000A 2

40:40− P a40 if both are alive = 39.49

12.4. A PRACTICAL APPLICATION – THE PURCHASE OF REVERSIONS 211

Lapse optionsOne should avoid having a negative prospective reserve at any duration, as this may lead to a lapseoption against the office. In particular, one should ensure that premiums cease as soon as there isno possibility of future benefits. (See Example 12.2.2 above.)

12.4 A Practical Application – The Purchase of Reversions

DefinitionsA reversion is a contract providing a sum of money payable on the death of a certain life, (y).An absolute reversion is a reversion in which the sum is paid under all circumstances, whilst in acontingent reversion the sum is paid only if certain other lives are (or are not) then alive.

Example 12.4.1. J. Brown(50) will receive £1,000,000 on the death of his mother (aged 80),provided that he is then alive. He wishes to sell his “interest”. What is it worth?

SolutionSuppose that a purchaser uses a certain rate of interest, i p.a., and assumes that a certain mortalitytable applies to both J. Brown and to his mother. (Note that the purchaser might assume differenttables for the lives). The interest is contingent since it requires (50) to be alive when (80) dies.Hence

M.P.V. = 1, 000, 000A50:

180

In practice, the purchaser will want to be sure of getting the money when (80) dies, and will“plug the gap” with an insurance policy (which pays out on the death of (80) if she dies second.)Suppose that the life office issuing the insurance policy uses the same mortality and interest basisas the purchaser, and also ignores expenses.

Single premium to buy insurance = 1, 000, 000A50:8

20

Hence purchase price = 1, 000, 000A80 − cost of premium for insurance policy

= 1, 000, 000[A80 − A

50:280

]

= 1, 000, 000A50:

180

(as found before)

Note:- The life office issuing the policy would examine the health (and any other risk factors) of(50) very carefully.


12.5 Extension to Three Lives

We briefly mention the following symbols and formulae:

∞q1xyz

= Pr{(x) dies first of (x), (y), (z)}

=∫ ∞

0tpxµx+t · tpy · tpz dt (12.5.1)

∞q2xy

1z

= Pr{(x) dies second, (y) having died first}

=∫ ∞

0tpxµx+t(1− tpy)tpz dt

= ∞q1xz−∞q1

xyz(12.5.2)

∞q2xyz

= Pr{(x) dies second of (x), (y), (z)}= ∞q2

xy1z

+∞q2xyz

1

(since either (y) or (z) dies first)

= ∞q1xy

+∞q1xz− 2∞q1

xyzusing formula (12.5.2)

Similar definitions and relationships apply for contingent assurances, e.g.

A2xyz

= m.p.v. of £1 payable immediately

on the death of (x) if he dies second

=A1xy

+ A1xz− 2A1

xyz

More complicated functions may be defined and evaluated, but we do not pursue this topic.

12.6. EXERCISES 213

Exercises

12.1 Which (if any) of the following statements are correct?A. Axx = 2A1

xx

B. Axx = 2Ax

C. Axx = 2Ax − A1xx

D. Axx = A1xx

+ A2xx

12.2 Adams (aged 40) and Brown (aged 50) are two business partners. Adams wishes to providefor the sum of £80,000 to be paid immediately on Brown’s death if Brown predeceases himwithin ten years, and effects a policy providing this benefit by single premium. The life officeissuing the contract employs the following basis:

Mortality (both lives) : A1967− 70 ultimate

Interest : 6%

Expenses: 2% of the single premium.

Using Simpson’s rule, or otherwise, estimate the single premium payable by Adams.

12.3 Estimate the value of ∞q174:84

on the basis of A1967− 70 ultimate mortality.

(Assume l108 = 0, so that the integral is over a range of 24 years. Break this into 3 sub-intervalsand use Simpson’s Rule over each.)

12.4 (i) Express A2xx

in terms of ax, axx and the rate of interest.

(ii) Smith and Jones are both aged 60. A life office has been asked to issue a special joint-lifeassurance policy providing £10,000 at the end of the year of death of the first to die of thesetwo lives. In addition, if Smith is the second to die, a further £5,000 will be payable at theend of the year of his death. The policy is to have annual premiums payable during the jointlifetime of Smith and Jones.

Calculate the annual premium on the following basis:

A1967− 70 ultimate mortality

4% interest

expenses are 5% of all premiums, with an additional initial expense of £100.

(iii) Write down (but do NOT evaluate) formulae for the reserve at duration 10 years (imme-diately before payment of the premium then due) on the premium basis, if

(a) both Smith and Jones are alive; and

(b) Jones has died but Smith is alive.

12.5 A policy providing the sum of £100,000 immediately on the death of (x) if she dies before (y)is to be issued by a life office to a group of trustees.(i) Ignoring expenses, write down an expression for the single premium in terms of an integral.

(ii) The trustees suggest that level annual premiums should be payable in advance until thedeath of the last survivor of (x) and (y).


(a) Ignoring expenses, give a formula for the annual premium.

(b) Would you advise the life office to issue the policy with premiums payable as suggested?Give reasons for your answer.

12.6 Estimate, by the trapezoidal rule or another suitable rule for approximate integration, thesingle premium for a temporary contingent assurance of £50,000 payable immediately on thedeath of Mrs Smith (aged 60), provided that this event occurs within 5 years and that herhusband (aged 50) is alive at the date of her death. Mrs Smith is subject to the mortality ofa(55) ultimate (females) and Mr Smith is subject to the mortality of A1967 − 70 ultimate.An interest rate of 7.5% p.a. is to be used, and allowance is to be made for expenses of 6% ofthe single premium.(Note A very accurate answer is not expected.)

12.7 Define the following functions in words, and give an expression for each of them in terms ofan integral.(i) ∞q1

xy

(ii) A2xy

(iii) A1x:y:n

12.8 The chief of a certain tribe holds that office until age 50 or earlier death, and may be succeededonly by a person aged from 36 to 45. (A person aged exactly 36 is eligible, but a person agedexactly 45 is not.) The customs of the tribe require that a chief’s successor be his oldesteligible brother; if there is no eligible brother, the position of chief is given to someone fromoutside the previous chief’s family, who are then permanently debarred from becoming chief.The present chief is aged exactly 47 and has two brothers, aged exactly 37 and 33 respectively.The chief and his brothers may be regarded as independent lives subject to the mortality ofa given table.Obtain an expression, in terms of quantities of the form npx, nq1

xyonly, for the probability

that (33) will become chief.12.9 Your life office has been asked to quote a single premium for a contingent assurance policy

providing £300,000 immediately on the death of a woman now aged 80 within 15 years,provided that at the date of her death a man now aged 60 has died. Your office uses thefollowing basis:

mortality : males - a(55)ultimate (males)females - a(55) ultimate (females)

interest : 8% per annumexpenses : 10% of the single premium.

(i) Assuming that the two lives are independent, write down a formula for the single premiumin terms of an integral.

(ii) State a suitable non-repeated rule of approximate integration for evaluating this integral.(You are NOT required to carry out the evaluation.)

(iii) Would you subject the male life to stringent underwriting procedures? Give brief reasonsfor your answer.

12.7. SOLUTIONS 215

Solutions

12.1 Only A is correct.

12.2 Let single premium be P .

P = 80000∫ 10

0

vttp50µ50+t · tp40 dt + 0.02P at 6% interest

=800000.98

∫ 10

0

f(t) dt where f(t) = vttp50µ50+t.tp40

' 800000.98

106

[f(0) + 4 · f(5) + f(10)] using Simpson’s rule

= £4, 686.

12.3

∞q174:84

=∫ ∞

0tp74µ74+t.tp84dt

=∫ 24

0

f(t) dt

where f(t) = tp74µ74+t.tp84

' 43[f(0) + 4f(4) + 2f(8) + 4f(12) + 2f(16) + 4f(20) + f(24)]

= 0.2907.

12.4(i) A2

xx=

12Axx = Ax − 1

2Axx

(or use A2xx

= Ax −A1xx

= Ax − 12Axx);

= (1− dax)− 12(1− daxx)

=12−

(i

1 + i

)(ax − 1

2axx)

(ii) Let the annual premium be P, and set x = 60.

0.95P axx = 10, 000Axx + 5, 000A2xx

+ 100

= 10, 000[1− daxx]

+ 5, 000[12− d(ax − 1

2axx)

]+ 100

Putting x = 60 gives

0.95P · (9.943)− 100 =10, 000[1− 0.04

1.04· 9.943

]

+ 5, 000[0.5− 0.041.04

(12.551− 0.5× 9.943)]

=6175.77 + 1042.40 = 7218.17

Hence P =7318.179.44585

= £774.75


(iii) (a) Reserve = 10, 000A70:70 + 5, 000A70:

270

− 0.95P a70:70

(b) Reserve = 5, 000A70

12.5 (i) 100, 000∫∞0

vttpxyµx+t dt

(ii) (a) Annual premium, P =100000

∫∞0

vttpxyµx+t dt

axy.

(b) No, because premiums should not continue after first death. If (x) dies first, the benefitis paid and the policy may be lapsed, and if (y) dies first there is no possibility of benefit andthe policy be lapsed.There is thus a lapse option on the first death.

12.6 Let single premium = P .Then

0.94P = 50, 000A1f

60:m50:5

at 712% interest

= 50, 000∫ 5

0

vttp f

60µ f

60+t.tpm

50dt

By the trapezoidal rule. ∫ 5

0

vttp f

60µ f

60+t.tpm

50dt ' 0.041726

Hence

P ' 50000× 0.0417260.94

= £2, 219

12.7 (i)∞q1

xy= Pr{(x) will die before (y)}

=∫ ∞

0tpxyµx+t dt

(ii)A2

x:y= m.p.v. of a contingent assurance of £1

payable immediately on death of (x),provided this occurs after the death of (y).

=∫ ∞

0


(iii)A1

x:y:n= m.p.v. of a temporary contingent assurance of £1 payable

immediately on death of (x), provided that this occurs withinn years, and before (y) dies.

=∫ n

0

vt.tpxyµx+t dt

12.7. SOLUTIONS 217

12.8 Let chief = (y) = (47), brothers (x1) = (37), (x2) = (33).(x2) becomes chief if and only if

(1) he survives for 3 years and (y) “retires” then, (x1) having died.

(2) (x1) becomes chief (by succeeding (y) on his death or “retirement”) and dies after time 3and before time 12, leaving (x2) alive.

Probability of event (1) is3py(1− 3px1

)3px2

Probability of event (2) isPr{(x1) dies between times 3 and 12, leaving (x2) alive}

=∫ 12

3tpx1

µx1+t · tpx2dt

=∫ 9

0r+3px1

µx1+r+3 · r+3px2dr

= 3px1x2

∫ 9

0rpx1+3:x2+3µx1+r+3 dr

= 3px1· 3px2

.9q 1x1+3:x2+3

12.9

(i) annual premium =300, 000

∫ 15

0vt

tp f

80µ f

80+t(1− tpm

60) dt

0.90

(ii) The three-eighths rule would be suitable, since it avoids evaluation of the integrand whent = 7 1

2 .

(iii) Yes. If (60) dies soon, there is a high chance that (80) will die before age 95 and so giverise to a claim. The office must check:

(1) the health of (60);

(2) whether he has any occupational or other risks, e.g. participation in a dangerous sport.

Note. The sum assured is large enough to justify the costs of a medical examination.


Chapter 13

REVERSIONARY ANNUITIES

13.1 Reversionary Annuities Payable Continuously

Consider an annuity of £1 p.a. payable continuously to (y) after the death of (x). The present valueof this reversionary annuity is

Z =

{aU − aT if U > T

0 if U ≤ T

whereT = future lifetime of (x), with p.d.f. = tpxµx+t (t > 0);U = future lifetime of (y), with p.d.f. = upyµy+u (u > 0).

Note. If (x) and (y)’s mortality rates follow different tables, this should be indicated.Define

ax|y = m.p.v. of Z

=∫∫

u>t

(au − at )(upyµy+u)(tpxµx+t) du dt

=∫ ∞

0

[∫ ∞

t

(au − at )upyµy+u du

]tpxµx+t dt (13.1.1)

Theoremax|y = ay − axy (13.1.2)

Proof It is easiest to proceed indirectly, as follows. Observe that

Z + amin(T,U)

=

{(aU − aT ) + aT if U > T

aU if U ≤ T= aU

Taking expected values on both sides gives

E(Z) + axy = ay

Another important formula

ax|y =∫ ∞

0

vttpxµx+ttpyay+t dt

219

220 CHAPTER 13. REVERSIONARY ANNUITIES

Proof

ax|y =∫ ∞

0

vttpy(1− tpx) dt

=∫ ∞

0

f(t)g′(t) dt

wheref(t) = 1− tpx = tqx (which is such that f ′(t) = tpxµx+t ) andg(t) = −t|ay = − ∫∞

tvr

rpy dr (which is such that g′(t) = vttpy )

Using integration by parts

ax|y = [f(t)g(t)]∞0 −∫ ∞

0

f ′(t)g(t) dt

= [(−t|ay)tqx]∞0 +∫ ∞

0tpxµx+t(t|ay) dt

=∫ ∞

0tpxµx+tv

ttpyay+t dt.

Evaluation of reversionary annuitiesUsing the Euler–Maclaurin formula we have

ax|y = ay − axy

' (ay +12)− (axy +

12)

= ay − axy

= ax|y

Example 13.1.1. Calculate an appoximate value of am65|

f

69on the a(55) tables at 8% interest.

Solution

am65|

f

69' a f

69− am

65:f

69

= 7.533− 5.877 (using interpolation)= 1.656

13.2 Reversionary Annuities Payable Annually or mthly

We first assume that payments are made mthly, the first payment being at the end of the 1m year

(measured from the issue date) following the death of (x), and use the symbol a(m)x|y to refer to this

case. Using the equation

a(m)x|y + a(m)

xy = a(m)y

it can be seen that

a(m)x|y = m.p.v. of reversionary annuity of £1 p.a.

payable monthly to (y) after the deathof (x)

= a(m)y − a(m)

xy

13.3. WIDOW’S (OR SPOUSE’S) PENSION ON DEATH AFTER RETIREMENT 221

When m = 1, it may be omitted, giving

ax|y = ay − axy

By Woolhouse’s formula

a(m)x|y ' (ay +

m− 12m

)− (axy +m− 12m

)

= ay − axy

= ax|y.

An alternative approach is to regard this reversionary annuity as a collection of “pure endow-ments” (payable at times 1

m , 2m , · · · ) which are paid if (x) has died and (y) is alive, i.e.

a(m)x|y =

∞∑t=1

1m

vtm Pr{(x) has died but (y) is alive at time

t

m}

=1m

∞∑t=1

vtm (1− t

mpx). t

mpy

= a(m)y − a(m)

xy (as before).

Suppose now that a mthly reversionary annuity (of £1 p.a.) begins immediately on the death of(x). Since the payments begin on average 1

2m year earlier than in the case discussed previously, them.p.v. is approximately

(1 + i)1

2m a(m)x|y (13.2.1)

An exact formula is∫ ∞

0

vttpxµx+t·tpya

(m)y+t dt

'∫ ∞

0

vttpxyµx+t

(ay+t +

12m

)dt

=ax|y +1

2mA1

xy

'ax|y +1

2mA1

xy(13.2.2)

In practice this gives results similar to (13.2.1).

13.3 Widow’s (or Spouse’s) Pension on Death after Retire-ment

Many pension schemes provide a spouse’s pension on the death of the member in service (D.I.S. =Death In Service) and/or on death after retirement (D.A.R. = Death After Retirement)

We consider D.A.R. only at this stage and suppose that a male employee retires at age 65.Now consider 2 different cases:-Case 1

A widow’s pension of £1 p.a. is payable on the death of (65) only if he was married to the samewoman at retirement. In this case, the m.p.v. at age 65 per married man is

am65|fy

(assuming pension is payable continuously)


where

y = average age of wife of member aged 65= 65− d, where d = age difference between husband and wife

(approximately 3 years in practice)

Hence the value of widow’s pension for each member retiring at age 65 (marital status unknown)is

h65am65|

f

65−d(13.3.1)

where hx = the probability that a man aged x is married.Note. We ignore the possibility of divorce (or assume that the ex-wife still gets pension).

Case 2 (now more common)The widow’s pension is payable to any widow: we use the “collective” approach.Suppose, for example, that the widow’s pension of £1 p.a. is payable monthly in advance, beginningimmediately on death of her husband.

The m.p.v. for a man retiring at age 65 (marital status then unknown) is∫ ∞

0

vttp

m65µ

m65+th65+ta

(12)f

65−d+t

dt (13.3.2)

where

a(12)f65−d+t = m.p.v. of widow’s annuity to the widow of a

man dying at age 65 + t, if widow isassumed to be d years younger than husband

Example 13.3.1. A life office sells “personal pensions” policies under which the benefits for menon retirement at age 65 consist of:

(a) a member’s pension (payable monthly in advance for 5 years certain and for life thereafter),and (for men married at age 65 only)

(b) a spouse’s pension (payable monthly in advance, beginning immediately on the death of themember) of half the member’s pension. The possibility of divorce of men aged over 65 may beignored, and post-retirement marriages do not give rise to spouse’s pension.

The member’s contributions are invested in certain unitised with-profits funds, and it is assumedthat, in respect of a certain Mr Brown’s policy, the fund available to purchase pension at age 65 willbe £100,000.

Suppose that the life office uses the following basis to calculate the amount of pension which maybe purchased at retirement in respect of a given fund:

mortality of males: a(55) males ultimatemortality of females: a(55) females ultimateinterest: 8% per annumexpenses: 1% of the fund (at age 65)

13.3. WIDOW’S (OR SPOUSE’S) PENSION ON DEATH AFTER RETIREMENT 223

Men who are not married at age 65 need not buy spouse’s pension, but married men must buythis. Calculate Mr Brown’s expected monthly pension if

(i) he is assumed to be single at age 65; and

(ii) he is assumed to be married at age 65, and his wife is 3 years younger.

Solution. (i) Let P be annual pension, payable monthly in advance.

0.99× 100, 000 = P (a(12)

5+5 |a(12)

65 ) on males’ table

= P (i

d(12)a5 + v5 l70

l65

[a70 − 11

24

])

= P (4.1637 + 0.68058× 0.86621× 6.8097)= 8.1782P

Hence P = £12, 105, so monthly pension = £1, 008.75

(ii) The equation of value is now

99, 000 ' P (8.1782) +12P (1.08)

124

(a f

62− am

65:f

62

)

' P (8.1782 + 1.2026)Hence P = £10, 553, so monthly pension = £879.42

Notes(1) There may be rules to try to exclude benefits for widows of “deathbed marriages”.(2) There may be problems if the man was married more than once (and his ex-wives are still

alive). In the U.K. the last wife receives all the pension.(3) If a wife is very much younger than her husband, there may be an “actuarial reduction” (see

later).(4) Widow’s pensions might cease on remarriage, but this rule is no longer common.(5) h65+t is sometimes assumed to be “piecewise continuous”, e.g.

h65+t =

0.85 for 0 ≤ t < 10.87 for 1 ≤ t < 20.89 for 2 ≤ t < 30.90 for t ≥ 3.

The integral in (13.3.2 ) may be evaluated approximately by a sum, i.e.∫ ∞

0

vttp

m65µ m

65+t(h65+t)a

(12)f

65−d+t

dt

' 0.85v12 qm

65a(12)

f

65−d+ 12

+ 0.87v1 12 1|qm

65a(12)

f

65−d+1 12

+ 0.89v2 12 2|qm

65a(12)

f

65−d+2 12

+ 0.9

[v3 1

2 ·3 |qm65

a(12)f65−d+3 1

2

+ v4 12 ·4 |qm

66a(12)f65−d+4 1

2

+ · · ·]

(assuming that the man dies on average half-way through each year of age).


13.4 Actuarial Reduction Factors

Suppose that, in a pension policy or scheme, the rules state that a reduction applies to the normalwidow’s pension if wife is more than 10 years younger than her husband. (This rule may also applyto widow’s D.I.S. pensions).Suppose a member dies aged x, leaving widow aged y (y < x− 10).

The value of this widow’s pension is calculated to be the same (actuarially speaking) as for awidow aged x − 10 (whose pension is not, of course, reduced.) Let the widow aged y get £R p.a.for each £1 p.a. of ‘normal’ widow’s pension. Then

(Full widow’s pension).Ra(12)fy

= (Full widow’s pension).a(12)f

x−10

This gives

R = actuarial reduction factor =a(12)

f

x−10

a(12)fy

13.5. EXERCISES 225

Exercises

13.1 (i) Define the following symbols in words, and give a formula in terms of an integral for eachof them:

(a) Ax

1y

(b) A2xy

(c) ay|x

(ii) Consider the following sets of payments:

(1) £1 immediately on the death of (y) if (y) dies before (x), and

(2) an income of £δ p.a. payable continuously to (x) after the death of (y), plus £1 immedi-ately on the death of (x) if this occurs after that of (y).

Prove that the present values (at force of interest δ p.a.) of (1) and (2) are equal. Hence writedown a relationship involving A

x1y, A2

xyand ay|x.

13.2 A special life policy on 2 lives aged x and y respectively provides cash sums of £10,000 and£20,000 immediately on the first and second deaths respectively. In addition, an annuityat the rate of £1,000 per annum will be paid continuously, commencing immediately on thefirst death and ceasing immediately on the second death. Obtain an expression for the meanpresent value of the benefits in terms of joint-life and single- life annuity functions and theforce of interest. Ignore expenses.

13.3 An office issues a policy on the lives of a woman aged 60 and her husband aged 64. Under thispolicy, level premiums are payable annually in advance for 20 years or until the first death ofthe couple, if earlier.

On the first death of the couple, the survivor will receive an annuity of £10,000 per annum,payable weekly, beginning immediately on the first death.

Calculate the annual premium if the office uses the basis given below:

Mortality males: a(55) males ultimatefemales: a(55) females ultimate

Expenses: 20% of the first premium5% of each premium after the first

Interest: 6% per annum.

13.4 A special annuity, payable yearly in arrear, is effected on the lives of a man aged x and hiswife aged y. The conditions of payment are:

(a) so long as both survive the rate of payment will be £3,000 per annum;

(b) if the wife dies first, the rate of payment will be £2,000 per annum until the man’s death;

(c) payments at the rate of £3,000 per annum will continue for six years certain after thedeath of the husband, the first payment being at the end of the year of his death, and will bereduced thereafter to £1,500 per annum during the lifetime of the wife.

Obtain an expression for the present value of this annuity in terms of single and joint-lifeannuity factors, life table and compound interest functions. Assume that the same (non-select) table of mortality is appropriate for the two lives.


13.5 A single-premium policy provides the following benefits to a husband and wife each aged 40.

(1) An annuity of £5,000 per annum, payable continuously, commencing on the husband’sdeath within 25 years, or on his survival for 25 years, and continuing so long as either husbandor wife is alive.

(2) A return of half the single premium without interest immediately on the death of thehusband within 25 years, provided that his wife has already died.

The office issuing the contract uses the following basis:

mortality : A1967-70 ultimateinterest : 4% per annum


Calculate the single premium.

13.6 A husband and wife, aged 70 and 64 respectively, effect a policy under which the benefits are(1) a lump sum of £10,000 payable immediately on the first death, and (2) a reversionaryannuity of £5,000 p.a. payable continuously throughout the lifetime of the surviving spouseafter the death of the first. Level premiums are payable annually in advance until the firstdeath.

Calculate the annual premium on the undernoted basis:

Males’ Mortality: a(55) males ultimateFemales’ Mortality: a(55) females ultimateInterest: 8% p.a.

Expenses: 10% of all premiums

Ignore the possibility of divorce.

13.6. SOLUTIONS 227

Solutions

13.1 (i) (a) The m.p.v. of £1 payable immediately on the death of (y), if this occurs before thatof (x).

Ax

1y

=∫ ∞

0

vttpxyµy+t dt

(b) The m.p.v. pf £1 payable immediately on the death of (x) if this occurs after that of (y).

A2xy

=∫ ∞

0


(c) The m.p.v. of a reversionary annuity of £1 p.a. payable continuously to (x) after thedeath of (y).

ay|x =∫ ∞

0

vttpyµy+t · tpx · ax+t dt

(other expressions also possible)

(ii) Consider payment (1). Suppose it is invested (at force of interest δ p.a.) to give incomeso long as (x) lives (assuming (y) dies first, otherwise there is no payment (1)). On the deathof (x), if after that of (y), the capital (£1) is paid immediately. Payments (2) thus have thesame present value as payment (1) (both being random variables depending on the futurelifetimes of (x) and (y)). Take means of these present values to get

Ax

1y

= δ.ay|x + A2xy

13.2Benefit = 10, 000Axy + 20, 000Axy + 1000(ax|y + ay|x)

= 10, 000(Axy + 2Ax + 2Ay − 2Axy) + 1000(ax + ay − 2axy)

= 10, 000(2Ax + 2Ay −Axy) + 1000(ax + ay − 2axy)= 10, 000(2− 2δax + 2− 2δay − 1 + δaxy) + 1000(ax + ay − 2axy)= (1000− 20, 000δ)(ax + ay) + (10, 000δ − 2000)axy + 30, 000

13.3 Let annual premium be P . Then0.95P am

64:f

60:20− 0.15P = 10000[am

64|f

60+ a f

60|m64]

= 10000[a f

60− am

64:f

60+ am

64− am

64:f

60]

' 10000[a f

60+ am

64− 2am

64:f

60]

= 42230

am64:

f

60:20= am

64:f

60− v20

lm84

lm64

·l f

80

l f

60

· am84:

f

80= 8.5709

Hence P =42230

0.95× 8.5709− 0.15= £5, 284 .

13.4 (a) value of benefit = 3000axy.

(b) value of benefit = 2000(ax − axy)


(c) value of benefit is value of(1) an annuity-certain (of 3000 p.a., for 6 years) beginning at end of year of death of (x), plus(2) 1500 p.a., payable at times t (t ≥ 7) if (y) alive and (x) dead 6 years previously

= 3000a6 ·Ax + 1500∞∑

t=7

vttpy(1− t−6px)

= 3000a6 (1− dax) + 1500v66py(ay+6 − ax:y+6)

Add (a), (b), (c) to find total value of the annuity.

13.5 Let single premium be P .

Consider benefit (1) with payment at rate £1 p.a. There are two cases:

(a) husband dies within 25 years;

(b) husband survives for 25 years.(Indicate “m,f” to clarify which life is which.)

In case (a), M.P.V. is ∫ 25

0

vttp

m40µ m

40+t·tpf40a f

40+tdt

In case (b), M.P.V. is

v2525p

m40

[25p f

40a

m65:

f

65

+ (1− 25p f

40)am

65

]

=v2525pm

40· 25p f

40(a f

65− am

65:f

65) + v25

25pm40

am65

=v2525pm

40· 25p f

40·∫ ∞

0

vttpm

65:f

65µ m

65+ta f

65+tdt + v25

25pm40

am65

Therefore total M.P.V. is ∫ ∞

0

vttpm

40:f

40µ m

40+ta f

40+tdt + v25

25pm40

am65

=am40|

f

40+ v25

25pm40

am65

Benefit (1) = 5000(a40|40) + 5000v2525p40a65

= 5000(a40 − a40:40) + 5000v2525p40a65

' 5000[a40 − a40:40 + v2525p40(a65 − 1

2)]

= 5000(1.842 + 3.1418) = 24, 919.

Benefit (2) =P

2

∫ 25

0

vttp40µ40+t(1− tp40) dt

=P

2[A 1

40:25−A 1

40:40:25]

=P

2[A 1

40:25− 1

2A 1z }| {

40 : 40:25]

=P

2(1.04)

12

[A40:25 − D65

D40− 1

2

(1− d.a40:40:25 − D65:65

D40:40

)]

=P

2(0.094993− 0.5× 0.17511) = 0.003719P

13.6. SOLUTIONS 229

Hence equation of value isP = 24919 + 0.003719P

Therefore P = £25, 012.

13.6 Let annual premium be P .Value of benefits = 10, 000Am

70:f

64+ 5000(am

70|f

64+ a f

64|m70)

= 10, 000(1− δam70:

f

64) + 5000(am

70+ a f

64− 2am

70:f

64)

' 10, 000[1− δ(am70:

f

64+

12)] + 5000[am

70+ a f

64− 2am

70:f

64]

= 5, 323.85 + 18, 450= 23, 773.85

Value of premiums less expenses = 0.9P am70:

f

64= 5.9184P .

Hence P =23773.855.9184

= £4, 017 .


Chapter 14

PROFIT TESTING FORUNIT-LINKED POLICIES

14.1 Unit-Linked Policies

Most of the money paid in premiums by the policyholders is used to purchase “units”: that is,money is placed in a unitised investment of some kind (U.K. equities, property, etc.). The assetsunderlying each policyholder’s units form the Unit Fund of the policy, the value of which may becalculated from the unit price and the number of units held. Unit prices are quoted at two levels,the “Bid Price” (at which the units can be sold) and the “Offer Price” (at which the units must bebought).

The proceeds on maturity (at policy duration n) or earlier surrender are usually equal to the bidvalue of the units. On death, the benefits are usually equal to the value of the units (at bid price),subject to a minimum death benefit.

In addition, the office holds a balancing account, called the Sterling Fund (or Sterling Reserves),into which are paid deductions from the premiums for expenses and fund management charges, andfrom which it pays the actual office expenses and death guarantee costs.

The office may also transfer profits/losses from the sterling reserves to the shareholders or with-profits policyholders (who may be considered to be “investing” in the sale of the unit-linked policies).They may not take money from the unit fund as this belongs entirely to the policyholders.

14.2 Mechanics of the Unit Fund

Unit prices are quoted at 2 prices, the bid price and the offer price, which is an artificial higherprice. Define

1− λ =Bid PriceOffer Price

where λ is called the Bid/Offer Spread (λ being perhaps 0.05).

The bid price is the “real” price and all valuation calculations use the bid price.We use the following notation:

231

232 CHAPTER 14. PROFIT TESTING FOR UNIT-LINKED POLICIES

Pt = the office premium in year t (t = 1, 2, ..., n) (which is actually paid at time t− 1)

at = the allocation proportion in year t

= the proportion of the premium Pt which is allocated to buying units(at the offer price)

The cost of allocation in year t is the money actually used by the office to buy units, that is

(1− λ)atPt.

Hence Pt − (1− λ)atPt may be transferred to the sterling reserves as a deduction for expenses.Define

ct = the fund management charge in year t

= a charge made at the end of year t, usually a percentageof the value of the unit fund, which is transferred to the sterling fund.

The accumulation of the unit fundLet iu be the assumed rate of growth per annum of the unit fund.Define

Ft =the value of the unit fund at time t (afterdeduction of the fund management charge,but before payment of any premiumthen due) asssuming that the policy isstill in force.

DefineF0 = 0

ThenFt = [Ft−1 + (1− λ)atPt] (1 + iu)− ct (14.2.1)

Suppose that the fund management charge is a proportion m of the unit fund; we have

ct = m [Ft−1 + (1− λ)atPt] (1 + iu) (14.2.2)

and thusFt = (1−m) [Ft−1 + (1− λ)atPt] (1 + iu) (14.2.3)

Note. In some cases ct may be a fixed sum rather than a proportion of the fund.

Example 14.2.1. A life office issues a large block of 3-year unit-linked endowment assurances underwhich 80% of the first year’s premium and 101% of subsequent premiums are invested in units atthe offer price. The bid price of the units is 95% of the offer price. The units are subject to anannual management charge of 0.75% of the bid value of the fund at the end of each policy year.

The annual premium is £1,000 and unit prices are assumed to grow at 9% per annum.Calculate the bid value of the units at the end of each year, according to the office’s projections.

14.3. THE STERLING FUND (OR STERLING RESERVES) 233

Solution(1) (2) (3) (4) (5)

Cost of Fund brought Fund at endAllocation forward from start of year

t Pt(1− λ)at Ft−1 + Pt(1− λ)at before F.M.C. F.M.C. Ft

1 760.00 760.00 828.40 6.21 822.192 959.50 1,781.69 1,942.04 14.57 1,927.473 959.50 2,886.97 3,146.80 23.60 3,123.20

(3) = (1.09)× (2)(4) = 0.0075× (3)(5) = (3)− (4).

14.3 The Sterling Fund (or Sterling Reserves)

Money is assumed to earn interest at rate is p.a. in the sterling reserves. Define

et = projected expenses for office in year t (payable at the start of year t)ct = fund management charge (as before)

(DG)t = death guarantee cost in year t

=

{qx+t−1(St − Ft) if St > Ft

0 if St ≤ Ft

where x = the age of the policyholder at the start of the policy,and St = the guaranteed minimum death benefit in year t. We suppose that the policy is oneof a large number of similar unit-linked contracts on lives whose mortality follows a specified table,random variations being ignored.

Define

(SCF )t = the expected net cash flow in the sterlingfund in year t per policy in force at the start of the year

= the “in force” net cash flow

The “initial” expected net cash flow in the sterling fund in year t; that is the expected net cashflow per policy sold, is given by

t−1px.(SCF )t (t = 1, 2, ..., n) (14.3.1)

The formula for (SCF )t is

(SCF )t = [Pt − (1− λ)atPt − et](1 + is) + ct − (DG)t (14.3.2)

Maturity BonusesIn some policies, there may be a maturity bonus in the form of a fixed sum, or a proportion of

the bid value of the fund at maturity. This money must come from the sterling fund, so we mustadjust (SCF )n as follows:

(SCF )n = “normal” (SCF )n (as in (14.3.2))− px+n−1 × (Maturity Bonus) (14.3.3)


The Profit Vector and Profit SignatureDefine

(PRO)t = the profit vector

= the expected net profit to the office inyear t per policy in force at the start of the year

σt = the profit signature

= the expected net profit to the office inyear t per policy sold=t−1 px.(PRO)t (14.3.4)

There are 2 cases to consider.Case 1If there is no need (or desire) to maintain sterling reserves at the end of each policy year, we

have(PRO)t = (SCF )t for t = 1, 2, ..., n

Case 2Suppose now that the office wishes to maintain sterling reserves of tV at the end of year t

(t = 1, 2, ..., n− 1). It is assumed that 0V = nV = 0.The calculations are very similar to those for conventional profit-testing.The profit vector is

(PRO)t = (SCF )t + is ·t−1 V − (IR)t (14.3.5)

where

(IR)t = increase in reserves= px+t−1 ·t V −t−1 V

Therefore

(PRO)t = (SCF )t + (1 + is)t−1V − px+t−1 ·t V (14.3.6)

In both cases the profit signature is found by equation 4.3.4, i.e.

σt =t−1 px · (PRO)t

Example 14.3.1. (continued from Example 14.2.1)Suppose that, in addition to the information in example 14.2.1, the death benefit, payable at the

end of the year of death, is the greater of twice the annual premium and the bid value of the units.The office expects to incur initial expenses on these policies of 20% of the first premium. Renewalexpenses are expected to be £20, payable at the beginning of each policy year after the first. Themortality rate at each age is assumed to be 0.003. Sterling reserves are assumed to earn interest at4% per annum. Ignore the possibility of withdrawal.

Assuming that the office holds zero sterling reserves at the end of each policy year, calculate theprofit signature.

14.4. THE ASSESSMENT OF PROFITS 235

Solution(1) (2) (3) (4) (5) (6)

Premium less cost Accumulation Death Guarantee In Forceof allocation Expenses of Sterling Costs FMC cash flow

t Pt − Pt(1− λ)at et Fund in year (DG)t ct (SCF )t

1 240.00 200 41.60 3.53 6.21 44.282 40.50 20 21.32 0.22 14.57 35.673 40.50 20 21.32 0 23.60 44.92

(3) = [(1)− (2)]× (1.04)(4) = 0.003(2000− Ft)(6) = (3)− (4) + (5)

(7) (8) (9)Profit Vector Profit signature

t (PRO)t t−1px σt

1 44.28 1 44.282 35.67 0.997 35.563 44.92 0.994 44.65

(7) = (6) (as no reserves held)

(8) = (0.997)t−1

(9) = (7)× (8)

14.4 The Assessment of Profits

The profit signature {σt} can be assessed in one or more of the following ways.(1) One could work out the Internal Rate of Return (or yield) by solving

n∑t=1

vtσt = 0 where v =1

1 + j

(the internal rate of return, j, being the solution of this equation.)

(2) The shareholders may value the net profits at a certain rate of interest, j per annum. Thisrate is called the Risk Discount Rate, and may reflect uncertainties in {σt}, with j normally higherthan is.

The net present value of the profits is thus

NPV (j) =n∑

t=1

vtσt at rate j.

(3) The Profit Margin is defined as

n.p.v. of profitsn.p.v. of premiums

, both at some rate of interest, im say

=∑n

t=1 vtσt∑n−1t=0 Pt+1 · tpxvt

, at rate im.


Example 14.4.1. Find the net present value of the profit signature in example 4.3.1 at a riskdiscount rate of 10%.

Solution

n.p.v. of profit =3∑

t=1

vtσt at 10%

= 44.58v + 35.56v2 + 44.65v3

= 103.46.

14.5 Zeroisation of the Profit Signature

Suppose that the profit signature, σt, with no provision for sterling reserves at the end of each year(i.e. tV = 0 for all t), is of the following form:

(σt) =

26.22−8.05−5.18−2.29

0.62

(14.5.1)

It may be considered undesirable for the shareholders to take a profit of £26.22 at time 1, asthere may not be enough money to cover the “negative profits” at times 2, 3 and 4. Thus the profittaken by the shareholders in year 1 should be reduced in order to avoid negative sterling fund profitsin years 2, 3 and 4. This process is called Zeroisation.

Let X be the sum needed at time 1 to pay the negative cash flows at times 2, 3 and 4 when theSterling Reserves earn interest at rate is = 4 1

2% p.a. (for example).Then

X = 8.05v + 5.18v2 + 2.29v3 at rate 412%

= £14.46

NoteX is the amount required at time 1, and hence the negative cash flows are discounted to time 1 (notthe start of the policy).

Hence the shareholders may take a profit at time 1 of

26.22− 14.46 = £11.76.

There will now be no need for capital injections (from the shareholders to the Sterling Reserves)at times 2, 3 and 4. The shareholders may still take a profit of £0.62 at time 5.

The zeroised profit signature {σ′t} is thus

(σ′t) =

11.76000

0.62

Notes


(1) This process may be carried out even if σ1 < 0.

(2) We may calculate the revised profit vector using the equation

(PRO)′t =σ′t

t−1px

(3) We may also calculate the sterling reserves, tV , implied by zeroisation.

Example 14.5.1. Using the profit signature (14.5.1), and supposing that qx+t−1 = 0.01 for t =1, 2, 3, 4, 5, calculate the sterling reserves in each year that are implied by zeroisation of the profitsignature.

Solution(1) (2) (3) (4) (5) (6)

Original profit Zeroised profit Remainder of Accum. Prob. thatsignature signature net-cash flow of (3) policy is in force

t σt σ′t σt − σ′t at is tpx tV

1 26.22 11.76 14.46 14.46 0.99 14.612 -8.05 0 -8.05 7.06 0.980 7.203 -5.18 0 -5.18 2.19 0.970 2.264 -2.29 0 -2.29 0 0.961 05 0.62 0.62 0 0 0.951 0

(6) =(4)(5)

Observe that column (4) gives the sterling fund per policy sold, and the probability of being inforce at time t is tpx. Thus

Funds needed at time t per policy sold = tV × Pr{ policy is still in force at time t}.

14.6 Withdrawals

So far we have ignored the possibility of surrender. Now assume that the surrender of a policy mayoccur only at the end of a policy year (just before payment of the premium then due).

Define

wt = the probability that a policy will besurrendered at the end of year t (t = 1, 2, ..., n− 1)

Assume that wn = 0 as the policyholder will receive the maturity benefit at that time.The chance that a policy in force at the start of year t is surrendered at the end of year t is

therefore,px+t−1 · wt

Define(SV )t = the surrender value at time t

which is usually equal to Ft, the bid value of the policy’s units, in some cases minus a surrenderpenalty of (say) £10 or 1% of Ft.


The revised profit vector, (PRO)′t, allowing for withdrawals is

(PRO)′t = (PRO)t + px+t−1 · wt[Ft − (SV )t +t V ] (14.6.1)

where (PRO)t is as before.If tV = 0 for all t, then obviously

(PRO)′t = (PRO)t + px+t−1 · wt[Ft − (SV )t]. (14.6.2)

Note that if there is no surrender penalty (i.e. (SV )t = Ft),

(PRO)′t = (PRO)t.

The profit signature, σt, must also be adjusted to allow for surrenders; we have

σ′t =t−1 p′x · (PRO)′t (14.6.3)

where

t−1p′x = Pr{ policy is in force at time t− 1, allowing for withdrawals}

=

{t−1px(1− w1)(1− w2)...(1− wt−1) if t ≥ 21 if t = 1

NoteEven if (SV )t = Ft, and hence (PRO)′t = (PRO)t, σ′t will still differ from σt as the probability ofthe policy still being in force in each year will be different, when t > 1.

The profit signature allowing for withdrawals, {σ′t}, may be zeroised in the same way as before.When calculating the reserves, tV , implied by zeroisation, the probability of the policy being in forceat time t should be changed from tpx to

tpx(1− w1)(1− w2)...(1− wt)

Example 14.6.1. A life office issues a large number of 3-year unit-linked endowment policies to menaged 65, under each of which level annual premiums of £1,000 are paid. 80% of the first premiumand 105% of each subsequent premium is invested in units at the offer price. There is a bid/offerspread in unit values, the bid price being 95% of the offer price.

A fund management charge of 0.5% of the bid value of the policyholder’s fund is deducted at theend of each policy year, before payment of any benefits then due.

The death benefit, which is payable at the end of the year of death, is £3,000 or the bid valueof the units if greater. The maturity value is equal to the bid value of the units.

The office incurs expenses of £100 at the start of the first year and £20 at the start of each ofthe second and third years.

Mortality is assumed to follow A1967-70 ultimate. It is assumed that, at the end of each of thefirst two policy years, 2% of the surviving policyholders withdraw. The withdrawal benefit is 98%of the bid value of the units, after deducting the management charge.


(a) Assuming that the growth in the unit value is 7% p.a. and that the office holds unit reservesequal to the bid value of units and zero Sterling Reserves at the end of each year, calculate theprofit emerging at the end of each policy year per policy sold. Sterling Reserves are assumed to earninterest at 6% p.a.

(b) Calculate the revised profit emerging at the end of each year if the office takes a smallerprofit in year 1 in order to ensure that the profit emerging in the second and third policy years iszero.

Solution(a) We first work out the unit fund, Ft, per policy sold, assuming it remains in force.

Policy Cost of Funds brought Funds on end Unit fundYear allocation forward from start before deduction F.M.C. at end of year

t Pt(1− λ)at Pt(1− λ)at + Ft−1 of charge ct Ft

1 760.00 760.00 813.20 4.07 809.132 997.50 1806.63 1933.09 9.67 1923.423 997.50 2920.92 3125.38 15.63 3109.75We now calculate (PRO)t = (SCF )t and then find (PRO)t + px+t−1wt(0.02Ft), where x = 65.

Policy Prem. less cost Accumulation D. G. In forceYear of allocation Expenses in Sterling Fund costs F.M.C. cash flow

t Pt − Pt(1− λ)at et at 6% interest (DG)t ct (SCF )t

1 240 100 148.40 -52.65 4.07 99.822 2.50 20 -18.55 -28.57 9.67 -37.453 2.50 20 -18.55 0 15.63 -2.92

(PRO)t px+t−1wt(0.02Ft) (PRO)′t t−1p′x σ′t

99.82 0.32 100.14 1 100.14-37.45 0.75 -36.70 0.95645 -35.10-2.92 0 -2.92 0.91245 -2.66where t−1p

′x =t−1 px(1− w1)...(1− wt−1) with w3 = 0.

(b) Let reduction in profit at time 1 be X. Then

X = 35.10v + 2.66v2 at is = 6%= 35.48.

Hence profit taken at time 1 is

100.14− 35.48 = £64.66


Exercises

14.1 A life office issues a three-year unit-linked endowment policy to a life aged exactly 60. Theannual premium is £2,000, payable at the start of each year. The allocation proportion is90% in year 1 and 97% thereafter. At the end of year of death during the term, the policypays the higher of £10,000 and the bid value of units allocated to the policy, after deductionof the fund management charge. A bonus of 2% of the (bid) value of the unit fund is payableat maturity. The life office makes the following assumptions in projecting future cash flows:

Mortality A1967-70 ultimateInitial expenses: £300Renewal expenses: £50, incurred at the start of the

second and the third yearsFund management charge: 2% per annum, taken at the end of each year

before payment of any benefitsSterling fund interest rate: 4% per annumBid/offer spread: 6%Unit fund growth rate: 10% per annum.

Construct tables to show the following:

(i) the growth of the unit fund;

(ii) the profit signature, assuming that no sterling reserves are held;

(iii) the profit signature after taking into account sterling reserves, given that the sterlingreserves per policy are to be £36.48 before receipt of the premium due at time 1 year and£78.64 before receipt of the premium due at time 2 years.

In each case, indicate clearly how you calculate your table entries. Ignore the possibility ofsurrenders.

14.2 (a) In the context of profit-testing of unit-linked business, define the following terms briefly:(i) the Unit Fund,(ii) the Sterling Reserves,(iii) the profit vector of a policy,(iv) the profit signature of a policy,(v) the risk discount rate, and(vi) zeroisation of Sterling Reserves.

(b) A life office issues a large number of identical 4-year annual premium unit-linked endow-ment assurances to lives aged 65. According to the office’s calculations, the profit vector perpolicy sold, ignoring withdrawals and assuming that no Sterling reserves are maintained atthe end of each year, is as follows (£):

191.12−111.45−3.2810.95

The office’s mortality basis is A1967-70 ultimate, and Sterling Reserves earn interest at 5%per annum.Calculate

(i) the profit signature per policy sold, ignoring any need to maintain SterlingReserves at the end of each year, and

14.7. EXERCISES 241

(ii) the profit signature per policy sold if Sterling reserves are zeroised.

(c) The office now wishes to make an allowance for surrenders. It assumes that, at the endof the first and the second policy years, 3% of the surviving policyholders will surrender (justbefore payment of the second and third annual premiums respectively.) Surrender values areequal to the value of the policyholder’s units (after deduction of fund management charges),with a surrender penalty of £10. Calculate

(i) the revised profit signature per policy sold, ignoring any need to maintain Sterling Reservesat the end of each year,

(ii) the revised profit signature per policy sold if the Sterling Reserves are zeroised, and

(iii) the net present value, at a risk discount rate of 15% per annum, of the revised profitsignature per policy sold, assuming that the Sterling Reserves are zeroised.

14.3 (a) If a profit test for a unit-linked policy reveals negative cash flows in the second or anysubsequent policy year, it is customary to eliminate these negative values by setting up sterlingreserves at the end of each year.Describe briefly the technique (“zeroisation”) by which these reserves are calculated.

(b) An office issues a 3-year unit-linked policy with a yearly premium of £500. The deathbenefit, payable at the end of the year of death, is £1,000 or the bid value of units if greater.The maturity value is the bid value of the units at maturity.95% of each premium is invested in units at the offer price. The bid price of units is 95% ofthe offer price. Management charges of 1

4% of the bid value of the units are deducted at theend of each year (before payment of death and maturity claims).The office expects to incur expenses of £75 at the start of the first year and £25 at the startof each subsequent year.

Using a profit testing analysis, calculate for a life aged 60 at entry

(i) the expected profit in each of the 3 years per policy in force at the beginning of the year,

(ii) the net present value at the issue date of the expected profit from one policy assuming arisk discount rate of 10% per annum.

Assume that the unit fund grows at 8% per annum (before deduction of management charges),that sterling reserves need not be maintained at the end of each year, and that the possibilityof surrender may be ignored. The mortality of policyholders follows A1967-70 ultimate andsterling reserves earn interest at 6% per annum during each policy year.

14.4 If a profit test for a unit-linked policy reveals negative cash flows in the second or any sub-sequent policy year, it is customary to eliminate these negative values by setting up sterlingreserves at the end of each year.

Calculate the sterling reserves required at the end of each policy year, per policy then in force,for a 3- year policy for which the profit signature (with no allowance for sterling reserves atthe end of each year) is (250, - 100, - 50), given that the rate of mortality is 0.01 per annumat each age and sterling reserves earn interest at 8% p.a.

14.5 An office issues a unit-linked endowment assurance with annual premium £400 and term fiveyears to a life aged 60 who is subject to A1967-70 ultimate mortality.

The sum assured, payable at the end of year of death or at the maturity date, is the bid valueof the units held, subject to a guaranteed minimum death benefit of £2,000. The allocationproportion is 70% for the first annual premium and 98% for all subsequent annual premiums.For units the bid/offer spread is 5% and the annual rate of management charge is 0.75%.

In determining the sterling reserves necessary for the policy the office makes the followingassumptions:


Initial expenses: £125

Renewal expenses (associated with the payment of the second and each subsequent premium)£20 increased by 7% p.a. compound from the outset of the policy.

Growth rate for units: 7% p.a.

Interest rate for sterling fund: 4% p.a.

(a) On this basis(i) construct a table showing the growth of the unit fund over the duration of the policy, and(ii) construct a table showing the growth of the sterling fund in the absence of reserves.

(b) Hence determine the sterling reserves which should be held by the office to eliminate thesterling fund negative cash flows in the second and subsequent years of the policy’s duration.

(c) Consider the unit-linked policy described above.

Suppose, however, that the growth rate for units will be 10% p.a., that the sterling fundinterest rate will be 6% p.a., and that the inflationary growth rate for renewal expenses willbe 4% p.a. (from the outset of the policy).

(i) Construct a table showing the growth of the unit fund over the duration of the policyand(ii) Construct a table showing the growth of the sterling fund in the absence of reserves.(iii)Assuming that the office sets up the sterling fund reserves found above, determine theresulting sterling fund profit vector and signature. Find also the internal rate of returncorresponding to the profit signature.

14.8. SOLUTIONS 243

Solutions

14.1 (i)(1)

t Pt(1− λ)at Pt(1− λ)at + Ft−1 (1)× 1.1 F.M.C. Ft

1 1692 1692 1861.20 37.22 1823.982 1823.60 3647.58 4012.34 80.25 3932.093 1823.60 5755.69 6331.26 126.63 6204.63

(ii)(2) (3) Maturity

t Pt − Pt(1− λ)at et [(2)− (3)]× 1.04 (DG)t FMC Cost (SCF )t

1 308 300 8.32 118.00 37.22 0 -72.462 176.40 50 131.46 97.17 80.25 0 114.543 176.40 50 131.46 67.37 126.63 121.89 68.83

(Maturity cost in year 3 = 0.02(6204.63)× p62.)

t (PRO)t t−1p60 σt

1 -72.46 1 -72.462 114.54 0.98557 112.893 68.83 0.96979 66.75

(iii) (PRO)t = (SCF )t + (1 + is)t−1V −t V.px+t−1.

t (SCF )t (1.04)t−1V p59+t.tV (PRO)t t−1p60 σt

1 -72.46 0 35.95 -108.41 1 -108.412 114.54 37.94 77.38 75.10 0.98557 74.023 68.83 81.79 0 150.62 0.96979 146.07

14.2 (a) Simple definition of each term required. Check definitions with text.

(b) (i)


1 191.12 1 191.122 -111.45 0.97597 -108.773 -3.28 0.95007 -3.124 10.95 0.92226 10.10

(ii) Let X be the amount withheld at time 1 to cover negative cash flows at times 2 and 3.X = 108.77v + 3.12v2 at 5% interest

= 106.42

Hence σt =

84.7000

10.10

(c) (i)

Notice that Ft − (SV )t = 10 for all t.Hence (PRO)′t = (PRO)t + 10wtp64+t


t (PRO)t 10wtp64+t (PRO)′t t−1p′65 σ′t

1 191.12 0.29 191.41 1 191.412 -111.45 0.29 -111.16 0.94669 -105.233 -3.28 0 -3.28 0.89392 -2.934 10.95 0 10.95 0.86775 9.50

Notice that wt =

{0.03 for t = 1, 20 for t = 3, 4

(ii) Retain X at time 1.X = 105.23v + 2.93v2 at 5% interest

= 102.88

Hence zeroised profit signature is

88.5300

9.50

(iii) NPV = 88.53v + 9.50v4 at 15% interest = £82.41.

14.3 (a) Zeroisation is the process whereby the profit in the first year is reduced to pay for anyfuture negative cash flows (as explained in text).

(b) (i)

(1)t Pt(1− λ)at Pt(1− λ)at + Ft−1 (1)× 1.08 FMC Ft

1 451.25 451.25 487.35 1.22 486.132 451.25 937.38 1012.37 2.53 1009.843 451.25 1461.09 1577.98 3.94 1574.04

(2) (3)t Pt − Pt(1− λ)at et [(2)− (3)]× 1.06 (DG)t FMC (SCF )t = (PRO)t

1 48.75 75 -27.82 7.42 1.22 -34.022 48.75 25 25.17 0 2.53 27.703 48.75 25 25.17 0 3.94 29.11

(ii)t (PRO)t t−1p60 σt

1 -34.02 1 -34.022 27.70 0.98557 27.303 29.11 0.96979 28.23

Hence net present value =3∑

t=1

σtvt at 10% interest

= −34.02v + 27.30v2 + 28.23v3

= £12.84.

14.4

σt =

250−100−50

14.8. SOLUTIONS 245

Let X be the sum retained in year 1 to cover the later negative cash flows. ThenX = 100v + 50v2 at 8% interest

= £135.46

Hence the zeroised profit signature is σ′t =

114.5400

(1) Accumulationt σt σ′t σt − σ′t of (1) tpx tV

1 250 114.54 135.46 135.46 0.99 136.832 -100 0 -100 46.30 0.980 47.243 -50 0 -50 0 0.970 0

14.5 (a) (i)(1)

t Pt(1− λ)at Pt(1− λ)at + Ft−1 (1)× 1.07 FMC Ft

1 266.00 266.00 284.62 2.13 282.492 372.40 654.89 700.73 5.26 695.473 372.40 1067.87 1142.62 8.57 1134.054 372.40 1506.45 1611.90 12.09 1599.815 372.40 1972.21 2110.26 15.83 2094.43

(ii)

(2) (3)t Pt − Pt(1− λ)at et [(2)− (3)]× 1.04 (DG)t FMC (SCF )t

1 134 125 9.36 24.79 2.13 -13.302 27.60 21.40 6.45 20.89 5.26 -9.183 27.60 22.90 4.89 15.37 8.57 -1.914 27.60 24.50 3.22 7.87 12.09 7.445 27.60 26.22 1.44 0 15.83 17.27


1 -13.30 1 -13.302 -9.18 0.98557 -9.053 -1.91 0.96979 -1.854 7.44 0.95257 7.095 17.27 0.93385 16.13

(b) Let X be the amount that has to be withheld at time 1 to pay for the negative cash flowsat times 2 and 3.

Hence X = 9.05v + 1.85v2 at 4% interest= £10.41

So the zeroised profit signature is

σ′t =

−23.7100

7.0916.13


(4) Accumulation oft σt − σ′t (4) at 4% interest tpx tV

1 10.41 10.41 0.98557 10.562 -9.05 1.78 0.96979 1.843 -1.85 0 0.95257 04 0 - 05 0 - 0

(c) (i)

t Pt(1− λ)at Pt(1− λ)at + Ft−1 (4)×1 · 1 FMC Ft

1 266.00 266.00 292.60 2.19 290.412 372.40 662.81 729.09 5.47 723.623 372.40 1096.02 1205.62 9.04 1196.584 372.40 1568.98 1725.88 12.94 1712.945 372.40 2085.34 2293.87 17.20 2276.67

(ii)

(5) (6)t Pt − Pt(1− λ)at et [(5)− (6)]× 1.06 (DG)t FMC (SCF )t

1 134.00 125.00 9.54 24.67 2.19 -12.942 27.60 20.80 7.21 20.44 5.47 -7.763 27.60 21.63 6.33 14.26 9.04 1.114 27.60 22.50 5.41 5.64 12.94 12.715 27.60 23.40 4.45 0 17.20 21.65

(iii)t (SCF )t tV (1.06)t−1V p59+t · tV (PRO)t t−1p60 σt

1 -12.94 10.56 0 10.41 -23.35 1 -23.352 -7.76 1.84 11.19 1.81 1.62 0.98557 1.603 1.11 0 1.95 0 3.06 0.96979 2.974 12.71 0 0 0 12.71 0.95257 12.115 21.65 0 0 0 21.65 0.93385 20.22

This uses(PRO)t = (SCF )t + (1 + is)t−1V − px+t−1.tV

Let the internal rate of return be j p.a. Then j solves5∑

t=1

σtvt = 0 where v =

11 + j

.

i.e. −23.35v + 1.60v2 + 2.97v3 + 12.11v4 + 20.22v5 = 0.

Solving this equation by trials and interpolation givesv = 0.87173.

Hence j = 0.147 = 14.7% p.a.

Chapter 15

MULTIPLE-DECREMENTTABLES

15.1 Introduction

Consider a body of lives subject to two “modes of decrement”, α and β. For example, considera group of bachelor employees of a large company, from which men can leave by either mode α,marriage, or by mode β, leaving the company (mortality being ignored).

Take a bachelor employee aged x, and let

T1 =time to marriage of (x), whetheror not he is still an employee of the company

and

T2 =time until (x) leaves the service of the company, whetherhe is then married or not

(At this point it is assumed that all bachelors eventually marry.)

Let T = min{T1, T2} =time until exit from the group ofbachelor employees, by either mode α

(marriage) or mode β (withdrawal from service),whichever comes first, of abachelor employee aged x

Note This is similar to the joint-life situation in Chapter 11.

Define

t(ap)x =the probability that (x) will “survive” for atleast t years with respect to both modesof decrement. (In the above example, he will notmarry or leave the service within t years.)

=Pr{T ≥ t}

247

248 CHAPTER 15. MULTIPLE-DECREMENT TABLES

and

t(aq)x =Pr{T < t}=Pr{(x) leaves before time t, either by

mode α or mode β}In the example above

t(aq)x = the probability that (x) gets marriedor leaves service (or does boththese things) within t years

We may proceed through the development of “life tables”, by merely putting “a” in front of thevarious functions. For example, the function, (al)x, x ≥ x0, is constructed to be such that

t(ap)x =(al)x+t

(al)x, (x ≥ x0, t ≥ 0)

When t = 1 it may be omitted, giving

(aq)x =(ad)x

(al)x=

(al)x − (al)x+1

(al)x

and

t|(aq)x =(ad)x+t

(al)x

Also, define the “force of exit” from the double-decrement table (by whichever mode occurs first)by

(aµ)x = limh→0+

h(aq)x

h

It follows as for ordinary life tables that

t(ap)x = exp[−

∫ t

0

(aµ)x+r dr

]

The probability density function of T is{

t(ap)x(aµ)x+t , t ≥ 00 , t < 0

and its distribution function is

t(aq)x =∫ t

0r(ap)x (aµ)x+r dr (t ≥ 0)

We also define

(am)x =central rate of “total decrement” (modesα and β together)

=(ad)x

(aL)x=

∫ 1

0(al)x+t(aµ)x+t dt∫ 1

0(al)x+t dt

15.2. THE ASSOCIATED SINGLE-DECREMENT TABLES 249

15.2 The Associated Single-Decrement Tables

Each mode of decrement may be thought of as possessing its own “life table”, denoted by, forexample, lαx or lβx . Thus

tpαx = Pr{(x) does not exit by mode α before time t,

whether exit by mode β happens or not}( = Pr{(x) does not marry within t years } in our example.)

tpβx = Pr{(x) does not exit by mode β before time

t, whether exit by mode α happens or not}( = Pr{(x) does not leave service within t years} in our example.)

Note thattp

αx = Pr{T1 ≥ t} and tp

βx = pr{T2 ≥ t}.

We also have

tqαx = Pr{(x) exits by mode α before time t,

whether exit by mode β happens or not}= Pr{T1 ≤ t}= 1− tp

αx .

Similarly

tqβx = Pr{T2 ≤ t}

= 1− tpβx .

The functions, tqαx , tq

βx are called the independent rates (or probabilities) of exit within t years

at age x in their respective single-decrement tables.

15.3 The Relationships between the Multiple-Decrement Ta-ble and its Associated Single-Decrement Tables

It is usually assumed that the variables T1 and T2 are independent. This gives

t(ap)x = Pr{T1 ≥ t and T2 ≥ t}= Pr{T1 ≥ t}Pr{T2 ≥ t}= tp

αx tp

βx (15.3.1)

Hence we may construct the function (al)x from lαx , lβx . In particular

t(ap)x =(al)x+t

(al)x=

lαx+t

lαx· lβx+t

lβxfor all x ≥ x0, t ≥ 0


Put x = x0, y = x0 + t to obtain

(al)y = k.lαy .lβy (y ≥ x0)

where k is a constant. If we choose the radix (al)x0 to be equal to lαx0.lβx0

, then

(al)y = lαy .lβy for all y ≥ x0( as k = 1)

Some Important Formulae

t(aq)x = 1− t(ap)x

= 1− (1− tqαx)(1− tq

βx), assuming independence of T1 and T2,

= tqαx + tq

βx − tq

αx .tq

βx

If t = 1 ,(aq)x = qα

x + qβx − qα

x qβx (15.3.2)

Also

(aµ)x = limh→0+

h(aq)x

h

= limh→0+

[hqα

x + hqβx − hqα

x .haβx

h

]

= limh→0+

hqαx

h+ lim

h→0+

hqβx

h+ lim

h→0+

(hqα

x

h

)(hqβ

x

h

)h

= µαx + µβ

x (similar to joint-life argument).

15.4 Dependent Rates of Exit

Define

t(aq)αx = Pr{(x) exits by mode α within time t,

exit by mode β not having previously occurred}

In terms of our example, this is the probability that (x) gets married within t years while stillan employee of the company.We have

t(aq)αx = Pr{T1 < t and T1 < T2}

where T1, T2 have p.d.f.’s

f1(t1) = t1pαxµα

x+t1 , t1 > 0,

f2(t2) = t2pβxµβ

x+t2 , t2 > 0, respectively .

So

15.4. DEPENDENT RATES OF EXIT 251

t(aq)αx =

∫∫t1<t2t1<t

f1(t1).f2(t2) dt1 dt2

=∫ t

0

∫ ∞

t1t1p

αxµα

x+t1 .t2pβxµβ

x+t2 dt2 dt1

=∫ t

0t1p

αxµα

x+t1

(∫ ∞

t1t2p

βxµβ

x+t2 dt2

)dt1

=∫ t

0t1p

αxµα

x+t1 .t1pβx dt1

Thus

t(aq)αx =

∫ t

0rp

αx .µα

x+r.rpβx dr (15.4.1)

=∫ t

0r(ap)xµα

x+r dr

When t = 1,

(aq)αx =

∫ 1

0tp

αxµα

x+t.tpβx dt.

(aq)αx , (aq)β

x are called the dependent rates (or probabilities) of exit at age x by mode α, β respec-tively. (The word ‘dependent’ indicates that (aq)α

x depends not only on mode α but also on mode β).

Example 15.4.1. A population is subject to 2 modes of decrement, α and β. For 30 ≤ x ≤ 32, itis known that

µαx = 0.05

and µβx = (35− x)−1

For an individual in this population at exact age 30, calculate(a) the probability that he will still be in the population at age 32.(b) the probability that he will leave by mode α before age 32.

Solution

(a) tpα30 = exp

(−

∫ t

0

µα30+r dr

)= e−0.05t

tpβ30 = exp

(−

∫ t

0

[35− (30 + r)]−1 dr

)

= exp(−

∫ t

0

15− r

dr

)

= exp[log(5− r)]t0 =5− t

5.

Thus 2(ap)30 =2 pα30.2p

β30 = e−0.1 · 3

5 = 0.5429


(b) 2(aq)α30 =

∫ 2

0tp

α30µ

α30+t.tp

β30 dt

=∫ 2

0

0.05e−0.05t

(5− t

5

)dt

= 0.05∫ 2

0

e−0.05t dt− 0.01∫ 2

0

t.e−0.05t dt

= (1− e−0.1) +25e−0.1 − 4(1− e−0.1) (using integration by parts)

= 0.07645.

Theorem If there is a Uniform Distribution of Decrements between ages x and x + 1 in eachsingle-decrement table, then

(aq)αx = qα

x

(1− 1

2qβx

)

and

(aq)βx = qβ

x

(1− 1

2qαx

)

Proof Under a Uniform Distribution of Decrements (U.D. of D.) for a life table,(a) lx+t is linear (0 ≤ t ≤ 1)(b) tqx = t.qx (0 ≤ t ≤ 1)(c) tpxµx+t = qx (0 ≤ t ≤ 1)

Hence

(aq)αx =

∫ 1

0tp

αx .µα

x+t.tpβx dt

=∫ 1

0

qαx (1− tq

βx) dt (using U.D. of D. for mode α)

= qαx

∫ 1

0

(1− t.qβx ) dt (using U.D. of D. for mode β)

= qαx

[t− t2

2qβx

]t=1

t=0

= qαx (1− 1

2qβx ).

NoteIt is often assumed that U.D. of D. holds approximately in each single-decrement table, so the aboveresults may be used as approximations.

Example 15.4.2. In a double-decrement table with 2 causes of decrement, death (d) and withdrawal(w), the central rates of decrement at a certain age x aremd

x = 0.01mw

x = 0.2Show that if the central rate of withdrawal was doubled, the dependent rate of mortality would bereduced by approximately 0.00076.

15.5. PRACTICAL CONSTRUCTION OF MULTIPLE-DECREMENT TABLES 253

Solution Assuming U.D. of D.,

qdx '

mdx

1 + 12md

x

= 0.009950

qwx ' mw

x

1 + 12mw

x

= 0.18182

Hence (aq)dx ' qd

x(1− 12.qw

x ) = 0.00905

Now suppose mwx is increased to 0.4.

Then

qwx ' 0.4

1 + 0.2= 0.3333.

Hence

(aq)dx ' qd

x(1− 12× 0.3333) = 0.00829

i.e. a reduction of 0.00076.

Getting qαx , qβ

x from (aq)αx , (aq)β

x

Assuming U.D. of D., we have shown that

(aq)αx = qα

x (1− 12qβx ), (i)

(aq)βx = qβ

x (1− 12qαx ) (ii)

Substituting qβx = (aq)β

x

1− 12 qα

xfrom (ii) into (i) gives a quadratic equation for qα

x , i.e.

(qαx )2 + qα

x [(aq)βx − (aq)α

x − 2] + 2(aq)αx = 0.

This can be solved (rejecting any solution outside the range 0 and 1) to find qαx , and hence qβ

x

may be found from (i) or (ii).

15.5 Practical Construction of Multiple-Decrement Tables

We define

(ad)αx =the expected number of exits by mode α

between ages x and x + 1 in amultiple-decrement table with(al)x lives at age x

=(al)x(aq)αx

Note also that

(aq)x = (aq)αx + (aq)β

x


To prove these results, we note that

t(aq)αx + t(aq)β

x = Pr{T1 ≤ t and T1 ≤ T2}+ Pr{T2 ≤ t and T2 ≤ T1}= Pr{min{T1, T2} ≤ t}= t(aq)x.

and hence (putting t = 1, which can be omitted)

(ad)x = (al)x(aq)x

= (ad)αx + (ad)β

x

Also,

t|(aq)αx = Pr{(x) leaves by mode α in the

multiple-decrement table betweenages x + t and x + t + 1}

= t(ap)x(aq)αx+t

=(ad)α

x+t

(al)x

ProcedureThere are 3 basic steps in the construction of a multiple-decrement table.

Step 1Choose a radix (al)x0 where x0 = the youngest age considered.(For example, (al)x0 may be 10,000 or 100,000).

Step 2Calculate qα

x , qβx (x = x0, x0 + 1, ...) and hence evaluate (aq)α

x , (aq)βx(x = x0, x0 + 1, ...), assuming

U.D. of D.Step 3

Calculate(ad)α

x0= (al)x0(aq)α

x0

and

(ad)βx0

= (al)x0(aq)βx0

Then find

(al)x0+1 = (al)x0 − [(ad)αx0

+ (ad)βx0

]= (al)x0 − (ad)x0

Repeat this procedure for x0 + 1, x0 + 2 and so on.

Example 15.5.1. A certain population is subject to 2 modes of decrement,

d = deathi = permanent disability

15.5. PRACTICAL CONSTRUCTION OF MULTIPLE-DECREMENT TABLES 255

The independent rates of mortality are in accordance with A67-70 ult., and there is an indepen-dent rate of disablement of 0.01 at each age x (60 ≤ x ≤ 62).Construct a multiple-decrement table for ages x = 60, 61, 62 (and include the number of survivorsat age 63.)

SolutionUse U.D. of D. to find

(aq)dx = qd

x(1− 12· qi

x)

and(aq)i

x = qix(1− 1

2· qd

x)

Choose a radix of (al)x0 = 100, 000 (for example)

x qdx qi

x (aq)dx (aq)i

x (al)x (ad)dx (ad)i

x

60 0.01443 0.01 0.01436 0.00993 100,000 1,436 99361 0.01601 0.01 0.01593 0.00992 97,571 1,554 96862 0.01775 0.01 0.01776 0.00991 95,049 1,679 94263 92,428

Applications of Multiple-Decrement Tables

1. The probabilities of various events can be calculated.

Example 15.5.2. Using the multiple-decrement table constructed in example 15.5.1, calculate(a) The probability that a person aged exactly 60 is alive and not disabled

at age 63.(b) The probability that a person aged 60 becomes disabled within 3 years.

Solution(a)

3(ap)60 =(al)63(al)60

=92428100000

= 0.92428

(b)

3(aq)i60 =

(ad)i60 + (ad)i

61 + (ad)i62

(al)60

=993 + 968 + 942

100, 000= 0.02903

2. Financial calculations can be carried out using multiple-decrement tables, as will be shown in thenext chapter.


15.6 Further Formulae

The Identity of the forcesDefine

(aµ)αx = lim

h→0+

h(aq)αx

h

Theorem(aµ)α

x = µαx for all x (15.6.1)

ProofAssume that µα

x and µβx are continuous.

Then

(aµ)αx = lim

h→0+

h(aq)αx

h

= limh→0+

∫ h

0 t(ap)xµαx+t dt

h

Now, by L’Hopital’s rule,

(aµ)αx = lim

h→0+

ddh [

∫ h

0 t(ap)xµαx+t dt]

ddh (h)

= limh→0+

h(ap)xµαx+h

1= µα

x (as limh→0+

h(ap)x = 1)

This result is called the identity of the forces. Note that

(aµ)x = limh→0+

h(aq)x

h

= limh→0+

[h(aq)α

x

h+ h(aq)β

x

h

]

= (aµ)αx + (aµ)β

x

= µαx + µβ

x

Central Rates of DecrementThe dependent central rate of decrement by mode α at age x is given by

(am)αx =

(ad)αx

(aL)x

=

∫ l

0(al)x+t(aµ)α

x+t dt∫ 1

0(al)x+t dt

=

∫ 1

0(al)x+tµ

αx+t dt∫ 1

0(al)x+t dt

(from (15.6.1))

' µαx+ 1

2

Compare this to the corresponding independent central rate of decrement:

15.7. GENERALIZATION TO 3 MODES OF DECREMENT 257

mαx =

∫ 1

0lαx+tµ

αx+t dt∫ 1

0lαx+t dt

' µαx+ 1

2

Therefore it is normally assumed that

(am)αx ' mα

x

NoteWe may also construct tables using select rates of decrement, using

qdx = q[x], q

dx+1 = q[x]+1, ...

for example.

15.7 Generalization to 3 Modes of Decrement

Suppose there are now 3 modes of decrement, α, β and γ. By generalising the results for 2 modes,it can be shown that

t(ap)x = tpαx .tp

βx .tp

γx

and

(aq)αx =

∫ 1

0tp

αxµα

x+t.tpβx .tp

γx dt

TheoremUnder U.D. of D. of each mode of decrement in its single-decrement table

(aq)αx = qα

x [1− 12(qβ

x + qγx) +

13qβx · qγ

x ]

Proof

(aq)αx =

∫ 1

0tp

αxµα

x+t.tpβx .tp

γx dt

=∫ 1

0

qαx (1− tq

βx)(1− tq

γx) dt (using U.D. of D. on α)

= qαx

∫ 1

0

(1− t.qβx )(1− t.qγ

x) dt (using U.D. of D. on β and γ)

= qαx

∫ 1

0

[1− t(qβ

x + qγx) + t2qβ

x .qγx

]dt

= qαx

[t− t2

2(qβ

x + qγx) +

t3

3qβx .qγ

x

]t=1

t=0

= qαx

[1− 1

2(qβ

x + qγx) +

13qβx .qγ

x

].

Note also that the identity of the forces remains true, and

(aµ)x = µαx + µβ

x + µγx.


In the practical construction of a multiple-decrement table, we find the values of (al)x, (ad)αx ,

(ad)βx , (ad)γ

x and use(al)x+1 = (al)x − [(ad)α

x + (ad)βx + (ad)γ

x].

Example 15.7.1. The members of a large company’s manual workforce are subject to three modesof decrement, death, withdrawal and promotion to supervisor. It is known that these workers’independent rates of mortality are those of English Life Table No. 12 - Males, the independentwithdrawal rate is 0.04 at each age, and their independent promotion rate is 0.02 at age 50 and 0.03at age 51.

(a) Draw up a service table for manual workers from age 50 to age 51 with a radix of 100,000at age 50, including the value of (al)52.

(b) Calculate the probability that a life aged exactly 50 will gain promotion within 2 years.

Solution

(a) Let d = death, p = promotion, w = withdrawal.x qd

x qwx qp

x (aq)dx (aq)w

x (aq)px (al)x (ad)d

x (ad)wx (ad)p

x

50 0.00728 0.04 0.02 0.00706 0.03946 0.01953 100,000 706 3946 195351 0.00823 0.04 0.03 0.00795 0.03924 0.02928 93,395 742 3665 273552 86,253

This table assumes U.D. of D., i.e.

(aq)dx = qd

x[1− 12(qw

x + qpx) +

13qwx .qp

x].

with similar formulae for withdrawal and promotion rates.

(b)

2(aq)p50 =

(ad)p50 + (ad)p

51

(al)50

=1953 + 2735

100, 000= 0.04688.

15.8 “Abnormal” Incidence of Decrement

Suppose that there are 2 modes of decrement, α and β operating between ages x and x+1. Assumemode β operates “smoothly” (i.e. µβ

x exists and is continuous) between ages x and x + 1. However,mode α only operates at a particular age, x + k (0 ≤ k ≤ 1).That is

tpαx =

{1 for t < k

1− qαx for t > k.

It follows that µαx+t is not defined for t = k and hence formulae such as

(aq)αx =

∫ 1

0tp

αx .tp

βxµα

x+t dt

15.8. “ABNORMAL” INCIDENCE OF DECREMENT 259

cannot be used.

To deal with this “abnormal” mode of decrement, we argue that from age x to x + k, mode βoperates by itself. Then, at age x + k, there is a chance qα

x of exit by mode α, so

(aq)αx = Pr{(x) survives to age x + k under mode β only}

× Pr{(x) leaves at age x + k by mode α, givensurvival until then}

= kpβx .qα

x

= qαx (1− kqβ

x)

If (as usually the case) U.D. of D. for mode β is assumed

(aq)αx = qα

x (1− k.qβx ) (15.8.1)

To obtain (aq)βx , we may use

(aq)βx = (aq)x − (aq)α

x

= [qαx + qβ

x − qαx .qβ

x ]− qαx (1− k.qβ

x )

So

(aq)βx = qβ

x [1− (1− k)qαx ] (15.8.2)

Special Cases1. If k = 1

2 , we obtain the familiar formulae

(aq)αx = qα

x (1− 12· qβ

x )

and(aq)β

x = qβx (1− 1

2· qα

x )

2. If k = 0(aq)α

x = qαx

and(aq)β

x = qβx (1− qα

x )

3. If k = 1, (i.e. we consider exits by mode α to occur just before reaching age x + 1)

(aq)αx = qα

x (1− qβx )

and(aq)β

x = qβx .

Example 15.8.1. Military personnel attend a training camp for an intensive course which involves3 weeks of continual exercises.

Each soldier remains in the camp for exactly 3 weeks, provided he is not hospitalised because


of injury or “failed” by one of the instructors and sent back to his base. The independent weekly

rates of decrement are as follows:

HospitalisedWeek through injury Being failed

1 0.078 0.1322 0.102 0.0923 0.058 0.043

(i) Of a group of 1,000 soldiers who start the course, calculate the number who will successfullycomplete the course, the number who will be hospitalised and the number who will be failed.

Assume a uniform distribution over each week of hospitalisation and failure. Anyone sent tohospital or failed leaves the camp immediately and does not return. There are no other modes ofdecrement.

(ii) Some time later the course is altered so that in the first week the instructors test the sol-diers only at the start of the sixth day, and the independent weekly rate of being failed in week 1alters to 0.160. During weeks 2 and 3, testing takes the form of continuous assessment as previously,with no change to the independent weekly rates of being failed.

Calculate the revised numbers of soldiers successfully completing the course, being hospitalisedand being failed, assuming nothing else changes.

Solution

(i) Consider “age” as time (in weeks) since entry to camp. Assume U.D. of D., which gives

(aq)hx = qh

x(1− 12qfx)

and (aq)fx = qf

x(1− 12· qh

x)

where h = hospitalised, f = failed.

x qhx qf

x (aq)hx (aq)f

x (al)x (ad)hx (ad)f

x

0 0.078 0.132 0.07285 0.12685 1000 73 1271 0.102 0.092 0.09731 0.08731 800 78 702 0.058 0.043 0.05675 0.04175 652 37 273 588

Hence number who complete the course = 588.

(ii) Here, k = 57 (test is at end of 5th day), so

(aq)hx = qh

x(1− 27· qf

x)

and(aq)f

x = qfx(1− 5

7· qh

x)

for x = 0 only.qf0 is now 0.16, but all the other values of qf

x and qhx are unchanged.

x qhx qf

x (aq)hx (aq)f

x (al)x (ad)hx (ad)f

x

0 0.078 0.16 0.07443 0.15109 1000 74 1511 0.102 0.092 0.09731 0.08731 775 75 682 0.058 0.043 0.05675 0.04175 632 37 263 570

Hence now the number who complete the course = 570.

15.8. “ABNORMAL” INCIDENCE OF DECREMENT 261

Application to Profit-TestingLet α = withdrawal, and β = mortality. It is assumed that withdrawals occur only at the end of apolicy year. That is, mode α only operates at time k = 1.In chapter 14, the following results were used:

Pr{ policy in force at time t− 1 will be surrendered at time t}= (1− qmortality

x+t−1 ).qwithdrawalx+t−1

where qwithdrawalx+t−1 was denoted by wt

andPr{ death occurs in policy year given it is in force at start of year} = qmortality

x+t−1

These formulae correspond exactly to those given in this section when k = 1.

Extension to 3 modes of decrementWith 3 modes of decrement, formulae similar to (15.8.1) and (15.8.2) may be awkward to derive.

These problems are best handled from first principles. Often one mode (α, say) operates at exactage x and the others (β and γ) operate uniformly over the year of age from x to x + 1.

This can be dealt with by defining (al)x = number of lives at age x before mode α operates, and(al)+x = number of lives at age x after mode α operates.

Then the numbers of exits by modes β and γ between ages x and x + 1 can be worked out as ifthere were no other modes of decrement. Note however that

(ad)βx = (al)+x (aq)β

x

where (aq)βx = qβ

x (1− 12qγx)

Similar calculations apply to “abnormal” exits at the end of the year.


Example 15.8.2. Among the employees of a certain firm retirement may take place at or after age57, but is compulsory at age 60. The independent rates of mortality of the employees are those ofA1967-70 ultimate. 20% of those attaining age 57 retire at once. Leaving aside these retirements,the central rates of retirement are as follows:

Central RateAge of Retirement57 0.0958 0.0859 0.05

There are no withdrawals or ill-health retirements after age 50.Construct a service table for employees from age 57 to age 60, with a radix of 100,000 employeesattaining age 57.

SolutionCreate a special mode of decrement, r′, to refer to retirement at exact age x. Then,

(ad)r′57 = 20, 000,

leaving (al)+57 lives who are subject to ‘normal’ retirements and death, both of which are approxi-mately “U.D. of D.”So construct a table of (aq)d

x and (aq)rx (ignoring mode r′).

x qdx qr

x ' mrx

1+ 12 mr

x(aq)d

x (aq)rx

57 0.01050 0.08612 0.01004 0.0856758 0.01169 0.07692 0.01124 0.0764759 0.01299 0.04878 0.01267 0.04846

Now construct the multiple-decrement table, noting that

(ad)rx = (al)+x (aq)r

x.

x (al)x (ad)r′x (al)+x (ad)r

x (ad)dx

57 100,000 20,000 80,000 6,854 80358 72,343 0 72,343 5,532 81359 65,998 0 65,998 3,198 83660 61,964 61,964 0

NoteWe may take (al)60 = (ad)r′

60 since all survivors at age 60 retire at once.

15.9. EXERCISES 263

Exercises

15.1 Let α1, α2 be the modes of decrement in a double-decrement table. Suppose that α1 isuniformly distributed over the year of age from x to x + 1 in its associated single- decrementtable, and µα2

x+t = c for 0 ≤ t ≤ 1.Find formulae for (aq)α1

x and (aq)α2x in terms of qα1

x and qα2x .

15.2 For a certain group of married women, for whom remarriage is not permitted, the dependent q-type rate of widowhood at each integer age x from 70 to 72 inclusive is twice the correspondingdependent q-type rate of mortality. The independent rates of mortality of wives follow a(55)ultimate (females).

(i) Using a radix of 100,000 and assuming a uniform distribution of each mode of decrement inits associated single-decrement table, construct a double-decrement table for married womenfrom age 70 to age 72 inclusive, giving also the value of (al)73, the number of wives at age 73.

(ii) Find the probabilities that a wife now aged 70 will(a) be alive and married at age 73; and(b) be widowed within 3 years.

15.3 In a certain country, widowed and divorced men are subject to the following independentq-type rates of decrement:mortality: English Life Table No. 12 - Malesremarriage: rates depend on the age at, and the duration since, the end of former marriage;the following table is an extract from these rates:

exactage at end duration durationof former 0 1marriage year year

50 0.050 0.02551 0.045 0.02352 0.042 0.020

Calculate the probability that a man aged exactly 50 whose marriage has just ended willremarry within 2 years.

15.4 A large industrial company recruits a constant number of school leavers aged exactly 18 yearson 1 July each year.

Upon joining, workers undergo training for one year. Of those who complete this period oftraining, ten per cent fail a final test of competence and are dismissed. Employees may alsoleave service voluntarily at any time. The central rate of voluntary withdrawal from serviceis 0.15 for trainees and 0.10 at each age for fully trained employees.

The occupation is hazardous and all workers, including trainees, are exposed to the risk ofinjury. The independent q-type rate of injury is 0.051219 at age 18 and 0.050030 at ages 19and above. An employee who is injured is transferred to alternative work with a subsidiarycompany, at a relocation cost of £1,000.

The mortality of all employees follows English Life Tables No.12 - Males. The number ofemployees attaining age 21 each year is 500.

(i) Construct a service table covering the first 3 years of employment with the original com-pany, distinguishing between those about to take the final test of competence and those whopass it. [Regard failing the test as a special mode of decrement ](ii) How many people are recruited on each 1st July?


Solutions

15.1 α1 is uniformly distributed, sotp

α1x µα1

x+t = qα1x for 0 ≤ t ≤ 1

µα2x+t = c, so

tpα2x = exp[−

∫ t

0

µα2x+t dt] = e−ct for 0 ≤ t ≤ 1

(aq)α1x =

∫ 1

0tp

α1x µα1

x+t.tpα2x dt

=∫ t

0

qα1x e−ct dt

= qα1x

[−1

ce−ct

]t=1

t=0

= qα1x

(1− e−c)c

= qα1x

[ −qα2x

log(1− qα2x )

]as c = − log(1− qα2

x )

(aq)α2x = (aq)x − (aq)α1

x

= [qα1x + qα2

x − qα1x qα2

x ]− qα1x

[ −qα2x

log(1− qα2x )

](from above)

15.2 (i) (aq)wx = 2(aq)d

x for x = 70, 71, 72wherew = widowhoodd = death.Using U.D. of D.

(aq)wx = qw

x (1− 12qdx) = 2qd

x(1− 12qwx ) = 2(aq)d

x

Hence qwx = 2qd

x

1+ 12 qd

x

Therefore (aq)dx = qd

x(1− 12 qd

x)

1+ 12 qd

x, and (aq)w

x = 2(aq)dx.

x qdx (aq)d

x (aq)wx (al)x (ad)d

x (ad)wx

70 0.02307 0.02254 0.04509 100,000 2,254 4,50971 0.02559 0.02494 0.04989 93,237 2,326 4,65172 0.02839 0.02760 0.05519 86,260 2,380 4,76173 79,119

(ii)(a) 3(ap)70 = 79119100,000 = 0.79119

(b) 3(aq)w70 = (ad)w

70+(ad)w71+(ad)w

72(al)70

= 0.1392.

15.3 Denote

d = death


r = remarriage

Notice that qr51 = q[50]+1, i.e. duration 1 year from age 50 at end of former marriage.

x qdx qr

x (aq)dx (aq)r

x (al)x (ad)dx (ad)r

x

50 0.00728 0.050 0.00710 0.04982 100,000 710 4,98251 0.00823 0.025 0.00813 0.02490 94,308 767 2,348

91,193

2(aq)r50 =

(ad)r50 + (ad)r

51

(al)50= 0.0733.

15.4 (i) Denoted = deathw = withdrawal (excluding those who fail test)f = those who fail testi = injury

Useqwx ' mw

x

1 + 12mw

x

and in view of U.D. of D. use formulae such as

(aq)dx = qd

x[1− 12(qw

x + qix) +

13qwx qi

x]

for deaths, injuries and withdrawals, and treat the test as an “abnormal” mode of exit.

x qdx qi

x qwx (aq)d

x (aq)ix (aq)w

x

18 0.00112 0.05122 0.13953 0.001016 0.04762 0.1358819 0.00117 0.05003 0.09524 0.001087 0.04762 0.0928020 0.00119 0.05003 0.09524 0.001105 0.04762 0.09280

Define(al)19 = number of people before test (at age 19 exactly)(al)+19 = number of people who pass test.with

(al)+19 = 0.9(al)19

Let (al)18 = 10, 000 (for example)

x (al)x (ad)fx (al)+x (ad)d

x (ad)ix (ad)w

x

18 10,000 0 10,000 10 476 135919 8,155 816 7,339 8 349 68120 6,301 0 6,301 7 300 58521 5,409

Notice that (ad)fx = (al)x − (al)+x

(ii) 10,000 entrants give 5,409 employees after 3 years. Hence 924 entrants are needed to get,on average, 500 employees aged exactly 21.


Chapter 16

FINANCIAL CALCULATIONSUSINGMULTIPLE-DECREMENTTABLES

16.1 Principles

One may value cash flows, calculate premiums and find reserves using the same ideas as are usedwhen there is just one mode of decrement (death).Notation is best considered from first principles, and there are no commutation functions (exceptfor pensions, which are covered later).

The procedure is generally as follows:Stage 1Construct a multiple-decrement table.Stage 2Write down an equation of value of the formm.p.v. of premiums = m.p.v. of benefits + m.p.v. of expenses

(This assumes that the expected present value of the profits to the office is zero. If it is not, adda suitable profit term to the equation.)

Then solve the equation of value for the unknown quantity (for example, the annual premium orthe sum assured payable on death)NoteIt should normally be assumed that exits occur, on average, in the middle of each policy year, andthat premiums cease if a life exits by any mode of decrement.

16.2 The Use of “Defective” Variables

A defective random variable, T , is such that

limt→∞

F (t) = k < 1

where F (t) is the distribution function of T . For example, letT = time of exit of (x) by mode α in a double-decrement table, the other mode of exit being β.Then

267

268CHAPTER 16. FINANCIAL CALCULATIONS USING MULTIPLE-DECREMENT TABLES

F (t) = Pr{T ≤ t}= t(aq)α

x

=∫ t

0rp

αxµα

x+r.rpβx dr

Since∞(aq)α

x +∞(aq)βx = 1,

we havelim

t→∞F (t) = lim

t→∞ t(aq)αx < 1

A defective variable T may have a probability density function, f(t) = F ′(t). In the aboveexample,

f(t) =

{tp

αxµα

x+t.tpβx , t > 0

0, t < 0

With defective variables, we still have results such as

E[g(T )] =∫ ∞

−∞g(t)f(t) dt

For example, consider a benefit of £1 payable immediately on exit by mode α to a life now agedx. The m.p.v. of this benefit is

E[vT ] =∫ ∞

0

vt.tpαxµα

x+t.tpβx dt

Notice the similarities to the joint-life and contingent assurance functions in chapters 11 and 12.

16.3 Evaluation of Mean Present Values

This can either be done by integrals or sums.

(A) IntegralsThe theory of ‘defective’ variables is used to value the m.p.v. of benefits on exit by a given mode ofdecrement. We also consider the m.p.v. of premium payments.(i) Consider a double-decrement table with exits by mode α and mode β, and suppose that there isa benefit of £S payable immediately on the exit of (x) by mode α within n years.The present value of this benefit, as a random variable, is

Z =

{SvT if T < n

0 if T > n.

where T = time to exit by mode α in the double-decrement table.T is a defective variable with p.d.f.

tpαx .µα

x+t.tpβx (t > 0).

Hence the m.p.v. of this benefit is

S

∫ n

0

vt.tpαxµα

x+t.tpβx dt (16.3.1)

16.3. EVALUATION OF MEAN PRESENT VALUES 269

(ii) Consider premiums of £P per annum, payable continuously for at most n years while (x) remainsa member of the group under consideration. The mean present value is

P

∫ n

0

vt · (al)x+t

(al)xdt = P

∫ n

0

vt.t(ap)x dt

= P

∫ n

0

vt.tpαx .tp

βx dt (16.3.2)

(assuming 2 modes of decrement, α and β)

Example 16.3.1. Suppose there are 2 modes of decrement, death (d) and withdrawal (w), andthere is a constant force of withdrawal of k per annum.Calculate the value of premiums of £P per annum payable continuously for at most n years while(x) remains a member of the group.

Solutionµw

x+t = k for all t.Hence tp

wx = e−kt for all t.

m.p.v. of premiums = P

∫ n

0

vt.tpdx.tp

wx dt

= P

∫ n

0

e−δt.tpdx.e−kt dt

(as vt = e−δt where δ is the force of interest)

= P

∫ n

0

e−(δ+k)t.tpdx dt

= P ax:n at force of interest δ′ = δ + k

(B) SumsSums can be used to value death and other benefits and premium payments either exactly (if financialtransactions occur at the end of policy years) or by approximating integrals (16.3.1) and (16.3.2) ifbenefits are payable immediately or premiums are paid continuously.(i) Consider a benefit of £S payable at the end of the year of exit of (x) by mode α within n years.The m.p.v. of this benefit is

S

[v(ad)α

x

(al)x+ v2 (ad)α

x+1

(al)x+ · · ·+ vn (ad)α

x+n−1

(al)x

].

If the benefit is payable immediately, then exits can be assumed to occur, on average, mid-waythrough year. The m.p.v. of the benefit is

S

[v

12(ad)α

x

(al)x+ v

32(ad)α

x+1

(al)x+ · · ·+ vn− 1

2(ad)α

x+n−1

(al)x

]

(ii) Consider an annual premium of £P per annum, payable in advance while (x) is still a memberof the group, for at most n years. The m.p.v. is

P

[1 + v

(al)x+1

(al)x+ v2 (al)x+2

(al)x+ · · ·+ vn−1 (al)x+n−1

(al)x

]


If the premiums are payable continuously, expression (16.3.2) can be approximated to a sum.The m.p.v. is now

P

[v

12(al)x+ 1

2

(al)x+ v

32(al)x+ 3

2

(al)x+ · · ·+ vn− 1

2(al)x+n− 1

2

(al)x

]

Note(al)x+ 1

2, (al)x+ 3

2, etc. must often be found by interpolation.

Example 16.3.2. A multiple-decrement table referring to mortality (d) and withdrawal (w) froma life assurance contract is

x (al)x (ad)wx (ad)d

x

60 10,000 64 12661 9,810 61 13162 9,618 48 13463 9,436

Suppose a 3-year term assurance is issued to a life aged 60, providing £20,000 at the end of theyear of death. If expenses consist of 5% of each premium, calculate the annual premium, P , payablein advance while a policyholder, for at most 3 years. Interest is at 4% per annum.

Solution

m.p.v. of benefits =20, 000(al)60

[v.(ad)60 + v2(ad)61 + v3(ad)62]

= 722.79

m.p.v. of premiums =P

(al)60[(al)60 + v(al)61 + v2(al)62]

= 2.8325P.

Hence P solves0.95× (2.8325P ) = 722.79

Therefore P = £268.61 per annum.

Benefits other than cash sumsIn some cases, the benefits on exit by a certain mode of decrement consists not of a cash sum butan annuity or some other benefit. In such cases, treat the m.p.v. of the annuity (or other benefit)at the date of exit as if it were paid out at that time. (The annuity will often have to be evaluatedby interpolation if it begins mid-way through a year of age.)

Example 16.3.3. A life office issues policies to lives aged under 60 providing the following benefits:(i) on becoming permanently disabled before age 60, an annuity of £2,000 per annum payable weeklyfor life and £20,000 immediately on death, and(ii) immediately on death before age 60 while not permanently disabled, £20,000.

Calculate the office annual premium, payable weekly and ceasing on death, on permanent disabilityor on reaching age 60, for a life aged 58 if the office uses the following basis:

16.3. EVALUATION OF MEAN PRESENT VALUES 271

Mortality: the independent rates of mortality of those not permanently disabled are those ofA1967-70 ultimate; the permanently disabled are subject to the mortality of English Life TableNo.12 - Males with the age rated up by 6 1

2 years;Permanent disability: a constant independent rate of 0.006;Interest: 4% per annum;Expenses: 2 1

2% of all office premiums, plus £50 at the issue date.

Solution

Construct service table with d = death and i = disability, assuming U.D. of D.x qd

x qix (aq)d

x (aq)ix (al)x (ad)d

x (ad)ix

58 0.01169 0.006 0.01165 0.00596 100,000 1,165 59659 0.01299 0.006 0.01295 0.00596 98,238 1,272 58560 96,381

Value of benefits:

(i) v12(ad)i

58

(al)58

[2, 000aih

58 12

+ 20, 000Aih58 1

2

]

+ v32(ad)i

59

(al)58

[2, 000aih

59 12

+ 20, 000Aih59 1

2

]

(ii) 20, 000[v

12(ad)d

58

(ad)58+ v

32(ad)d

59

(al)58

]= 468.34

For benefit (i)

aih58 1

2= a65 on E.L.T. 12 - Males = 8.918

Aih58 1

2= A65 on E.L.T. 12 - Males = 0.65023

Similarly aih59 1

2= a66 = 8.587

Aih59 1

2= A66 = 0.66323.

So value of (i) = 180.24 + 167.89 = 348.13.

Let P be the annual premium. The value of premiums less expenses is

0.975P

[v

12(al)58 1

2

(al)58+ v

32(al)59 1

2

(al)58

]− 50

(al)58 12' 1

2[(al)58 + (al)59] = 99119

(al)59 12' 97310

Hence value of premiums less expenses is 1.8422P − 50.Therefore

1.8422P − 50 = 468.34 + 348.13

.Consequently P = £470.35.


16.4 Benefits on Death by a Particular Cause

Sometimes a policy will provide death benefits if (x) dies from a particular cause. One may writeµα

x+t = force of mortality from cause α, andµβ

x+t = force of mortality from all other causes.

When these are added together to give (aµ)x+t, this will often be the force of mortality for agiven life table.

Example 16.4.1. A national newspaper recently advertised “free insurance for subscribers”, wherebya benefit of £10,000 would be paid immediately onaccidental death (mode α) within n years. Assume µα

x+t = 0.0005 for t ≥ 0 (for all x) and mortalitydue to all causes follows A1967-70 ult.

Assuming a given rate of interest and ignoring expenses, calculate the annual premium payablecontinuously by the newspaper to provide this “free” policy.

SolutionLet the annual premium be P .

m.p.v. of benefits = 10000∫ n

0

vt.tpαxµα

x+t.tpβx dt

= 10000× 0.0005∫ n

0

vt.tpA67−70ultx dt

= 5ax:n on A67-70 ult.m.p.v. of premiums = Pax:n

Thus P = £5 per annum (or £0.42 per month.)

Note:Observe that P does not depend on the mortality table (for all causes), nor the rate of interest, northe term of n years. Also notice that £0.42 is a net premium, whereas an office premium (allowingfor expenses) would be larger.

16.5 Extra Risks Treated as an Additional Mode of Decre-ment

Suppose that in addition to “normal” mortality (mode α), a group of lives are subject to certainadditional hazards (mode β). Mode β may refer to:

(a) Certain occupational hazards;(b) Risks associated with a leisure activity, such as motor racing;(c) Some medical conditions.

One may wish to calculate premiums for policies which pay out only on death from the “addi-tional” cause, in which case the formulae in Section 16.3 may be used. If the death benefit is paidon any cause of death, the problem can be treated as an “extra risk” question, as covered in theActuarial Subject A2.

Example 16.5.1. A certain life office’s premium basis for policies accepted at normal rates is:

16.5. EXTRA RISKS TREATED AS AN ADDITIONAL MODE OF DECREMENT 273

A1967-70 select,4% interest,expenses are ignored.

A proposer, aged 45, for temporary assurance ceasing at age 65 is subject to an extra occupationalhazard which is considered to be equivalent to an addition of 0.009569 to the force of mortality atall ages. The sum assured, which is payable immediately on death, is £10,000.

(a) Calculate the level annual premium, payable throughout the term of the policy.(b) The proposer requests that, in the event of death occurring as a result of the special occupa-

tional hazard, the sum assured should be doubled, and offers to pay an additional single premiumat the outset for this extra cover.

Calculate this single premium.

Solution(a) Consider α = normal mortality, β = extra occupational mortality

Value of benefits is

10000∫ 20

0

vttp

α[45].tp

β[45](µ

α[45]+t + µβ

[45]+t) dt

= 10, 000∫ 20

0

vte−kt.tp[45](µ[45]+t + k) dt (where k = 0.009569)

= 10, 000A∗1[45]:20

where * indicates normal plus extra mortality

= 10, 000(1.04)12 A∗1

[45]:20

= 10, 000(1.04)12 [(1− d.a∗

[45]:20)−A∗

[45]:1

20

]

Now use the rule (as in extra risks) that the rate of interest (in annuity and pure endowmentfunctions) may be altered to allow for the addition to the force of mortality.

a∗[45]:20

is at force of interest δ′ = δ + k

where δ = log(1.04) = 0.03922 (rate of interest = 4%)So δ′ = 0.048790 and hence, i′ = eδ′ − 1 = 0.05 = 5%.

Hence m.p.v. of benefits is

10, 000(1.04)12 [1− d.a[45]:20 0.05 −A

[45]:1

20 0.05]

= £2, 102.87 (using A[45]:

1

20 0.05=

l65l[45]

v200.05)

So annual premium =2102.87a∗[45]:20

=2102.87

a[45]:20 0.05

= £167.78

(b) The m.p.v. of this benefit is found by considering exit by mode β only. This gives a single


premium of

10, 000∫ 20

0

vt.tpα[45].tp

β[45].µ

β[45]+t dt

= 10, 000∫ 20

0

e−δt.e−kt.tp[45].k dt (k = 0.009569)

= 10, 000ka∗[45]:20

= 10, 000ka[45]:20 0.05

' 95.69[a[45]:20 0.05 −

12(1− l65

l[45]v200.05)

]

' 95.69× 12.1899 = £1, 166.45

16.6 Calculations Involving a Change of State

Often there is interest in the probabilities of survival for a life who leaves a multiple-decrement tableand continues in another state. These problems may often be treated from first principles, ratherthan by the construction of a multi-state model.

Example 16.6.1. Manual workers between the ages of 50 and 60 are subject to 2 modes of decre-ment, d = death, and p = promotion to foreman.

Foremen are subject to the mortality of another table, T ′, with functions l′x, etc...Find the probabilities that

(a) A manual worker aged 50 will be alive and a foreman at age 52.(b) A manual worker aged 50 will be promoted to foreman but then die before age 52.

Solution(a) The exact formula is ∫ 2

0tp

d50.tp

p50µ

p50+t

(l′52

l′50+t

)dt

(the expression in brackets giving survival as a foreman to age 52), whichapproximates to

(ad)p50

(al)50· l′52l′50 1

2

+(ad)p

51

(al)50l′52l′51 1

2

(b) The probability is approximately

(ad)p50

(al)50

[1− l′52

l′50 1

2

]+

(ad)p51

(ad)50

[1− l′52

l′51 1

2

]

16.7. EXERCISES 275

Exercises

16.1 The following is an extract from a multiple-decrement table referring to mortality and with-drawal from certain life assurance contracts, these modes being referred to as ‘d’ and ‘w’respectively.

age,x (al)x (ad)w

x (ad)dx

50 15,490 24 5151 15,415 21 5852 15,336 15 60

(i) What is the probability that a policyholder aged 51 will withdraw before attaining age 53?

(ii) Suppose that a single-premium three-year term assurance contract is to be issued to a lifeaged 50, subject to the mortality and withdrawal rates shown in the above table. The sumassured is £50,000, payable immediately on death, and the benefit on withdrawal is zero. Onthe basis of rate of interest of 6% p.a., and allowing for expenses of 10% of the single premium,calculate the single premium for the above policy.

(iii) Suppose now that the office issuing the policy of (ii) above wishes to introduce surrendervalues for these three-year contracts. The surrender value is to be equal to 0.5 per cent ofthe single premium for each week between the date of withdrawal and the end of the term.Assuming that surrenders in each policy year take place on average half-way through the year,write down an equation of value from which you could calculate the revised single premium.(Do NOT proceed to the evaluation of this premium.)

16.2 Employees of a certain company are given the opportunity of early retirement immediatelyafter they complete a 3-year overseas assignment.

Those who undertake this assignment effect a 3-year policy providing the following benefitspayable at the end of the year of claim:

(a) on death or ill-health retirement during the term, the sum of £6,000;

(b) on survival as an employee of the company to the end of the term, a lump sum;

(c) on withdrawal from the company during the second year an amount equal to 1 14 times the

annual premium, and on withdrawal from the company during the third year an amount equalto 2 1

2 times the annual premium. No benefit is payable on withdrawal from the company inthe first year.

The following multiple-decrement table is applicable to employees going overseas at exact age47.(d = death, w = withdrawal, i = ill-health retirement)


x (ad)wx

47 100,000 853 13,059 23,00748 63,081 616 11,604 8,46849 42,393 478 9,035 1,87550 31,005

Using an interest rate of 4% p.a., calculate the lump sum for an employee going overseas atexact age 47 who pays level annual premiums of £2,000 in advance. Ignore expenses.

16.3 An organization recruits new employees at exact age 20. The employees serve an apprentice-ship during the first two years of employment. The central rate of withdrawal of apprentices


at ages 20 and 21 is 0.05 per annum, and withdrawals are uniformly spread over the yearof age (in the single-decrement table for withdrawals). At exact age 22, apprentices join thepermanent staff, who retire at exact age 60. The force of withdrawal of permanent staff is0.0094787 per annum. Both apprentices and permanent staff experience mortality accordingto A1967-70 ultimate. The employer pays an immediate benefit of £2,000 on withdrawal and£8,000 on death in service to permanent staff. The death benefit (but not the withdrawalbenefit) is paid to apprentices.

(a) Construct a double-decrement table for apprentices between ages 20 and 22 with a radixof 100,000 at age 20.

(b) Calculate, using an interest rate of 5% per annum, the value of the benefits in respect ofa new employee aged 20.

16.4 A life office sells policies to lives aged 63 which provide a benefit of £50 per week, ceasing atage 65 or earlier death, to those becoming permanently disabled before age 65. The followingbasis is used to calculate the single premium for this contract:

Independent rates ofmortality of policyholders

(not disabled): A1967-70 select (at entry)Force of disablement: 0.01 p.a.Mortality of disabled lives: A1967-70 ultimate, rated up by 7 yearsRate of interest: 4% p.a.Expenses are ignored.

Weekly benefits may be taken as payable continuously.

Evaluate the single premium for this contract.

16.5 In a certain country, married men are subject to the following independent q-type rates ofdecrement:


Widowhood: wives of married men are subject to A1967-70 ultimate

Divorce: 0.02 per annum at all agesWidowed and divorced men are subject to the following independent q-type rates of decrement:

Mortality: English Life Table No. 12 - Males;

Remarriage: rates depend on the age at, and the duration since, the end of the formermarriage; the following table is an extract from these rates:

age at end duration durationof former 0 1Marriage year year

50 12 0.050 0.025

51 12 0.045 0.023

52 12 0.042 0.020

53 12 0.040 0.018

Calculate the probability that a married man aged 50, whose wife is also aged 50, will attainage 52 as a widower who has not remarried between ages 50 and 52.

16.8. SOLUTIONS 277

Solutions

16.1 (i)

2(aq)w51 =

(ad)w51 + (ad)w

52

(al)51= 0.002335

(ii) Let single premium be P . Then P solves

0.9P = 50, 000

[(ad)d

50v12 + (ad)d

51v1 1

2 + (ad)d52v

2 12

(al)50

]at 6% interest

= 50, 000× 0.009977= 498.86

Hence P = £554.29.

(iii) Let P ′ be the premium allowing for surrender values. There are on average 52.18 weeksin 1 year. So, the surrender value in year 1 is on average 52.18× 0.005× 2.5P ′.The average surrender value in year 2 is 52.18× 0.005× 1.5P ′, and so on.The equation of value is

0.9P ′ = value of death benefit (as above)

+52.18× 0.005

(al)50P ′[v

12 (ad)w

50(2.5) + v1 12 (ad)w

51(1.5) + v2 12 (ad)w

52(0.5)]

16.2 Let the lump sum be X.

m.p.v. premiums = 2000[1 +

(al)48(al)47

v +(al)49(al)47

v2

]

= 3996.99.

m.p.v. benefits:(a)

6, 000(al)47

[v((ad)d47 + (ad)i

47) + v2((ad)d48 + (ad)i

48) + v3((ad)d49 + (ad)i

49)]

= 1987.92

(b) X.v3 (al)50(al)47

= 0.27563(X)

(c) 2000(al)47

[1.25(ad)w48.v

2 + 2.5(ad)w49.v

3]= 279.07

So X solves3996.99 = 2266.99 + 0.27563(X)

Hence X = £6276.53

16.3 (a)

qw20 '

mw20

1 + 12mw

20

=0.051.025

= 0.04878 = qw21


Using (aq)dx = qw

x (1− 12qd

x), the double-decrement table is,

x qdx qw

x (aq)dx (aq)w

x (al)x (ad)dx (ad)w

x

20 0.000889 0.04878 0.000868 0.04876 100,000 87 487621 0.000841 0.04878 0.000821 0.04876 95,037 78 463422 90,325

(b) Consider value of benefits for employee attaining age 22. Let k = 0.0094787.Value of death benefit is

8000∫ 38

0

vte−kt.tpd22µ

d22+t dt

= 8000A∗122:38

at force of interest δ′ = δ + k = 0.058269

= 8000A 122:38 0.06

= 8000(1.06)12

[A22:38 0.06 −

l60l22

v380.06

]

= 206.80

Value of withdrawal benefit is

2000∫ 38

0

vte−kt.tpd22.k dt

= 2000.ka22:38 ,0.06

= 2000× 0.0094787[a22:38 − 1

2(1− l60

l22v380.06)

]

= 285.64

Value of benefit in first 2 years, valued at age 20, is8000(al)20

[v12 (ad)d

20 + v1 12 (ad)d

21]

= 12.59

Hence value of benefits for new employee is

12.59 + v2 (al)22(al)20

[206.80 + 285.64]

= £416.03

16.4 Let d = death and i = disablement.Note that qd

63 = q[63] and qd64 = q[63]+1 on A67-70 tables.

Also, qi63 = qi

64 = 1− e−0.01 = 0.00995

x qdx qi

x (aq)dx (aq)i


x

63 0.00839 0.00995 0.00835 0.00991 100,000 835 99164 0.01245 0.00995 0.01239 0.00989 98,174 1216 97165 95,987

m.p.v. benefits ' 50× 52.18(al)63

[v

12 (ad)i

63.a∗63 1

2 :1 12

+ v1 12 (ad)i

64a∗64 1

2 : 12

]

where

16.8. SOLUTIONS 279

a∗63 1

2 :1 12

= a70 1

2 :1 12

on A67-70 ultimate

' 12[a70:2 + a71:1 ]

=12[1.84904 + 0.96010] = 1.4046

a∗64 1

2 :1 12

= a71 1

2 : 12on A67-70 ultimate

' 12(0.96010) = 0.48005

Hence single premium = £47.08.

16.5 Denoteα = mortality of married menβ = mortality of wivesγ = divorce.

Use formula of the form(aq)α

x = qαx [1− 1

2(qβ

x + qγx) +

13qβxqγ

x ]

x qαx qβ

x qγx (aq)β

x

50 0.004789 0.004789 0.02 0.00473051 0.005377 0.005377 0.02 0.005309

A multi-decrement table can be used, but short cuts can be taken.(ap)50 = (1− qα

50)(1− qβ50)(1− qγ

50) = 0.97064.

So, Pr{ married man aged 50 becomes widower between ages 51 and 52}= 1|(aq)β

50 = (ap)50(aq)β51 = 0.005154

Pr{ widower aged 5012

attains age 52 without remarrying}

=l52l50 1

2

(1− 0.050)(1− 12× 0.025) (using E.L.T. 12-Males)

= 0.92674. (i)

Pr{ widower aged 5112

attains age 52 without remarrying}

=l52l51 1

2

(1− 12× 0.045) = 0.97322. (ii)

Hence required probability is(aq)β

50 × (i) + (ap)50(aq)β51 × (ii)

= 0.004730× 0.92674 + 0.005154× 0.97322= 0.009399


Chapter 17

MULTIPLE-STATE MODELS

17.1 Two Points of View

There are essentially two approaches to the theory of multiple-decrement tables:(a) Traditionalist - uses the underlying single-decrement tables;(b) Modernist - uses continuous-time stochastic processes with a finite number of states.Both view-points have advantages and disadvantages, and both produce the same practical re-

sults. In chapters 15 and 16, the traditional approach and notation was used. This chapter shallbriefly take a more modernist approach, and show the differences between the two methods.

A development of the modernist approach for multiple-state models is covered in ActuarialSubject D2.

17.2 Kolmogorov’s Forward Equations

Suppose there are n possible states. For example, n = 3 in the following system, which refers to adouble-decrement table:

Define

pij(x, x + t) =Pr{(x) will be in state j at age x + t

given that he is in state i at age x}

Figure 17.2.1: a double-decrement model

281

282 CHAPTER 17. MULTIPLE-STATE MODELS

Note that the initial conditions are

pij(x, x) =

{1 if i = j

0 if i 6= j.

The forces of transition, µij(y), are defined as

µij(y) = limh→0+

pij(y, y + h)h

(i 6= j)

Also,

µi(y) =∑

i 6=j

µij(y)

= the force of transition from state i

to any other state at age y

NoteIf all these forces are constant, this is a homogeneous chain. If the forces of transition vary withage, it is called inhomogeneous.

Kolmogorov’s Forward Equations (for the general case in which there are n states) give thefollowing system for each fixed x:

d

dtpij(x, x + t) = −µj(x + t)pij(x, x + t)

+∑

ν 6=j

µνj(x + t)piν(x, x + t) (1 ≤ i, j ≤ n, t ≥ 0) (17.2.1)

The three-state model of Figure 17.2.1 is such that

µ1(y) = µ12(y) + µ13(y)µ2(y) = 0µ3(y) = 0 (as no one leaves states 2 or 3)

so Kolmogorov’s Forward Equations give

d

dtp12(x, x + t) = p11(x, x + t)µ12(x + t) (17.2.2)

and

d

dtp13(x, x + t) = p11(x, x + t)µ13(x + t) (17.2.3)

Note also that

p11(x, x + t) = 1− p12(x, x + t)− p13(x, x + t) (17.2.4)

and hence

d

dtp11(x, x + t) = − d

dtp12(x, x + t)− d

dtp13(x, x + t)

17.3. LIFE TABLES AS STOCHASTIC PROCESSES 283

Adding equations (17.2.2) and (17.2.3) gives

d

dt[p12(x, x + t) + p13(x, x + t)] = p11(x, x + t)[µ12(x + t) + µ13(x + t)]

= p11(x, x + t)µ1(x + t)

And hence, using equation (17.2.4),

d

dtp11(x, x + t) = −p11(x, x + t)µ1(x + t) (17.2.5)

(Equation (17.2.5) also follows directly from equation (17.2.1) with i = j = 1.)

This is a straightforward first-order differential equation. Using the initial condition p11(x, x) = 1,(17.2.5) can be solved to give the unique solution

p11(x, x + t) = exp[−∫ t

0

µ1(x + r) dr] (17.2.6)

This corresponds exactly to a result derived in chapter 15, i.e.

t(ap)x = exp[−∫ t

0

(aµ)x+r dr]

= exp[−∫ t

0

(µαx+r + µβ

x+r) dr]

where µαx+r corresponds to µ12(x + r), and µβ

x+r corresponds to µ13(x + r) .

Returning to the differential equation (17.2.2) and solving with the initial condition p12(x, x) = 0gives

p12(x, x + t) =∫ t

0

p11(x, x + r)µ12(x + r) dr

Again this corresponds to a “traditional” formula in chapter 15, i.e.

t(aq)αx =

∫ t

0r(ap)xµα

x+r dr

17.3 Life Tables as Stochastic Processes

Life tables just contain one mode of decrement (death) and hence the continuous-time model isNote that

µ12(y) = the force of mortality at age y

= µy in “traditional” notation

Proceeding as in section 17.2, but with only one mode of decrement, we find that Kolmogorov’sForward Equations may be solved to give

p11(x, x + t) = exp[−

∫ t

0

µ12(x + r) dr

]

In “traditional” notation, this is just the familiar result that

tpx = exp[−

∫ t

0

µx+r dr

]


Figure 17.3.1: the life table model

RemarkIn the stochastic processes approach, it is assumed that the Chapman–Kolmogorov equations hold.That is,

pij(x, z) =n∑

k=1

pik(x, y)pkj(y, z) (17.3.1)

for all x ≤ y ≤ z, and for the n states in the continuous-time model.In Figure 17.3.1 there are only 2 states (alive and dead), and clearly

p21(x1, x2) = 0 for all x1 ≤ x2

andp22(x1, x2) = 1 for all x1 ≤ x2

Equation (17.3.1) with i = 1 and j = 1 is just a re-statement of the following axiom of thetraditional life table:

z−xpx =y−x px.z−ypy

17.4 Sickness Models

The Continuous Mortality Investigation Bureau (C.M.I.B.) has used the following multiple-statemodel for sickness rates:

Define

σy = the force of sickness at age y,

py,z = the force of recovery at age y and duration of sickness z,µy = the force of mortality of healthy lives at age y,

νy,z = the force of mortality of sick lives at age y and duration of sickness z

The various forces of transition have been graduated using Permanent Health Insurance data.Suppose one wishes to calculate probabilities such as

p12(x, x + t) = Pr{ a healthy life aged x will be sick at time t}.

The existence of the parameter z complicates the multiple-state model: Kolmogorov’s ForwardEquations have to be modified, as the forces of recovery and mortality of sick lives depend on

17.4. SICKNESS MODELS 285

Figure 17.4.1: a sickness model

duration of sickness as well as age. This leads to “semi-Markov” processes which are more difficultto handle than the cases so far discussed.

A further complication is caused by the need (in practical applications) to consider “deferredperiods”. For example, one may wish to calculate

Pr{ a healthy life aged x will be sick at time t and has been sick forat least d weeks (d is the deferred period)}

A possible simplificationOne possible way to simplify this model is to initially ignore mortality and to consider only forwardtransitions between “healthy” and “sick” states. Define T1, T2, T3, ... to be the times (in years) spentas

1. a healthy life (who has not been sick since age x),2. a sick life (who has been sick exactly once),3. a healthy life (who has been sick exactly once),and so on.A pictorial representation of this is as follows:In this set-up, one can work out

Pr{ a healthy life aged x will be sick at age x + t}=

∑

j=2,4,6,...

Pr{ a life in state 1 at age x will be in state j at time t}

Each term may be evaluated, using the joint density function f(t1, t2, ..., tj) of the variable(T1, T2, ..., Tj). That is,

Pr{ a life aged x in state 1 will be in state j at time t}=Pr{T1 + ... + Tj−1 < t, but T1 + ... + Tj > t}= a multiple integral (we omit the details)

Advantages of this method:1. The mathematics is not so difficult to follow.2. The deferred period may be easily handled by modifying the range of the multiple integrals.

Disadvantages:1. There can be difficulties in evaluating the multiple integrals for large values of j.


Figure 17.4.2: sick and healthy states

In practice, the Manchester Unity System is often used to calculate sickness functions; this willbe covered in Chapter 18.

Chapter 18

SICKNESS FUNCTIONS

18.1 Rates of Sickness

The force of sickness, zx, at age x is the probability that a life aged exactly x is “sick” (accordingto the rules of the scheme).

In practice, sickness benefit must be claimed for short periods (e.g. days or weeks) during whichtime the life is either “sick” or “not sick”.Consider these periods to be of length h years.Define

z(h)x = Pr{(x) is entitled to sickness benefit for the time period until age x + h}

It is supposed that, uniformly on any bounded age interval,

limh→0+

z(h)x = zx

If benefit is paid at the rate of £1 per week, this amounts to £52.18 per year on average. Supposethat benefit is payable in advance over intervals of length h years, where h = 1

n , i.e. each year isdivided into n intervals. The expected cash to be paid in sickness benefit between ages x and x + 1for a life now aged x is

n−1∑

j=0

jhpx(52.18h)z(h)x+jh

Letting n →∞ (or h → 0+) gives

sx = expected cash paid out in sickness benefit

= 52.18∫ 1

0tpxzx+t dt (18.1.1)

sx is called the annual rate of sickness at age x: it is the expected number of weeks of sicknessbetween ages x and x + 1 for a life now aged x.

We also define

zx = the central rate of sickness at age x

=52.18

∫ 1

0 tpxzx+t dt∫ 1

0 tpx dt(18.1.2)

287

288 CHAPTER 18. SICKNESS FUNCTIONS

Equations (18.1.1) and (18.1.2) give the approximations:

sx ' 52.18. 12pxzx+ 1

2(18.1.3)

zx ' 52.18zx+ 12

(18.1.4)

and hencesx ' 1

2p

xzx (18.1.5)

“Formulae and Tables for Actuarial Examinations” makes use of the Manchester Unity Expe-rience 1893-97, Occupational Groups AHJ. Fuller details of this experience can be found in theAppendix to this chapter.

Sickness by DurationBenefits may depend on the duration of sickness. Define

zm/nx = Pr{(x) is “sick” and has been sick for more than m weeks

but not more than m + n weeks.}

Similar modifications are used to define zm/nx and s

m/nx . For example,

z13/13x = Pr{(x) is “sick” with duration of sickness more

than 13 weeks but less than 26 weeks}.Note that

z13x = z0/13

x = Pr{(x) is “sick” and has been sick for less than 13 weeks}Also,

z104/allx = Pr{(x) is sick with duration of sickness greater than 104 weeks, or 2 years}.

The Tables give the following values:z13x , z

13/13x , z

26/26x , z

52/52x , z

104/allx

Furthermore,zx = zall

x = z13x + z

13/13x + z

26/26x + z

52/52x + z

104/allx

Some More Definitions(a) The Deferred Period

This is the time between falling sick and being able to claim sickness benefits.In Permanent Health Insurance, one encounters the notation D1, D4, etc. which refers to the lengthof the deferred period in weeks. For example, in a D1 policy, a member must be sick for 1 weekbefore being allowed to receive sickness benefit.

(b) The Waiting PeriodThis is the time between joining a friendly society or sickness benefit scheme and being able toclaim sickness benefits. This time-period is often 6 months or a year.

(c) The Off-PeriodThis is the minimum time that must elapse between 2 bouts of sickness in order for them not to beconsidered as the same bout of sickness for benefit calculations.

18.2. VALUING SICKNESS BENEFITS 289

If the sickness benefit falls with duration of sickness, there may be a temptation for people totemporarily “recover” and then to start claiming benefit again at the higher initial rate. To preventthis, an off-period is specified so that if 2 spells of sickness are separated by less than the off-period,the later spell is treated, for benefit purposes, as a continuation of the first spell. In the ManchesterUnity experience, the off-period is 1 year. If benefits do not fall as duration of sickness increasesthere is no need for an off-period rule.

18.2 Valuing Sickness Benefits

Consider a life aged x, subject to a certain mortality table, and suppose that sickness benefit willbe payable at the rate of £1 (per week) during all sickness within the next T years. Consider theage-range from x to x + T to be split into nT short intervals, each of length h years, and supposethat sickness benefit is payable in advance on sickness during any of these short-age intervals. Themean present value of the sickness benefit is thus

nT−1∑

j=0

vjhjhpx(52.18h)z(h)

x+jh

Letting n →∞ gives

m.p.v. of sickness benefit = 52.18∫ T

0

vt.tpxzx+t dt (18.2.1)

This can be approximated by a sum, i.e.

m.p.v. of sickness benefit 'T−1∑t=0

vt+ 12 .t+ 1

2p

x(52.18zx+t+ 1

2)

=T−1∑t=0

vt+ 12 .t+ 1

2pxzx+t (18.2.2)

Notes1. If sickness benefits continue throughout life (although this is unusual as it is not easy to define“sickness” among the very old), we may let T →∞, giving

m.p.v. of sickness benefit = 52.18∫ ∞

0

vt.tpxzx+t dt (18.2.3)

'∞∑

t=0

vt+ 12 .t+ 1

2pxzx+t (18.2.4)

2. If the sickness benefit is payable for sickness of duration greater than m weeks but less than m+n

weeks, replace zx and zx by zm/nx and z

m/nx , respectively.

The M.P.V. of a sickness benefit may be evaluated by direct evaluation of a sum, by approximateintegration or by commutation functions.

Commutation functionsDefine

Hx = 52.18∫ 1

0

vx+tlx+tzx+t dt (18.2.5)


This can be approximated to give

Hx ' 52.18vx+ 12 lx+ 1

2zx+ 1

2(18.2.6)

and so

Hx ' Dx+ 12zx (18.2.7)

where Dx = lxvx, as usual.We also define

Kx =∞∑

t=0

Hx+t (18.2.8)

so that the m.p.v. of a sickness benefit of £1 per week for life to (x) is approximately

∞∑t=0

Dx+t+ 12

Dxzx+t (from (18.2.4))

=∑∞

t=0 Hx+t

Dx

=Kx

Dx(18.2.9)

NoteKx = Kall

x = K13x + K

13/13x + K

26/26x + K

52/52x + K

104/allx , where H13

x and K13x , etc., are defined by

replacing zx by z13x , etc., in the formulae for Hx and Kx.

Commutation functions are available in “Formulae and Tables” on the following basis only:

• English Life Tables, No.12 - Males

• 4% interest

• Manchester Unity 1893-97 (A, H, J)

Temporary and Deferred benefitsIf sickness benefit ceases at a certain age, say 65, then the m.p.v. of sickness benefit of £1 per weekto (x) is

Kx −K65

Dx(18.2.10)

If benefits are deferred for n years, we obtain

Kx+n

Dx

rather than

Kx

Dx

This is used in connection with waiting periods. If there is a waiting period of 6 months for sicknessbenefits, replace Kx by Kx+ 1

2(interpolate in the Tables.)

18.2. VALUING SICKNESS BENEFITS 291

Example 18.2.1. A life office is proposing to issue 3-year sickness benefit policies to lives aged 30.The benefits are £50 per week during sickness within the next three years. There is no waitingperiod and the off-period is as in the Tables provided. Find the single premium on each of thefollowing bases:

mortality: English Life Table No.12 - Malesinterest: (i) 4% p.a.,

(ii) 5% p.a.sickness: Manchester Unity 1893-97 (AHJ)expenses: none.

Solution(i)

m.p.v. of benefits = 50 · K30 −K33

D30

= 50[1784760− 1706624

29372

]

= £133.01

(ii) Calculate value of benefits from first principles. The value is

50l30

[v

120.05l30 1

2z30 + v

1 12

0.05l33 12z31 + v

2 12

0.05l32 12z32

]

The value of l30 12, l31 1

2, l32 1

2can be found by interpolation using English Life Tables, No.12 -

Males, to give

m.p.v. of benefits =50

95265× 249682

= £131.05

If sickness benefit is payable for sickness lasting more than m weeks but less than m + n weeks,replace Kx and Hx by K

m/nx and H

m/nx respectively.

For example,

K13/allx −K

13/all65

Dx= m.p.v. of a sickness benefit of £1 per week

payable up to age 65 on sickness lasting more than 13 weeks

Since K13/allx is not given directly in the Tables, one must use

K13/allx = K

13/13x + K

26/26x + K

52/52x + K

104/allx

Example 18.2.2. A friendly society issued a policy providing the following benefits to a man agedexactly 25 at entry:

(a) on death at any time before age 60, the sum of £4,000 payable immediately;(b) on survival to age 60, an annuity of £8 per week payable weekly in advance for as long as he

survives;(c) on sickness, an income benefit to be payable during sickness of £32 per week for the first 6

months reducing to £16 per week for the next 18 months and to £8 per week thereafter. Sickness


benefit is not payable after age 60. There is no waiting period.Premiums are payable monthly in advance for at most 35 years, and are not waived during periodsof sickness.

The society uses the following basis to calculate premiums. Find the monthly premium.

mortality: English Life Table No.12 - Malessickness: Manchester Unity Sickness Experience 1893-97, Occupation Group AHJinterest: 4% p.a.expenses: none

SolutionValue of benefits:

(a) 4000A 125:35

= 4000(

M25−M60D25

)= 268.05

(b) 52.18× 8 · N60D25

= 926.65

(c) 32(

K2625−K26

60D25

)+ 16

(K

26/7825 −K

26/7860

D25

)+ 8

(K

104/all25 −K

104/all60

D25

)

= 660.92Let P = annual premium payable monthly. Then the equation of value is

P a(12)

25:35= 268.05 + 926.65 + 660.92

= 1855.62

a(12)

25:35= a

(12)25 − D60

D25a(12)60

= (a25 +124

)− D60

D25

(a60 +

124

)= 18.495.

Hence P = £100.33 and monthly premium = £8.36.

18.3 Various Other Points

(a) A possible adjustmentConsider a policy providing a sickness benefit of £10 per week on sickness lasting more than 2 yearsfor an entrant aged x, with all benefits ceasing at age 65. One might want to adjust the formula

10

[K

104/allx −K

104/all65

Dx

](18.3.1)

to

10

[K

104/allx+2 −K

104/all65

Dx

](18.3.2)

since the benefit cannot be received in the first 2 years.

This point is discussed in the Appendix to this Chapter, and it is concluded that neither formula isexactly right, so the use of adjusted version (18.3.2) is usually optional.

The only exception (in which (18.3.2) is accurate and (18.3.1) is not) occurs when the term ofthe policy is very short. For example when x = 63, formula (18.3.2) gives zero, which is correct.

18.3. VARIOUS OTHER POINTS 293

(b) Waiver of Premium BenefitsSuppose that a policy has a weekly premium of £P but this is waived during sickness (or duringsickness of a certain duration).

This is handled by assuming that all premiums are paid by the policyholder, but there is anadditional sickness benefit of £P per week.

Also, as a general rule, expenses will apply even when premiums are waived. If this is not thecase, one should adjust the equation of value appropriately.

Example 18.3.1. A friendly society issues sickness insurance policies which provide income duringperiods of sickness as follows:

(a) £100 per week while a sickness has duration in excess of 13 weeks but less than 1 year;(b) £75 per week while a sickness has duration in excess of 1 year but less than 2 years;(c) £50 per week while a sickness has duration in excess of 2 years.

All benefits cease at age 65. Premiums are payable weekly until age 65 and are waived whensickness benefit is being paid. There is no waiting period.

Calculate the weekly premium for a life aged 38 at entry. Basis:mortality: E.L.T. No.12 (Males)sickness: Manchester Unity Sickness Experience 1893-97, Occupation Group

AHJ.interest: 4% p.a.expenses: 50% of all premiums payable in the first year, plus 10% of all

premiums payable after the first year.

SolutionLet the weekly premium be P .

m.p.v. of (premiums-expenses) is0.9× 52.18P a38:27 − 0.4× 52.18Pa38:1

= 713.98P (i)

m.p.v. of benefits and premiums waived is

100[K

13/3938 −K

13/3965

D38] + 75[

K52/5238 −K

52/5265

D38] + 50[

K104/all38 −K

104/all65

D38]

+ P [K

13/all38 −K

13/all65

D38]

= 784.80 + 267.63 + 543.00 + P × 22.276= 1595.43 + 22.276P (ii)

Setting (i) = (ii) givesP (713.98− 22.276) = 1595.43Hence P = £2.31

(c) ReservesThese are calculated prospectively or retrospectively (usually the former) in a way similar to thatused for life policies.

If the premium and reserve bases agree, the prospective and retrospective reserves are equal. Ifthe bases differ, one must specify how to find the reserve.


Example 18.3.2. Consider a policy issued to (x) providing a sickness benefit of £10 per weekceasing at age 65.

(a) Give a formula for the weekly premium, P , ceasing at age 65 (ignoring expenses).(b) Calculate the reserve at duration t years by

(i) the prospective method, and(ii) the retrospective method,

on the premium basis.

Solution(a) 52.18P ax:65−x = 10

[Kx−K65

Dx

]

(b) (i) tV = 10[

Kx+t−K65Dx+t

]− 52.18P ax+t:65−x−t

(ii) tV = Dx

Dx+t

[52.18P ax:t − 10

(Kx−Kx+t

Dx

)]

Note that (i) and (ii) are equal, by an argument similar to that used for life policies.

(d) Understanding the Tables for Actuarial Exams1. Pages 82-83 give

z13x , z13/13

x , z26/26x , z52/52

x , z104/allx , zx(= zall

x )

2. Pages 84-85 give

K13x

Dx,K

13/13x

Dx,K

26/26x

Dx,K

52/52x

Dx,K

104/allx

Dx,Kx

Dx

3. Pages 86-87 give

Dx,K13x , K13/13

x ,K26/26x ,K52/52

x ,K104/allx

18.4. EXERCISES 295

Exercises

18.1 (i) In a combined sickness and mortality table

Kx+1 = 554, 405zx = 1.129

Dx = 24, 510Dx+1 = 23, 425

Estimate Kx.

(ii) An office offers an optional waiver of premium benefit on sickness of any duration in respectof a 25-year with or without profits endowment assurance policy with weekly premiums payablefor 25 years or until earlier death. There is a waiting period of 12 months for the waiver ofpremium benefit, and, during the second year of the policy, only half the premium (includingthe extra premium for the waiver benefit) is waived. The sum assured under the endowmentpolicy is payable immediately on death, or on survival until the end of the term.

Using the basis given below, calculate the percentage by which the normal weekly premium(i.e. the premium for a policy without the waiver benefit) for a life aged exactly 30 at entryshould be increased in order to provide the waiver benefit.

mortality: English Life Table No.12-Males;sickness: Manchester Unity 1893-97, Occupation Group AHJ;interest: 4% per annum;expenses: 15% of each extra premium for the waiver benefit.

18.2 A policy issued by a life office to a male life aged exactly 35 is subject to level weekly premiumsceasing at exact age 65. If the man has been sick for 6 months or more when a premium fallsdue, the premium is waived. The policy provides the following benefits:

(a) on survival to exact age 65, an annuity of £5,000 per annum payable monthly in advance,

(b) on death before age 65, a return of all premiums paid (including those waived duringsickness) together with compound interest at 4% per annum to the date of death.

There is no waiting period and the off periods are the same as those underlying the tables inFormulae and Tables for Actuarial Examinations. Calculate the weekly premium.

Basis: English Life Table No.12-Males, Manchester Unity Sickness Experience 1893-97, Oc-cupation Group AHJ, interest 4% per annum, no expenses.

18.3 (a) Ten years ago, a man then aged exactly 30 effected an insurance policy providing sicknessbenefits of £100 per week for the first six months of sickness, £50 per week for the remainderof the first year and £30 per week thereafter, with benefit ceasing at age 60. Calculate theweekly premium payable to age 50 on the following basis:

Mortality: English Life Table No. 12 - Males;

Sickness: Manchester Unity 1893-97, Occupation Group AHJ;

Interest: 4% per annum;

Expenses: 10% of each premium.

(b) The man now wishes to alter his policy so that premiums will in future be waived duringall periods of sickness. Calculate the revised premium payable assuming that the alterationbasis follows the premium basis above.Note. Expenses are incurred even when premiums are waived.


18.4 A certain friendly society recruits only married men aged under 55. The society provides thefollowing benefits:

(i) immediately on the death of the member at any age, £2,000,

(ii) during the first 10 years of membership, immediately on the death of the member’s wifebefore her husband, £500,

(iii) on survival of the member to age 65, an annuity of £10 per week for life, and

(iv) during any spell of sickness of the member before age 65, £10 per week reducing to £6per week after 3 months’ sickness.The basis for all calculations is:

mortality of members: E.L.T. No.12 - Males

mortality of wives: E.L.T. No. 12 - Males with an age-deduction of 5 years


sickness: Manchester Unity 1893-97 AHJ

expenses: 5% of all contributions, including those waived during sickness

Wives of members are taken as being of the same age as their husbands; there is no benefiton the death of wives of marriages taking place after entry to the society, and the possibilityof divorce is to be ignored. There is no waiting period and the off-period is as in the Tablesprovided. The possibility of withdrawal is ignored.

(a) Calculate the weekly contribution rate, waived during sickness and ceasing at age 65 orthe previous death of the member, for an entrant aged 25.

(b) Calculate the reserve for a member aged 35 who joined at age 25.Note Use Simpson’s rule for approximate integration, i.e.∫ b

a

f(t) dt ' b− a

6[f(a) + 4f(

a + b

2) + f(b)]

18.5 A Friendly Society issues policies providing the following benefits:

(i) A sickness benefit of £25 per week for the first 13 weeks of sickness and £12.50 per weekthereafter, benefit ceasing at age 60. Contributions are waived during sickness.

(ii) On death before age 60, a lump sum of £1,000 plus a return of contributions (includingany waived) without interest.

(iii) On survival to age 60, a lump sum of £2,000.

Contributions are payable by level weekly amounts until age 60. There is a six-month waitingperiod for the sickness benefit (including the premium waiver) and the off-period may beassumed to be the same as that underlying the tables in “Formulae and Tables for ActuarialExaminations”.

(a) Calculate the weekly contribution payable by a new member aged 35 on the basis givenbelow.

(b) Calculate the reserve (on the basis given below) to be held for this member five years afterhe joins the Society.

Basis: English Life Table No. 12 - Males

Manchester Unity Sickness Experience 1893-97, Occupation Group AHJ

interest 4% per annum.

expenses are 5% of all premiums (including those waived).

18.4. EXERCISES 297

18.6 A friendly society provides the following benefits:

(i) On sickness, £40 per week for the first 26 weeks and £50 per week for the next 26 weeks.No sickness benefit is payable after age 65.

(ii) On attaining age 65 or immediately on earlier death, the sum of £3,000.

Members contribute £1 per week, ceasing at age 65 or earlier death, but this is waived duringperiods of sickness (whether benefit is payable or not.) Calculate the reserve which should beheld for a member aged 50, using the following basis:

English Life Table No.12 - Males,4% per annum interest,Manchester Unity Sickness Experience 1893-97 (AHJ).There is no waiting period, and off-periods are as in “Formulae and Tables for ActuarialExaminations”.

18.7 On 1 January 1995 a friendly society had in force a large group of policies providing certainsickness and life assurance benefits. These policies had all been issued 10 years ago to livesnow aged 45. The benefits cease when the policyholders reach age 65, and the policies havelevel weekly premiums payable throughout the term of the policy. Each policy provides anincome of £20 per week during any period of sickness which has lasted more than 6 months,and a lump sum of £20,000 immediately on death before age 65. Premiums are waived duringsickness (even when no benefit is payable), and the off-period is as in the Manchester Unitysickness tables. The basis for the calculation of premiums and reserves is:

Mortality: E.L.T. 12 - Males

Interest: 4% p.a.

Sickness: Manchester Unity Experience 1893-97 (Occupation Group AHJ)

Expenses: 20% of all premiums, including those waived

(a) Calculate the weekly premium for each of these policies.

(b) Calculate the reserve per policy in force on 1 January 1995.


Solutions

18.1 (i)Kx = Hx + Kx+1

' Dx+ 12.zx + Kx+1

' 12(Dx + Dx+1)zx + Kx+1

= 581, 464

(ii) Let weekly premium for the basic policy be P , and let k.P be the extra premium for thewaiver benefit.The equation of value is

52.18× 0.85kP a30:25 = P (1 + k)[ 1

2 (K31 −K32) + (K32 −K55)D30

]

(note the waiting period of 1 year).Hence 690.07k = (1 + k)20.81Therefore k = 0.0311 = 3.11%.

NoteOne does not need to calculate the actual premium P in this example.

18.2 Let weekly premium be P .The equation of value is

52.18P.a35:30 = 5000D65

D35a(12)65 +

∫ 30

0

vt.tp35µ35+t(52.18P st|) dt

+ P[K

26/all35 −K

26/all65

]/Dx (i)

Note that∫ 30

0

vt.tp35µ35+t(52.18P st|) dt

= 52.18P

∫ 30

0

(1− vt

δ

)tp35µ35+t dt

=52.18P

δ

[30q35 − A 1

35:30

]

Hence (i) gives869.13P = 9, 994.55 + 203.26P + 16.8815PHence P = £15.40.

18.3 (a) Let weekly premium be P . The equation of value is

52.18× 0.9P a30:20 = 100(

K2630 −K26

60

D30

)+ 50

(K

26/2630 −K

26/2660

)

+ 30

(K

52/all30 −K

52/all60

D30

)

Hence 640.51P = 1711.35 + 125.21 + 242.74and so

P = £3.25 per week.

18.5. SOLUTIONS 299

(b) Let extra premium be E per week. The equation of value for E is

(E + 3.25)(

K40 −K50

D40

)= 52.18× 0.9Ea40:10

Therefore (E + 3.25)× 13.077 = 382.67Eand so

E = 0.115, say £0.12.

Hence revised premium is £3.37 per week.

18.4 (a) Let weekly contribution be P .

m.p.v. benefits:

(i) 2000A25 = 377.64

(ii)

500∫ 10

0

vt.tp20µ20+t.tp25 dt ' 500× 0.008638

= 4.319 (using Simpson’s Rule)

(iii) 10× 52.18 · D65D25

a65 = 693.28

(iv) (6 + P )(

K25−K65D25

)+ 4

(K13

25−K1365

D25

)= 254.23 + 31.853P

Hence the equation of value is52.18× 0.95P a25:40 = 377.64 + 4.319 + 693.28 + 254.23 + 31.853P

Hence 927.54P = 1329.46and so, P = £1.43.

(b)

10V = 2000A35 + 52.18× 10N65

D35+ (6 + 1.43)

(K35 −K65

D35

)

+ 4(

K1335 −K13

65

D35

)− 0.95× 1.43× 52.18a35:30

= 537.38 + 1038.18 + 270.88 + 62.55− 1180.71= £728.28

The reserve may also be calculated by the retrospective method.

18.5 (a) Let the weekly contribution be P . The equation of value is0.95× 52.18P a35:25 = 1, 000A 1

35:25+ 52.18P (IA) 1

35:25

+ 2000D60

D35+ (12.5 + P )

(K35 1

2−K60

D

)

+ 12.5

(K13

35 12−K13

60

D35

)(i)

Use,

(IA) 135:25

' (IA) 135:25

− 12A 1

35:25

and find K35 12

and K1335 1

2by linear interpolation.


Equation (i) gives759.52P = 86.28 + 71.38P + 625.57 + (12.5 + P )28.066 + 165.92

and soP = £1.86 per week

(b)5V = (1000 + 5× 52.18P )A1

40:20+ (52.18× 1.86)(IA)1

40:20

+ 2000D60

D40+ (12.5 + P )

(K40 −K60

D40

)+ 12.5

(K13

40 −K1360

D40

)

− 0.95× 52.18P a40:20 (Prospectively)

Using P = 1.86, this gives5V = 142.35 + 113.99 + 768.11 + 413.06 + 59.53− 1222.68

= £374.36

18.6 Using the prospective method,

Reserve = 40(

K2650 −K26

65

D50

)+ 50

(K

26/2650 −K

26/2665

D50

)+

(K50 −K65

D50

)

+ 3000[A1

50:15+

D65

D50

]− 52.18a50:15

= 681.85 + 209.36 + 39.08 + 1775.84− 542.88= £2163.25

18.7 (a) Let the weekly premium be P . The equation of value is

0.8× 52.18P a35:30 = 20000A135:30

+ 20

(K

26/all35 −K

26/all65

D35

)

+ P

(K35 −K65

D35

)

This gives 695.30P = 2472.44 + 337.63 + 36.46P

Hence P = £4.27 per week.

(b) Using the prospective method

10V = 20000A145:20

+ 20

(K

26/all45 −K

26/all65

D45

)+ P

(K45 −K65

D45

)

− 0.8× 52.18P a45:20

With P = 4.27, this gives10V = 3176.82 + 416.27 + 168.42− 2285.91

= £1475.60

18.5. SOLUTIONS 301

Appendix: The Manchester Unity Experience 1893-97

1. Until fairly recently, when the C.M.I. published statistics concerning P.H.I. (Permanent HealthInsurance), the principal published set of sickness rates were those derived from the experiencein 1893-97 of the Independent Order of Oddfellows, Manchester Unity (a large “affiliatedorder” friendly society, i.e. an association of lodges with some measure of central control.)The members were all male.

2. Friendly societies existed largely to protect wage-earners in times of adversity, there beingno State benefits before 1908 (unless one counts the Workmen’s Compensation Acts of 1897,which covered injuries at work.) There was (and to some extent still is) a wide variety ofsocieties ranging in size and financial stability, and not all societies provided sickness benefits.From 1911 to 1946, sickness benefits for wage-earners were provided by Approved Societies(friendly societies or life offices approved under the National Insurance Act, 1911). A partof their income was provided by a Government grant, and they had to use a valuation basisspecified by the Government Actuary. Societies with favourable experience were able to pro-vide additional (in practice, usually dental and ophthalmic) benefits. Since 1946 the State hasprovided welfare benefits (including sickness benefits) directly, either in return for nationalinsurance contributions or under a means test. The friendly societies have generally declinedin importance; many have disappeared, and others concentrate on social activities. A certainnumber (particularly of centralized societies) continue, and indeed prosper, in modern con-ditions: under the Friendly Societies Act 1992, they are able to conduct a greater variety ofbusiness.

3. The Manchester Unity 1893-97 investigation was carried out by Alfred W. Watson, and pub-lished in 1903. A calendar year system was used, lives being classified according to age nearestbirthday at the start of the calendar year. The unadjusted sickness rates are of the form

zx =no. of weeks of sickness at age x last birthday

Ecx

These rates were graduated by an adjusted-average formula to produce the published rates, zx.(The statistical basis of the graduation of sickness rates is more complex than for mortalityrates, and we do not attempt a discussion.) The rates were also subdivided by periods ofsickness, giving z13

x , z13/13x , z

26/26x , z

52/52x , z

104/allx . The off-period assumed in the investigation

was 1 year, this being the actual off-period for most of the lodges.

4. Members were also subdivided according to the following occupational groups:

A: Agricultural Workers

B: Outdoor Tradesmen and Labourers

C: Railwaymen

D: Seamen and Fishermen

E: Quarry Workers

F: Iron and Steel Workers

G: Miners

H: Rural workers not included in A-G

J: Urban workers not included in A-G

Sickness rates were calculated for the four groups AHJ, BCD, EF, G, it being found thatsickness rates were lowest in the first group and increased as one moved from one group to the


next. The experience of occupation group AHJ is given in Tables for Actuarial Examinations.Since the middle classes and well-to-do did not (in general) join friendly societies providingsickness benefits, this represents the experience of wage-earners in 1893-97, excluding those inthe more hazardous occupations. (In some cases, however, sickness rates were high becausethe jobs demanded a high level of physical fitness.)

5. Mortality investigations were also conducted, with separate tables for 3 geographical areasand for urban and rural areas, but these have long been out of date. It is usual to combinethe M.U. sickness rates with a more modern mortality table (e.g. English Life Table No. 12-Males, as used in Tables for Actuarial Examinations.)

6. The sickness rates of M.U. (whole society) were found to be remarkably similar to those foremployed men in the national insurance scheme in 1953-58.

7. Strictly speaking, sickness rates should be “select”, i.e. they should depend on duration ofmembership as well as on attained age, since new entrants are not normally accepted unlessthey are reasonably healthy. Another point is illustrated by the example of a new entrant agedx who is to receive £20 per week for the first 26 weeks of sickness, £10 for the second 26 weeksand £5 per week for the remainder of sickness, subject to a waiting period of 6 months. Thevalue of the benefits might be adjusted from

20K26x+ 1

2+ 10K

26/26

x+ 12

+ 5K52/all

x+ 12

Dx(i)

to20K26

x+ 12

+ 10K26/26x+1 + 5K

52/allx+1.5

Dx(ii)

to allow for the fact that no “26/26” benefit can be received before age x + 1 and no “52/all”benefit can be received before age x + 1.5. But formula (ii) is not quite right; consider, forexample, the “26/26” benefits, which are now valued as

10[Dx+1.5z26/26x+1 + Dx+2.5z

26/26x+2 + ...]

Dx

Now z26/26x+1 is based on an experience (M.U.) in which the exposed to risk included some recent

entrants who could not claim “26/26” benefit because they had not been eligible for sicknessbenefit for more than 6 months. That is, z

26/26x+1 should be slightly adjusted upwards relative

to that of M.U. Similarly, z26/26x+2 should be slightly increased, so formula (ii) understates the

expected value of the benefits. It is therefore better to use the formula

10[0.5Dx+0.5z26/26x + Dx+1.5z

26/26x+1 + ...]

Dx

since the “extra” first term allows approximately for the fact that the later terms are too low.This argument leads us to use sickness rates without deferment (except for the waiting period),as in formula (i), except perhaps when benefits cease soon after the entry age.In practice, the uncertain definition of “sickness” and other factors are likely to be of muchgreater importance in the estimation of sickness rates. But when comparing the actual weeksof sickness claim at the later durations in a given experience with that expected on the basis ofManchester Unity (say), one must bear in mind any differences in the proportions of memberswhose durations of membership are so short that they cannot claim. Thus, for example, anexperience consisting mainly of recent entrants may be expected to have low rates of sicknessat the later durations.

Chapter 19

PENSION FUNDS

19.1 General Introduction

Pension schemes may be described (in broad terms) as either:(a) defined - benefit schemes,

or(b) defined - contribution schemes.

Defined - benefit schemesThese are pension schemes whereby the pension and other benefits are set out in the rules of the

scheme. Most schemes of this type provide benefits which depend on the ‘final salary’. For example,in a typical U.K. public sector final salary scheme, the annual pension is 1

80× the final salary peryear of service. There may be various other benefits such as death benefits and a lump sum onretirement.

The benefits are paid for by a combination of(i) the employee’s contributions (e.g. a fixed percentage of salary)

and(ii) the employer’s contributions, which may vary from time to time according to actuarial advice.

Remarks1. A “non-contributory” pension scheme is one in which the employees do not contribute (e.g. U.K.civil service).2. An “insured” pension scheme is one in which benefits are secured by contracts with a life office.

Defined - contribution schemesThese are also called “money purchase” schemes; the contributions of both employees and em-

ployer are fixed (often as percentages of the salary). The pension benefit is what these combinedcontributions can buy, usually after investment (either directly by the scheme or by life office con-tracts) until retirement age. Obviously, the annual pension depends on the rate of return obtainedon the employer’s and member’s contributions and the terms on which pension may be purchasedat retirement (unless pension is purchased directly when the contributions are received).

19.2 Valuation Principles

Only defined-benefit schemes shall be considered, and the mean present value of the future benefitsand future contributions of an “active” member aged x will be calculated. (“Active” refers to a

303

304 CHAPTER 19. PENSION FUNDS

member who has not yet retired).The reserve for each member is calculated prospectively. That is,

reserve = mean present value of future benefits - mean presentvalue of future contributions (of both employee and employer)

Notes1. The rate of interest, i per annum, used in valuing the benefits and contributions is normally gross(free of tax). i is called the valuation interest rate.2. Expenses are usually ignored, as they are paid for separately by the employer.3. A service table is required. This is a multiple-decrement table with various modes of decrement:death, ill-health retirement, etc.4. If benefits or contributions depend on the employee’s salary, a salary scale is required to estimatefuture salaries from current salaries.5. Sometimes one may require to find the employer’s contribution rate by setting the reserve equalto zero at entry to the scheme of a new entrant, or for a group of new entrants.

19.3 Service Tables

The following notation is used, as in “Formulae and Tables in Actuarial Examinations”:lx = number of members at exact age xdx = number of deaths at age x last birthdayix = number of “ill-health” retirements at age x last birthdayrx = number of “age” retirements at age x last birthdaywx = number of withdrawals at age x last birthday

Notes1. “Age” retirements are the retirements at or above the minimum normal retirement age (NRA) ofthe scheme. “Age” retirements may be concentrated at an exact age (for example 60 or 65) or maybe spread uniformly between ages 60 and 65, or both.2. Members often have to retire by a certain age (often 65). In the “Formulae and Tables”, “age”retirements occur only from age 60 onwards, and all members must retire at age 65 at the latest.3. To calculate the value of benefits one may require separate mortality tables for “age” retirementpensioners and ill-health retirement pensioners. We may also have to deal with benefits for peoplewith “deferred” pensions in the scheme who no longer work for the company, so a mortality tablefor them is sometimes required (see later.)

19.4 Salary Scales

Assume that salaries are revised continuously.Define {sx}, the salary scale function, to be such that the salary rate per annum at age x + t of alife now aged x with current salary rate £(SAL) per annum is

(SAL)×(

sx+t

sx

)(19.4.1)

Also define

sx =∫ x+1

x

sy dy =∫ 1

0

sx+t dt (19.4.2)

19.4. SALARY SCALES 305

Using approximate integration in (19.4.2), we have

sx ' sx+ 12' 1

2[sx + sx+1] (19.4.3)

“Formulae and Tables” gives values of sx (not sx), so we find sx by linear interpolation:

sx ' 12[sx−1 + sx] ' sx− 1

2(19.4.4)

NoteTo estimate s65, one must use s64 1

2' s64 and s63 1

2' s63 and then employ linear extrapolation: see

Example 19.4.1 below.The salary to be earned by (x) between ages x + t and x + t + 1, given that he is an active

member during this age-interval, is estimated as:

(SAL)sx+t

sx(19.4.5)

where SAL is the current salary rate (in £) earned by (x).

We also find that:

(SAL)sx+t

sx= assumed salary rate per annum

at exact age x + t (19.4.6)

Adjustments1. If SAL refers to the earnings received in the past year (i.e. between ages x− 1 and x), adjust thedenominator from sx to sx−1.2. If SAL refers to the expected earnings in the coming year (i.e. between ages x and x + 1), adjustthe denominator from sx to sx.

Example 19.4.1. (a) Consider a person now aged exactly 25 whose annual salary rate is currently£9,192. Estimate(i) his annual salary rate at exact age 53,(ii) his earnings between exact ages 64 and 65,(iii) the average amount earned by him each year between exact ages 60 and 65,(iv) his annual salary rate at exact age 65.If he dies at age 57 last birthday, determine the average values of(v) his annual salary rate at the moment of death,(vi) the total amount earned over his last year of life.

(b) Calculate revised answers for (i) to (vi) assuming that the person aged exactly 25 is expectedto earn £9,192 in the coming year.

Assume that salaries are revised continuously, and use the pension table in “Formulae and Tablesfor Actuarial Exams” with 4% p.a. interest.

Solution(a) (i) 9192

s25· s53 ' 9192

s24 12

s52 12

= 9192× 4.771.80 = £24, 359

(ii) 9192s25

s64 ' 9192s24 1

2

s64 = £27, 576

(iii) 9192s25

[s60+s61+s62+s63+s64

5

] ' £27, 198.


(iv) Annual salary rate at 64 is 9192 s64s25

' 27, 499

Annual salary rate at 64 12 is 9192

s64 12

s25' 27, 576

Hence by linear extrapolation the annual salary rate at age 65 is approximately27, 576 + (27, 576− 27, 499) = £27, 653(v) On average he will die at exact age 57 1

2 . Hence answer is

9192s25

· s57 12' 9192

s25· s57 = £25, 891

(vi) 9192s25

s56 12' £25, 738.

(b) Adjusting the denominator from s25 to s25 results in multiplying each answer by a factor of

s25

s25'

s24 12

s25' 0.962567.

This gives answers of:(i) £23, 447 (ii) £26, 544 (iii) £26, 180(iv) £26, 618 (v) £24, 922 (vi) £24, 775

Estimation of sx and sx

sx is usually estimated by a product of 2 factors, one to allow for future inflationary increases andone to take account of “career progression”. The inflation factor is usually of the form (1 + e)x,where e is the assumed annual rate of future salary escalation.

It is often assumed that

i− e ' the “real” rate of interest (assumingwage inflation and priceinflation are about the same)

'2% (or some similar figure)

where i is the valuation interest rate and future salary inflation is assumed to be e per annum.

19.5 The Value of Future Contributions

There are 2 main ways of calculating contributions:(a) contributions are at the rate of k% of salary;(b) contributions are at a fixed rate of £F per annum (payable continuously).

(a) The mean present value of the future contributions (of employee, employer or both) at rate k%of salary for a member age x with current salary rate of £SAL per annum is

k

100· SAL

sx

∫ 65−x

0

vt lx+t

lxsx+t dt

This is usually approximated to

k

100· SAL

sx

64−x∑t=0

vt+ 12lx+t+ 1

2

lxsx+t (using sx+ 1

2' sx) (19.5.1)

(Note the summation is up to “64− x”).Now define

Dx = vxlx, (as previously defined)

19.5. THE VALUE OF FUTURE CONTRIBUTIONS 307

Dx =∫ 1

0Dx+t dt ' Dx+ 1

2' 1

2 [Dx + Dx+1]sDx = sxDx

andsNx =

∑64−xt=0

sDx+t

One can now evaluate (19.5.1) by means of commutation functions, giving a mean present valueof

k

100· SAL

sx

64−x∑t=0

vx+t+ 12 .lx+t+ 1

2

vxlxsx+t

=k

100SAL

sx

64−x∑t=0

Dx+t+ 12

Dxsx+t

' k

100SAL

sxDx

64−x∑t=0

sDx+t

=k

100SAL

sxDx· sNx.

So, the m.p.v. of future contributions is

k

100SAL

sxDx· sNx (19.5.2)

Notes

1. If SAL refers to the past year, change sx to sx−1 in the denominator.

2. If SAL refers to the coming year, change sx to sx in the denominator.

Example 19.5.1. Consider a life aged 35 with current salary rate £10,000 who contributes 5% ofhis salary to a pension scheme. Using the Examination Tables, calculate the mean present value ofthe employee’s future contributions.

Solutionm.p.v. of contributions is

5100

· 10000s35D35

·s N35 ' 500× 417, 22412 (2.98 + 3.08)× 7232

= £9520.

(b) Suppose contributions are independent of salary and that these are a fixed annual sum of £F,


payable continuously. Then the mean present value of future contributions is

F

∫ 65−x

0

vt lx+t

lxdt

' F

64−x∑t=0

vt+ 12lx+t+ 1

2

lx

= F

64−x∑t=0

Dx+t+ 12

Dx

' F · 1Dx

64−x∑t=0

Dx+t

= F

(Nx

Dx

)(19.5.3)

where Nx =∑64−x

t=0 Dx+t.

Example 19.5.2. If a life aged 35 contributes £500 each year to his pension scheme, calculate thevalue of his future contributions.

Solution

m.p.v. = 500 · N35

D35= £7, 096.

Suppose now that contributions are k% of the “pensionable salary”, where pensionable salaryis defined as “salary - A”, A being a fixed amount. Assuming that the current salary rate, SAL,already exceeds A, the m.p.v. of these contributions is

k

100· SAL

sx

(sNx

Dx

)− k

100·A

(Nx

Dx

)(19.5.4)

Example 19.5.3. Employees contribute to a pension scheme at a rate of 6% of “pensionable salary”,where “pensionable salary” equals actual annual salary rate less £2000 (to allow for other pensionarrangements). Find the present value of the future contributions payable by a member aged exactly30 whose current annual salary rate is £15,500.

Solution

m.p.v. ' 15500× 0.06s29 1

2·D30

·s N30 − 2000× 0.06N30

D30

= 18, 377.7− 1, 606.5 = £16, 771

19.6 The Value of Pension Benefits

Pension benefits are usually of one of the following forms:(a) A fixed pension of £P per year of service;(b) A fraction of the average salary per year of service;

19.7. FIXED PENSION SCHEMES 309

(c) A fraction of the final salary per year of service.

Notes1. Years of service normally include fractions counting pro rata.2. The benefits on the date of age retirement or ill-health retirement are valued by multiplying

the annual pension by an appropriate annuity function. For example, the benefit on age retirementat age 65 is

annual pension × ar65

where r indicates age retirement mortality rates.Similarly, ai

x refers to a life retiring at age x due to ill-health. Ill-health retirement mortality isusually heavier than that for age retirements.

3. Although the symbol ar65 is usually used, the benefit may be payable in various ways, e.g.

monthly in advance, and is possibly subject to a guarantee that at least 5 years’ payments will bemade. In this example,

ar65 = a

(12)

5+ 5|a(12)

65

on a suitable mortality table.

19.7 Fixed Pension Schemes

Consider a fixed pension of £P per year of service, including fractions pro rata, for a life now agedx with n years’ past service.

Ill-health and age retirements will be considered separately.

Ill-health retirementsThe m.p.v. of the benefit is

64−x∑t=0

vt+ 12ix+t

lx(n + t +

12)P ai

x+t+ 12

This can be divided into 2 terms:1. The value of the Past Service Pension (P.S.P.), which is the pension earned in respect of pastservice with the employer; in this case the P.S.P. is nP ;2. The value of the Future Service Pension (F.S.P.), which is the pension which will be earned infuture; in this case the F.S.P. is (t + 1

2 )P if retirement occurs (due to ill-health) at age x + t + 12 .

So the m.p.v. is

nP

64−x∑t=0

vt+ 12ix+t

lxai

x+t+ 12

+ P

64−x∑t=0

vt+ 12ix+t

lx(t +

12)ai

x+t+ 12

(19.7.1)

Define the following commutation functions:

Ciax = vx+ 1

2 ixaix+ 1

2,

M iax =

64−x∑t=0

Ciax+t

M iax = M ia

x − 12Cia

x ,

and

Riax =

64−x∑t=0

M iax+t


It follows that

Riax =

64−x∑t=0

(t +12)Cia

x+t

Proof.

Riax = M ia

x + M iax+1 + ... + M ia

64

= (12Cia

x + Ciax+1 + ... + Cia

64)

+ (12Cia

x+1 + Ciax+2 + ... + Cia

64)

+ ...

+12Cia

64

=12Cia

x + 112Cia

x+1 + ... + (64− x +12)C64

=64−x∑t=0

(t +12)Cia

x+t.

These commutation functions can now be used to calculate (19.7.1).The value of the P.S.P. is

nP

64−x∑t=0

vx+t+ 12

vxlxix+ia

ix+t+ 1

2

= nP · 1Dx

64−x∑t=0

Ciax+t

= nP

(M ia

x

Dx

)

The value of the F.S.P. is

P

64−x∑t=0

vx+t+ 12 ix+t

vxlx(t +

12)ai

x+t+ 12

= P · 1Dx

64−x∑t=0

(t +12)Cia

x+t

= P

(Ria

x

Dx

)

Hence the value of all ill-health pension benefits is

P

[n.M ia

x + Riax

Dx

](19.7.2)

Age retirementsThe valuation of benefits caused by age retirements is very similar to that for ill-health retirements,but with a final term corresponding to age retirement at exact age 65. The m.p.v. of age retirementbenefits is

64−x∑t=0

vt+ 12

(rx+t

lx

)(n + t +

12)P ar

x+t+ 12

+ v65−x r65

lx(n + 65− x)P ar

65 (19.7.3)

19.8. AVERAGE SALARY SCHEMES 311

Again, this may be separated into the P.S.P. and the F.S.P. terms.Define the commutations

Crax =

{vx+ 1

2 rxarx+ 1

2, x < 65

v65r65ar65 , x = 65

Mrax =

∑65−xt=0 Cra

x+t (note the summation is up to “65− x”)

Mrax = Mra

x − 12Cra

x

and

Rrax =

64−x∑t=0

Mrax+t =

64−x∑t=0

(t +12)Cra

x+t + (65− x)Cra65

(using a very similar proof to that for ill-health retirements).Then the value of the P.S.P. is

nP · 1Dx

[64−x∑t=0

vx+t+ 12 rx+ta

rx+t+ 1

2+ v65r65a

r65

]

= nP ·(

Mrax

Dx

)

The value of the F.S.P. is

P · 1Dx

[64−x∑t=0

(t +12)vx+t+ 1

2 rx+tarx+t+ 1

2+ (65− x)v65r65a

r65

]

= P · 1Dx

[64−x∑t=0

(t +12)Cra

x+t + (65− x)Cra65

]

= P

(Rra

x

Dx

)

Hence the value of all age retirement pension benefits is

P

[n.Mra

x + Rrax

Dx

].

So if one defines M(i+r)ax = M ia

x + Mrax , and similarly for other commutation functions, then the

value of a fixed pension of £P per annum payable for any retirement is

P

[n.M

(i+r)ax + R

(i+r)ax

Dx

](19.7.4)

for a member aged x with n years’ past service.

19.8 Average Salary Schemes

Suppose that the annual pension on retirement is

160× (total salary in service of company)

=160× (total past salary, T.P.S.)

+160× (total future salary).


Thus the benefits can be separated into the Past Service Pension and the Future Service Pension.Value of P.S.P.

This is just the value of a fixed pension of 160 (T.P.S.) so, using the functions defined in the

previous section,

m.p.v. of past service benefits =T.P.S.

60

(M

(i+r)ax

Dx

)(19.8.1)

Value of F.S.P.Consider ill-health retirement first. The salary to be earned between ages x + t and x + t + 1

2 isestimated as

12(SAL)

sx+t

sx

The m.p.v. of benefits is therefore

160

64−x∑t=0

vt+ 12ix+t

lx

SAL

sx(sx + sx+1 + ... + sx+t−1 +

12sx+t)ai

x+t+ 12

=SAL

60sxDx

64−x∑t=0

(sx + sx+1 + ... + sx+t−1 +12sx+t)Cia

x+t (19.8.2)

Define the commutation functionssM ia

x = sxM iax

andsRia

x =64−x∑t=0

sMiax+t

The ‘summation’ in (19.8.2) can be re-written as

sx(12Cia

x + Ciax+1 + ... + Cia

64)

+ sx+1(12Cia

x+1 + Ciax+2 + ... + Cia

64)

+ ...s64

(12Cia

64

)

= sxM iax + sx+1M

iax+1 + ... + s64M

ia64

So the value of the F.S.P. is

SAL

60sxDx

64−x∑t=0

sx+tMiax+t

=SAL

60sxDx

64−x∑t=0

sMiax+t

=SAL

60sxDx· sR

iax

Consider age retirements now. The m.p.v. of the future service pension is very similar to thesummation in (19.8.2), with all ‘i’ terms being replaced by ‘r’ terms and with an extra term relatingto retirements at exactly age 65; this means that a final term of

SAL

60sxDx· (sx + sx+1 + ... + s64)Cra

65

19.9. FINAL SALARY SCHEMES 313

has to be added.On defining the commutation functions

sMrax = sxMra

x and sRrax =

∑64−xt=0

sMrax+t

and using a similar argument to that used for ill-health retirements, the value of the F.S.P. isfound to be

SAL

60sxDx· sR

rax

Thus the value of all benefits for an average salary scheme is

TPS

60

(M

(i+r)ax

Dx

)+

SAL

60sx

(sR

(i+r)ax

Dx

)(19.8.3)

Example 19.8.1. A pension scheme provides each member who retires (whether for “age” or “ill-health” reasons) with an annual pension of 1

60 th of his average annual income over a member’sentire service, for each year of service. Fractions of years of service are included when calculatingthe amount of pension payable.

If contributions are paid entirely by the employer, calculate the appropriate contribution rate(as a percentage of salary) for a new entrant aged 20.

SolutionLet k% be the contribution rate. Then k must solve

k

100· SAL

s20

(sN20

D20

)=

SAL

60s20

(sR

(i+r)a20

D20

)

Hence, k =sR

(i+r)a20

0.6sN20= 5.25%.

19.9 Final Salary Schemes

Suppose that the annual pension is given by the formula

180× “final salary” per year of service.

“Final salary” is usually defined as the average annual salary in the last m years of service.Define

zx =1m

(sx−m + sx−m+1 + ... + sx−1).

In “Formulae and Tables”, m = 3, and hence

zx =13(sx−3 + sx−2 + sx−1)

The method of valuing benefits is similar to that in section 19.7, allowing for salary factors.

Ill-health retirementsThe m.p.v. of benefits for a member aged x with n years past service is

64−x∑t=0

SAL

80sxzx+t+ 1

2(n + t +

12)vt+ 1

2ix+t

lxai

x+t+ 12

(19.9.1)


Define the commutation functionszCia

x = zx+ 12Cia

x ,

zM iax =

64−x∑t=0

zCiax+t,

zM iax = zM ia

x − 12

zCiax ,

andzRia

x =64−x∑t=0

zMiax+t =

64−x∑t=0

(t +12)zCia

x+t

(note similarity to definitions in section 19.7)

Expression (19.9.1) becomes

SAL

80sxDx

64−x∑t=0

(n + t +12)zCia

x+t

and, following the usual steps, we find that the m.p.v. of the benefits is equal to

SAL

80sx

n.zM iax

Dx+

SAL

80sx·

zRiax

Dx(19.9.2)

This shows the terms for past and future service pensions separately.

Age retirementsAgain this is similar to the fixed pension case, but with salary factors inserted. The new commutationfunctions are

zCrax =

{zx+ 1

2Cra

x if x < 65,

z65Cra65 if x = 65,

zMrax =

65−x∑t=0

zCrax+t (note the upper limit of 65− x)

zMrax = zMra

x − 12

zCrax

zRrax =

64−x∑t=0

zMrax+t

We find that the m.p.v. of benefits due to age retirements is

SAL

80sxDx

[64−x∑t=0

(n + t +12)zCra

x+t + (n + 65− x)zCra65

]

=SAL

80sx

n.zMrax

Dx+

SAL

80sx

zRrax

Dx(19.9.3)

Thus the value of all pension benefits for a final salary scheme is

SAL

80sxDx

[n.zM (i+r)a

x + zR(i+r)ax

](19.9.4)

19.9. FINAL SALARY SCHEMES 315

Example 19.9.1. Three members of a pension scheme whose age nearest birthday is 45 have thefollowing annual rates of salary and exact numbers of years of past service:A : £15,000 20 yearsB : £12,000 10 yearsC : £14,000 5 years

For these members find the present value of a pension, payable on age-retirement or on ill-healthretirement, of 1

100 th of the average salary in the final 3 years before retirement for each year of service,including fractions. Give separately the values of the past-service and future-service benefits.

Solution

Value of P.S.P. =1

100

∑3i=1 ni(SAL)i

s45D45· zM

(i+r)a45

' 1100

490000(zM ia45 + zMra

45)12 (s44 + s45)D45

= 29, 071

Value of F.S.P. =1

100·∑3

i=1(SAL)i

s45D45· zR

(i+r)a45

' 1100

4100012 (s44 + s45)D45

(zRia45 + zR

ra45)

= 42, 929.

Hence total benefits have m.p.v. £72,000.

Remarks1. If retirement is possible within m years, the factor zx should be adjusted to allow for the actualpast salary progression. Also, if someone retires with less than m years’ service, the final salarymust be redefined as the average over a shorter period. (In calculations these points are sometimesignored.)2. If “final salary” means the salary rate at the date of retirement, then we change zx to sx. Also,if m = 1, zx = sx−1. But one must not change zR

(i+r)ax to sR

(i+r)ax , as the latter refers to average

salary schemes.3. If fractions of a year are not included when calculating the pension value, it is usually sufficientlyaccurate to adjust the n years’ past service to n− 1

2 . If there is no past service, consider the memberas having “− 1

2” year past service. So one must substitute

(n− 12).zM (i+r)a

x for n.zM (i+r)ax

in formula (19.9.4) (even when n = 0).4. Benefits may depend on a “pensionable salary” equal to salary minus some fixed sum, say £A. Ifso, then the m.p.v. of the pension benefit in a final salary scheme is now

SAL

80sxDx

(n.zM (i+r)a

x + zR(i+r)ax

)− A

80

(n.M

(i+r)ax − R

(i+r)ax

Dx

)

5. It may be the case that the number of years of service are restricted to, say, 40 for pensionpurposes. This will affect lives joining at ages under 25, assuming a latest retirement age of 65.

There are 2 cases to consider:


(a) If n < 40, then leave the P.S.P. unchanged but restrict the F.S.P. . In the final salary schemementioned above, the value of the F.S.P. is altered to

SAL

80sxDx

(zR(i+r)a

x −z R(i+r)ax+40−n

)(19.9.5)

(b) If n ≥ 40, then restrict the past service to 40 years and make the F.S.P. zero. Hence thevalue of pension benefits is

4080· SAL

sxDx

zM (i+r)ax (19.9.6)

Example 19.9.2. An executive pension scheme provides a pension of 145 of final salary for each

year of scheme service, with a maximum of 23 of final salary, upon retirement due to age between

the ages of 60 and 65.Final salary is defined as salary in the 3 years prior to retirement.A director, now aged 47 exactly has 14 years of past service and expects to earn £80,000 over

the coming year.Using the symbols defined in the Formulae and Tables for Actuarial Examinations, what is the

expected present value of the future service pension on age retirement for this member?

Solution As the maximum pension is 23 = 30

45 of the final salary, service is restricted to a maximumof 30 years for pension purposes. Hence, as n is 14,

m.p.v. of F.S.P. =80000

45·

zRra47 − zR

ra63

s47D47

(notice that £80,000 is the salary next year and not the salary rate, and hence we put s47 in thedenominator).

19.10 Lump Sums on Retirement

Suppose there is a cash payment on retirement of 3 times the annual pension, where the pension is180× final salary per year of service (as in section 19.9).

Define the commutation functions

Cix = ixvx+ 1

2

zCix = zx+ 1

2Ci

x,

and so on, changing the annuity factor in all commutation functions (both “age” or “ill-health”) to1.

These functions are tabulated (on the same basis as in “Formulae and Tables”) in the Supplementto this book.

Using nearly the same arguments as before, the m.p.v. of the lump sum on retirement is

3 · SAL

80sxDx(n.zM i+r

x + zRi+rx ) (19.10.1)

If the annual pension is a fixed sum of £P per year of service, then a lump sum of 3P onretirement has m.p.v.

3P

(n.M i+r

x + Ri+rx

Dx

)(19.10.2)

19.11. DEATH AND WITHDRAWAL BENEFITS 317

19.11 Death and Withdrawal Benefits

The benefits on death in service usually consist of one or more of:1. a fixed sum, or a certain multiple of salary;2. a return of the employee’s contributions, accumulated at a rate of interest

j (where j may be zero);3. a spouse’s pension.

We begin by considering the first type of benefit.Suppose the death benefit is 2× the salary rate at date of death. The benefit for a person aged x,with current salary rate SAL, has m.p.v.

2(SAL)64−x∑t=0

vt+ 12dx+t

lx

sx+t

sx(19.11.1)

Define the commutation functions

Cdx = vx+ 1

2 dx,sCd

x = sxCdx

Mdx =

64−x∑t=0

Cdx+t

sMdx =

64−x∑t=0

sCdx+t

Expression (19.11.1) can be written in the form

2SAL

sxDx

64−x∑t=0

sx+tCdx+t

and hence the value of the benefit is

2SAL

sx·

sMdx

Dx(19.11.2)

If the benefit on death is just a fixed amount, £B, then the m.p.v. is

B

64−x∑t=0

vt+ 12dx+t

lx

B ·(

Mdx

Dx

)(19.11.3)

WithdrawalsSimilar formulae may be developed (replacing the ‘d’ with ‘w’) but withdrawal benefits are usuallyin the form of a return of contributions or a deferred pension. (We shall discuss the formulae forvaluing these benefits later.)

NoteThe commutation functions Md

x , Mwx , etc. are tabulated in the Supplement to this book. Note that

Mdx corresponds to jMd

x when j = 0 (page v of the Supplement) and Mwx corresponds to jMw

x whenj = 0 (page vi of Supplement).


Example 19.11.1. A company has a pension scheme which provides a pension upon retirementof 1

80 th of the final salary per year of service. In addition the sum of £10,000 is paid on death inservice of a member. If all contributions are paid by the employer, find the contribution rate as apercentage of salary required for a new entrant aged 40 with a salary rate of £10,000.Use the basis of “Formulae and Tables” (with the supplement); final salary is the average annualsalary in the 3 years prior to retirement.

SolutionLet k be the contribution rate per cent. Then k solves

0.01k · 10, 000s40D40

· sN40 =10, 000

80s40D40× zR

(i+r)a40 + 10, 000

Md40

D40

Thus,

0.01k × 172628.7 = 14827.60 + 895.6

Hence k = 9.11%.

19.12 Return of Contributions on Death or Withdrawal

Suppose that the employee’s contributions are returned with compound interest at rate j per annumif the member leaves service.

Let TPC be the member’s total past contributions to the scheme.The following additional commutation functions are defined:

jCwx = (1 + j)x+ 1

2 vx+ 12 wx,

jMwx =

64−x∑t=0

jCw

x+t

jMwx = jM

w

x −12

jCw

x

sjMwx = sx

jMw

x

sjRwx =

64−x∑t=0

(1 + j)−(x+t+ 12 )sjM

w

x+t

Similar functions for returns of contributions on death are defined by substituting ‘d’ for ‘w’.Suppose that the employees contribute at rate k% of salary. Then the m.p.v. of the return of

contributions on withdrawal for a member aged x is

(TPC)(1 + j)−x ·jMw

x

Dx+

k

100· SAL

sxDx· sjR

w

x (19.12.1)

Proof. See the Appendix to this Chapter.

NoteThe values of jM

wx , sjR

wx , etc. are given for j = 0.03 and j = 0 in the Supplement to this book.

If the employee’s contributions are fixed at £A per annum, payablecontinuously, the m.p.v. of the return of contributions on withdrawal is

(TPC)(1 + j)−xjMw

x

Dx+ A ·

jRwx

Dx(19.12.2)

19.12. RETURN OF CONTRIBUTIONS ON DEATH OR WITHDRAWAL 319

(see the Appendix to this Chapter.)NoteWhen j = 0 one may omit the ‘j’ from the commutation functions, so if contributions are returnedwithout interest their mean present value is

(TPC)Mw

x

Dx+

k

100SAL

sxDx· sR

wx (19.12.3)

if the employee contributes k% of salary, and

(TPC)Mw

x

Dx+ A · Rw

x

Dx(19.12.4)

if he contributes at the fixed rate of £A p.a.

If employee’s contributions are returned on death (with or without interest), formulae (19.12.1)to (19.12.4) are changed by just substituting ‘d’ for ‘w’.

Furthermore, if contributions are returned on either death or withdrawal, one may adjust formula(19.12.1) to

(TPC)(1 + j)−xjMd+w

x

Dx+

k

100· SAL

sxDx· sjR

d+w

x (19.12.5)

where jMd+wx = jM

dx + jM

wx

and sjRd+wx = sjR

dx + sjR

wx , as expected.

(Similar remarks apply to formulae (19.12.2) to (19.12.4).)

Example 19.12.1. You are actuary to a pension scheme which provides the following benefits:(a) a pension of 1

80× final salary for each year of service (including fractions pro rata) on retire-ment for “age” or “ill-health” reasons.

(b) a lump sum on retirement of 3 times the annual pension.

Your actuarial basis is that given in the “Formulae and Tables”, and you calculate reserves prospec-tively, ignoring expenses. Final salary is the average annual salary in the 3 years before retirement.

Members pay contributions at the rate of 2% of salary. The total contribution rate is assessedfor each member separately, and is the proportion of salary which will pay for the benefits, whenthe member joins, i.e. the prospective reserve at the entry date is zero.

1. Find the employer’s contribution rate for a member joining at age 20.2. Find the reserve held for a member aged 45, with salary at the rate of £20,000 per annum,

who joined the employer at age 20.3. Your pension scheme accepts transfer values from other pension schemes. Suppose that a life

aged 45 joins the scheme, bringing a transfer value of £10,000. The member’s current salary rate is£15,000 p.a., and he asks for his transfer value to be used to give him “added years” of service, i.e.to credit him with n years of past service. Find n, given that, in the event of death in service, thetransfer value will be returned with compound interest at 3% p.a.

NoteThere are no withdrawals over age 45 in the Tables, so we need not specify what happens to thetransfer value in that event.


Solution1. Let k be total contribution rate per cent for a new member aged 20. Then

k

100SAL

s20D20·s N20 =

SAL

80s20D20[zR(i+r)a

20 + 3.zRi+r20 ]

Thus

k =10080

[zR

(i+r)a20 + 3.zRi+r

20sN20

]= 7.443.

So employer contributes at rate 7.443%− 2% = 5.443% of salary.2. Reserve is

20, 00080s45D45

[25.zM(i+r)a45 +z R

(i+r)a45 + 3(25.zM i+r

45 +z Ri+r45 )]− 20, 000

s45D45(0.0744)sN45

= £59, 585.

3. Let n years of past service be credited. The equation of value is

10000 =15000n

80s45D45(zM

(i+r)a45 + 3.zM i+r

45 )

+ 10000(1.03)−45

(jMd

45

D45

)where j = 0.03

Hence 10, 000 = 1420.47.n + 1394.42Therefore n = 6.06 years

(The transfer value is treated as if it were the T.P.C.)

19.13 Spouse’s Benefits

A pension scheme may provide the following benefits (among others):(a) a spouse’s death in service (D.I.S.) pension,

and/or(b) a spouse’s death after retirement (D.A.R.) pension.

If spouse’s pension ceases on remarriage, a double-decrement table (for death and remarriage)should be constructed and used in place of the mortality table for spouses.

(a) Spouse’s D.I.S.Consider (for example) a male member aged x of a pension scheme providing an annuity payable

(say) monthly in advance to his widow if he dies in service. As in chapter 13, we definehx = probability that a man is married at age x

and d = the average age difference (in years) between husband and wife, i.e. average value of(husband’s age - wife’s age).

The m.p.v. of this benefit is

64−x∑t=0

vt+ 12dx+t

lxhx+t+ 1

2× (spouse’s pension p.a.)a(12)

f

x+t+ 12−d

(19.13.1)

19.13. SPOUSE’S BENEFITS 321

The size of spouse’s pension may depend on the member’s salary at or near the date of deathand the number of years of service.

Note. Commutation functions are complicated and are usually ignored.

(b) Spouse’s D.A.R.One must consider the following 2 cases:

(i) Widow’s pension is only payable if the widow was married to the scheme member when heretired. (Post-retirement marriages do not count for benefit purposes.)

(ii) Any widow may receive the pension.The spouse’s D.A.R. pension is valued by first considering the appropriate value at the retirement

date of the member, and then allowing for survivorship and interest before retirement.For example, consider a man retiring at age 65 in normal health. Assuming that spouse’s pensions

are payable continuously, the value at that age of the spouse’s D.A.R. benefit is as follows:Case (i) : (annual widow’s pension).h65am

65|f

65−d(19.13.2)

Case (ii) :(annual widow’s pension).

∫∞0

vt.tpm65µ m

65+th65+ta f

65−d+tdt (19.13.3)

Example 19.13.1. A pension fund provides the following benefits for widows of male members:(a) on death of the member in service, a widow’s pension of annual amount equal to one-third

of the member’s salary rate at the time of his death; and(b) on death of the member after age or ill-health retirement, a widow’s pension of 1

120 th of themember’s average salary in the 3 years immediately preceding retirement for each year of service,fractions of a year counting pro rata.

Develop formulae for valuing the widow’s benefits for a male member aged x with n years’ pastservice (including fractions). You may assume that all age-retirements take place at exact age 60,between ages 60 and 65, or at exact age 65, and that ill-health retirements may take place at any agebetween 35 and 60. Widows’ pensions on death after retirement are payable to any widow (not justto a widow who was married to the member when he retired.) Widows’ pensions are payable monthlyin advance and do not cease on remarriage. You may assume that a service table has already beenconstructed, and that the member’s current salary rate per annum is SAL. Commutation functionsneed not be developed.

Solution(a) Define sx, sx, hx as previously. Let

zx =13(sx−3 + sx−2 + sx−1) as before.

Let

y = average age of wife of (x)= x− d

Then value of benefit is

SAL

3

64−x∑t=0

vt+ 12

(dx+t

lx

)sx+t+ 1

2

sxhx+t+ 1

2a(12)

y+ft+ 1

2


(b) Consider ill-health retirement first.

m.p.v. =SAL

120

59−x∑t=0

vt+ 12

(ix+t

lx

)zx+t+ 1

2

sx(n + t +

12)aIH

x+t+ 12

where

aIHx+t+ 1

2=

∫ ∞

0

vr.rpihx+t+ 1

2µih

x+t+ 12+rhx+t+ 1

2+ra(12)

y+t+ 12+r

dr

(“ih” indicates that a mortality table for men retiring due to ill-health is employed).For age retirements, there are terms for retirement (i) between ages 60 and 65, (ii) at exact ages

60 and 65.(i) value =

SAL

120

64−x∑t=60−x

vt+ 12

(rx+t

lx

)zx+t+ 1

2

sx(n + t +

12)aNH

x+t+ 12

where aNHx+t+ 1

2is as for aIH

x+t+ 12, but with a mortality table for age retirements.

(ii) value =SAL

120v60−x

(r′60lx

)z60

sx(n + 60− x)aNH

60

+SAL

120v65−x

(r′65lx

)z65

sx(n + 65− x)aNH

65

(r′x refers to retirements at exact ages 60, 65; note that l65 = r′65).

19.14 Preserved Pensions on Leaving Service

Suppose that (for example) “early leavers” get a deferred annual pension of160× final salary per year of service, vesting at age 65. ‘Final’ salary refers to the average salary inthe 3 years before leaving service.

In practice, this pension may escalate between the date of leaving and age 65; suppose thatescalation is at rate j per annum compound. Also, one must specify a mortality table for the earlyleavers: let us use the notation lx for their life table. The m.p.v. of this withdrawal benefit for a lifeage x is

SAL

60

64−x∑t=0

v65−x wx+t

lx(n + t +

12)zx+t+ 1

2

sx(1 + j)65−(x+t+ 1

2 ) l65

lx+t+ 12

· a65 (19.14.1)

where a65 is the annuity factor, with the pension possibly payable monthly, etc.To evaluate (19.14.1) by commutation functions, one needs to define suitable ‘new’ commutation

functions. That is, the expression (19.14.1) is equal to

SAL

60sxDx

64−x∑t=0

(n + t +12)zCwa

x+t

where

zCwax = l65a65v

65

[wxzx+ 1

2(1 + j)65−x− 1

2

lx+ 12

]

19.14. PRESERVED PENSIONS ON LEAVING SERVICE 323

We now define

zMwax =

64−x∑t=0

zCwax+t,

zMwax = zMwa

x − 12

zCwax

and

zRwax =

64−x∑t=0

zMwax+t

(Note the similarities with the definitions used in previous sections.) The present value of thedeferred pension is

SAL

60sxDx

[n.zMwa

x + zRwax

](19.14.2)

Note The notation used here is not standard, but is an example of “do-it-yourself” commutationfunctions.


Exercises

19.1 Contributions to a pension scheme by employees are made at a rate of 5% of salary when agedunder 35, 6% between ages 35 and 45, and 7 1

2% when aged 45 or over. Calculate the presentvalue of the future contributions payable by a member aged exactly 30 who in the past yearhas received a total salary of £12,718.

19.2 A company pension scheme provides the following benefits for all members:

(1) a pension on retirement (on grounds of ill-health or of age) of one-eightieth of final pen-sionable salary for each year of service (including fractions),

(2) a lump sum on retirement of 3 times the annual pension,

(3) on death in service, a lump sum of £30,000,

(4) on withdrawal from service, a return of the employee’s contributions, accumulated at 3%per annum compound.

Final pensionable salary is defined as the average annual salary in the three years immediatelybefore retirement. All members who reach age 65 retire immediately.

Employees contribute to the scheme at the rate of 3% of salary, payable continuously. Salariesare revised continuously. The employer’s contribution rate is assessed for each member sep-arately, and is such that the prospective reserve for each new entrant is zero. Expenses areignored.

(i) (a) Derive a formula, in terms of suitable commutation functions, for valuing benefit (1)above in respect of a new entrant aged x with annual salary rate SAL. (You need not definethe service table functions.)

(b) In respect of a new entrant aged x with annual salary rate SAL, give formulae for valuingbenefits (2), (3) and (4) above, using suitable commutation functions. (You need not derivethe formulae.)

(c) Hence find a formula for the employer’s contribution rate for a new member aged x and astarting salary rate of £10,000 p.a.

(ii)(a) Using the basis given in the pension fund section of the Formulae and Tables (and thesupplement), find the value of each of the benefits (1), (2), (3) and (4) for a new entrant aged45 with salary rate £10,000 per annum.

(b) Hence or otherwise find the employer’s contribution rate for this new member.

19.3 The pension scheme of a certain company provides an annual pension on retirement (for‘age’ or ‘ill- health’ reasons) of amount equal to one per cent of the member’s total earningsthroughout his service. The pension is payable weekly. In addition, in the event of a memberdying in service there is payable at the time of death a lump sum of £30,000. There is nobenefit on withdrawal.The company pays a constant percentage of all the members’ salaries into the pension fund.The percentage is that which will exactly cover the cost of benefits for a new entrant to thefund at age 30 with an initial salary rate of £10,000 per annum. Contributions are payablecontinuously, and the employees do not contribute to the scheme. Expenses are negligible.

(a) Calculate the contribution rate paid by the company, assuming the last retirement age is65.

(b) A valuation of the fund is to be conducted. For each active member of the scheme thereis recorded

19.15. EXERCISES 325

(i) the age nearest birthday (which is regarded as the member’s exact age) at the valuationdate,

(ii) the annual salary rate at the valuation date, and

(iii) the total past earnings in service (prior to the valuation date.)

For each age, the totals of (ii) and (iii) are recorded and the following is an extract from thedata.

Age x No. of members Total past earnings Total of annual salaryaged x for members aged x rates for members aged x

£ £25 11 302,100 70,100

Assuming that the basis of the Tables provided is appropriate, find the liability at the valuationdate for the benefits payable to the members aged 25, and determine whether the futurecontributions payable in respect of these members are more or less than sufficient to cover thebenefits.

19.4 A pension scheme provides each member who retires (for any reason) with annual pensionequal to 1

60× final salary per year of service. Final salary is the average income over the last3 years of service, and fractions of a year of service are not included when calculating thepension.Assuming that equal contributions are payable by the member and his employer, that in theevent of death in service a benefit is payable equal to the return without interest of both themember’s and the employer’s contributions, and that in the event of withdrawal from servicea return without interest is made of the member’s contributions, calculate the appropriatecontribution rate payable by both the member and his employer in respect of a new entrantaged 40.

19.5 You are consulting actuary to a small pension scheme, which has just been established. Youhave decided to use the pension fund tables in Formulae and Tables for Actuarial Examinationsas the basis for all calculations. The scheme provides the following benefits to employees:

(i) on retirement (for ill-health reasons or otherwise), a pension of 180 th of annual pension-

able salary, averaged over the previous three years, for each year of future service includingfractions;

(ii) on withdrawal or death in service, a return of the member’s contributions, accumulatedat 3% per annum compound interest.

Employees contribute 2% of salary to the scheme. Pensionable salary is defined as salary less£4000. Salaries are revised continuously, and contributions are made continuously.

Details of the current membership are as follows.member age current salary rate (£)

1 45 30,0002 45 20,0003 35 10,0004 35 10,000

(a) The employer has decided to contribute the proportion of total salaries which, togetherwith the employees’ contributions, will exactly pay for the benefits. Calculate the employer’scontribution rate.

(b) A new employee, aged 35 and with current salary rate £8,000, is about to be hired.Calculate the surplus or deficiency in the pension fund after this new member joins.


19.6 It is desired to set up a pension scheme for the group of employees described below. For eachmember there is recorded his exact age, his exact length of service with the company, and hisannual rate of salary.

Past Rate ofAge Service Salary

(Years) p.a.£

25 4 8,80025 6 8,50025 5 8,50025 3 8,60025 1 8,600


(Years) p.a.£

35 12 12,40035 18 12,50035 15 12,80035 5 12,50035 10 12,400


(Years) p.a.£

45 25 15,00045 15 14,00045 10 13,00045 20 13,20045 5 13,000

The scheme will provide pensions of 160 th of “pensionable salary” for each year of service

(fractions of a year being included) and on death or withdrawal from service a return will bemade of the member’s contributions with 3% compound interest. All members will contributeat the same rate and the employer will contribute the same amount as each member. Thecontribution rate will be such as to provide exactly the benefits for future service. The basisof “Formulae and Tables” is to be used.

(a) Assuming that pensionable salary is the average annual earnings over the three-year periodending on the retirement date, calculate the contribution rate payable by each member.

(b) Calculate also the total liability for past service benefits. The employer wishes to meetthis liability by paying additional contributions proportional to future salary payments. Atwhat rate should these additional contributions be made?

(c) Immediately after the scheme is set up as described above the 45 year old member with5 years of service withdraws. Is the position of the fund improved or worsened by this with-drawal?

19.7 The pension scheme of a large company provides the following benefits, among others:

(1) on death in service of married members: a spouse’s pension of one-third of the member’sannual salary at the date of death;

(2) on death after normal retirement of the member at age 65 or after ill-health retirementat an earlier age: a spouse’s pension of 1% of annual salary at the date of retirement foreach year of scheme membership (including fractions of a year); no pension is payable if themarriage has taken place after the member’s retirement date.

Assume that a service table has been constructed, and that the proportion of members mar-ried at exact y is hy. Assume further that spouses are the same age as members, and thata unisex mortality table is used for all calculations, with the age rated up by 5 years onill-health retirement. You are given the age nearest birthday, current salary rate and yearsof past service (including fractions) of each member. Spouse’s pension is payable monthly inadvance, beginning immediately on the death of the member.

(i) Using the rate of interest i per annum, find formulae for the mean present value of eachof the benefits (1) and (2) above for a member aged x (x < 65). You are NOT required toconstruct commutation functions.

(ii) Suppose that the scheme’s rules are to be changed so that benefit (2) is to be payable toany surviving spouse. Show how to modify the formulae of (i) to accommodate this change.


Solutions

19.112, 718

s29 ×D30{0.05sN30 + 0.01sN35 + 0.015sN45}= 12, 718× 1.30162= £16, 554

19.2 (i)(a)

m.p.v. =SAL

80sxDx

[64−x∑t=0

((t +

12)zCia

x+t + (t +12)zCra

x+t

)+ (65− x)zCra

65

]

=SAL

80sx

zR(i+r)ax

Dx

(b)(2): 3.SAL80sx

· zRi+rx

Dx

(3) 30000 · Mdx

Dx

(4) 0.03 · SALsxDx

sjRw

x where j = 0.03

(c) Let employer’s contribution rate be k for this member. Then

(k + 0.03)SAL

sxDx

sNx = above benefits.

So

k + 0.03 =1

sN x

[180

zR(i+r)ax +

380

zRi+rx + 3sxMd

x + 0.03.sjRwx

]

(ii)(a) Values of benefits are:(1) 13,088(2) 3,642 (using Supplement)(3) 3,016(4) 0

Hence total value of benefits = £19,746.

(b) value of contributions = 142580(k + 0.03)Hence k + 0.03 = 19746

142580and hence k = 10.85%.

19.3 (a) Let k be the contribution rate. Then

k · SAL

s30D30·s N30 =

SAL

100s30·

sR(i+r)a30

D30+ 30000 · Md

30

D30

where SAL = 10, 000.

Hence k =s30

sN30

[0.01

sR(i+r)a30

s30+ 3.Md

30

]

= 0.0535 = 5.35%


(b) Value of benefits for the group is302100

100· M

(i+r)a25

D25+

70100100s25

·sR

(i+r)a25

D25+ 11× 30000

Md25

D25

= 2977.37 + 54730.83 + 11811.67 = £69, 528

Value of future contributions is0.0535× 70100

s25D25

sN25

= £76, 718.

Hence contributions are more than sufficient.

19.4 Let k% of salary be the contribution rate for both the member and the employer. Then if themember has a salary rate of £SAL,

2k

100SAL

s40D40·s N40 =

SAL

60s40D40

[zR

(i+r)a40 − 1

2zM

(i+r)a40

]

+k

100· SAL

s40D40·s Rw

40 +2k

100· SAL

s40D40·s Rd

40

Hence 6342.42k = 35513.07 + 13.74k + 562.35k.Therefore k = 6.16%

19.5 (a) Construct the following table, using j = 0.03:

Member age salary SAL80sx

· zR(i+r)ax

Dx

400080 · R(i+r)a

x

Dx0.02SAL

sx

sjRd+wx

Dx

1 45 30,000 39,264 3,999 8992 45 20,000 26,176 3,999 5993 35 10,000 15,189 3,498 5494 35 10,000 15,189 3,498 549

95,818 14,994 2,596

Let the employer contribute k% of salary, so m.p.v. of contributions is

(k + 0.02)[

50000s45D45

sN45 +20000s35D35

·s N35

]

= 1093716(k + 0.02)

Therefore k + 0.02 = 95818−14994+25961093716

and hence k = 5.627%.

(b) Reserve for new member is800080s35

·zR

(i+r)a35

D35− 4000

80· R

(i+r)a35

D35

+ 0.02× 8000s35D35

.sjRd+w35

− 8000s35D35

× 0.07627sN35

= 12, 151− 3, 498 + 440− 11, 617= −2, 524.

Hence there is now a surplus of £2,524.

19.6 (a) For each age x,

Value of Past Service Pension =P

(n×SAL)60sx

· zM(i+r)ax

Dx


Value of Future Service Pension =P

SAL60sx

· zR(i+r)ax

Dx

Value of return of contributions =P

SAL100sx

· sjRd+wx

Dx(where j = 0.03) per 1% of salary

return ofx

∑SAL

∑(n× SAL) P.S.P. F.S.P. contributions (per 1% of salary)

25 43,000 163,100 7,776 76,783 2,223.735 62,600 752,300 55,421 126,780 1,720.545 68,200 1,044,000 103,230 119,012 1,022.7

166,427 322,575 4,966.9

The value of the contributions is, per 1% of contributions,∑SAL

100sx

sNx

Dx(1)

x Value of (1)25 8,791.235 11,919.145 9,724.0

30,434.3

Let members’ contribution be k%. The equation of value is thus2k × 30, 434.3 = 322, 575 + 4, 966.9k

Hence k = 5.77%.

(b) The past service liability is £166,427. Let the contribution rate per cent needed to payfor this be p. Then p solves,

p× 30, 434.3 = 166, 427Hence p = 5.468%.

(c) The reserve for the withdrawn member is13, 000

60s45D45·z R

(i+r)a45 +

5× 1300060s45D45

zM(i+r)a45 + 0.0577× 13000

s45D45·sj Rd+w

45

− (2× 0.0577 + 0.05468) · 13000s45D45

·s N45

= 30, 238− 31, 526= −1, 288

So the position of the pension fund has worsened by £1,288.

19.7 (i) Benefit (1) has valueSAL

3

64−x∑t=0

vt+ 12dx+t

lxhx+t+ 1

2a(12)

x+t+ 12

sx+t

sx

Benefit (2):On normal retirement the value is

SAL

100sx· v65−x.s65(n + 65− x)× r65

lx

[h65a

(12)65|65(1 + i)

124

]

On ill-health retirement, the value isSAL

100

64−x∑t=0

vt+ 12 (

sx+t

sx)ix+t

lx(n + t +

12)[hx+t+ 1

2a(12)

x+t+ 12+5|x+t+ 1

2(1 + i)

124 ]


(ii) Benefit (1): no changeBenefit (2):For normal retirement change

h65a(12)65|65(1 + i)

124

to ∫ ∞

0

vt.tp65µ65+th65+ta(12)65+t dt

For ill-health retirement, changehx+t+ 1

2a(12)

x+t+ 12+5|x+t+ 1

2(1 + i)

124

to

∫ ∞

0

vr.rpx+t+ 12+5µx+t+ 1

2+5+rhx+t+ 12+ra

(12)

x+t+ 12+r

dr


Appendix: Formulae for valuing a return of contributions

Suppose employee’s contributions are to be returned on withdrawal with interest at rate j perannum compound. Consider a member aged x whose current salary rate is SAL and whose totalpast contributions, accumulated at rate j p.a., are TPC.

Assume that the employee will in future contribute k% of salary. The value of the return offuture contributions is

k

100SAL

sx

64−x∑t=0

vt+ 12wx+t

lx[sx(1 + j)t + sx+1(1 + j)t−1 + ... + sx+t−1(1 + j) +

12sx+t]

=k

100

(SAL

sxlx

){v 1

2 wx × 12sx + v1 1

2 wx+1[sx(1 + j) +12sx+1]

+ v2 12 wx+2[sx(1 + j)2 + sx+1(1 + j) +

12sx+2]

+ ...

+ v64 12−xw64[sx(1 + j)64−x + sx+1(1 + j)64−x−1 + ... + s63(1 + j) +

12s64]}

Collecting the coefficients of sx, sx+1, ... gives

k

100

(SAL

sxlx

){sx[

12v

12 wx + v1 1

2 wx+1(1 + j) + ... + v64 12−xw64(1 + j)64−x]

+ sx+1[12v1 1

2 wx+1 + v2 12 wx+2(1 + j) + ... + v64 1

2−xw64(1 + j)64−x−1]

+ ... + s64[12v64 1

2−xw64]}

Define

jCwx = (1 + j)x+ 1

2 vx+ 12 wx

jMwx =

64−x∑t=0

jCw

x+t

jMwx = jM

w

x −12

jCw

x

sjMwx = sx.jMw

x

sjRwx =

64−x∑t=0

(1 + j)−(x+t+ 12 ).sjMw

x+t

The value of the return of future contributions can be written as

k

100

(SAL

sxvxlx

){sx(1 + j)−(x+ 1

2 )

[12(v(1 + j))x+ 1

2 wx + (v(1 + j))x+1 12 wx+1 + ... + (v(1 + j))64

12 w64

]

+ sx+1(1 + j)−(x+1 12 )

[12(v(1 + j))x+1 1

2 wx+1 + (v(1 + j))x+2 12 wx+2 + ... + (v(1 + j))64 frac12w64

]

+ ... + s64(1 + j)−64 12

[12(v(1 + j))64

12 w64

]}

=k

100

(SAL

sxDx

).sjRw

x


The value of the return of past contributions with interest at rate j is

(TPC)64−x∑t=0

vt+ 12 (1 + j)t+ 1

2wx+t

lx

=(TPC)

Dx

64−x∑t=0

(1 + j)−x(v(1 + j))x+t+ 12 wx+t

= (TPC)(1 + j)−xjMw

x

Dx

Hence the total value is

(TPC)(1 + j)−xjMw

x

Dx+

k

100SAL

sxDx

sjRw

x

Suppose now that the employee’s contributions do not depend on salary, but are instead £A eachyear, payable continuously. The value of the return of future contributions is obtained in a similarway to the previous derivation, with the salary scale function taken as 1. This gives a present valueof

A ·jRw

x

Dx

wherejRw

x =64−x∑t=0

(1 + j)−(x+t+ 12 ).jMw

x+t.

Appendix A

Some notes on examination technique

Each candidate should spend some time going over his or her notes, the textbook and (especially)past examination papers, making a note of subjects and formulae of particular importance. For thispurpose a set of postcards/computer cards is useful: each topic may be summarised on one postcard,enabling the candidate to learn the most important facts and then have a “revision aid”. (He or sheshould ask a friend to choose a card at random, to give the title and to ask for a description of thetopic.)

In actuarial examinations (particularly in life contingencies) it is important to separate the basicideas or “rationale” of the solution from

(i) technical aspects, such as the correct use of actuarial tables and commutation functions, and

(ii) arithmetic.

It is sometimes necessary to omit (ii) or even (i) under extreme time-pressure, but one shouldleave space to return to the question later if time permits. On the other hand, it is satisfying to givea complete answer, and this should be attempted if time is not short. One should keep the left-handpages of the examination book free for rough working and arithmetic.

On first seeing the examination paper, one should tick the questions about which one is mostconfident - especially bookwork. Leave the less familiar questions until later. One should ensure thatan approximate time-scheme (say, 15 minutes per 10-mark question) is adhered to, within reasonablelimits. Routine arithmetic (or even looking up the tables) may sometimes be left till near the end,when one is too tired to think out new ideas but can still do (or check) arithmetic. As in musicalperformances, it is essential to get off to a good start, by being in the right frame of mind andpicking the most suitable questions to attempt first.

333

Appendix B

Some technical points about the tables used in examinations

1. Avoid excessive reliance on commutation functions.In theoretical/statistical questions, it is nearly always best to avoid the use of commutation func-tions.

2. In the A1967-70 section of “Formulae and Tables” commutation functions are given only at 4%interest. For other interest rates one most use the more limited tables provided, e.g.

ax:n at 6% = ax − vn lx+n

lxax+n

(There is no point writing vn lx+n

lxas

Dx+n

Dxsince Dx and Dx+n are not given at 6% in the tables.)

3. Remember to make use of the functions maturing at ages 60, 65, etc. This saves time.

4. If only limited tables are available, one must proceed directly, using suitable approximations. For

example, suppose that one is asked to evaluate A 150.5:2.5 at 10% interest on A1967-70 ultimate.

This may be done by the trapezoidal rule as follows:∫ 2.5

0

vttp50.5µ50.5+tdt + 2.5

2

[µ50.5 + v2.5 l53µ53

l50.5

]

where µ50.5 and l50.5 are estimated by linear interpolation.

5. Remember that for a(55) there are two tables - male and female (each with a select period of 1year.) Assume that males and females are subject to the table of the appropriate sex (unless thequestion states otherwise.)

6. Be careful (especially in A1967-70, but also in a(55)) that you distinguish between “select” and“ultimate” tables and use the correct rate of interest.

334

Appendix C

Some common mistakes

WRONG CORRECT

1. Ax:n ; (1 + i)12 Ax:n Ax:n ; (1 + i)

12 A1

x:n + A 1x:n

=(1 + i)

12 (Mx −Mx+n) + Dx+n

Dx

2. ax + i

δax ax ; ax + 1

2 or ax − 12

3. a(m)x:n ; i

d(m)ax:n a(m)

x:n ; ax:n −(

m− 12m

) (1− Dx+n

Dx

)

4. a1x:n , a 1

x:n do not exist

5.∫ t

0

crdr =[cr

]t

0or

[cr+1

r + 1

]t

0

∫ t

0

crdr =[

cr

log c

]t

0

(Note that cr = er log c)

6. Mean present value = vT

Present value (as random variable) = Ax

M.P.V. = Ax

P.V. = vT

7. var(P aT − vT )

= var(P aT )− var(vT )

or var(P aT ) + var(vT )

var(

P(1− vT )

δ− vT

)

= var[−

(P + δ

δ

)vT +

P

δ

]

=(

P + δ

δ

)2

var(vT )

8. ax:n = 1 + ax:n ax:n = 1 + ax:n−1

9. µx =l′xlx

µx = − l′xlx

10. tpx = exp(∫ t

0

µx+rdr

)tpx = exp

(−

∫ t

0

µx+rdr

)

11. ax:n =Nx −Nx+n

Dxax:n =

Nx+1 −Nx+n+1

Dx

335

Appendix D

Some formulae for numerical integration

1. Mid-point rule

∫ b

a

f(x)dx + (b− a)f(

a + b

2

)

2. Trapezoidal rule∫ b

a

f(x)dx +(

b− a

2

)[f(a) + f(b)]

3. Simpson’s rule∫ b

a

f(x)dx +(

b− a

6

)[f(a) + 4f

(a + b

2

)+ f(b)

]

4. Three-eighths rule∫ b

a

f(x)dx +(

b− a

8

)[f(a) + 3f

(2a + b

3

)+ 3f

(a + 2b

3

)+ f(b)

]

336

336

SUPPLEMENT

agex Ci

x M ix R

i

xzCi

xzM i

xzR

i

x

20 0.0 188.8 6835.5 0.0 904.4 33753.7 TABLES FOR21 0.0 188.8 6646.7 0.0 904.4 32849.3 LUMP SUM22 0.0 188.8 6457.9 0.0 904.4 31944.9 BENEFITS ON23 0.0 188.8 6269.1 0.0 904.4 31040.5 RETIREMENT24 0.0 188.8 6080.3 0.0 904.4 30136.1

25 0.0 188.8 5891.5 0.0 904.4 29231.7 Ill-health26 0.0 188.8 5702.8 0.0 904.4 28327.3 retirement functions27 0.0 188.8 5514.0 0.0 904.4 27422.928 0.0 188.8 5325.2 0.0 904.4 26518.529 0.0 188.8 5136.4 0.0 904.4 25614.1

30 0.0 188.8 4947.6 0.0 904.4 24709.731 0.9 188.8 4758.8 2.2 904.4 23805.332 0.8 187.9 4570.5 2.2 902.2 22902.033 0.8 187.1 4383.0 2.2 900.0 22000.934 1.6 186.3 4196.3 4.4 897.8 21101.9

35 1.5 184.7 4010.8 4.4 893.5 20206.336 1.4 183.2 3826.8 4.3 889.1 19315.037 1.8 181.8 3644.3 5.8 884.8 18428.138 1.8 180.0 3463.5 5.7 879.0 17546.239 1.7 178.2 3284.4 5.7 873.3 16670.1

40 2.0 176.5 3107.1 7.0 867.6 15799.641 2.0 174.4 2931.6 6.9 860.6 14935.542 2.3 172.5 2758.1 8.2 853.7 14078.343 2.2 170.2 2586.8 8.1 845.5 13228.744 2.4 168.0 2417.6 9.4 837.3 12387.3

45 2.4 165.6 2250.8 9.2 828.0 11554.746 2.6 163.2 2086.4 10.4 818.7 10731.347 2.9 160.7 1924.5 12.2 808.3 9917.848 3.1 157.7 1765.3 13.3 796.2 9115.549 3.3 154.6 1609.1 14.3 782.9 8326.0

50 3.9 151.3 1456.2 17.1 768.6 7550.351 4.2 147.4 1306.9 19.2 751.5 6790.252 5.0 143.2 1161.6 22.9 732.4 6048.353 5.4 138.2 1020.9 25.3 709.5 5327.454 6.3 132.8 885.4 29.8 684.1 4630.6

55 6.9 126.5 755.7 33.5 654.3 3961.356 7.7 119.6 632.6 38.0 620.8 3323.757 8.7 111.9 516.9 43.3 582.8 2721.958 10.0 103.2 409.4 50.3 539.5 2160.859 11.4 93.2 311.2 58.3 489.2 1646.5

60 13.6 81.8 223.7 70.2 430.8 1186.561 13.7 68.2 148.7 71.5 360.6 790.862 15.3 54.4 87.4 80.7 289.1 465.963 18.0 39.1 40.7 95.5 208.4 217.164 21.1 21.1 10.6 112.9 112.9 56.5

337

SUPPLEMENT

agex Cr

x Mrx R

r

xzCr

xzMr

xzR

r

x

20 0.0 1524.0 65671.0 0.0 8048.7 347094.1 TABLES FOR21 0.0 1524.0 64147.0 0.0 8048.7 339045.4 LUMP SUM22 0.0 1524.0 62623.1 0.0 8048.7 330996.8 BENEFITS ON23 0.0 1524.0 61099.1 0.0 8048.7 322948.1 RETIREMENT24 0.0 1524.0 59575.2 0.0 8048.7 314899.5

25 0.0 1524.0 58051.2 0.0 8048.7 306850.8 Age retirement26 0.0 1524.0 56527.2 0.0 8048.7 298802.2 functions27 0.0 1524.0 55003.3 0.0 8048.7 290753.528 0.0 1524.0 53479.3 0.0 8048.7 282704.929 0.0 1524.0 51955.4 0.0 8048.7 274656.2

30 0.0 1524.0 50431.4 0.0 8048.7 266607.631 0.0 1524.0 48907.4 0.0 8048.7 258558.932 0.0 1524.0 47383.5 0.0 8048.7 250510.333 0.0 1524.0 45859.5 0.0 8048.7 242461.634 0.0 1524.0 44335.6 0.0 8048.7 234413.0

35 0.0 1524.0 42811.6 0.0 8048.7 226364.336 0.0 1524.0 41287.6 0.0 8048.7 218315.737 0.0 1524.0 39763.7 0.0 8048.7 210267.038 0.0 1524.0 38239.7 0.0 8048.7 202218.439 0.0 1524.0 36715.8 0.0 8048.7 194169.7

40 0.0 1524.0 35191.8 0.0 8048.7 186121.141 0.0 1524.0 33667.8 0.0 8048.7 178072.442 0.0 1524.0 32143.9 0.0 8048.7 170023.843 0.0 1524.0 30619.9 0.0 8048.7 161975.144 0.0 1524.0 29096.0 0.0 8048.7 153926.5

45 0.0 1524.0 27572.0 0.0 8048.7 145877.846 0.0 1524.0 26048.0 0.0 8048.7 137829.247 0.0 1524.0 24524.1 0.0 8048.7 129780.548 0.0 1524.0 23000.1 0.0 8048.7 121731.849 0.0 1524.0 21476.2 0.0 8048.7 113683.2

50 0.0 1524.0 19952.2 0.0 8048.7 105634.551 0.0 1524.0 18428.2 0.0 8048.7 97585.952 0.0 1524.0 16904.3 0.0 8048.7 89537.253 0.0 1524.0 15380.3 0.0 8048.7 81488.654 0.0 1524.0 13856.3 0.0 8048.7 73439.9

55 0.0 1524.0 12332.4 0.0 8048.7 65391.356 0.0 1524.0 10808.4 0.0 8048.7 57342.657 0.0 1524.0 9284.5 0.0 8048.7 49294.058 0.0 1524.0 7760.5 0.0 8048.7 41245.359 0.0 1524.0 6236.5 0.0 8048.7 33196.7

60 378.5 1524.0 4712.6 1952.7 8048.7 25148.061 212.4 1145.4 3377.9 1107.4 6096.0 18075.762 112.3 933.0 2338.7 591.0 4988.6 12533.463 93.8 820.7 1461.8 498.0 4397.5 7840.464 77.7 726.9 688.1 415.5 3899.6 3691.8

65 649.2 649.2 3484.1 3484.1

338

SUPPLEMENT

Functions for valuing (at interest rate i per annum) refunds on death of contributions withcompound interest at interest rate j per annum

i = .04j = 0.03

x jMdx

jRd

xsjR

d

x

=∑64

y=xjC

dy =

∑64y=x(1 + j)−(y+.5).jM

d

y =∑64

y=x(1 + j)−(y+.5).sjMd

y

20 3108.18 31393.46 87740.9825 2878.57 23700.42 76568.1530 2700.59 17508.94 63315.7635 2524.59 12500.67 49478.9740 2338.12 8479.14 36324.4645 2120.85 5291.34 24305.7150 1804.15 2861.89 13941.8055 1333.41 1177.45 6005.0260 655.64 243.17 1286.3361 490.11 147.36 784.2962 350.45 79.12 423.2963 225.80 33.70 181.2064 108.85 8.09 43.67

Notes.(1) jCd

x = (1 + j)x+.5.(1 + i)−(x+.5).dx

(2) jMdx =

∑64−xt=0

jCdx+t

(3) jMd

x = jMdx − 1

2 · jCdx

(4) sjMd

x = sx · jMd

x

339

SUPPLEMENT

Functions for valuing (at interest rate i per annum) refunds on withdrawal of contributions withcompound interest at interest rate j per annum

i = .04j = 0.03

x jMwx

jRw

xsjR

w

x

=∑64

y=xjC

wy =

∑64y=x(1 + j)−(y+.5).jM

w

y =∑64

y=x(1 + j)−(y+.5).sjMw

y

20 43547.50 119414.16 207941.0825 19045.16 42216.97 98759.3230 8041.95 13393.17 38299.3935 3006.98 3281.50 10746.7536 2399.53 2334.90 7831.2237 1875.23 1608.25 5520.4938 1426.97 1063.28 3732.9739 1049.20 666.53 2391.9540 735.82 388.85 1425.6341 482.26 204.89 767.0342 284.70 92.43 353.1843 140.12 31.95 124.5844 46.19 6.20 24.67

Notes.(1) jCw

x = (1 + j)x+.5.(1 + i)−(x+.5).wx

(2) jMwx =

∑64−xt=0

jCwx+t

(3) jMw

x = jMwx − 1

2 · jCwx

(4) sjMw

x = sx · jMw

x

340

SUPPLEMENT

Functions for valuing (at interest rate i per annum) refunds on death of contributions withcompound interest at interest rate j per annum

i = .04j = 0

x jMdx

jRd

xsjR

d

x

=∑64

y=xjC

dy =

∑64y=x(1 + j)−(y+.5).jM

d

y =∑64

y=x(1 + j)−(y+.5).sjMd

y

20 792.85 18575.75 56093.0525 674.16 14932.38 50789.9430 594.98 11769.88 44005.8035 527.62 8965.83 36247.4340 466.07 6483.42 28117.4545 404.38 4303.49 19889.5950 326.91 2467.10 12048.6955 227.54 1071.57 5469.3360 104.08 232.42 1229.6261 76.40 142.17 756.7562 53.72 77.11 412.5763 34.07 33.21 178.5964 16.18 8.09 43.67

Notes.(1) jCd

x = (1 + j)x+.5.(1 + i)−(x+.5).dx

(2) jMdx =

∑64−xt=0

jCdx+t

(3) jMd

x = jMdx − 1

2 · jCdx

(4) sjMd

x = sx · jMd

x

341

SUPPLEMENT

Functions for valuing (at interest rate i per annum) refunds on withdrawal of contributions withcompound interest at interest rate j per annum

i = .04j = 0

x jMwx

jRw

xsjR

w

x

=∑64

y=xjC

wy =

∑64y=x(1 + j)−(y+.5).jM

w

y =∑64

y=x(1 + j)−(y+.5).sjMw

y

20 20579.08 103387.17 180950.8925 7847.29 37164.60 87256.7530 2916.96 12076.18 34605.9935 971.53 3051.10 10002.5436 758.81 2185.93 7337.8237 580.57 1516.24 5208.2038 432.61 1009.65 3546.5939 311.55 637.57 2288.9640 214.05 374.77 1374.4141 137.46 199.01 745.2042 79.53 90.52 345.9443 38.36 31.58 123.1444 12.40 6.20 24.67

Notes.(1) jCw

x = (1 + j)x+.5.(1 + i)−(x+.5).wx

(2) jMwx =

∑64−xt=0

jCwx+t

(3) jMw

x = jMwx − 1

2 · jCwx

(4) sjMw

x = sx · jMw

x

342

Documents

48044934 Life Assurance Mathematics W F Scott