MATH 3286

Mathematics of Finance

Alex Karassev

COURSE OUTLINE• Theory of Interest

1. Interest: the basic theory

2. Interest: basic applications

3. Annuities

4. Amortization and sinking funds

5. Bonds

• Life Insurance6. Preparation for life contingencies

7. Life tables and population problems

8. Life annuities

9. Life insurance

Chapter 1

INTEREST: THE BASIC THEORY

• Accumulation Function

• Simple Interest

• Compound Interest

• Present Value and Discount

• Nominal Rate of Interest

• Force of Interest

1.1 ACCUMULATION FUNCTION

• The amount of money initially invested is called the principal.

• The amount of money principal has grown to after the time period is called theaccumulated value and is denoted byA(t) – amount function. t ≥0 is measured in years (for the moment)

• Define Accumulation function a(t)=A(t)/A(0)• A(0)=principal• a(0)=1• A(t)=A(0)∙a(t)

Definitions

Natural assumptions on a(t)

• increasing

• (piece-wise) continuous

(0,1)t

a(t)

(0,1)

a(t)

t(0,1)

a(t)

t

Note: a(0)=1

Definition of Interest and

Rate of Interest

• Interest = Accumulated Value – Principal:Interest = A(t) – A(0)

• Effective rate of interest i (per year):

• Effective rate of interest in nth year in:

a(t)) A(0) A(t) (since

A(0)

A(0)A(1)

a(0)

a(0)a(1)1a(1)

i

1)-a(n

1)-a(na(n)

1)-A(n

1)-A(nA(n)

ni

Example (p. 5)

• Verify that a(0)=1

• Show that a(t) is increasing for all t ≥ 0

• Is a(t) continuous?

• Find the effective rate of interest i for a(t)

• Find in

a(t)=t2+t+1

Two Types of Interest

• Simple interest: – only principal earns interest– beneficial for short term (1 year)– easy to describe

• Compound interest: – interest earns interest– beneficial for long term– the most important type of accumulation

function

( ≡ Two Types of Accumulation Functions)

1.2 SIMPLE INTEREST

a(t)=1+it, t ≥0

(0,1)t

a(t) =1+it

1

1+i

•Amount function:A(t)=A(0) ∙a(t)=A(0)(1+it)

•Effective rate is i

•Effective rate in nth year:

)1(1

ni

iin

Example (p. 5)

Jack borrows 1000 from the bank on January 1, 1996 at a rate of 15% simple interest per year. How much does he owe on January 17, 1996?

a(t)=1+it

Solution

A(0)=1000

i=0.15

A(t)=A(0)(1+it)=1000(1+0.15t)

t=?

How to calculate t in practice?

• Exact simple interest number of days 365

• Ordinary simple interest (Banker’s Rule) number of days 360

t =

t=

Number of days: count the last day but not the first

Number of days (from Jan 1 to Jan 17) = 16

• Exact simple interest t=16/365 A(t)=1000(1+0.15 ∙ 16/365) = 1006.58

• Ordinary simple interest (Banker’s Rule) t=16/360 A(t)=1000(1+0.15 ∙ 16/360) = 1006.67

A(t)=1000(1+0.15t)

1.3 COMPOUND INTEREST

Interest earns interest

• After one year:a(1) = 1+i

• After two years:a(2) = 1+i+i(1+i) = (1+i)(1+i)=(1+i)2

• Similarly after n years:a(n) = (1+i)n

COMPOUND INTEREST Accumulation Function a(t)=(1+i)t

1+it

(0,1)

t

a(t)=(1+i)t

1

1+i

•Amount function:A(t)=A(0) ∙a(t)=A(0) (1+i)t

•Effective rate is i

•Moreover effective rate in nth year is i (effective rateis constant):

iii

iii

n

nn

n

11)1(

)1()1(1

1

How to evaluate a(t)?

• If t is not an integer, first find the value for the integral values immediately before and after

• Use linear interpolation• Thus, compound

interest is used for integral values of t and simple interest is used between integral values

1

t

a(t)=(1+i)t

1

1+i

2

(1+i)2

Example (p. 8)

Jack borrows 1000 at 15% compound interest.

a) How much does he owe after 2 years?

b) How much does he owe after 57 days, assuming compound interest between integral durations?

c) How much does he owe after 1 year and 57 days, under the same assumptions asin (b)?

d) How much does he owe after 1 year and 57 days, assuming linear interpolation between integral durations

e) In how many years will his principal have accumulated to 2000?

a(t)=(1+i)t

A(t)=A(0)(1+i)t

A(0)=1000, i=0.15

A(t)=1000(1+0.15)t

1.4 PRESENT VALUE AND DISCOUNT

The amount of money that will accumulate to the principal over t years is called the

present value t years in the past

PRINCIPAL ACCUMULATEDVALUE

PRESENTVALUE

-t t

Calculation of present value

• t=1, principal = 1

• Let v denote the present value• v (1+i)=1

• v=1/(1+i)

In general:

• t is arbitrary

• a(t)=(1+i)t

• [the present value of 1 (t years in the past)]∙ (1+i)t = 1

• the present value of 1 (t years in the past) = 1/ (1+i)t = vt

v=1/(1+i)

tt

t ii

v )1(1

1

(0,1)

t

a(t)=(1+i)t

a(t)=(1+i)t

gives the valueof one unit(at time 0)at any time t,past or future

If principal is not equal to 1…

present value = A(0) (1+i)t

PRINCIPALA (0)

ACCUMULATEDVALUE

A(0) (1+i)t

PRESENTVALUE

A(0) (1+i)t

t < 0 t = 0 t > 0

Example (p. 11)

The Kelly family buys a new house for 93,500 onMay 1, 1996.How much was this house worth on May 1, 1992 if real estate prices have risen at a compound rate for 8 % per year during that period?

Solution

a(t)=(1+i)t

• Find the present value ofA(0) = 93,500

• 996 - 1992 = 4 yearsin the past

• t = - 4, i = 0.08

• Present value = A(0) (1+i)t = 93,500 (1+0.8) -4 = 68,725.29

If simple interest is assumed…

• a (t) = 1 + it

• Let x denote the present value of one unit t years in the past

• x ∙a (t) = x (1 + it) =1

• x = 1 / (1 + it)

NOTE:

In the last formula, t > 0

Thus, unlikely to the case of compound interest, we cannot use the same formula for present value and accumulated value in the case of simple interest

1

a(t) =1+it

1 / (1 + it)

1

t

a(t) =1+it

1 / (1 - it)

Discount

• We invest 100

• After one year it accumulates to 112

• The interest 12 was added at the end of the term

Alternatively:

• Look at 112 as a basic amount

• Imagine that 12 were deducted from 112 at the beginning of the year

• Then 12 is amount of discount

Rate of Discount

Definition Effective rate of discount d

d = accumulated value after 1 year – principal accumulated value after 1 year

= A(1) – A(0) A(1)

= A(0) ∙a(1)– A(0) A(0) ∙a(1)

= a(1) – 1 a(1)

i = accumulated value after 1 year – principal principal = a(1) – 1

a(0)

Recall:

In nth year…

)(

)1()(

na

nanadn

Identities relating d to i and v

i

i

i

i

a

aad

11

1)1(

)1(

)0()1(

i

id

1Note: d < i

vii

ii

i

id

1

1

1

)1(

111

vd 1

d

di

1

Present and accumulated values in terms of d:

• Present value = principal * (1-d)t

• Accumulated value = principal * [1/(1-d)t]

ivd

1

11

If we consider positive and negative values of t then:

a(t) = (1 - d)-t

Examples (p. 13)

1. 1000 is to be accumulated by January 1, 1995 at a compound rate of discount of 9% per year.

a) Find the present value on January 1, 1992

b) Find the value of i corresponding to d

2. Jane deposits 1000 in a bank account on August 1, 1996. If the rate of compound interest is 7% per year, find the value of this deposit on August 1, 1994.

1.5 NOMINAL RATE OF INTEREST

Note: t is the number ofeffective interest periodsin any particular problem

Example (p. 13)A man borrows 1000 at an effective rate of interest of 2% per month. How much does he owe after 3 years?

More examples… (p. 14)

• You want to take out a mortgage on a house and discover that a rate of interest is 12% per year. However, you find out that this rate is “convertible semi-annually”. Is 12% the effective rate of interest per year?

• Credit card charges 18% per year convertible monthly. Is 18% the effective rate of interest per year?

In both examples the given ratesof interest (12% and 18%) werenominal rates of interest

…yet another example

• You have two credit card offers:

– 17% convertible semi-annually

– 16% convertible monthly

• Which is better?

Definition

• Suppose we have interest convertible m times per year

• The nominal rate of interest i(m) is

defined so that i(m) / m is an effective rate of interest in 1/m part of a year

Note:

If i is the effective rate of interest per year, it follows that

mm

m

ii

)(

11

Equivalently:

1]1[ /1)(

mm

im

i

In other words,i is the effective rate of interestconvertible annually which is equivalent to the effective rate of interest i(m) /m convertible mthly

Examples (p. 15)

1. Find the accumulated value of 1000 after three years at a rate of interest of 24 % per year convertible monthly

2. If i(6)=15% find the equivalent nominal rate of interest convertible semi-annually

Formula that relatesnominal rates of interest

nnmm

n

ii

m

i

)()(

111

Nominal rate of discount

• The nominal rate of discount d(m) is

defined so that d(m) / m is an effective rate of discount in 1/m part of a year

• Formula:

mm

m

dd

)(

11

Formula relating nominal rates of interest and discount

nn

n

dd

)(

11

1)1(1

1

11

di

id

mm

m

ii

)(

11

nnmm

n

d

m

i

)()(

11

Example

• Find the nominal rate of discount convertible semiannualy which is equivalent to a nominal rate of interest of 12% convertible monthly

nnmm

n

d

m

i

)()(

11

1.6 FORCE OF INTEREST

• What happens if the number m of periods is very large?

• One can consider mathematical model of interest which is convertible continuously

• Then the force of interest is the nominal rate of interest, convertible continuously

Definition

]1)1[( /1)( mm imiNominal rate of interest equivalent to i:

Let m approach infinity:]1)1[(limlim

/1)(

m

m

m

mimi

We define the

force of interest δequal to this limit: ]1)1[(limlim

/1)(

m

m

m

mimi

Formula

• Force of interest δ = ln (1+i)

• Therefore eδ = 1+i

• and a (t) = (1+i)t =eδt

• Practical use of δ: the previous formula gives good approximation to a(t) when m is very large

Example

• A loan of 3000 is taken out on June 23, 1997. If the force of interest is 14%, find each of the following:

– The value of the loanon June 23, 2002

– The value of i

– The value of i(12)

Remark

)(

)(

)1(

])1[(

)1()1ln()1(])1[(

ta

ta

i

i

iiii

t

t

ttt

The last formula shows that it is reasonable to define forceof interest for arbitrary accumulation function a(t)

Definition

)(

)(

ta

tat

Note:

1) in general case,force of interest depends on t

2) it does not depend on t ↔ a(t)= (1+i)t !

The force of interest corresponding to a(t):

Example (p. 19)

• Find in δt the case of simple interest

• Solution

it

i

it

it

ta

tat

11

)1(

)(

)(

How to find a(t)

if we are given by δt ?

Consider differential equation

in which a = a(t) is unknown function:

)(

)(

ta

tat

We have:

at

a

Since a(0) = 1 its solution is given by

t

rdr

eta 0)(

Applications

• Prove that if δt = δ is a constant then

a(t) = (1+i)t for some i

• Prove that for any amount function A(t) we have:

• Note: δt dt represents the effective rate of

interest over the infinitesimal “period of time” dt .

Hence A(t)δt dt is the amount of interest earned

in this period and the integral is the total amount

)0()()(0

AnAdttAn

t

Remarks

• Do we need to define the force of discount?

• It turns out that the force of discount coincides with the force of interest!(Exercise: PROVE IT)

• Moreover, we have the following inequalities:

• and formulas:

iiiddd mmmm )()1()1()(

midid mm

111 and 1

11)()(

Chapter 2INTEREST: BASIC APPLICATIONS

• Equation of Value

• Unknown Rate of Interest

• Time-Weighted Rate of Return

2.1 Equation of Value

• Four numbers:• principal A(0)• accumulated value A(t) = A(0) ∙ a(t)• period of investment t (determine

effective period to find t)• rate of interest i

• Time diagram

• Bring all entries of the diagram to the same point in time and writethe equation of value

Examples

1. Find the accumulated value of 500 after 173 months at a rate of compound interest of 14% convertible quarterly (p. 30)

2. Alice borrows 5000 from FF Company at a rate of interest 18% per year convertible semi-annually. Two years later she pays the company 3000. Three years after that she pays the company 2000. How much does she owe seven years after the loan is taken out? (p. 31)

3. Eric deposits 8000 on Jan 1, 1995 and 6000 on Jan 1, 1997 and withdraws 12000 on Jan 1, 2001. Find the amount in Eric’s account on Jan 1, 2004 if the rate of compound interest is 15% per year (p. 31)

More Examples…

Unknown time

John borrows 3000 from FFC. Two years later he borrows another 4000. Two years after that he borrows an additional 5000. At what point in time would a single loan of 12000 be equivalent if i = 0.18 ? (p. 32)

Unknown rate of interest

Find the rate of interest such that an amount of money will triple itself over 15 years (p. 32)

2.2 UNKNOWN RATE OF INTEREST

• We need to find the rate of interest i

• Set up equation of value and solve it for i

• Very often the resulting equation is a polynomial

equation in i of degree higher than 2

• In general, there is no formula for solutions of equation of degree ≥ 5 (and the formulas for degrees 3 or 4 are very complicated)

• Use approximations (numerical methods)

Examples

1. Joan deposits 2000 in her bank account on January 1, 1995, and then deposits 3000 on January 1, 1998. If there are no other deposits or withdrawals and the amount of money in the account on January 1, 2000 is 7100, find the effective rate of interest.

2. Obtain a more exact answer to the previous question

2.3 TIME-WEIGHTED RATE OF RETURN

• Let B0, B1, … , Bn-1, Bn denote balances in a fund such that precisely one deposit or withdrawal (denoted by Wt) is made immediately after Bt starting from t=1

• Let W1, … , Wn-1 denote the amounts of deposits (Wt > 0) or withdrawals (Wt > 0) and let W0 = 0

• Determine rate of interest earned in the time period between balances: 1

11

tt

tt WB

Bi

The time-weighted rate of return is defined by

i = (1+i1 ) (1+i2 ) … (1+in ) - 1

Example (p. 35)

On January 1, 1999, Graham’s stock portfolio is worth 500,000. On April 30, 1999, the value has increased to 525,000. At that point, Graham adds 50,000 worth of stock to this portfolio. Six month later, the value has dropped to 560,000, and Graham sells 100,000 worth of stock. On December 31, 1999, the portfolio is again worth 500,000. Find the time-weighted rate of return for Graham’s portfolio during 1999.

Chapter 3

ANNUITIES

• Basic Results

• Perpetuities

• Unknown Time andUnknown Rate of Interest

• Continuous Annuities

• Varying Annuities

3.2 Basic Results

• Definition: Annuity is a series of payments made at regular intervals

• Practical applications: loans, mortgages, periodic investments

• Basic model: consider an annuity under which payments of 1 are made at the end of each period for n equal periods

present valueof annuity

FormulasSeries of n payments

of 1 unit

0 1 32 n

1 1 1 1

…..

an|

accumulated valueof annuity

The present (accumulated) value of the series of payments is equalto the sum of present (accumulated) values of each payment

n

nvvvva 32

i

v

i

v

v

v

v

vv

vvvvvvvva

nnnn

nn

n

1

11

1

11

1

1

1

)1( 1232 i

va

n

n

1

n

n

nais )1(

i

i

i

ivi

ii

vias

nnnn

nn

n

nn

1)1()1()1(

)1(1

)1(

i

is

n

n

1)1(

sn|

Examples (p. 46)

1. (Loan) John borrows 1500 and wishes to pay it back with equal annual payments at the end of each of the next ten years. If i = 17% determine the size of annual payment

2. (Mortgage) Jacinta takes out 50,000 mortgage. If the mortgage rate is 13% convertible semiannually, what should her monthly payment be to pay off the mortgage in 20 years?

3. (Investments) Eileen deposits 2000 in a bank account every year for 11 years. If i = 6 % how much has she accumulated at the time of the last deposit?

One more example… (p. 47)

• Elroy takes out a loan of 5000 to buy a car. No payments are due for the first 8 months, but beginning with the end of 9th month, he must make 60 equal monthly payments. If i = 18%, find:

a) the amount of each payment

b) the amount of each payment if there is no payment-free period

Annuity-immediate and Annuity-dueAnnuity-immediate

(payments made at the end)

0 1 32 n

1 1 1 1

…..

sn|

Annuity-due (payments made at the beginning)

1 32 n

1 1 1 1

…..

än| .. sn|

n+1

)1( iaann

d

vvi

i

viaa

n

i

i

nn

nn

11)1(

1)1(

1

d

va

n

n

1d

is

n

n

1)1(

)1( issnn

an|

Example (p. 50)

• Henry takes out a loan of 1000 and repays it with 10 equal yearly payments, the first one due at the time of the loan. Find the amount of each payment ifi = 16%

3.3 Perpetuities

• Definition: Perpetuity is an annuity whose payments continue forever

• Practical applications: perpetual bonds

• Basic model: consider perpetuity under which payments of 1 are made forever

Formulas

0 1 32

1 1 1

…..

a∞|

present value ofperpetuity

1

32 ...k

kn vvvvva

ii

vaa

n

nnn

11limlim

ia

1

di

ii

iiaa

11)1(

1)1(

da

1

3.4 Unknown Time and Unknown Rate of Interest

1. A fund of 5000 is used to award scholarships of amount 500, one per year, at the end of each year for as long as possible. If i=9% find the number of scholarships which can be awarded, and the amount left in the fund one year after the last scholarship has been awarded

2. A trust fund is to be built by means of deposits of amount 5000 at the end of each year, with a terminal deposit, as small as possible, at the end of the final year. The purpose of this fund is to establish monthly payments of amount 300 into perpetuity, the first payment coming one month after the final deposit. If the rate of interest is 12% per year convertible quarterly, find the number of deposits required and the size of the final deposit

Examples – Unknown Time

3. At what effective yearly rate of interest is the present value of 300 paid at the end of every month, for the next 5 years, equal to 15,000?

• 1st method: linear interpolation

• 2nd method: successive approximations

Example – Unknown Rate of Interest

3.5 Continuous Annuities

Series of n∙m paymentsof 1/m

0 1/m 3/m2/mn = nm(1/m)

1/m 1/m 1/m 1/m

…..

Let effective period be 1/m part of the year and i(m)/mbe the effective rate of interest: (1+ i(m)/m)m = 1 + i

)(

|

m

na )(

|

m

ns

Formula:

)()(

|

1m

nm

n i

va

)()(

|

1)1(m

nm

n i

is

)()(

)(

)()(

|

)(

|

1)1()1(

1

)1(1

1

m

nn

m

n

nm

nmnmm

n

m

n

i

ii

i

v

ii

v

m

ias

0 1/m 3/m2/mn = nm(1/m)

1/m 1/m 1/m 1/m

…..

)(

|

m

na )(

|

m

ns

Let m approach infinity…

• Present valueof continuous annuity

• Accumulated valueof continuous annuity

)(

||lim m

nmnaa

)(

||lim m

nmnss

Formulas:

nn

nnnnt

mn

t

mt

m

tmmn

tm

m

nmn

v

i

v

i

v

i

v

v

vdtv

vmm

i

maa

1

1ln

1

1ln

1

11

ln

1

ln

1

1lim1

1limlim

0

1

/)(

1

)(

||

n

n

va

1|

1)1(|

n

n

is

3.6 Varying Annuities

• Arithmetic annuities– increasing

– decreasing

• Arithmetic increasing perpetuities

Arithmetic annuity

• Definition: Arithmetic annuity is a series of payments made at regular intervals such that– the first payment equals P– payments increase by Q every year

• Thus payments formarithmetic progression:

P, P+Q, P+2Q, …, P+(n-1)Q

Formulas

Series of n paymentsk-th payment = P+ (k-1)Q

k=1,2,…, n

0 1 32 n

P P+Q P+2Q P+(n-1)Q

…..

present valueA

accumulated valueS

i

nvaQPaA

n

n

n

i

nsQPsS

n

n

1) Increasing annuity with P = 1, Q = 1

Series of n paymentsk-th payment = k

k=1,2,…, n

0 1 32 n

1 2 3 n

…..

present value

(Ia)n|

accumulated value

(Is)n|

i

nvaQPaA

n

n

n

i

nsQPsS

n

n

Two Special CasesWe have:

i

nvaaIa

n

n

nn)(

i

nssIs

n

nn)(

P = 1Q = 1

i

nvaIa

n

n

n

)(i

nsIs

n

n

)(

2) Decreasing annuity with P = n, Q = -1

Series of n paymentsk-th payment = n – k + 1

k=1,2,…, n

0 1 32 n

n n-1 n-2 1

…..

present value

(Da)n|

accumulated value

(Ds)n|

i

nvaQPaA

n

n

n

i

nsQPsS

n

n

We have:

i

nvanaDa

n

n

nn)(

i

nsnsDs

n

nn)(

P = nQ = -1 i

anDa

n

n

)(

i

sinDs

n

n

n

)1()(

Increasing perpetuity

k-th payment = kcontinues forever

0 1 32

1 2 3

…..

present value

(Ia)∞|

i

nvdv

i

nvaIaIa

nn

n

n

n

nnn

1

limlim)(lim)(

2

11)(

iiIa

Examples (p. 62 - 65)

1. Find the value, one year before the first payment, of a series of payments 200, 500, 800,… if i = 8% and the payments continue for 19 years

2. Find the present value of an increasing perpetuity which pays 1 at the end of the 4th year, 2 at the end of the 8th year, 3 at the end of the 12th year, and so on, if i = 0.06

3. Find the value one year before the first payment of an annuity where payments start at 1, increase by annual amounts of 1 to a payment of n, and then decrease by annual amounts of 1 to a final payment of 1

4. Show both algebraically and verbally that(Da)n| = (n+1)an| - (Ia)n|

More examples: Geometric annuities

1. Geometric annuityAn annuity provides for 15 annual payments. The first is 200, each subsequent is 5% less than the one preceding it. Find the accumulated value of annuity at the time of the final payment if i = 9%

2. Inflation and real rate of interestIn settlement of a lawsuit, the provincial court ordered Frank to make 8 annual payments to Fred. The first payment of 10,000 is made immediately, and future payments are to increase according to an assumed rate of inflation of 0.04 per year. Find the present value of these payments assuming i = .07

Chapter 4

AMORTIZATION AND SINKING FUNDS

• Amortization

• Amortization Schedule

• Sinking Funds

• Yield Rates

4.1 Amortization

• Amortization method: repay a loan by means of instalment payments at periodic intervals

• This is an example of annuity

• We already know how to calculate the amount of each payment

• Our goal: find the outstanding principal

• Two methods to compute it:

– prospective

– retrospective

Two Methods

• Prospective method:outstanding principal at any point in time is equal to the present value at that date of all remaining payments

• Retrospective method:outstanding principal is equal to the original principal accumulated to that point in time minus the accumulated value of all payments previously made

• Note: of course, this two methods are equivalent. However, sometimes one is more convenient than the other

Examples (p. 75-76)

• (prospective) A loan is being paid off with payments of 500 at the end of each year for the next 10 years. If i = .14, find the outstanding principal, P, immediately after the payment at the end of year 6.

• (retrospective) A 7000 loan is being paid of with payments of 1000 at the end of each year for as long as necessary, plus a smaller payment one year after the last regular payment. If i = 0.11 and the first payment is due one year after the loan is taken out, find the outstanding principal, P, immediately after the 9th payment.

One more example… (p. 77)

• (Renegotiation) John takes out 50,000 mortgage at 12.5 % convertible semi-annually. He pays off the mortgage with monthly payments for 20 years, the first one is due one month after the mortgage is taken out. Immediately after his 60th payment, John renegotiates the loan. He agrees to repay the remainder of the mortgage by making an immediate cash payment of 10,000 and repaying the balance by means of monthly payments for ten years at 11% convertible semi-annually. Find the amount of his new payment.

4.2 Amortization Schedule

• Goal: divide each payment (of annuity) into two parts – interest and principal

• Amortization schedule – table, containing the following columns:

– payments

– interest part of a payment

– principal part of a payment

– outstanding principal

Example

Duration Payment Interest PrincipalRepaid

Outstanding Principal

0 5000.00

1 1387.05 600.00 787.05 4212.95

2 1387.05 505.55 881.50 3331.45

3 1387.05 399.77 987.28 2344.17

4 1387.05 281.30 1105.75 1238.42

5 1387.05 148.61 1238.44 0

5,000 at 12 % per year repaid by 5 annual payments

Amortization schedule:

Example

t - 1 t

PaymentX

Outstanding principal

P

• Interest earned during interval (t-1,t) is iP• Therefore interest portion of payment X is iP and principal portion is X - iP

A 1000 loan is repaid by annual payments of 150, plus a smaller final payment. If i = .11, and the first payment is made one year after the time of the loan, find the amount of principal and interest contained in the third payment

Recall: in practical problems, the outstanding principal P can be found by prospective or retrospective methods

General method

0 1 t2 n

1 1 1 1

…..

an|

present value = outst. principal at 0

…..

an-t|

outstanding principal at t

interest portion of (t+1)-st payment = i a n-t| = 1 – vn-t

t+1

1

principal portion of (t+1)-st payment = 1 – (1 – vn-t ) = vn-t

If each payment is X theninterest part of kth payment = X (1 – vn-k+1 )principal part of kth payment = X∙vn-k+1

Example (p. 79)

• A loan of 5000 at 12% per year is to be repaid by 5 annual payments, the first due one year hence. Construct an amortization schedule

General rules to obtain an amortization schedule

I. Take the entry from “Outs. Principal” of the previous row, multiply it by i, and enter the result in “Interest”

II. “Payment” – “Interest” = “Principal Repaid”

III. “Outs. Principal” of prev. row - “Principal Repaid” = “Outs. Principal”

IV. Continue

i = 12 %Duration Payment Interest Principal

RepaidOutstanding

Principal

0 5000.00

1 1387.05 600.00 787.05 4212.95

2 1387.05 505.55 881.50 3331.45

3 1387.05 399.77 987.28 2344.17

4 1387.05 281.30 1105.75 1238.42

5 1387.05 148.61 1238.44 0

Example (p. 80)

A 1000 loan is repaid by annual payments of 150, plus a smaller final payment. The first payment is made one year after the time of the loan and i = .11. Construct an amortization schedule

4.3 Sinking Funds

• Alternative way to repay a loan – sinking fund method:

– Pay interest as it comes due keeping the amount of the loan (i.e. outstanding principal) constant

– Repay the principal by a singlelump-sum payment at some point in the future

0 1 2 n

interestiL iL iL

…..

Loan L

lump-sum payment L

• Lump-sum payment L is accumulated by periodic deposits into a separate fund, called the sinking fund

• Sinking fund has rate of interest j usually different from (and usually smaller than) i

• If (and only if) j is greater than i then sinking fund method is better (for borrower) than amortization method

Examples (p. 82)• John borrows 15,000 at 17% effective annually. He agrees to

pay the interest annually, and to build up a sinking fund which will repay the loan at the end of 15 years. If the sinking fund accumulates at 12% annually, find– the annual interest payment– the annual sinking fund payment– his total annual outlay– the annual amortization payment which would pay off this

loan in 15 years• Helen wishes to borrow 7000. One lender offers a loan in

which the principal is to be repaid at the end of 5 years. In the mean-time, interest at 11% effective is to be paid on the loan, and the borrower is to accumulate her principal by means of annual payments into a sinking fund earning 8% effective. Another lender offers a loan for 5 years in which the amortization method will be used to repay the loan, with the first of the annual payments due in one year. Find the rate of interest, i, that this second lender can charge in order that Helen finds the two offers equally attractive.

4.4 Yield Rates• Investor:

– makes a number of payments at various points in time

– receives other payments in return

• There is (at least) one rate of interest for which the value of his expenditures will equal the value of the payments he received (at the same point in time)

• This rate is called the yield rate he earns on his investment

• In other words, yield rate is the rate of interest which makes two sequences of payments equivalent

• Note: to determine yield rate of a certain investor, we should consider only payments made directly to, or directly by, this investor

Examples

• Herman borrows 5000 from George and agrees to repay it in 10 equal annual instalments at 11%, with first payment due in one year. After 4 years, George sells his right to future payments to Ruth, at a price which will yield Ruth 12% effective– Find the price Ruth pays.– Find George’s overall yield rate.

• At what yield rate are payments of 500 now and 600 at the end of 2 years equivalent to a payment of 1098 at the end of 1 year?

Chapter 5

BONDS

• Price of a Bond

• Book Value

• AmortizationSchedule

• Other Topics

5.1 Price of a BondBonds:

• certificates issued by a corporation or government

• are sold to investors

• in return, the borrower (i.e. corporation or government) agrees:

– to pay interest at a specified rate (the coupon rate) until a specified date (the maturity date)

– and, at that time, to pay a fixed sum (the redemption value)

• Usually:

– the coupon rate is a nominal rate convertible semiannually and is applied to the face (or par) value stated on the front of the bond

– the face and redemption values are equal (not always)

• Thus we have:

– regular interest payments

– lump-sum payment at the end

Example of a bond

• Face amount = 500

• Redemption value = 500

• Redeemable in 10 years with semiannual coupons at rate 11%, compounded semiannually

• Then in return investor receives:

• 20 half-yearly payments of (.055)(500) = 27.50 interest

• a lump-sum payment of 500 at the end of the 10 years

Notations• F = the face value (or the par value)

• r = the coupon rate per interest period (we assume that the quoted rate will be a nominal rate 2r convertible semiannually)

• Note: the amount of each interest payment (coupon) is Fr

• C = the redemption value (often C = F, i.e. bond “redeemable at par”)

• i = the yield rate per interest period

• n = the number of interest periods until the redemption date (maturity date)

• P = the purchase price of a bond to obtain yield rate i

0 1 2 n

coupon (interest)

Fr Fr Fr

…..

purchase price P

redemption value C Note: time ismeasured inhalf-years

Equation to find yield rate i

nin

iCaFrP )1()(|

Note: often C = F

Examples (p. 94 – p. 96)• A bond of 500, redeemable at par after 5

years, pays interest at 13% per year convertible semiannually. Find the price to yield an investor– 8% effective per half-year– 16% effective per year

• Remarks– P < C since the yield rate is higher than the

coupon rate, i > r– therefore the investor is buying the bond at a

discount– otherwise (if i < r) we would have P > C and

then the investor would have to buy the bond at a premium of P - C

• A corporation decides to issue 15-years bonds, redeemable at par, with face amount of 1000 each. If interest payments are to be made at a rate of 10% convertible semiannually, and if George is happy with a yield of 8% convertible semiannually, what should he pay for one of these bonds?

• A 100 par-value 15-year bond with coupon rate 9% convertible semiannually is selling for 94. Find the yield rate.

5.2 Book Value• The book value of a bond at a time t is an analog of an

outstanding balance of a loan

• The book value Bt is the present value of all future payments

• At time t where the tth coupon has just been paid we have:

tntnt CvaFrB

|)(

• Remarks– Usually C = F

– In the last formula, an-t and v are computed using the yield rate i

– P = B0 < Bt < Bt+1< Bn = C or P = B0 > Bt > Bt+1 > Bn = C

Examples (p. 96 - 97)

1. Find the book value immediately after the payment of 14th coupon of a 10-year 1,000 par-value bond with semiannual coupons, if r=.05 and the yield rate is 12% convertible semiannually.

2. Let Bt and Bt+1 be the book values just after the tth and (t+1)th coupons are paid. Show that Bt+1 = Bt (1+i) – Fr

3. Find the book value in 1) exactly 2 months after the 14th coupon is paid.

How do we find the book value between coupon payment dates?

Assumesimple interest

at rate i per period between adjacent coupon payments

ExampleFind the book value exactly 2 months after the 14th coupon is paid of a 10-year 1,000 par-value bond with semiannual coupons, if r=.05 and the yield rate is 12% convertible semiannually.

Alternative approach

• Since Bt+1 = Bt (1+i) – Fr we can view Bt (1+i) as the book value just before next (i.e. (t+1)th) coupon is paid

• Book value calculated using simple interest between coupon dates is called the flat price of a bond

• Using linear interpolation between Bt+1 and Bt we obtain the market price (or the amortized value) of the bond

• Clearly market price ≤ flat price at any given moment

t t+1

Bt Bt+1

(1+i) Bt

flat p

rice

market price

Fr

Example (p. 98)

Find the market price exactly 2 months after the 14th coupon is paid of a 10-year 1,000 par-value bond with semiannual coupons, if r =.05 and the yield rate is 12% convertible semiannually.

5.3 Bond Amortization Schedule

• Goal: trace changes of the book value

• Bond amortization schedule – table, containing the following columns:

– time

– coupon

– interest

– principal adjustment

– book value

Example1000 par value two-year bond which pays interest at 8% convertible semiannually; yield rate is 6% convertible semiannually

Time Coupon Interest Principaladjustment

Book Value

0 1037.17

1 40 31.12 8.88 1028.29

2 40 30.85 9.15 1019.14

3 40 30.57 9.43 1009.71

4 40 30.29 9.71 1000.00

Algorithm

• Book value at time t is Bt

• Amount of coupon at time t+1 is Fr

• The amount of interest contained in this coupon is iBt

• Fr – iBt represents the change in the book value between these dates

Time Coupon Interest Principaladjustment

Book Value

0 1037.17

1 40 31.12 8.88 1028.29

2 40 30.85 9.15 1019.14

3 40 30.57 9.43 1009.71

4 40 30.29 9.71 1000.00

Example (p. 99)

• Consider 1000 par-value 10-year bond with semiannual coupons, r = .05 and the yield rate i = 0.06 effective semiannually. Find the amount of interest and change in book value contained in the 15th coupon of the bond.

Example (p. 99)

• Construct a bond amortization schedule for a 1000 par-value two-year bond which pays interest at 8% convertible semiannually, and has a yield rate of 6% convertible semiannually

5.4 Other Topics

• Different frequency of coupon payments

• Increasing or decreasing coupon payments

• Different yield rates

• Callable bonds

Examples (p. 101 – p. 102)

• (Different frequency) Find the price of a 1000 par-value 10-year bond which has quarterly 2% coupons and is bought to yield 9% per year convertible semiannually

• (Increasing coupon payments) Find the price of a 1000 par-value 10-year bond which has semiannual coupons of 10 the first half-year, 20 the second half-year,…, 200 the last half-year, bought to yield 9% effective per year

• (Different yield rates) Find the price of a 1000 par-value 10-year bond with coupons at 11% convertible semiannually, and for which the yield rate is 5% per half-year for the first 5 years and 6% per half-year for the last 5 years

Callable bonds

• A borrower (i.e. corporation, government etc.) has the right to redeem the bond at any of several time points

• The earliest possible date is the call date and the latest is the usual maturity date

• Once the bond is redeemed, no more coupons will be paid

Purchase Date Call Date Maturity Date

possible redemption

Examples (p. 103 – p. 105)

• Consider a 1000 par-value 10-year bond with semiannual 5% coupons. Assume this bond can be redeemed at par at any of the last 4 coupon dates. Find the price which will guarantee an investor a yield rate of

– 6% per half-year

– 4% per half-year

• Consider a 1000 par-value 10-year bond with semiannual 5% coupons. This bond can be redeemed for 1100 at the time of the 18th coupon, for 1050 at the time of the 19th coupon, or for 1000 at the time of 20th coupon. What price should an investor pay to be guaranteed a yield rate of

– 6% per half-year

– 4% per half-year

Chapter 6

PREPARATION FOR LIFE CONTINGENCIES

• Introduction

• ContingentPayments

6.1 Introduction

• Ideal situation: all payments are made

• Real-life situations:

– failure to make a payment

– default on a loan

– bad credit ratings

– life contingencies and life insurance

• Contingent payments (we need to combine the theory of interest and elementary probability)

6.3 Contingent Payments

• Assume that for each payment of the loan (annuity, bond etc.) there is a probability that this payment is made

• Finding the present value of such sequence of payments we need to take into account these probabilities

• Example Henry borrows 1000 from Amicable Trust and agrees to repay the loan in one year. If payment were certain, the company would charge 13% interest. From prior experience, however, it is determined that there is a 5% chance that Henry will not repay any money at all. What should Amicable Trust ask Henry to repay?

Examples (p. 119 – p. 122)

• (Loan) The All-Mighty Bank lends 50,000,000 to a small Central American country, with the loan to be repaid in one year. It is felt that there is a 20% chance that a revolution will occur and that no money will be repaid, a 30% chance that due to inflation only half the loan will be repaid, and a 50% chance that the entire loan will be repaid. If payments were certain, the bank would charge 9%. What rate of interest should the bank charge?

• (Payments contingent upon survival) Mrs. Rogers receives 1000 at the end of each year as long as she is alive. The probability is 80% she will survive one year, 50% she will survive 2 years, 30% she will survive 3 years, and negligible that she will survive longer than 3 years. If the yield rate is 15%, what should Mrs. Rogers place on these payments now?

• (Life insurance) An insurance company issues a policy which pays 50,000 at the end of the year of death, if death should occur during the next two years. The probability that a 25-year-old will live for one year is .99936, and the probability he will live for two years is .99858. What should the company charge such a policyholder to earn 11% on its investments?

• (Annuity) Alphonse wishes to borrow some money form Friendly Trust. He promises to repay 500 at the end of each year for the next 10 years, but there is a 5% chance of default in any year. Assume that once default occurs, no further payments will be received. How much can Friendly Trust lend Alphonse if it wishes to earn 9% on its investments?

• Redo the last example, without the restriction that once default occurs, no further payments will be received

• (Bond) A 20-year 1000 face value bond has coupons at 14% convertible semiannually and is redeemable at par. Assume a 2% chance that, in any given half-year, the coupon is not issued, and that once default occurs, no further payments are made. Assume as well that a bond can be redeemed only if all coupons have been paid. Find the purchase price to yield on investor 16% convertible semiannulllay.

Chapter 7

LIFE TABLES AND POPULATION PROBLEMS

• Introduction

• Life Tables

• The StationaryPopulation

• Expectation of Life

7.1 Introduction

• How do we find probabilities?

• Data obtained from practice

• Data required to findprobabilities of survivingto certain ages (or, equivalently, of dying before certain ages) are contained in life tables

7.2 Life Tables

• lx – number of lives survived to age x

• Thus l0 = 1,000,000 is a starting population

• Survival function S(x) = lx / l0 is the probability of surviving to age x

• dx = lx – lx+1 – number of lives who died between (x, x+1)

• qx = dx /lx – probability that x – year-old will not survive to age x + 1

Age lx dx 1000 qx

0 1,000,000 1580 1.58

1 998,420 680 68

2 997,740 485 .49

3 997,255 435 .44

Examples (p. 129 – p. 130)

• Find– the probability that a newborn will live to age 3– the probability that a newborn will die between

age 1 and age 3

Age lx dx 1000 qx

0 1,000,000 1580 1.58

1 998,420 680 68

2 997,740 485 .49

3 997,255 435 .44

• Find an expression for each of the following:

– the probability that an 18-year-old lives to age 65

– the probability that a 25-year-old dies between ages 40 and 45

– the probability that a 25-year-old does not die between ages 40 and 45

– the probability that a 30-year-old dies before age 60

• There are four persons, now aged 40, 50, 60 and 70. Find an expression for the probability that both the 40-year-old and the 50-year-old will die within the five-year period starting ten years from now, but neither the 60-year-old nor the 70-year-old will die during that five-year period

Note:

• If we are already given by probabilities qx and

starting population l0 we can construct the whole

life table step-by-step since dx = qx lx and lx+1 = lx -

dx

Example

• Given the following probabilities of deaths

q0 = .40, q1 = .20, q2 = .30, q3 = .70, q4 = 1

and starting with l0 = 100 construct a life table

More notations…• qx = dx / lx – probability that x – year-old will not survive

to age x + 1

• px = 1- qx – probability that x – year-old will survive to age x + 1

• Note: qx = (lx – lx+1) / lx and px = lx+1 / lx

•nqx – probability that x – year-old will not survive to age x + n

•npx = 1- nqx – probability that x – year-old will survive to age x + n

• Note: nqx = (lx – lx+n) / lx and n px = lx+n / lx

• Formulas:lx – lx+n = dx + dx+1 + …+ dx+n

n+mpx = mpx ∙ npx+m

What is mPx when m is not integer?

•tPx where 0 < t <1

• Assuming that deaths are distributed uniformly during any given year we can use linear interpolation to find tpx :

x

xx

x

xxx

x

xxxt l

tllt

l

lltl

l

tdlp 11 )1()(

Examples (p. 132 – p. 133)

• 30% of those who die between ages 25 and 75 die before age 50. The probability of a person aged 25 dying before age 50 is 20%. Find 25P50

• Using the following life table and assuming a uniform distribution of deaths over each year, find:– 4/3P1

– The probability that a newborn will survive the first year but die in the first two months thereafter

Age lx dx 1000 qx

0 1,000,000 1580 1.58

1 998,420 680 68

2 997,740 485 .49

3 997,255 435 .44

7.4 The Stationary Population• Assume that in every given year (or, more precisely, in any

given 12-months period) the number of births and deaths is the same and is equal to l0

• Then after a period of time the total population will remain stationary and the age distribution will remain constant

• px and qx are defined as before

• lx denote the number of people who reach heir xth birthday during any given year

• dx = qx lx represent the number of people who die before reaching age x+1

• Also, dx represent the number of people who die during any given year between ages x and x + 1

• Similarly lx – lx + n represent the number of people who die during any given year between ages x and x + n

Number of people aged x

• Let Lx denote the number of people aged x (last birthday) at any given moment

• Note: Lx ≠ lx

• Assuming uniform distribution of deaths we obtain:

• More precisely:

)(2

1)(

2

1

2

111 xxxxxxxx llllldlL

1

0

dtlL txx

Number of people aged x and over

• Let Tx denote the number of peopleaged x and over at any given moment

• Then

• Assuming uniform distribution we obtain:

• More precisely:

101 2

1)(

2

1

iixx

iixixx llllT

0

dtlT txx

0iixx LT

Example (p. 139)• An organization has a constant total membership. Each

year 500 new members join at exact age 20.20% leave after 10 years, 10% of those remaining leave after 20 years, and the rest retire at age 65. Express each of the following in terms of life table functions:– The number who leave at age 40 each year– The size of the membership– The number of retired people alive at any given time– The number of members who die each year

7.5 Expectation of Life

• What is the average future life time ex of a person aged x now?

• The answer is given by expected value (or mathematical expectation) and is called the curtate expectation:

132

4332211

1 )(3)(2)(1)(

txtxxx

xxxxxxt

xtxtx

pppp

ppppppppte

Complete Expectation

1t

xtx pe(1)

dtpxtxe

0

Complete expectation: (4)

Approximation:

2

1

xxee

(2)

0

dtlT txx

x

txxt l

lp (3)

(2)&(3)&(4) imply:

x

xx l

Te

12

1

iixxx llT

Since

we get

Remarks

x

xx l

Te

xxx elT

Thus Tx can be viewed as the total number of years of future life of those who form group lx

Note: this interpretation makes sense in any population (stationary or not)

Average age at death

• Average age at death of a person currently aged x is given by

• Letting x = 0 we obtain the average age at death for all death among l0 individuals

x

xx l

Txx e

0

00 l

Te

Examples (p. 142 – 143)

• If tp35 = (.98)t for all t, find e35 and eo35

without approximation. Compare the value for eo

35 with its approximate value

• Interpret verbally the expressionTx-Tx+n – nlx+n

• Fin the average age at death of those who die between age x and age x+n

Chapter 8

LIFE ANNUITIES

• Basic Concepts

• Commutation Functions

• Annuities Payable mthly

• Varying Life Annuities

• Annual Premiums and Premium Reserves

8.1 Basic Concepts

• We know how to compute present value of contingent payments

• Life tables are sources of probabilities of surviving

• We can use data from life tables to compute present values of payments which are contingent on either survival or death

Example (pure endowment), p. 155

• Yuanlin is 38 years old. If he reaches age 65, he will receive a single payment of 50,000. If i = .12, find an expression for the value of this payment to Yuanlin today. Use the following entries in the life table: l38 = 8327, l65 = 5411

Pure Endowment

• Pure endowment: 1 is paid t years from now to an individual currently aged x if the individual survives

• Probability of surviving is t px

• Therefore the present value of this payment is the net single premium for the pure endowment, which is:

t Ex = (t px ) (1 + t) – t = v t t px

Example (life annuity), p. 156

• Aretha is 27 years old. Beginning one year from today, she will receive 10,000 annually for as long as she is alive. Find an expression for the present value of this series of payments assuming i = .09

• Find numerical value of this expression ifpx = .95 for each x

Life annuity

Series of payments of 1 unitas long as individual is alive

x x + 1 x + 2 x + n

1 1 1

…..

present value(net single premium)

of annuity ax

age

probability px 2px npx

…..

xtt

txn

nxxxx pvpvpvpvvpa

1

33

22

Temporary life annuity

Series of n payments of 1 unit(contingent on survival)

x x + 1 x + 2 x + n

1 1 1

…..

present value

ax:n|

age

probability px 2px npx

xt

n

t

txn

nxxxnx

pvpvpvpvvpa

1

33

22

|:

last payment

n - years deferred life annuity

Series of payments of 1 unit as long as individual is alivein which the first payment is at x + n + 1

x x + 1 x + 2 x + n +1

1

…

present value

n|ax

age

probability n+1px

11

33

22

11|

nsxs

s

txtn

tn

xtntn

xnn

xnn

xnxn

pvpv

pvpvpvpva n

first payment

x + n + 2

1

n+2px

x + n …

Note: |:|

nxxxn aaa

Life annuities-due

x x + 1 x + 2 x + n

1 1 1

…..

äx

px 2px npx

…

x x + 1 x + 2 x + n

1 1

…..

äx:n|

px 2px

x x + 1 x + 2 x + n +1

1

…

n|äx

n+1px

x + n + 2

1

n+2px

x + n …

1

1

xtt

txx pvaa

1

11

xt

n

t

tnxnx

pvaa

1

1|1:|:

11

xnxn aa || 11

x + n-1

1

n-1px

npx

Note

|1:|:1

nxnxaa

but

|:|:)1(

nxnxaia

|:1|:1 nxxnxavpa

8.2 Commutation Functions

Recall: present value of a pure endowment of 1 to be paid n years hence to a life currently aged x

Denote Dx = vxlx

Then nEx = Dx+n / Dx

xx

nxnx

x

nxnxn

nxn lv

lv

l

lvpvE

Life annuity and commutation functions

11 t

xtxtt

tx Epva Since nEx = Dx+n / Dx

we have

3211

1xxx

t xx

txx DDD

DD

Da

Define commutation

function Nx as follows:

00 ttx

tx

ttxx lvDN

Then:x

xx D

Na 1

Identities for other types of life annuities

x

nxxn

t x

txnx D

NN

D

Da 11

1|:

x

nxxn D

Na 1|

x

nxxnx D

NNa

|:

temporary life annuity

n-years delayed l. a.

temporary l. a.-due

Accumulated values of life annuities

x

nxxnx D

NNa 11

|:

temporary life annuity

similarly for temporary life annuity-due:

x

nxxn D

DE

|:|: nxxnnxsEa

since and

|:|: nxxnnxsEa

we have

nx

nxxnx D

NNs

11|:

and

nx

nxxnx D

NNs

|:

Examples (p. 162 – p. 164)

• (life annuities and commutation functions) Marvin, aged 38, purchases a life annuity of 1000 per year. From tables, we learn that N38 = 5600 and N39 = 5350. Find the net single premium Marvin should pay for this annuity

– if the first 1000 payment occurs in one year

– if the first 1000 payment occurs now

• Stay verbally the meaning of (N35 – N55) / D20

• (unknown rate of interest) Given Nx = 5000, Nx+1=4900,

Nx+2 = 4810 and qx = .005, find i

Select group

• Select group of population is a group with the probability of survival different from the probability given in the standard life tables

• Such groups can have higher than average probability of survival (e.g. due to excellent health) or, conversely, higher mortality rate (e.g. due to dangerous working conditions)

Notations• Suppose that a person aged x is

in the first year of being in the select group

• Then p[x] denotes the probability of survival for 1 year and q[x] = 1 – p[x] denotes the probability of dying during 1 year for such a person

• If the person stays within this group for subsequent years, the corresponding probabilities of survival for 1 more year are denoted by p[x]+1, p[x]+2, and so on

• Similar notations are used for life annuities:a[x] denotes the net single premium for a life annuity of 1 (with the first payment in one year) to a person aged x in his first year as a member of the select group

• A life table which involves a select group is called a select-and-ultimate table

Examples (p. 165 – p. 166)

• (select group) Margaret, aged 65, purchases a life annuity which will provide annual payments of 1000 commencing at age 66. For the next year only, Margaret’s probability of survival is higher than that predicted by the life tables and, in fact, is equal to p65 + .05, where p65 is taken from the standard life table. Based on that standard life table, we have the values D65 = 300, D66 = 260 and N67 = 1450. If i = .09, find the net single premium for this annuity

• (select-and-ultimate table) A select-and-ultimate table has a select period of two years. Select probabilities are related to ultimate probabilities by the relationships p[x] = (11/10) px and p[x]+1 = (21/20) px+1. An ultimate table shows D60 = 1900, D61 = 1500, and ä 60:20| = 11, when i = .08. Find the select temporary life annuity ä[60]:20|

• The following values are based on a unisex life table: N38 = 5600, N39 = 5350, N40 = 5105, N41 = 4865,N42 = 4625.It is assumed that this table needs to be set forward one year for males and set back two years for females. If Michael and Brenda are both age 40, find the net single premium that each should pay for a life annuity of 1000 per year, if the first payment occurs immediately.

8.3 Annuities Payable mthly

• Payments every mth part of the year

• Problem: commutation functions reflect annual probabilities of survival

• First, we obtain an approximate formula for present value

• Assume for a moment that the values Dy are also given for non-integer values of y

Usual life annuity

x x + 1 x + 2 x + n

1 1 1

…..

ax

age …..

Annuity payable every 1/m part of the year

x x + 1/m

x + 2/m

x + (m-1)/m

1 1 1

…..

ax

age …..x + n

1