Acteduk Ct6 Hand Qho v04

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CT6 Online Classroom Questions Handout Page 1

The Actuarial Education Company IFE: 2014 Examinations

Subject CT6

Statistical Methods

CT6 Online Classroom Questions Handout

Decision theory

1 ActEd

For each of the zero-sum game payoff matrices below determine the minimax solution:

(i)

A

I II

B1 1- 52 0 6

[2]

(ii)

A

I II III

B

1 1 5 2

2 0 3 1

3 2 4 5

[2]

(iii)

A

I II

B1 1 2

2 3 4- [4]

[Total 8]


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Page 2 CT6 Online Classroom Questions Handout

IFE: 2014 Examinations The Actuarial Education Company

2 Subject 106, April 2003, Question 4

The loss function under a decision problem is given by:

1Q 2Q 3Q

1D 23 34 16

2D 30 19 18

3D 23 27 20

4D 32 19 19

(i) State which decision can be discounted immediately and explain why. [2]

(ii) Determine the minimax solution to the problem. [2]

(iii) Given the following distribution 1( ) 0.25PQ = , 2( ) 0.15PQ = , 3( ) 0.60PQ = ,determine the Bayes criterion solution to the problem. [2]

[Total 6]

3 ActEd

A statistician is trying to decide whether a coin is fair or biased towards tails. The

decision is to be made on whether a tail is obtained on onetoss of the coin.

(i) List the 4 possible decision functions. [1]

If the coin is biased, the probability of obtaining a tail is 0.75. The loss function is:

nature

loss biased fair

statisticianbiased 1- 2

fair 1 0

(ii) (a) Determine the risk function for each of the 4 decision functions.

(b) Which decision function is inadmissible? [5]

(iii) (a) Determine the solution using the minimax criterion.

(b) Determine the solution using Bayes criterion, if (fair coin) 0.8P = . [3]

[Total 9]


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4 Subject CT6, April 2007, Question 4

A casino operator moving into a country for the first time must apply to the casino

regulator for a licence. There are three types of licence to choose from slots, dice and

cards each with different running costs. The casino operator has to pay a fixed

amount annually (1,300,000) to the regulator, plus a variable annual licence cost.

The variable licence cost and expected revenue per customer for each type of game are

as follows:

Variable

Licence cost

Expected

Revenue

per customer

Slots 250,000 60

Dice 550,000 120

Cards 1,150,000 160

The casino operator is uncertain about the number of customers and decides to prepare a

profit forecast based on cautious, best estimate and optimistic numbers of customers.

The figures are 14,000, 20,000 and 23,000 respectively.

(i) Determine the annual profits under each possible combination. [2]

(ii) Determine the minimax solution for optimising the profits. [2]

(iii) Determine the Bayes criterion solution based on the annual profit given the

probability distribution:

(cautious) 0.2=P (best estimate) 0.7=P

and (optimistic) 0.1=P . [2][Total 6]


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Bayesian statistics

1 Subject C2, September 1995, Question 3 (adapted)

The number of claims arising in a year from a group of policies follows a Poisson

distribution with mean m. The prior distribution for m is gamma with parameters

10a= and 2l= .

Given that 8 claims arose in the last year, determine the posterior distribution for . [4]

2 Subject 106, September 2001, Question 8

The number of claims per month are independent Poisson random variables with meanl, and the prior distribution for lis exponential with mean 0.2.

(i) Determine the posterior distribution for lgiven the observed values 1, , nx x

of the number of claims in each of n months. [2]

(ii) Determine the Bayesian estimator of l

(a) under quadratic loss

(b) under all-or-nothing loss [3]

(iii) If 5n= and5

1

1ix = , calculate to 2 significant figures the Bayesian estimate of

lunder absolute error loss. [4]

[Total 9]

3 Subject C2, April 1995, Question 3 (adapted)

The number of claims registered per week has a Poisson distribution for which the

mean, l, is either 1 or 2. The prior distribution for lis given by:

( 1) 0.4P l= = ( 2) 0.6P l= =

Given that three claims are registered in a particular week, calculate the Bayesian

estimate of lunder squared error loss, and under zero-one loss. [4]


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4 Subject C2, September 1999, Question 12 (corrected)

The number of claims arising each month from a general insurance portfolio has a

Poisson distribution, with unknown Poisson parameter l. Claims are monitored over a

period of 50 months, and an average of 210 claims per month are observed.

(i) It is suggested, based on knowledge gained from similar portfolios, that a

suitable prior distribution for l has mean 250 and variance 45. Using the

conjugate prior distribution, determine the posterior distribution of l and the

Bayesian estimate of lunder quadratic loss. [6]

(ii) An alternative suggestion for estimating l is to use the number of claims

occurring on a single day, which is assumed to have a Poisson distribution with

mean / 30l . It is suggested that the following prior distribution for l shouldbe used:

230 with probability 0.2



l

=

If 7 claims were recorded on the most recent day for which data are available,

determine the posterior distribution for l, and hence find the Bayesian estimate

of lunder quadratic loss. [6]

(iii) Discuss briefly the differences between the estimators in (i) and (ii), indicating

which you think is preferable. [2]

[Total 14]


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Estimation

1 Subject 106, April 2002, Question 10 (part)

The most recent ten claims under a particular class of insurance policy were:

35 111 201 309 442 617 843 1,330 2,368 4,685

(i) Assuming that the claims came from a lognormal distribution with parameters

and s , derive the formula for the maximum likelihood estimates of these

parameters and estimate the parameters based on the observed data. [5]

(ii) Assuming that the claims come from a Pareto distribution with parameters a

and l, use the method of moments to estimate these parameters. [3]

(iii) Assuming that the claims come from a Weibull distribution with parameters c

and g , use the method of percentiles (based on the 25th and 75th percentiles) to

estimate these parameters. [5]

[Total 13]

2 ActEd

In a portfolio of motor policies, the annual number of claims for a single policy has a

Poisson distribution with parameter l. The parameter lis not the same for all policies

in the portfolio, but is modelled as a random variable with density:

5( ) 25 , 0g e ll l l-= >

(i) Show that the probability of n claims in a year, on a policy picked at random

from the portfolio, is given by:

2( 2) 1 5( ) , 0,1,2,3,

( 1) (2) 6 6

nnP N n n

n

G + = = = G + G [3]

(ii) Hence, obtain the mean and variance of the number of claims per annum. [2]

[Total 5]


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Reinsurance

1 ActEd

Claims from a certain portfolio have a Pareto distribution with parameters = 3 and

= 500 . A retention limit of 400 is in force, with the excess of this amount on any

claim being paid by a reinsurer.

(i) What proportion of claims involve the reinsurer? [2]

(ii) What is the mean amount paid by the reinsurer on all claims? [4]

(iii) What is the mean amount paid by the reinsurer on all claims in which it is

involved? [2][Total 8]

2 ActEd

(i) An insurer effects excess of loss reinsurance with retention limit . Obtain the

distribution of the claim amounts paid by the reinsurer given that they are

referred to the reinsurer if claims amounts have an exponential distribution with

parameter l. [2]

(ii) An insurer introduces a policy excess (deductible) of E. Obtain the distribution

of the claim amounts now paid by the insurer if the total claim amounts have a

Pareto distribution with parameters a and l. [2]

[Total 4]


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3 Subject C2, April 1999, Question 12 (modified)

(i) The random variable X has the lognormal distribution with density function

( )f x and parameters and s . Show that for any positive integer k:

[ ]2 2( ) ( ) ( )

Uk k k

k k

L

x f x dx e U Lm s+= F - F

whereln

k

LL ks

s

-= - and

lnk

UU k

ms

s

-= - . [6]

(ii) The amount, X,of a claim, in thousands of pounds, from an insurance portfolio

has the lognormal distribution with mean 12.2 and standard deviation 16.

Consider an excess of loss reinsurance policy with a retention of 28,000 so thatthe claim paid by the insurer (000) is given by XI, where:

28

28 28I

X XX

X

= >

(a) Determine the probability that a claim involves the reinsurer.

(b) Calculate the mean and variance of the claims paid by the insurer.

(c) Given that a claim is referred to the reinsurer, what is the conditional

expected value paid by the reinsurer? [14]

[Total 20]


The loss severity distribution for a portfolio of household insurance policies is assumed

to be Pareto with parameters 3.5a= , 1,000l= .

Next year, losses are expected to increase by 5%, and the insurer has decided to

introduce a policyholder excess of 100.

Calculate the probability that a loss next year is borne entirely by the policyholder. [3]


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A specialist motor insurer writes policies with individual excesses of 500 per claim.

The insurer has taken out a reinsurance policy whereby the insurer pays out a maximum

of 4,500 in respect of each individual claim, the rest being paid by the reinsurer. Theindividual claims, gross of reinsurance and the excess, are believed to follow an

exponential distribution with parameter l.

Over the last year, the insurer has gathered the following data:

There were 5 claims which were not processed because the loss was less than the

excess.

There were 11 claims where the insurer paid out 4,500 and the reinsurer the remainder.

There were 26 other claims in respect of which the insurer paid out a total of 76,457.

Derive the loglikelihood function of l. [6]


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Credibility theory

1 Subject 106, September 2001, Question 7 (adapted)

Claim amounts under a particular insurance portfolio are believed to follow a Normal

distribution with variance21s and an unknown mean q . The insurer observes a sample

of n policies that have given rise to a claim for which the mean amount is a . The prior

distribution of q is assumed to be Normal with mean mand variance 22s .

(i) (a) State the posterior mean for q .

(b) Show that the posterior mean of q can be expressed as a weighted

average of the prior mean and the sample mean, and derive an expressionfor the weight placed on the sample mean. [3]

(ii) An insurer believes that individual claim amounts follow a Normal distribution

with an unknown mean and a standard deviation of 210. Prior information

suggests that the mean should be assumed to follow a Normal distribution with

mean = 155 and standard deviation = 84. Over the previous year, the insurer

collected data from 15 claims where the total amount paid out was 2,456.

Calculate the credibility factor placed on the sample mean and hence, given that

the insurer wishes to add a 30% loading for profit and expenses, calculate thepremium the insurer should charge. [2]

(iii) Determine the limiting value of the credibility factor as each of n ,21s and

22s

increases and briefly describe how it is affected by the insurers assumptions and

the observed data. [3]

[Total 8]


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2 Subject 106, September 2003, Question 7 (adapted)

In a portfolio of property insurance policies, let q denote the proportion of policies on

which claims are made in the year. The value of q is unknown and is assumed to have

a beta prior distribution with parameters a and b .

Let x be the number of policies on which claims are made from a sample of n policies.

(i) Show that the posterior mean of q can be expressed as:

.( ) (1 ).Z x n Z m+ -

where is the mean of the beta prior distribution and express Zas a function

of ,a b and n . [3]

A claims analyst estimates that the mean and standard deviation of the prior distribution

of q are 0.20 and 0.25 respectively.

(ii) Determine the values of the parameters, a and b . [3]

From a random sample of 50 policies, a claim is made on 24% of them during the year.

(iii) (a) Calculate the value of Zin this case and explain what it represents.

(b) Without performing any further calculations, explain how you would

expect the value of Zto change if:

(1) The analyst now believes the standard deviation, s , of the prior

distribution to be 0.50.

(2) The sample size, n , was 400.

(c) State the limiting value of Zas s and n increase and explain what thismeans. [4]

[Total 10]


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EBCT

1 Subject 106, April 2004, Question 7 (part)

The total amount claimed for a particular risk in a portfolio is observed for each of 5

consecutive years. An insurer decides to use Empirical Bayes Credibility, Model 1,

where the credibility premium combines the mean for the particular risk with the

estimated value of ( ( ))E mq . Data from 3 risks in this portfolio over 5 years are

available. Let ijX be the claim for Risk i in year j . The table shows various

summary statistics for the observed data.

iX

52

1

( )ij ij

X X

=

-

Risk 1 122 2,848

Risk 2 164 1,628

Risk 3 106 1,887

Calculate the estimated credibility factor, and calculate the credibility premium for

Risk 1. [4]

2 ActEd

The table below shows the total aggregate claims, ijY , and the corresponding risk

volumes, ijP , over the past six years for the three types of pet insurance offered by a

small company:

Risk

6

1

ij

j

Y=

6

1

ij

j

P=

Cat 54,278 362

Dog 68,861 485

Stick insect 1,689 78

(i) Using EBCT Model 2, the credibility premium per unit of risk volume for the

coming year for cats is 148.357. Calculate the credibility premium per unit of

risk volume for the coming year for dogs and stick insects. [7]

(ii) Comment on the values of the credibility factors. [2]

[Total 9]


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3 Subject 106, September 2002, Question 9 (part (iii))

An insurance company has insured a fleet of cars for the last four years. For year j (

1,...,4j= ), letj

Y andj

P be the total amount claimed and the number of cars in the

fleet, respectively. Let /j j jX Y P= be the average amount claimed per car in year j .

Assume that the distribution of jX depends on a risk parameter q and that the

conditions of Empirical Bayes Credibility Theory Model 2 are satisfied.

The company has insured ten similar fleets over the last four years. Using the data from

these years, [ ( )]E mq , 2[ ( )]E s q and [ ( )]V mq are estimated to be 62.8, 106.32 and 5.8

respectively.

(a) Calculate next years credibility premium for a fleet of cars with claims over the

last four years given below, if the fleet will have 16 cars next year.

Year

1 2 3 4

Total amount claimed 1,000 1,200 1,500 1,400

Number of cars 15 16 18 15

(b) Explain how and why the credibility factor would be affected if the estimate of

[ ( )]V mq increases, and comment on the effect on the credibility premium. [5]

[Total 8]


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4 Subject C2, April 1997, Question 13 (part (ii))

For the past five years an insurance company has insured 15 different chains of

newsagents shops against damage to their premises and stock from any cause. For

chain i i, , , ,= 1 2 15 , and year j j, , , ,= 1 2 5 , the random variable Yij represents the

annual claims and Pij represents the number of shops in the chain. The sequence

Y Pij ij ji

;n s=

=

RSTUVW1

5

1

15

satisfies all the assumptions for Empirical Bayes Credibility Theory

Model 2. The data for the first three chains in this collective are shown in the table

below. Also shown for each of the first two chains is the credibility premium per shop

for the coming year.

Y Pij ij

; Credibility

premium per

shopChain j = 1 2 3 4 5

1 450; 2 220; 2 3700; 2 250; 2 380; 2 750

2 2500; 3 1140; 4 3600; 4 3900; 4 860; 5 733

3 4950; 9 39600; 9 14850; 11 29700; 12 9900; 14

(a) Calculate the credibility premium per shop for the coming year for Chain

number 3.

(b) Explain carefully why the credibility premium per shop for the coming year ishigher for Chain 1 than for Chain 2 even though the average annual claim per

shop is lower for Chain 1 than for Chain 2. [11]

[Total 17]


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Risk models 1


The number of claims arising from a hurricane in a particular region has a Poisson

distribution with mean l. The claim severity distribution has mean 0.5 and variance 1.

(i) Determine the mean and variance of the total amount of claims arising from a

hurricane. [2]

(ii) The number of hurricanes in this region in one year has a Poisson distribution

with mean . Determine the mean and variance of the total amount claimed

from all the hurricanes in this region in one year. [3]

[Total 5]

2 Subject 106, April 2004, Question 4 (part (ii))

A portfolio consists of two types of policies. For type 1, the number of claims in a year

has a Poisson distribution with mean 1.5 and the claim sizes are exponentially

distributed with mean 5. For type 2, the number of claims in a year has a Poisson

distribution with mean 2 and the claim sizes are exponentially distributed with mean 4.

Let Sbe the total amount claimed on the whole portfolio in one year. All policies are

assumed to be independent.

Derive the MGF of Sand show that Shas a compound Poisson distribution. [4]


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3 Subject CT6, April 2008, Question 10 (corrected)

A bicycle wheel manufacturer claims that its products are virtually indestructible in

accidents and therefore offers a guarantee to purchasers of pairs of its wheels. There are

250 bicycles covered, each of which has a probability p of being involved in an

accident (independently) in a year. Despite the manufacturers publicity, if a bicycle is

involved in an accident, there is in fact a probability of 0.1 for each wheel

(independently) that the wheel will need to be replaced at a cost of 100. Let Sdenote

the total cost of replacement wheels in a year.

(i) Show that the MGF of Sis given by:

250200 10018 81

( ) 1100

t t

S

pe pe p

M t p

+ +

= + - [4]

(ii) Show that ( ) 5,000E S p= and 2var( ) 550,000 100,000S p p= - . [6]

Suppose instead that the manufacturer models the cost of replacement wheels as a

random variable Tbased on a portfolio of 500 wheels, each of which (independently)

has a probability of 0.1p of requiring replacement.

(iii) Derive expressions for ( )E T and var( )T in terms of p . [2]

(iv) Suppose 0.05p= .

(a) Calculate the mean and variance of Sand T.

(b) Calculate the probabilities that Sand Texceed 500.

(c) Comment on the differences. [5]

[Total 17]


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Risk models 2

1 Subject CT6, September 2007, Question 8

The total claim amount, S, on a portfolio of insurance policies has a compound Poisson

distribution with Poisson parameter 50. Individual loss amounts have an exponential

distribution with mean 75. However, the terms of the policies mean that the maximum

sum payable by the insurer in respect of a single claim is 100.

(i) Find ( )E S and var( )S . [7]

(ii) Use the method of moments to fit as an approximation to S:

(a) a normal distribution(b) a log-normal distribution [3]

(iii) For each fitted distribution, calculate ( 3,000)P S> . [3][Total 13]


The total claims arising from a certain portfolio of insurance policies over a given

month is represented by

1

if 0=

0 if = 0

N

i

i

X NS

N

=

>

where Nhas a Poisson distribution with mean 2 and 1 2, , , NX X X is a sequence of

independent and identically distributed random variables that are also independent ofN.

Their distribution is such that ( =1) =1/ 3 and ( = 2) = 2/ 3i iP X P X . An aggregatereinsurance contract has been arranged such that the amount paid by the reinsurer is

3S- (if S> 3) and zero otherwise.

The aggregate claims paid by the direct insurer and the reinsurer are denoted by

andI RS S , respectively.

Calculate ( ) and ( )I RE S E S . [8]


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3 Subject 106, September 2004, Question 9 (part)

A general insurance company has a portfolio of fire insurance policies, which offer

cover for just one fire each year.

Within the portfolio, there are three types of buildings for which the average cost of a

claim and probability of a claim are given in the table below.

Type of

building

Number of Risks

Covered

Average Cost

of a Claim

(000s)

Probability

of a Claim

Small 147 12.4 0.031

Medium 218 27.8 0.028

Large 21 130.3 0.017

It is assumed that the cost of a claim has an exponential distribution, and that all the

buildings in the portfolio represent independent risks for this insurance cover.

(i) Show that the mean and standard deviation of annual aggregate claims from this

portfolio of insurance policies are 272,715 and 150,671, respectively and

obtain the cumulant generating function. [4]

(ii) Using a normal distribution to approximate the distribution of annual aggregate

claims, calculate the premium loading factor necessary such that the probabilitythat annual aggregate claims exceed premium income is 0.05. [3]

(iii) Market conditions dictate that the insurer can only charge a premium which

includes a loading of 25%. Calculate the amount of capital that the insurer must

allocate to this line of business in order to ensure that the probability that annual

aggregate claims exceed premium income and capital is 0.05 (again using a

normal approximation). [2]

(iv) Comment on the assumption of independence and the use of a normal

approximation, in relation to your answers to (ii) and (iii). [4][Total 13]


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Ruin theory

1 ActEd

( )S t is a compound Poisson process with Poisson parameter 40, and claim size

distribution which is log (5, 4)N .

(i) Find [S(10)]E and var[ (10)]S . [3]

(ii) The initial surplus is 400,000, and the rate of premium income is 41,000 per unit

time. Assuming that ( )U t can be approximated by a normal distribution, find

the probability that (10) 0U < . [3][Total 6]


Aggregate annual claims on a portfolio of insurance policies have a compound Poisson

distribution with parameter l. Individual claim amounts have an exponential

distribution with mean 1.

The insurer calculates premiums using a loading of a (so that the annual premium is

1 a+ times the annual expected claims) and has initial surplus U.

(i) Show that if the first claim occurs at time t, the probability that this claim causes

ruin is (1 )U te e a l- - + . [3]

(ii) Show that the probability of ruin on the first claim is2

Ue

a

-

+. [4]

(iii) Show that if the insurer wishes to set a such that the probability of ruin at the

first claim is less than 1% then it must choose 100 2U

ea -

> - . [2][Total 9]


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3 ActEd

The aggregate daily claims (000s) on a certain portfolio of policies occur according to

a Poisson process with parameter 30. Individual claim sizes (000s) have a gamma

distribution with mean 40 and variance 800. The insurer adds a premium loading factor

of 50%.

(i) Calculate the value of the adjustment coefficient. [4]

(ii) Hence calculate the minimum amount of capital needed to ensure that the

probability of ultimate ruin is less than 0.1%. [1]

(iii) Calculate the expected daily profit on this portfolio of policies. [1]

[Total 6]

4 Subject 106, September 2004, Question 8 (part (ii))

Claims arrive as a Poisson process rate l, and the premium loading factor is 25%.

(a) Determine to one significant figure the adjustment coefficient R* if each claim

is exactly 100.

(b) If claims are exponentially distributed with mean 100, the adjustment

coefficient expR is 0.002. Compare R* with expR and comment. [3]


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5 ActEd

Claims arrive as a Poisson process with rate . Individual claim sizes are exponentially

distributed with mean 100. The insurer uses a premium loading factor of 0.2.

A proportional reinsurance arrangement has been proposed, with a retained proportion

of . The reinsurer uses a security loading of 0.4.

(i) State the range of possible values of a such that the probability of ruin is less

than 1. [2]

(ii) (a) Show that the adjustment coefficient, R , is given by:

2 1

100 (7 1)R a

a a-= -

(b) Hence, find the value of a that maximises R .

(c) Determine the expected profit per unit time and the upper bound for the

probability of ultimate ruin for the value of a calculated in part (ii)(b).

[10]

(iii) Compare the profit and the probability of ultimate ruin in part (ii)(c) with those

where no reinsurance is effected. [2][Total 14]


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Generalised linear models

1 ActEd

(i) Show that each of the following distributions are members of the exponential

family:

(a) ~ ( )i iY Poi m

(b) ~ ( )i iY Exp l

(c) 2~ ( , )i iY N m s

(d) i iY Z n= where ~ ( , )i iZ Bin n m [8]

(ii) For each distribution in part (i), use the properties of exponential families to

determine their mean, variance and variance function. [8]

[Total 16]

2 Subject CT6, April 2007, Question 10 (part)

(i) The gamma distribution with mean mand variance 2m a has density function:

1 /( )( )

yf y y ea

a a ma

a

m a

- -=G

( )0y>

(a) Show that this may be written in the form of an exponential family.

(b) Use the properties of exponential families to confirm that the mean and

variance of the distribution are mand

2

m a . [9]


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The next six questions refer to the following generalised linear model:

the number of claims iY on policy i ( 1, ,15)i= has a Poisson distribution with

unknown mean im .

The number of claims observed in the last year on 15 policies were as follows:

i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

iy 0 1 0 1 0 0 1 0 0 0 1 3 0 3 1

3 ActEd

(i) Show that the log-likelihood function for this model, based on observations

{ }: 1,...,15iy i= is given by:

{ }15

1

ln ( ) ln where is a constanti i i ii

L y c cm m m=

= - + . [1]

(ii) Identify the canonical link function associated with the Poisson model from the

Tables. [1]

4 ActEd

The insurance company proposes a simple model (Model A) with linear predictor:

for 1,...,15i ih a= =

(i) Describe in words what this model represents. [1]

(ii) Use the canonical link function to show that the log-likelihood in terms of a is

given by:

{ }15

1

ln ( ) ii

L y e caa a=

= - + [1]

(iii) Hence, show that the maximum likelihood estimate for a is

( )151 ln 15 0.3102ii ya == = - . [1]

(iv) State the fitted values, i , for this model. [1]

[Total 4]


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5 ActEd

The insurance company proposes a second model (Model B) with linear predictor:

for 1,...,10

for 11,...,15i

i

i

ah

b

===


(ii) Write down the log-likelihood function for Model B and derive maximum

likelihood estimates for a and b . [2]

(iii) State the fitted values, im , for this model. [1]

[Total 4]

6 ActEd

The insurance company proposes a third model (Model C) with linear predictor:

for 1,...,15i i ih a= =


(ii) Derive the maximum likelihood estimates for ia for 1, ,15i= . [2]

(iii) State the fitted values, im , for this model. [1]

[Total 4]


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7 ActEd

(i) Describe what is meant by thesaturatedmodel. [1]

(ii) Write down the log-likelihood function of the saturated model for the data in the

previous question in terms of i only. [1]

(iii) (a) Define the scaled deviance and describe what it measures.

(b) Show that the scaled deviances for Models A, B and C are:

20.01 12.89 0

(Assume that log 0 for 0i i iy y y= = ). [7]

(iv) By performing a chi-squared test, determine whether:

(a) Model B is a significant improvement over Model A

(b) Model C is a significant improvement over Model B. [2]

[Total 11]

8 ActEd

(i) Show that the deviance residual for 11 :

(a) under Model A is 0.2949

(b) under Model B is 0.5099- . [2]

(ii) Show that the Pearson residual for 11 :

(a) under Model A is 0.3114

(b) under Model B is 0.4743- . [2]

(iii) In Model B, the maximum likelihood estimate of*b b a= - is * 1.674b = and

the estimated standard error of the corresponding estimator is 0.677. Interpret

this result. [1]

[Total 5]


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9 ActEd

A statistician is using generalised linear modelling to try to estimate the probability of

lives in Country C developing a particular medical condition. He believes that the

relevant covariates are the region in which a person lives, socio-economic group and

age.

For the purposes of the investigation, the country has been divided into 4 regions.

There are 5 socio-economic groups.

The statistician is considering a linear predictor of the form:

i jh a b g = + +

where i denotes region ( 1,...,4i= ), j denotes socio-economic group ( 1,...,5j= ) andx denotes age. He has fitted a binomial model to a set of relevant data and has obtained

the following estimates of the parameters.

1 2.3975a = - 2 2.3118a = - 3 2.7375a = - 4 2.6562a = -

1 0b = 2 0.1242b = 3 0.3894b = 4 0.4665b = 5 0.6616b =

0.0012g =

(i) Explain why the statistician has chosen a binomial model and write down the

canonical link function. [2]

(ii) Use the canonical link function to predict the probability of each of the

following lives developing the condition.

(a) Life X, who is aged 40, lives in Region 1 and belongs to socio-economic

group 2

(b) Life Y, who is aged 42, lives in Region 3 and belongs to socio-economicgroup 1. [4]

(iii) Comment on the relative levels of risk for Life X and Life Y. [2]

[Total 8]


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10 Subject 106, April 2004, Question 9 (part (ii))

Let ijY be the number of accidents on a particular motorway in the j th quarter of

year i , 1,2,3=i , 1, ,4=

j . Suppose that ijY has a Poisson distribution with mean mij.

Three models are shown below.

DevianceDegrees of

Freedom

Model 1 log m=ij 266.35 11

Model 2 logm a=ij i 202.19 9

Model 3 logm a b= +ij i j 10.68 6

(a) Interpret each of these models.

(b) Determine which model you would recommend, giving your reasons. [7]

11 ActEd

A generalised linear model is being used to estimate the expected future lifetime of

individuals aged exactly 25. The following covariates are used:

C average number of cigarettes smoked per day

A average number of units of alcohol consumed per week

iS sex ( i= male, female)

jO occupation ( 1,2,3,4j= representing stress and risk levels)

(i) Write the parameterised form of the linear predictor for the following models:

(a) cigarettes (b) 2(cigarettes) (c) occupation [3]

(ii) Write the parameterised form of the linear predictor for the following models:

(a) 2cigarettes (cigarettes)+ (b) cigarettes alcohol+ (c) cigarettes sex+ (d) alcohol occupation+ (e) sex occupation+ (f) cigarettes sex occupation+ + [6]

[Total 9]


28/58



12 ActEd

(i) What is an interactive effect between two covariates? [1]

(ii) The average effect on life expectancy of smoking 20 cigarettes a day and/or

drinking 21 units a week are shown in the table below:

Covariates Reduction in life expectancy

Smoker only 5 years

Drinker only 2 years

Smoker and drinker 10 years

Explain in words what the following represent and relate that to a reduction in

life expectancy:

(a) smoker drinker (b) smoker drinker* [2]

(iii) Using the covariates from 11, write the parameterised form of the linear

predictor for the following models:

(a) cigarettes alcohol cigarettes alcohol+ + (b) cigarettes sex cigarettes sex+ + (c) sex occupation sex occupation+ + [3]

(iv) Using the covariates from 11, write the parameterised form of the linear

predictor for the following models:

(a) cigarettes alcohol sex* + (b) cigarettes sex alcohol* + (c) sex occupation cigarettes* * [3]

[Total 9]


29/58



13 Subject CT6, April 2007, Question 10 (part)

(ii) Explain the difference between a continuous covariate and a factor. [3]

(iii) A company is analysing its claims data on a portfolio of motor policies, and uses

a gamma distribution to model the claim severities. The company uses three

rating factors:

policyholder age (as a continuous variable)

policyholder gender

vehicle rating group (as a factor).

(a) Write down the form of the linear predictor when all rating factors are

included as main effects.

(b) State how the linear predictor changes if an interaction between

policyholder age and gender is included. [4]

[Total 7]


30/58



Run off triangles

1 ActEd

The figures below give the claim payments made during the calendar years 2007-2009

for a certain portfolio of general insurance policies:

Claim payments made

during year (000)

Development year

0 1 2

Accident

year

2007 320 460 110

2008 350 410

2009 400

Estimates of past and future inflation are given below:

Annual claim inflation

rate (past)

Annual claim inflation

rate (future)

2007/08 5% 2009/10 3%

2008/09 4% 2010/11 4%

Use the inflation adjusted chain ladder method to estimate the total amount outstanding

for future claims arising from accident years 2008 and 2009. [6]


31/58




The table below shows cumulative paid claims and premium income on a portfolio of

general insurance policies.

Underwriting

year

Development YearPremium

income0 1 2

2002 38,419 77,112 91,013 120,417

2003 31,490 78,504 117,101

2004 43,947 135,490

(i) Assuming an ultimate loss ratio of 93% for underwriting years 2003 and 2004,

calculate the Bornhuetter-Ferguson estimate of outstanding claims for this

triangle. [8]

(ii) State the assumptions underlying this estimate. [2]

[Total 10]


32/58




The delay triangles given below relate to a portfolio of motor insurance policies.

The cost of claims settled during each year is given in the table below:

(Figures in 000s)

Accident

Year

Development Year

0 1 2

2004 4,144 694 183

2005 4,767 832

2006 5,903

The corresponding number of settled claims is as follows:

Accident

Year

Development Year

0 1 2

2004 581 75 28

2005 626 71

2006 674

Calculate the outstanding claims reserve for this portfolio using the average cost per

claim method with grossing-up factors, and state the assumptions underlying your

result. [7]


33/58



Time series (part 1)

Time series processes

A time series is astochastic process, { : }tX t J , with a continuous state space, tX S,

and discrete time set J. A time series can be univariate or multivariate.

Covariances refresher

Give formulas for, or simplify the following:

1. cov( , )X Y

2. ( )cov ,X c 3. ( )cov 2 ,3X Y

4. cov(2 1, 5 3 )+ -X Y 5. cov( , )X X

6. cov( , )+ +aX b cY d 7. cov( , )+X Y Z 8. corr( , )X Y

If { }tX denotes a time series defined at integer times and { }tZ is white noise with

variance 2s (and mean 0), calculate each of the following:

1. 2 3cov( , )Z Z

2. 3 3cov( , )Z Z

3. 2 3cov( , )X Z

ie what is the connection between future white noise 3( )Z and the current

observed time series 2( )X ?

4.1 2 1 3

cov( , )

+ +Z Z Z Z

5. 1 2 2 3cov(0.5 , 0.5 )+ +Z Z Z Z

6. 1 2var( )+Z Z


34/58



Backward shift operator B

Shifts a process, tX , backwards:

1-=t tBX X

Note: B has no effect on constants, eg m=B

Example:

1 23 2- -- + =t t t t X X X e

Strict and weak stationarity

A time series is strictly stationary if all the statistical properties are unchanged over

time.

For a weakly stationary time series, we require that ( )tE X and var( )tX are constant,

and that the covariances are the same for the same lags:

cov( , ) cov( , )+ +=i j i jt t t k t k X X X X

Why do we want a time series to be stationary?

Indeterminism

A time series is purely indeterministic if knowledge of 1, , nX X is less useful in

predicting X as N . This excludes purely deterministic processes like 0,1,0,1,

What happens to corr( , )+t t kX X as k ? How does this help us?

Note: If an exam question states stationary time series it means weakly stationary and

purely indeterministic.


35/58



Invertibility

A time series is invertible if we can calculate the white noise terms (residuals) from

observed data values by inverting the formula for the process.

Examples:

10.8 -= +t t tX X e

10.6 -= +t t tX e e

Why do we want a time series to be invertible?

Markov property

A time series has the Markov property if we can predict the future development of a

time series from its present state alone. In the course this is formerly expressed as the

process probabilities depend only on the most recent value:

1 1( | , , ) ( | ) = = = =

m mt t t m t t mP X A X x X x P X A X x

Examples:

10.8 -= +t t tX X e

1 20.8 0.3- -= - +t t t t X X X e

10.6 -= +t t tX e e

Characteristic polynomials

1. Write the formula for the time series in terms of the backwards shift operators:

( ) ( )f q=t tB X B e

2. Replace B s by ls to obtain the 2 characteristic polynomials

( )f l and ( )q l

Testforstationarity

3a. If allroots of ( ) 0f l = satisfy 1l> then stationary

Testingforinvertibility

3b. If allroots of ( ) 0q l = satisfy 1l> then invertible


36/58



Examples:

Check stationarity and invertibility of:

1 2 1 212 10 2 12 11 2- - - -= - + - +t t t t t t X X X e e e

10.8 -= +t t tX X e

10.6 -= +t t tX e e

AutocovarianceFunction

Covariance between tX s of a stationary random process:

0 var( )

cov( , )

gg +

==

t

k t t k

X

X X

Note: if process is not stationary, then covariances ( )gk t would depend on tand k.

AutocorrelationFunction(ACF)

Correlation between tX s of a stationary random process:

corr( , ) 1 1r r+= - k t t k k X X

Note: 0r k as k for purely indeterministic processes.

What is rkexpressed in terms of gk?

PartialAutocorrelationFunction(PACF)

Conditional correlation of +t kX with tX given 1 1, ,+ + -t t kX X , ie how much the

variance of +t kX is due tojust tX .

22 1

1 1 2 21

, ,1

r rf r f f

r

-= =

- kgiven on page 40 of Tables


37/58



MA(q)

A moving average process of order q is a weighted average of the past q white noise

terms (plus a new white noise term):

1 1b b- -= + + +t t t q t qX e e e (zero mean)

1 1m b b- -= + + + +t t t q t qX e e e (mean m)

Always stationary

rkcuts off for >k q

fkdecays as k

Not Markov Do the test for invertibility

AR(p)

An autoregressive process of order p depends on the previous p values (plus just one

white noise term):

1 1a a- -= + + +t t p t p t X X X e (zero mean)

1 1( ) ( )m a m a m

- -= + - + + - +

t t p t p t X X X e (mean m)

Always invertible

rkdecays as k

fkcuts off for >k p

Not Markov unless 1=p

Do the test for stationarity


38/58



1 Subject C2, April 1999, Question 6 (adapted)

tY is a moving average process given by:

1 30.5 0.1 0.4- -= + -t t t t Y Z Z Z

where { }tZ are independent random variables, each with mean 0 and variance2s .

(i) Calculate the first three autocorrelations, 1 2,r r and 3r . [5]

(ii) State, without calculation, the values of rk, 3>k . [1][Total 6]


(i) (a) Calculate the autocovariance function { : 0}g k k and autocorrelation

function { : 0}r k k of a first-order Moving Average process:

1 1b -= + +t t tX e e

where { : 0}te t is a sequence of uncorrelated, zero-mean randomvariables with common variance 2se .

(b) State the conditions on the values of the parameters such that the process

is invertible. [5]

(ii) A sequence of observations 1 2, , , nx x x has sample variance 0 14.5g = , sample

lag-1 autocovariance 1 5.0g = . Show that there is more than one first-order

moving average process which can be fitted to these data, but verify that only

one of the fitted processes is invertible. [4]

[Total 9]


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A stationary stochastic process { }: 0,1,= tY t satisfies the relationship:

( ) ( )1 20.8 0.4m m m- -= + - - - +t t t t Y Y Y e

where { }: 0,1,= te t is a sequence of independent, zero-mean Normal random

variables with common variance 2s .

(i) Calculate the autocorrelation function, rk, and the partial autocorrelation

function, fk, of Yfor 1k= and 2. [5]

(ii) State, without performing additional calculations, what you would expect to find

if you were to calculate rkand fkfor larger values of k. [2]

[Total 7]

4 Subject C2, September 1996, Question 2 (adapted)

An autoregressive process of order 2 is defined by:

20.8 -- =t t tY Y Z

where { }tZ is a white noise process. Obtain the values of the autocorrelation

coefficients 1r and 2r . [3]

5 Subject 103, September 2004, Question 7 (part, adapted)

Consider the second order autoregressive process:

1 20.6 0.3t t t t X X X e- -= + +

where { : 1}te t is white noise process with variance2s .

(a) Determine whether the process can be stationary.

(b) State, with a reason, whether the process possesses the Markov property.

(c) Show that

2700

0 169g s= and2600

1 169g s= , and find the value of 2g . [8]


40/58



AR(p) stationarity explored

We can express an (1)AR as a summation of white noise terms:

21 2 1 2 1

1

0

0

( )

...

t t t t t t t t t

tt j

t j

j

X X e X e e X e e

X e

a a a a a

a a

- - - - -

-

-=

= + = + + = + +

=

= +

From this we can see that:

2

2 20 0 21( ) ( ) var( ) var( ) 1

t

t tt tE X E X X X aa a sa-= = + -

If we assume the time series has infinite history then we obtain:

0

jt t j

j

X ea

-=

=

with ( ) 0tE X = and, if | | 1a < then

2

2var( ) 1tX

s

a= - .

ARMA(p, q)

A process that is a combination of an ( )AR p and an ( )A q :

1 1

1 1

a a

b b

- -

- -

= + + +

+ + +

t t p t p t

t q t q

X X X e

e e (zero mean)

1 1

1 1

( ) ( )a m a m

b b

- -

- -

= + - + + - +

+ + +

t t p t p t

t q t q

X X X e

e e (mean )

rkdecays as k

fkdecays as k

Not Markov unless 1=p and 0=q

Do the test for stationarity

Do the test for invertibility


41/58




tY , 1,2,3,t= is a time series defined by:

1 10.8 0.2t t t t Y Y Z Z - -- = +

where tZ , 0,1,t= is a sequence of independent zero-mean variables with common

variance2s . Derive the autocorrelation kr , 0,1,2,k= . [6]

Difference operator

Finds the 1st difference of a process, tX :

1(1 ) - = - = -t t t t X B X X X

Note: 1 = -B

Examples:

1 2 33 3- - -= - + +t t t t t X X X X Z

1 2 3 46 17 17 7- - - -= - + - +t t t t t t X X X X X Z

Integrated ( )I d

A process, tX , is integrated of order d, if:

tX not stationary and = d

t tY X is stationary

ARIMA(p, d, q)

A series, such that the dth difference:

= dt tY X is astationary ( , )ARMA p q


42/58



Example:

Classify the following processes:

1 2 3 10.6 0.3 0.1 0.25- - - -- - - = -t t t t t t X X X X e e

1 2 3 4 22 7 9 5- - - - -= - + - + +t t t t t t t X X X X X e e

1 2 3 46 17 17 7- - - -= - + - +t t t t t t X X X X X Z

Multivariate Time Series

A time series that depends on more than one variable. This is as opposed to a univariate

time series which is a time series with only one variable in it (white noise doesnt count

as a variable). For example ( ), ( ), ( , )AR p MA p ARMA p q and ( , , )ARIMA p d q are all

univariate. Multivariate time series can be written in vector (VARMA) form:

Example:

1 1

1 1

0.7 0.1

0.2 0.3

- -

- -

= - +

= - +

t t t t

t t t t

X X Y e

Y X Y e

Is this example time series:

a) stationary (eigenvalues < 1 in magnitude)?

b) invertible?

c) Markov?

Cointegrated

This means that the individual processes are not stationary but when combined in some

way, they do produce a stationary series.

Two time series processes, tX and tY are cointegrated if:

,t tX Yare both (1)I

a b+t tX Y is stationary, where ( ) 0a b , where ( )a b is called the

cointegrating vector.


43/58



Bilinear models

( ) ( )1 1 1 1a m m b m - - - -+ - = + + + -t t t t t t X X e e b X e

Threshold AR models

1 1 1

2 1 1

( ) if =

( ) if

a mm

a m

- -

- -

- + + - + >

t t tt

t t t

X e X dX

X e X d

Random coefficient AR models

( )1 , randomm a m a-= + - +t t t t t X X e

ARCH models

( )2

0

1

m a a m-=

= + + -p

t t k t k

k

X e X

Note: ARCH has now been tested twice (Sept 2007 and Sept 2008).


44/58



Time series (part 2)

Box Jenkins

Tentative identification of a model from the ARIMA class

Estimation of parameters in the identified model

Diagnostic checks

Identify

p, dand q

remove trends and cycles

linear trend

differencing

least squares trend

removal (apply the

model = + +t tx a bt y )

exponential trend take logs

seasonal/periodic

method of moving

averages

seasonal differencing

method of seasonal

means

identify d

difference to minimise2sd (the sample

variance of the process( )d

z )

if kr tends towards 0 slowly, go to the next

difference

( )A q kr cuts off (ie within 95% CI) when >k q ,

fkdecays

( )AR p fk cuts off (iewithin 95% CI) when >k p ,

kr decays

( , )ARMA p q

if not ( )A q or ( )AR p then ( , )ARMA p q

start with (1,1)ARMA and work up to more

complicated models if fit poor

AIC only add new parameters if relative

reduction in sum of squares of residuals2-

ne

Estimate the

parameters

method of moments solve r=k kr

method of least squares minimise 2 te method of maximum

likelihood

need to assume a distribution so may not be

easy


45/58


46/58



(i) Explain which feature of the Figures indicates that differencing is not required in

order to obtain a stationary series. [1]

(ii) On the basis of the sample ACF, kr , the companys analyst decides to fit a first-

order autoregressive model to the data. State, with reasons, whether you

consider this to be a reasonable decision and indicate what additional plot you

would require in order to make a firmer recommendation. [3]

(iii) The model is fitted and the residuals calculated. The sample ACF of the

residuals is shown in Figure 1c. State what conclusions you would draw from

the plot. [1]

Figure 1c

[Total 5]

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 1 2 3 4 5 6 7

Autocorrelation,

rk

Lag, k

Autocorrelation Function for Ratio

-0.5

-0.4

-0.3

-0.2

-0.10

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5 6 7

Autocorrelation,rk

Lag, k

Autocorrelation Function for Residuals


47/58



2 Subject CT6, September 2008, Question 10 (part)

From a sample of 50 consecutive observations from a stationary process, the table

below gives values for the sample autocorrelation function (ACF) and the sample partial

autocorrelation function (PACF):

Lag ACF PACF

1 0.854 0.854

2 0.820 0.371

3 0.762 0.085

The sample variance of the observations is 1.253.

(i) Suggest an appropriate model, based on this information, giving your reasoning.

[2]

(ii) Consider the (1)AR model:

1 1t t tY a Y e-= +

where te is a white noise error term with mean zero and variance2s .

Calculate method of moments (Yule-Walker) estimates for the parameters of 1a and 2s on the basis of the observed sample. [4]

[Total 6]


48/58



3 ActEd

The monthly sales (in kilolitres) of red wine by Australian winemakers from January

1980 through to October 1991 are shown in the graph below:

(i) (a) Give two features which indicate that the time series is non-stationary.

(b) Describe how each of these features can be removed. [3]

The data is logged and seasonally adjusted and the statistician tries to fit an( , , )RIMA p d q model to the adjusted data:

tX tX 2 tX 3

tX

sample

variance0.0791 0.0159 0.0471 0.1565

SAC

F

1r 0.8805 0.4774- 0.6672- 0.7453-

2r 0.8541 0.0433 0.1918 0.2998

3r 0.8245 0.0374- 0.0010 0.0115- 4r 0.8025 0.0920- 0.1148- 0.1316-

5r 0.8110 0.1810 0.1967 0.1985

6r 0.7815 0.1284- 0.1409- 0.1456-

(ii) Use the data in the table to choose the most appropriate value for d. State your

reasons clearly. [2]

0

500

1000

1500

2000

2500

3000

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

Sales

Month


49/58



After being differenced an appropriate number of times, the statistician examines the

SACF and SPACF:

(iii) Using the principle of parsimony, suggest appropriate values of p and q . [2]

(iv) The statistician fits an (0,1,1)ARIMA model and examines the residuals:

Ljung-Box statistic p -value = 0.76836

Turning points test p -value = 0.59190

Interpret these results. [2]

[Total 9]

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

1 2 3 4 5 6 7 8 9 10

SACF

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

1 2 3 4 5 6 7 8 9

SPACF

Lag

Lag


50/58



Forecasting (Box Jenkins)

replace all unknown parameters by their estimated values

replace the random variables 1,..., nX X by their observed values 1,..., nx x

replace the random variables 1 1,...,+ + -n n kX X by their forecast values

(1),..., ( 1)-n nx x k

replace the innovations 1,..., ne e by the residuals 1 ,..., ne e

replace the random variables 1 1,...,+ + -n n ke e by 0 (their expectations)

( )nx k is the estimate of the expected value of +n kX (given the observations up to nX ) .

Example:

1 1 2 2 1 1a a b- - -= + + +n n n n nX X X e e

Forecasting (Exponential smoothing)

21 2 (1) (1 ) (1 )a a a- - = + - + - + n n n nx x x x

This is a weighted average of the past values but there is less emphasis on older values.

Rearrangements: 1 (1) (1 ) (1)a a -= + -n n nx x x or [ ]1 1 (1) (1) (1)a- -= + -n n n nx x x x


51/58



4 Subject CT6, April 2007, Question 8 (corrected, extended)

A modeller has attempted to fit an ARMA ( , )p q model to a set of data using the Box-

Jenkins methodology. The plot of residuals based on this proposed fit is shown below.

(i) Under the assumptions of the model, the residuals should form a white noiseprocess.

(a) By inspection of the chart, suggest two reasons to suspect that the

residuals do not form a white noise process.

(b) Define what is meant by a turning point.

(c) Perform a significance test on the number of turning points in the data

above. (There are 100 points in the data and 59 turning points.) [6]

(ii) On your suggestion, the original fitted model is discarded, and re-parameterisedto:

2 1 2= 5 0.9( 5) 0.5n n n nX X e e+ + ++ - + +

The sample autocorrelations of the 100 residuals at lags 1, 2, 3, 4 and 5 were

calculated to be:

0.10 +0.20 +0.05 +0.04 0.05

Carry out a portmanteau (Ljung and Box) test and state your conclusion. [3]

Residuals based on fitted model

-80

-60

-40

-20

0

20

40

60

80

100

120

1 6 11 16 2 1 26 3 1 36 41 4 6 51 5 6 61 66 71 7 6 81 8 6 91 9 6

Time


52/58



(iii) Given the following observations:

99 100 99 100 2 7 0.7 1.4x x e e= = = - =

(a) Use the Box-Jenkins methodology to calculate the forward estimates

100 (1)x , 100 (2)x and 100 (3)x .

(b) The simplest form of exponential smoothing used at time 99 gave a

forecast for 100x of 5.2. Assuming the smoothing parameter is equal to

0.2, find the forecast for 101. [6]

[Total 15]


53/58



Monte Carlo simulation

Inverse transform method (continuous random variable)

1. Generate a random number u from (0,1)U

2. Return:

1( )-=x F u

Inverse transform method (discrete random variable)

1. Generate a random number u from (0,1)U

2. Return i such that:

1( ) ( )- < i iF x u F x

where the discrete random variable, X , can take only the values 1 2, , ,x x x , (

1 2< < s go back to step 1, otherwise return:

1 1

2ln-=

Sz v

Sand 2 2

2ln-=

Sz v

S


54/58



Acceptance/rejection method

Scales up ( )h x , so that the area under ( )Ch x includes all of the area under ( )f x :

all

( )max

( )=

x

f xC

h x

1. Generate a random number 1u from (0,1)U

2. Use 1u to generate a random variate x from ( )h x

3. Generate a random number 2u from (0,1)U

4. If ( ) ( )

( )2< =

f xu g x

Ch xthen return x otherwise repeat.


55/58



1 ActEd

A random variable, X , has the density function:

2 71 28 3 24

( ) , 1 3= - + - <


56/58



Errors

Using1

q= ixn

to estimate ( )q=E X :

absolute error q q= -

relative errorq q

q

-=

Number of simulations required

Using

~ (0,1)

q q

t

-N

n, the number of simulations required, n , to ensure:

( )2

22 2

1 at

q q e a e

- < = - fi >P n z

22

2 2 2

1 a

q q te a

q e q

- < = - fi >

P n z

where xq= , [ ]q=E X and 2 2

1

1 ( )1

t q=

= --

n

k

k

xn

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2 ActEd

Monte Carlo simulation is being used to model the life expectancy of a particular group

of annuitants. The standard deviation of the life expectancy has been estimated as 6.52years and a previous estimate of the mean life-expectancy was 74.13 years. Calculate

how many simulations should be performed to ensure that, with 95% probability, the:

(i) absolute error when estimating the mean life expectancy is less than 0.05 [2]

(ii) relative error when estimating the mean life expectancy is less than 0.01%. [2]

[Total 4]


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Preparing for the exam

As the exam gets closer, we recommend that you focus on question practice in order to prepare for the

exam. If you have purchased the CMP, dont forget that you have the Series X Assignments and

Question and Answer Bank, which contain many exam-style questions as well as other questions todevelop your understanding of the material.

In addition, you may find some of the following study material very helpful in the run-up to the exam:

ASET ActEds solutions to exams set from 2010 to 2013, with useful exam technique

comments. In the summer, Mini-ASET will also be available covering the April 2014

paper only.

Mock Exam There is a 100-mark mock exam paper (Mock Exam A) available. You can purchase

the mock exam with or without mock exam marking.

AMP The Additional Mock Pack (AMP) consists of two further 100-mark mock exampapers. This pack is ideal for students who are retaking and have already sat Mock

Exam A, or for those who just want some extra question practice. You can have either

or both mock exams marked by submitting the relevant mock exam with a Marking

Voucher (purchased separately).

Revision Notes These are a set of conveniently-sized A5 spiral-bound booklets perfect for revising

on the train or tube to work. Each booklet contains relevant Core Reading with a set

of integrated short questions to develop your bookwork knowledge, relevant past

exam questions with concise solutions, detailed analyses of key past exam questions

(selected for their difficulty) and other useful revision aids.

Flashcards These A6-sized cards cover the key points of the course that most students like to

commit to memory and are an excellent supplement to our other study materials.

Each flashcard has questions on one side and the answers on the reverse.

For more information about these products, please see the Student Brochure or visit our website at

www.ActEd.co.uk. Please order online at www.ActEd.co.uk/estoreor complete an order form and send it

to us by post or fax.

If you have queries in this subject then please use the Questions forum that is within the Online

Classroom.


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