past exams subject 106 2000-2004 (1)

Faculty of Actuaries Institute of Actuaries

EXAMINATIONS

13 April 2000 (am)

Subject 106 — Actuarial Mathematics 2

Time allowed: Three hours

INSTRUCTIONS TO THE CANDIDATE

1. Write your surname in full, the initials of your other names and yourCandidate’s Number on the front of the answer booklet.

2. Mark allocations are shown in brackets.

3. Attempt all 9 questions, beginning your answer to each question on aseparate sheet.

Graph paper is not required for this paper.

AT THE END OF THE EXAMINATION

Hand in BOTH your answer booklet and this question paper.

In addition to this paper you should have availableActuarial Tables and an electronic calculator.

Faculty of Actuaries106—A2000 Institute of Actuaries

106—2

1 The profit per client-day made by a privately owned health centre depends onthe variable costs involved. Variable costs, over which the owner of the healthcentre has no control, take one of three levels θ1 = high, θ2 = most likely, θ3 =low. The owner has to decide at what level to set the number of client-days forthe coming year. Client-days can be either d1 = 16, d2 = 13.4 or d3 = 10 (each in000s). The profit (£) per client-day is as follows:

θ1 θ2 θ3

d1 85 95 110d2 105 115 130d3 125 135 150

(i) Determine the Bayes criterion solution based on the annual profits,given the probability distribution p(θ1) = 0.1, p(θ2) = 0.6, p(θ3) = 0.3. [3]

(ii) Determine both the minimax solution and the maximin solution to thisproblem. [2]

[Total 5]

2 An insurance company operates a No Claims Discount system for its motorinsurance business, with discount levels 0%, 15%, 30% and 50%. The fullannual premium is £500. The rules for moving between discount levels are:

• If no claims are made during a year, the policyholder moves to the nexthigher level of discount or remains at the maximum discount level.

• If one or more claims are made during a year, a policyholder at the 30% or50% discount level moves to the 15% discount level and a policyholder atthe 0% or 15% discount level moves to, or remains at, the 0% discountlevel.

When an accident occurs, the distribution of the loss is exponential with mean£1,000. In the event of an accident, a policyholder will claim only if the loss isgreater than the total extra premiums that would have to be paid over thenext three years, assuming that no further accidents occur.

For each discount level, calculate:

(i) the smallest loss for which a policyholder will make a claim. [3]

(ii) the probability of a claim being made in the event of an accidentoccurring. [3]

[Total 6]

106—3 PLEASE TURN OVER

3 (i) Loss amounts from a particular type of insurance have a Paretodistribution with parameters α and λ. If the company applies a policyexcess, E, find the distribution of claim amounts paid by the insurer. [3]

(ii) Assuming that α = 4 and λ = 15, calculate the mean claim amount paidby the insurer

(a) with no policy excess (E = 0),

(b) with an excess of 10 (E = 10). [2]

(iii) Using your answers to (ii), comment on the effect of introducing a policyexcess. [2]

[Total 7]

4 The proportion, θ, of staff working in a particular office who have access to theinternet at home is to be estimated. Of a sample of 50 people questioned, 29have access to the internet at home.

(i) Using a suitable uniform distribution as the prior distribution, calculatean estimate of θ under the quadratic loss function. [3]

(ii) Using instead a beta distribution, with parameters α = 4 and β = 4, asthe prior distribution for θ, calculate the Bayesian estimator for θunder the “all-or-nothing” loss function. [4]

[Total 7]

5 The following table shows incremental claims relating to the accident years1997, 1998 and 1999. It is assumed that claims are fully run-off by the end ofdevelopment year 2. Estimate total outstanding claims using the chain-laddertechnique, ignoring inflation.

Incremental Claims

Development Year

Accident Year 0 1 2

1997 2587 1091 2511998 2053 12981999 3190 [7]

106—4

6 A generalised linear model (GLM) has independent Poisson responses {Yix},with

E(Yix) = mix , Var(Yix) = mix .

The linear parameterised predictor ηix is linked to the mean response by thelog function, such that

log mix = ηix .

(i) (a) Write down an expression for the variance function V(.) for thisGLM.

(b) Evaluate the integral expression

dix = 2( )

ˆ

ixix

ix

yy t

dtV t

m

−∫

to determine an expression for the general component dix of thedeviance of this GLM in term of the observed responses yix andfitted values ˆ ixm .

(c) Write down expressions for the deviance residual and Pearsonresidual for this model. [5]

(ii) The actual deaths aix , with matching exposures rix to the risk of death,based on policy counts for UK female assured lives in 1987–90, aremodelled as the independent responses Aix of a Poisson GLM with

E(Aix) = mix = rix µix

where µix is the targeted unknown force of mortality, and

log mix = ηix = log rix + log µix .

The structure of the linear predictor is generated by the two covariates:

policy duration, denoted by the factor D(i), i = 1, 2, 3, whichrepresent the three levels 0, 1, or 2+ years respectively; and

age, denoted by the variable x, coded in 5 yearly bands,x = (mid age band − 17.5) / 5.


The term log rix is a known offset and log µix takes various linearparameterised forms as implied by the following computer output:

Fit 1:scaled deviance = 16829, residual df = 41

Add in straight-line age effects:scaled deviance = 338.83 (change = 16400),residual df = 40 (change = 1)

Add in quadratic age effects:scaled deviance = 306.71 (change = 32.12),residual df = 39 (change = 1)

Add in cubic age effects:scaled deviance = 306.51 (change = 0.1976),residual df = 38 (change = 1)

Add in duration effects:scaled deviance = 46.055 (change = 260.5),residual df = 36 (change = 2)

estimate s.e. parameter

−9.504 0.1382 10.3370 0.05627 x0.01011 0.007445 x2

−0.0001376 0.0003055 x3

0.3663 0.06365 D(2)0.6682 0.04979 D(3)

(a) List the various fitted parameterised formulae.

(b) Interpret the output and state your conclusions.

(c) What additional relevant computer output would be useful?[8]

[Total 13]

106—6

7 (i) Claims on a group of policies of a certain type arise as a Poisson processwith parameter λ1. Claims on a second, independent, group of policiesarise as a Poisson process with parameter λ2. The aggregate claimamounts on the respective groups are denoted S1 and S2 .

Using moment generating functions (or otherwise), show that S (thesum of S1 and S2) also has a compound Poisson distribution and hencederive the Poisson parameter for S. [4]

(ii) An insurance company offers accident insurance for employees. A totalof 650 policies have been issued split between two categories ofemployees. The first category contains 400 policies, and claims occur oneach policy according to a Poisson process at a rate of one claim per 20years, on average. In this category all claim amounts are £3,000. In thesecond category, claims occur on each policy according to a Poissonprocess at a rate of one claim per 10 years, on average. In this category,the claim amount is either £2,000 or £3,000 with probabilities 0.4 and0.6, respectively. All policies are assumed to be independent.

Let S denote the aggregate annual claims from the portfolio.

(a) Calculate the mean, variance and coefficient of skewness of S. [4]

(b) Using the normal distribution as an approximation to thedistribution of S, calculate Y such that the probability of Sexceeding Y is 10%. [3]

(c) The insurance company decides to effect reinsurance cover withaggregate retention £100,000, so that the insurance companythen pays no more than this amount out in claims each year. Inthe year following the inception of this reinsurance, the numbersof policies in each of the two groups remains the same but,because of changes in the employment conditions of which thecompany was unaware, the probability of a claim in group 2 fallsto zero. Using the normal distribution as an approximation tothe distribution of S, calculate the probability of a claim beingmade on the reinsurance treaty. [3]

[Total 14]

8 The aggregate claims process for a risk is a compound Poisson process withparameter λ. Individual claim amounts X are independent and identicallydistributed with density function

f(x) = 1 23

x+ e−x, x > 0.

The insurer’s premium is paid continuously at a constant rate and iscalculated so that the premium loading factor θ = 3

8 .


(i) Define the adjustment coefficient for such a process and calculate itsvalue. [11]

(ii) Define the probability of ruin ψ(u) with initial capital of u(> 0). [1]

(iii) For the detailed process described above it is possible to show that

ψ(u) = / 5 15 /113 14 14 .u ue e− −−

(a) Sketch this function.

(b) State Lundberg’s inequality and hence find an upper bound forψ(u). Comment on the result.

[3]

(iv) A company insuring the above risk is offered proportional reinsurancefrom another company which charges a premium loading factor of 50%.If the reinsurer pays 16% of each claim, derive the equation whichdetermines the adjustment coefficient for the direct insurer, but DONOT solve it.

[5][Total 20]

9 The aggregate claims X each year, from a portfolio of insurance policies, areassumed to have the normal distribution with unknown mean θ and knownvariance τ2. Prior information is such that θ is assumed to have a normaldistribution with known mean µ and known variance σ2. Independentaggregate claims over the last n years are denoted by x1 , x2 , ..., xn .

(i) (a) Derive the posterior distribution of θ.

(b) Using the answer in (a), write down the Bayesian point estimateof θ under a quadratic loss function.

(c) Show that the estimate in (b) can be expressed in the form of acredibility estimate, and derive the form of the credibility factor.

(d) Determine the limiting form for this credibility estimate as thenumber of years of data increases. [9]

(ii) The following denote the annual aggregate claims for two companiesover five years

Company: A B

Year: 1 217 1962 250 2393 249 2334 239 2225 265 244

106—8

(a) Determine the Bayes credibility estimate of the risk premiumbased on the modelling assumptions of part (i) above, in the twoseparate cases:

Case (1) Case (2)

Company: A B Company: A B

τ2 400 400 τ2 400 400µ 270 260 µ 270 260σ2 2500 2500 σ2 225 225

(b) Using the answer to (ii)(a), comment on the effect of the value ofσ2.

(c) Determine the empirical Bayes credibility estimate of the riskpremium for each of the two companies.

[12][Total 21]


EXAMINATIONS

April 2000


EXAMINERS’ REPORT

� Faculty of Actuaries

� Institute of Actuaries

Subject 106 (Actuarial Mathematics 2) — April 2000 — Examiners’ Report

Page 2

1

(i)

Annual profits

�1 �2 �3Expected profit

d1

d2

d3

1360

1407

1250

1520

1541

1350

1760

1742

1500

1576

1587.9 �1385

p(�) 0.1 0.6 0.3

decision: choose d2

(ii) minimax

max

d1

d2

d3

1760 �1742

1500

decision: choose d1

maximin

min

d1

d2

d3

1360

1407�1250

decision: choose d2

2 (i) Premiums in next three years:

Claim No claim Smallest loss forwhich will claim

0%15%30%50%

500, 425, 350500, 425, 350425, 350, 250425, 350, 250

425, 350, 250350, 250, 250250, 250, 250250, 250, 250

250425275275

(ii) 0% level: P(cost > 250) = e�250/1000 = 0.779

15% level: P(cost > 425) = e�425/1000 = 0.654

30% and 50% levels: P(cost > 275) = e�275/1000 = 0.760


Page 3

3 (i) Insurer pays X � E if X > E

0 if X � E

Define Y = X � E � X > E.

fY(y) = ( )

( )

xf y E

P X E

�

�

= 1

1

( )

( )E

E y

x dx

� ��

��

��

��

= 1( )

( )

E y

E

� ��

� ��

��

� � �

= �(� + E)� (� + E + y)��1

Y is Pareto, parameters � and � + E

(ii) E(Y) = 1

E� �

� �

(a) E(Y) = 15

4 1� = 5

(b) E(Y) = 15 10

3

� = 1

38

(iii) Small claims are excluded, so the mean claim amount increases.

A misprint occurred in part (ii) of this question. Many candidates spotted this andhighlighted it in their answer. However, the misprint was taken into account in markingscripts to ensure that nobody was disadvantaged by it.

4 (i) Let X = No. of people with access to the internet at home in a sample

of 50.

X ~ B(50, �)

f(x��) = 50� ��

�x (1 ��)50�x

f(�) = 1

f(��x) �x(1 � �)50�x

��x ~ beta(x + 1, 50 � x + 1)

x = 29 � ��x ~ beta (30, 22)


Page 4

Quadratic loss function estimate is posterior mean = 30

52.

(ii) f(x��) = 50� ��

�x (1 � �)50�x

f(�) = �3 (1 � �)3

f(��x) �x+3 (1 ��)53�x

x = 29 � f(��x) �32 (1 ��)24

“All-or-nothing” loss function estimate is posterior mode.

log f(��x) = 32 log � + 24 log (1 � �)

d

d� log f (��x) =

32 24

(1 )�

� � �

= 32(1 ) 24

(1 )

� � � �

� � �

Equate to 0 for maximum (mode)

32(1 ��) = 24�

� = 32

56

5 Cumulative claims:

2587 3678 39292053 33513190

�2 =

7029

4640 = 1.5149 , �

3 =

3929

3678 = 1.0682

Forecast cumulative claims:

3579.74832.4 5162.2

Total outstanding claims = 228.7 + 1972.2 = 2200.9


Page 5

6 (i) (a) V(m) = m

(b) d = ˆ

2y

m

y tdt

t

�

�

= � �ˆ

2 logy

my t t�

= � �ˆ2 logˆ

yy y m

m

� ��

� ��

(c) deviance residual sign ˆ( ) .y m d�

Pearson residual ˆ

ˆ

y m

m

�

(ii) (a) Formulae:

log �ix =

0

0 1

2

0 1 2

2 3

0 1 2 3

2 3

0 1 2 3 1( 0)

i

x

x x

x x x

x x x

��

��

(b) The term in x3 is not supported by the change in the deviance. Italso renders the parameter estimates

3�̂ (and

2�̂ ) non-significant

as measured by their respective t�statistics estimate

s.e.

� ��

.

Conclusion: Select the formula log �ix = �0 + �j + �

1 x + �

2 x2

(c) Additional output:

� parameter estimates and standard errors for the selected model

� residual plots of various types (e.g. vs fitted values; vs age)

A significant number of candidates did not attempt this question although those that didgenerally scored fairly well, particularly on part (ii).


Page 6

7 (i) MS(t) = E[etS]

= 1 2tS tSE e

��

= 1 2.

tS tSE e E e� � � ��

= 1 21 2( ( ) 1) ( ( ) 1)X XM t M t

e e� � � �

= 1 2 1 21 2[ ( ) ( )] ( )X XM t M t

e� ��

= [ ( ) 1]YM te� �

where � = �1 + �

2 , and MY(t) = 1 2

1 2( ) ( )

X XM t M t� � �

�.

Hence S has a compound Poisson distribution, with Poisson parameter

� = �1 + �

2 .

(ii)

(a) E[S] = 400 � 0.05 � 3,000 + 150 � 0.1 � 3,000 + 100 � 0.1 � 2000

= 125,000

V[S] = 400 � 0.05 � 3,0002 + 150 � 0.1 � 3,0002 + 100 � 0.1 � 20002

= 355,000,000

E[(S � E(S))3] = 20 � 30003 + 15 � 30003 + 10 � 20003

= 1.025 � 1012

Coefficient of skewness = 12

3 / 2

1.025 10

355,000,000

�

= 0.153

(b) S ~ N(125,000, 355,000,000)

P(S > Y) = 125,000 125,000

18,841 18,841

Y SP

� ��

� = 0.1

� Y = 12,500 + 18,841 � 1.2816 = 149,000


Page 7

(c) Now, E(S) = 60,000

V(S) = 180,000,000

S ~ N(60,000; 180,000,000)

P(S > 100,000)= P[N(0, 1) > 2.98]

= 0.0014

8 (i) The adjustment coefficient R is the solution of

�MX(r) � � � cr = 0

in the positive region for which MX(r) = E(erX) exists.

Note also, here c = (1 + �) E(X) � so look at

MX(r) ��1 ��(1 + �) E(X) r = 0

Get

MX(t) = 0

1 2

3

x�

�

� e�(1�t)x dx so require t < 1

= (1 ) (1 )1 23 3

0 0

t x t xe dx xe dx

� �

� � � �

��

But y = (1 � t) x, dy = (1 � t) dx

MX(t) = 1 2

3 3 2

1 1(1) (2)

1 (1 )t t

� � ��

MX(t) = 1 21 2

3 3(1 ) (1 )t t

� �

� � �

= 2

3

3(1 )

t

t

�

�

.

Also ( )X

M t� = 2 31 2

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

� �

� � � � � � �

� (0)X

M � = E(X) = 5

3


Page 8

We require the root of

2

3 11 51 .

8 33(1 )

r

r

r

�

� �

�

= 0

s.t. 0 < r < 1.

Look at

2

2

8(3 ) (24 55 )(1 )

24(1 )

r r r

r

� � � �

�

= 2

2

(55 86 15)

24(1 )

r r r

r

� � �

�

= 2

(11 15)(5 1)

24(1 )

r r r

r

� � �

�

so R = 0.2 (other roots are 0, 15

1)11

�

(ii) Ut = u + ct � St

St = loss (compound Poisson process)

u = initial capital

c = constant rate

Then �(u) = P(� t > 0 s.t. Ut < 0�U0 = u).

(iii) (a)

(b) Lundberg’s upper bound is e�Ru = 5u

e

�

.

This is clearly greater than �(u) �u given the specific nature of

�(u).


Page 9

(iv) Look at XI = �X with � = 0.84

E(XI) = �E(X) = 0.84 � 5

3 = 1.4 � E(XR) = E(X) � E(XI)

= 5

1.43� = 0.2667

the reinsurance premium is 1.5 � 0.2667 � � = 0.4� .

Recall that the insurer’s premium before reinsurance was

(1 + �) E(X) � = 11 5

8 3� � =

55

24�

We also require

( )I

XM t = ( )ItX

E e = E(et�X) = MX(�t) = 2

3

3(1 )

t

t

� �

� �

.

Hence the required equation is

( )I

XM r� � � �

550.4

24r

� ��

� � = 0

or

2

3 0.84

3(1 0.84 )

r

r

�

�

� 1 ��1.8917r = 0

9 (i)

(a) Given

f(�) 2

2 2

( ) 1exp exp

2 2

� � ��

� � (�2 � 2��)

and

p(x��) 2

2

2 2

11

( ) 1exp exp ( 2 )

2 2

n n

i

i

ii

x

x

��

� ��

� ��

2

2

1exp ( 2 )

2n nx� � � �

�1

=

n

i

i

x nx

�

� �� ,


Page 10

we want

p(��x) p(x��) p(�)

2

2 2 2 2

1exp

2 2

n nx� ��

2 2 2 2

2

2 2 2 2exp 2

2

n nx

n

� ��

22 2 2 2

2 2 2 2exp

2

n nx

n

� ��

� ��x ~ 2 2 2 2

2 2 2 2 2 2,

nN x

n n n

� ��

� � � � � � � � �

(b) The posterior mean (the point estimator under quadratic loss) is

( )E x�� = 2 2

2 2 2 2

n

x

n n

� ��

� � � � � � .

(c) ( )E x�� = (1 � Z) � + Zx

where

Z = 2

2 2

n

n

�

� � �

= 2

2

n

n

��

�

is the credibility factor. Hence ( )E x�� can be expressed in the form of a

credibility estimate.

(d) As n �, Z = 2

2

2

limn

n

��

�

��

1 and ( )E x x��

(ii) (a) A B

Wanti

x 244 226.8 with n = 5.

(1) A B (2) A B

�2 400 400 400 400

� 270 260 270 260

�2 2500 2500 225 225


Page 11

For both companies Z = 2

2

2

n

�

��

= 0.969 Z = 0.7377 (0.738)

Credibility premiums 0.969 0.031i

x � � 0.738 0.262i

x � �

A 244.8 250.8B 227.8 235.5

(b) With larger � in case (1), prior knowledge is more vague. This is reflected

in the lower weighting given to � relative to i

x .

(c) A B

Wanti

x 244 226.8 � x = 235.4 with N = 2, n = 52

is 314 363.7

21( )

1i

i

x xN

�

�

� = 147.92

Then 2ˆ ( ( ))E S � = 21i

i

SN� = 338.85

ˆVar( ( ))m � = 2 21 1 1( )

1i i

i i

x x SN n N

� �

�

� � = 147.92 �338.85

5 = 80.15

and Z = 80.15

338.8580.15

5�

= 0.5418

Credibility premiums are 0.5418 0.4582 = 0.5418 107.86i i

x x x� �

A 240.1B 230.7


EXAMINATIONS

14 September 2000 (am)




1. Write your surname in full, the initials of your other names and yourCandidate’s Number on the front of the answer booklet.






In addition to this paper you should have available, ActuarialTables and an electronic calculator.

� Faculty of Actuaries106—S2000 � Institute of Actuaries

106—2

1 An actuarial student is considering his post-examination holiday plans. Histravel agent tells him that there will definitely be vacancies in either Hotel A orHotel B but not at both. He must book his flight now to take advantage ofcurrent huge discounts on flight prices. Flights to airport 1 cost £110 and flightsto airport 2 cost £160. Accommodation at Hotel A costs £890 and accommodationat Hotel B costs £490. If he chooses to fly to airport 1, a taxi to Hotel A will cost£10 and a taxi to Hotel B will cost £70. If he chooses to fly to airport 2, a taxi toHotel A will cost £65 and a taxi to Hotel B will cost £5. The student believesthere is a 90% chance of the vacancies being at Hotel B. To which airport shouldhe book a flight in order to minimise the expected cost? [4]

2 In the context of Generalised Linear Models, consider the exponentialdistribution with density function f(x), where

f(x) = /1 xe− µ

µ(x > 0).

(i) Show that f(x) can be written in the form of an exponential family ofdistributions. [1]

(ii) Show that the canonical parameter, θ, is given by θ =1−µ

. [1]

(iii) Determine the variance function and the dispersion parameter. [3][Total 5]

3 The table below shows annual aggregate claim statistics for 3 risks over 4 years.Annual aggregate claims for risk i, in year j are denoted by Xij .

Riski

414

1

=i ijj

X X=�

42 21

31

= ( )i ij ij

s X X=

−�

1 2,517 4,121,2802 7,814 7,299,1753 2,920 3,814,001

(i) Calculate the value of the credibility factor for Empirical Bayes Model 1.[3]

(ii) Using the numbers calculated in (i) to illustrate your answer, describe theway in which the data affect the value of the credibility factor. [4]

[Total 7]


4 In a large portfolio of non-life policies involving a new product, let θ denote theproportion of policies on which claims are made in the first year. The value of θis unknown and is assumed to have a beta prior distribution with parameters αand β and mean µ0.

(i) If a random sample of n such policies gives rise to x claims in the firstyear, shown that the posterior mean of θ is given by

wn µ0 + ( 1 − wn)xn

expressing the weight wn as a function of α, β and n. [3]

(ii) Two alternative assessments A and B, of the prior probability densityfunction of θ are made as follows:

fA(θ) = 3(1 − θ)2 0 ≤ θ ≤ 1,

fB(θ) = 4θ3 0 ≤ θ ≤ 1.

81 claims subsequently arise during the year from 1,000 randomlyselected policies.

(a) Sketch the two prior densities and comment briefly on the natureof these two sets of prior beliefs.

(b) Determine the posterior Bayes estimate of θ for each priorassessment based on the squared error loss function, and commentbriefly on these posterior estimates.

[5][Total 8]

106—4

5 An insurance company has a portfolio of policies with a per-risk excess of lossreinsurance arrangement with a deductible of M(>0). Claims made to the directinsurer, denoted by X, have a Pareto distribution with cumulative distributionfunction

F(x; α) = 1 − 200.

200 x

α� �� +� �

There were a total of n claims from the portfolio. Of these, l were for amountsless then the deductible. The claims less than the deductible are

{xi : i = 1, 2, ..., l). The value of the statistic1

log(200 )l

ii

x=

+� = y is given.

(i) Show that the maximum likelihood estimate of α, α̂ , is given by

α̂ =( ) log(200 ) log 200

ln l M n y− + − +

[5]

(ii) From last year’s experience we have the following information:

M = 600, n = 500, l = 400, y = 2209.269

(a) Use the expression derived in (i) to verify that the maximumlikelihood estimate of α based on these figures is α̂ = 1.75.

(b) Assuming that α = 1.75, estimate the average amounts paid by theinsurer and the reinsurer on a claim made during the year. [6]

[Total 11]

6 The tables below show the cumulative cost of incurred claims and the number ofclaims reported each year for a certain cohort of insurance policies. The claimsare assumed to be fully run-off at the end of development year 2.

Cumulative cost of incurred claims:

DevelopmentAccident Year 0 1 2

012

288465773

634980

893

The numbers of claims reported in each year are:

DevelopmentAccident Year 0 1 2

012

110167285

85113

55

Given that the total amount paid in claims to date, relating to accident years 0, 1and 2, is £2,750, calculate the outstanding claims reserve using the average costper claim method. [11]


7 (i) Define the adjustment coefficient for a compound Poisson process, R, anddescribe how it can be used to assess reinsurance arrangements on thebasis of security. [4]

(ii) Claims occur on a portfolio of insurance business according to a Poissonprocess. Individual claims are a fixed amount of £500 and the insurancecompany uses a premium loading of 0.15.

A reinsurer offers excess of loss reinsurance for which it uses a premiumloading of 0.2, and the insurance company is considering purchasing thisreinsurance, with a retention of m (0 ≤ m ≤ 500).

(a) Derive an equation for the adjustment coefficient, R, in terms of m.

(b) By differentiating this equation with respect to m, show that Rsatisfies the following differential equation.

1.2R + (1.2m − 25)dRdm

= mRdRR m e

dm� �+� ��

[Hint: It should be noted that R is a function of m.]

(c) Determine the optimal value of m by settingdRdm

= 0 and

substituting for R in the adjustment coefficient equation derivedin (a).

(d) Compare the values of R and the expected profit with noreinsurance with the values with the optimal value of m, andcomment.

[Note: It is not necessary to calculate the value of R when there isno reinsurance.] [13]

[Total 17]

106—6

8 An insurance company operates a no claims discount (NCD) system with threecategories: 0%, 30% and 60% discount. The rules for moving between thesecategories are as follows:

If a policyholder makes no claims in a year, they move to the next higherdiscount level (or stay at the highest level) next year.

If a policyholder makes one or more claims, they move to the 0% level next year(or stay there).

Suppose that the probability that a policyholder does not make a claim(irrespective of discount level) is p, and that the premium in the 0% discountlevel is c.

(i) Write down the transition matrix. [2]

(ii) Derive the steady state distribution of policyholders in each discount level,in terms of p. [5]

(iii) (a) Calculate the average premium paid in the steady state, A, interms of p and c.

(b) By considering the value of A when p = 0.9 (A0.9), and p = 0.8 (A0.8),comment on the effectiveness of the NCD system.

(c) The company is considering changing the NCD system so that thediscount levels are 0%, 100d% and 200d%. (The present NCDsystem has d = 0.3.) Determine the value of d such thatA0.8 = 1.5A0.9, and comment on this. [10]

[Total 17]

106—7

9 The claim amount in the coming year from a specific type of risk is modelled asthe random variable

X = IB

where I is an indicator random variable and B is a random variable representingthe amount of the claim given that the claim occurs. Let µ and σ2 denote themean and variance of B respectively. The indicator random variable is such thatP(I = 1) = q, where I = 1 indicates that a claim takes place in the year, otherwiseP(I = 0) = 1 − q. For this type of risk, at most one claim can be made in the year.

(i) (a) By deriving an expression for the moment generating function of Xin terms of the moment generating function of B, or otherwise,show that the mean and variance of X are given by

E[X] = qµ , Var(X) = qσ2 + q(1 − q) µ2

(b) Suppose that B = a + bZ (£000s; a, b > 0) where Z has a Poissondistribution with parameter λ, so that B takes the values a, a + b,a + 2b, .… Determine expressions for the mean, variance andprobability distribution of X in terms of a, b, λ and q. [9]

(ii) Consider further two types of the above risk:

Risk Type Claim Amount inComing Year

Parameters

AB

XY

a = 2, b = 1, λ = 1.5, q = 0.1a = 3, b = 2, λ = 1.0, q = 0.2

(a) Given that the two types of risk are independent, calculate theprobability that X + Y is greater than 5 (£000), without resorting toany approximations.

(b) A portfolio consists of 100 independent type A risks and 50independent type B risks. Calculate the probability that theaverage claim amount per risk in the coming year is greater than2 (£000), by using an appropriate approximation. [11]

[Total 20]


EXAMINATIONS

September 2000


EXAMINERS’ REPORT

� Faculty of Actuaries� Institute of Actuaries

Subject 106 (Actuarial Mathematics 2) — September 2000 — Examiners’ Report

Page 2

1 θ1 = There are vacancies at Aθ2 = There are vacancies at Bα1 = Student chooses to fly to airport 1α2 = Student chooses to fly to airport 2

Then

α1 α2

θ1 1010 1115θ2 670 655

Expected costs:

α1 = 10%. 1010 + 90%. 670 = 704α2 = 10%. 1115 + 90%. 655 = 701

and so he should choose airport 2.

2 (i) f(x) =1

x

e−

µ

µ= exp log

x� �− − µ� �µ� �

which is in the form of an exponential family

(ii) The canonical parameter is θ =1−µ

(iii) b(θ) = log µ =1

log � �−� �θ� �= − log (−θ)

b ′(θ) =1

( 1)− × −−θ

= 1−θ

b ′′(θ) =2

1θ

= µ2

Hence V(µ) = µ2

The dispersion parameter is 1.


Page 3

3 (i) E[s2(θ)]: 5078152V[m(θ)]: 7425771

z =2

4=

5078152[ ( )] 47425771[ ( )]

nE s

nV m

θ ++θ

= 0.854

(ii) z depends on n, E[s2(θ)] and V[m(θ)].

E[s2(θ)] measures the variation of the data within risks and V[m(θ)]measures the variation between the risks. It can be seen that thevariation between risks is relatively large, which means that more weightis given to the individual risks (z is close to 1). If n were larger, z wouldalso be larger because there would be more information on each individualrisk.

4 (i) Given p(θ) α θα−1 (1 − θ)β−1 with µ0 = E(θ) =α

α + β

We observe p(xθ) α θx(1 − θ)n−x

and we want p(θx) α p(xθ) p(θ) α θα+x−1 (1 − θ)β+n−x−1

so that µ = E(θx) = = =x x x

x n x n n nα + α + α +

α + + β + − α + β + α + β + α + β +

=n x

n n nα + β α +

α + β + α + β α + β +

= wn µ0 + (1 − wn)xn

where wn =n

α + βα + β +


Page 4

(ii) (a)

(b) With a squared error loss function, look at the posterior means.

Get

A: α = 1, β = 3 � wn =4

4 n+= 0.003984

E(θx) = 0.003984 × 14 + 0.996016 × 81

1000= 0.0817

B: α = 4, β = 1 � wn =5

5 n+= 0.004975

E(θx) = 0.004975 × 45 + 0.995025 × 81

1000= 0.0846

Results are nearly identical.

The sample is large, so prior beliefs are dominated by theinformation from the data.

5 (i) f(x) = F ′(x) = 1

200(200 )x

α

α+α

+

The likelihood, L(α) is given by:

L(α) =( )

11

200 200200(200 )

n ll

i i mx

α −α

α+=

� �α × � �++ � �∏

= 1( )

1

200(200 )

(200 )

ll n

in li

xm

α−α−

α −=

α+

+ ∏

0 1 θ

BA

A: α = 1, β = 3, E(θ) = 14

B: α = 4, β = 1, E(θ) = 45

priors are “opposites” with Aplacing emphasis on a low valueof θ and B placing emphasis on ahigh value of θ. Note for examplethe prior means.


Page 5

l(α) = log L(α) = l log α + αn log 200 − α(n − l) log(200 + M)

− (α + 1)1

log(200 )l

ii

x=

+�

l∂∂α

=lα

+ n log 200 − (n − l) log(200 + M)

−1

log(200 )l

ii

x=

+�

2

2

l∂∂α

=2

l−α

which is negative and so this is a maximum

l∂∂α

= 0 when α = α̂ , where

α̂ =

1

( ) log(200 ) log(200) log(200 )l

ii

l

n l M n x=

− + − + +�

(ii) (a) α̂ =400

100 log 800 500 log 200 2209.269− +=

400228.57

= 1.75

(b) We require E(Y) =0

( ) ( )M

xf x MP X M+ >�

= E(X) − ( ) ( )M

x M f x dx∞

−�

E(X) =1

λα −

=2000.75

= 266.67

Also ( ) ( )M

x M f x dx∞

−� =0

( )zf z M dz∞

+�

= 10

200 800800 (800 )

zdz

z

α α∞

α α+α

+�

=14 1

α λ� �� α −� �

=1 8004 1

α� �� α −� �

=800

11.3137 0.75×

= 94.28

Thus E(Y) = 266.67 − 94.28 = 172.39.


Page 6

6 Note that it is acceptable to use either the grossing-up method or the chain-laddermethod to calculate the projected number of claims.

Cumulative number of claims reported:

110 195 250167 280285

Average incurred cost per claim:

ultimate2.618 3.251 3.572 3.57273.29% 91.01% 100%

2.784 3.500 3.84672.39% 91.01%

2.712 3.72372.84%

Projected ultimate average cost per claim in bold above.

Calculate projected numbers of claims:

ultimate110 195 250 25044.00% 78.00% 100%

167 280 35946.52% 78.00%

285 63045.26%

Projected total claims:

3.572 × 250 = 8933.846 × 359 = 13813.723 × 630 = 2345

Total 4619

less claims paid −2750

o/s claims reserve 1869


Page 7

7 (i) R is defined as the smallest positive solution to:

λ + cR = λMX(R)

If there is no positive solution, then R = 0.

where c is the rate of premium income received by the insurerλ is the claim frequencyMX(R) is the MGF of the individual claim size distribution

Security can be assumed by the probability of ruin ψ(u). By Lundberg’sinequality, ψ(u) ≤ e−Ru, we see that a large value of R is good for security.

(ii) (a) c = 500 × 1.15λ

=

0=

I

R

X X X mm X m

X mX

X m X m

≤�� >�

≤�� − >�

Hence XI = m and XR = 500 − m

∴ cR = 1.2 × (500 − m) λ

and ( )IXM R = emR

The adjustment coefficient equation is

λ + (500 × 1.15λ − 1.2 × (500 − m) λ) R = λemR

i.e. 1 + (575 − 600 + 1.2m) R = emR

i.e. 1 + (1.2m − 25) R = emR

(b) 1 + (1.2m − 25) R(m) = emR(m)

Differentiate wrt m:

1.2R(m) + (1.2m − 25)( )dR m

dm= ( )( )

( ) mR mmdR mR m e

dm� �+� ��


Page 8

(c) Put( )dR m

dm= 0:

1.2R(m) = R(m) emR(m)

Hence 1.2 = emR(m)

∴ R(m) = log(1.2) / m

Substituting into the adjustment coefficient equation:

1 + (1.2m − 25)log 1.2

m= elog 1.2 = 1.2

∴ 1 + 1.2 log 1.2 − 25 log 1.2m

= 1.2

∴ m =25 log 1.2

1.2 log 1.2 0.2−= £242.6 £243�

(d) When there is no reinsurance, the value of R will be lower thanwhen m = £243. The expected profit is

No reinsurance: 0.15 × 500 × λ = 75λ

With reinsurance: 500 × 1.15λ − 1.2 × 257.4λ − 242.6λ

= 23.5λ

So security is higher, but expected profit is lower when m = £243.It is a trade-off between these two factors.

8 (i) The transition matrix is

1 01 01 0

p pp pp p

−� �� −� �� −� �

(ii) πp = π

(π0 π1 π2)1 01 01 0

p pp pp p

−� �� −� �� −� �

= (π0 π1 π2)


Page 9

(1 − p) (π0 + π1 + π2) = π0

pπ0 = π1

p(π1 + π2) = π2

Since (π0 + π1 + π2) = 1, π0 = 1 − p

∴ π1 = p(1 − p)

and pπ1 = (1 − p) π2

∴π2 =1

pp−

π1 = p2

The steady state distribution is

(1 − p, p(1 − p), p2)

(iii) (a) A = ( 1 − p) c + p (1 − p) × 0.7c + p2 × 0.4c

= c(1 − p + 0.7p − 0.7p2 + 0.4p2)

= c(1 − 0.3p − 0.3p2)

= c(1 − 0.3p (1 + p))

(b) A0.9 = 0.487c

A0.8 = 0.568c

Although P(claim) is twice as much in A0.8 as in A0.9 , the averagepremium is only slightly higher. The NCD system is not effective.

(c) A = (1 − p) c + p (1 − p) × (1 − d) c + p2 (1 − 2d) c

= c (1 − p + p − pd − p2 + p2 d + p2 − 2p2 d)

= c (1 − pd − p2 d)

= c (1 − p (1 + p) d)

c (1 − 0.8 (1.8) d) = 1.5c (1 − 0.9 (1.9) d)

1 − 1.44d = 1.5 − 2.565d

1.125d = 0.5

d = 0.444


Page 10

To get a premium 50% higher, the discount categories have to be44% and 89%. Clearly this is not practical. It might be moreappropriate to have the premium 100% higher, but this wouldmake the discount categories even more unrealistic.

9 (i) (a) MX(t) = E(etX) = E{E(etXI)}

= E(etXI = 0) P(I = 0) + E((etXI = 1)) P(I = 1)

= 1 . (1 − q) + MB(t) . q

Then

( )XM t′ = ( )BqM t′ � E(X) = qE(B) = qµ

and

( )XM t′′ = ( )BqM t′′ � E(X2) = qE(B2)

so that

E(X) = qµ

Var(X) = E(X2) − {E(X)}2

= q{Var(B) + {E(B)2} − q2µ2

= q{σ2 + µ2} − q2µ2 = qσ2 + q(1 − q) µ2

(b) B = a + bZ with Z ~ Poi(λ) ∴ E(Z) = Var(Z) = λ

so µ = E(B) = a + bλ ∴ E(X) = q(a + bλ)

σ2 = Var(B) = b2λ ∴ Var(X) = qb2λ + q(1 − q) (a + bλ)2

Also

MB(t) = E(etB) = E(et(a+bZ)) = eat MZ(bt) = eat ( 1)bteeλ −


Page 11

Then

MX(t) = qeat ( 1)bteeλ − + 1 − q

= qe−λ eat bteeλ + 1 − q

= qe−λ( )

0

1!

i a ib t

i

eq

i

∞ +

=

λ+ −�

X takes values 0, a, a + b, a + 2b, ...

so selecting coefficients of ext gives

P(X = 0) = 1 − q

P(X = a + b(x−1)) =1

( 1)!

xqex

−λ −λ−

; x = 1, 2, 3, ...

(ii) (a) X takes values 0, 2, 3, 4, 5, ...

Y takes values 0, 3, 5, 7, ...

We can compute P(X + Y ≤ 5) as a convolution:

x pX(x) y pY(y)

023456

0.90.0223130.0334700.0251020.012551....

0357

0.80.0735760.073576....

X + Y sample points(x, y)

probability

02345

(0, 0)(2, 0)(0, 3); (3, 0)(4, 0)(0, 5); (2, 3); (5, 0)

0.7200000.0178500.0929940.0200820.0779010.928827

∴ P(X + Y > 5) = 0.071173


Page 12

(b) E(X) = 0.1 (2 + 1.5) = 0.35;

V(X) = 0.1 × 1 × 1.5 + 0.1 × 0.9 × 3.52 = 0.15 + 1.1025 = 1.2525

E(Y) = 0.2 (3 + 2) = 1

V(Y) = 0.2 × 4 × 1 + 0.2 × 0.8 × 52 = 0.8 + 4 = 4.8

c =100 50

1 1

1150 i j

i j

X Y= =

� �+� ��

� ��

( )E c =1

150(100 E(X) + 50 E(Y)) = 0.5667

( )V c =2

1150

(100 Var(X) + 50 Var (Y)) = 0.016233

Under the normal modelling distribution, we require

1 −2 0.5670.12741

−� �Φ � ��

= 1 − Φ (11.25) = 0


EXAMINATIONS

5 April 2001 (am)

Subject 106 � Actuarial Mathematics 2



1. Write your surname in full, the initials of your other names and yourCandidate�s Number on the front of the answer booklet.







� Faculty of Actuaries106�A2001 � Institute of Actuaries

106 A2001�2

1 The triangle below shows incremental claims for a portfolio of general insurancepolicies. The data have already been adjusted to take account of inflation effects.Calculate the basic chain ladder development factors, and the implied grossing-up factors.

2,541 1,029 2172,824 7901,981 [5]

2 A generalised linear model has independent Binomial responses Z1 , �, Zk withE(Zi) = nµ, Var(Zi) = nµ(1 − µ) for 0 < µ < 1.

(i) Show that Yi = Zi /n belongs to an exponential family. [2]

(ii) Identify the natural parameter and the canonical link function, and derivethe variance function. [4]

[Total 6]

3 The aggregate claims process for a particular risk is a compound Poisson processwith λ = 20. Individual claim amounts are £100 with probability ¼, £200 withprobability ½, or £250 with probability ¼. The initial surplus is £1,000. Using aNormal approximation, calculate approximately the smallest premium loadingfactor θ such that the probability of ruin at time 3 is at most 0.05. [7]

4 (i) Let S = X1 + � + XN , where X1 , X2 , � are independent, identicallydistributed random variables, and N is a random variable independent ofthe Xi�s. Derive an expression for the moment generating function of S interms of the probability generating function of N and the momentgenerating function of Xi . [2]

(ii) In a group of policies, the monthly number of claims for a single policy hasa Poisson distribution with parameter λ, where λ is a random variablewith density f(λ) = 2e−2λ, λ > 0.

(a) Show that the probability of n claims on a policy picked at random

from the group is 1

23n+ , n = 0, 1, 2, �

(b) Find the moment generating function for the aggregate claimsdistribution if the claims have a gamma distribution with mean 2and variance 2. [6]

[Total 8]

106 A2001�3 PLEASE TURN OVER

5 A company is considering setting up a new class of insurance. All risks will be atone of four possible levels of intensity, I1 , I2 , I3 and I4. The company has todecide at what level to set the premium, which will attract differing amounts ofbusiness as follows:

Annual premium £85 £81 £79No. of policies (�000s) 100 150 200

The company has annual fixed costs of £1.5 million plus annual per policyexpenses of £18. Under each level of intensity the company expects to pay out anaverage in claims per policy of:

Level of intensity I1 I2 I3 I4

Average cost (£) 40 45 57 60

(i) Determine the minimax solution based on annual profits. [7]

(ii) Given the probability distribution p(I1) = 0.1, p(I2) = 0.4, p(I3) = 0.3,p(I4) = 0.2, determine the Bayes criterion solution based on annual profits.

[2][Total 9]

6 An insurance company is monitoring the length of time staff take to pick uptelephones after they first ring. It is assumed that the time follows anexponential distribution with parameter θ.

10 calls are monitored at random and the average response time is calculated as3.672 seconds.

(i) (a) Show that the gamma distribution is the conjugate priordistribution for θ .

(b) Assuming that the prior distribution for θ has mean 0.315 andstandard deviation 0.251, derive the posterior distribution of θ andcalculate the Bayesian estimator of θ under quadratic loss. [7]

(ii) A further 70 calls are monitored and have the same average response timeof 3.672 seconds. Calculate the Bayesian estimator of θ under quadraticloss using all the data collected. [2]

(iii) Comment on your answers in (i) and (ii) [2][Total 11]

106 A2001�4

7 A no claims discount system has four levels, 0%, 25%, 50% and 75%. The rulesfor moving between levels are as follows:

If no claims are made in one year, the policyholder moves to the nexthigher level, or remains at the 75% level;

If one claim is made in one year, the policyholder moves down one level, orremains at the 0% level;

If two or more claims are made, the policyholder moves straight down to,or remains at, the 0% level.

Policyholders at different levels are found to experience different rates ofclaiming. The number of claims made per year follow a Poisson distribution withparameter λ as follows:

Level 0% 25% 50% 75%λ 0.29 0.22 0.18 0.10

(i) Derive the transition matrix. [6]

(ii) Calculate the proportions at each different level when the system reachesa steady state. [9]

[Total 15]

8 For each of m independent policies, the probability of one claim in a year isθ (0 < θ < 1) and the probability of no claims in a year is 1 − θ. The total numberof claims in one year is a random variable X. Independent observations x1 , �, xn

of X are available. The prior distribution of θ has density f(θ) ∝ {θ(1 − θ)}β−1,0 < θ < 1, for some constant β > 0.

(i) (a) Derive the posterior distribution of θ given x1 , �, xn .

(b) Derive the maximum likelihood estimate of θ, ( ).g x�

(c) Derive the Bayesian estimate of θ under quadratic loss, and showthat it takes the form of a credibility estimate,

( )Zg x�

+ (1 − Z) µ

where µ is a quantity you should specify in terms of the priordistribution of θ.

(d) Explain what happens to Z as the number of data points increases.[11]

106 A2001�5

(ii) Calculate the Bayesian estimate of θ and the value of Z if n = 6, m = 10and x1 = 1, x2 = 4, x3 = 2, x4 = 1, x5 = 1, x6 = 3, when

(a) β = 1

(b) β = 4

By considering the prior variance, comment on the effect on Z ofincreasing β, and relate this effect to the quality of the prior informationabout θ in each case. [6]

[Total 17]

9 (i) A random variable X has the lognormal distribution with density functionf(x) and parameters µ and σ. Show that for a > 0

∞ � �� σ − µ − σµ + − Φ� �� σ� � � ��

2 2log( ) = exp 1

2a

axf x dx

where Φ is the cumulative distribution function of the standard normaldistribution. [4]

(ii) Claims under a particular class of insurance follow a lognormaldistribution with mean 9.070 and standard deviation of 10.132 (figures in£000s). In any one year 20% of policies are expected to give rise to aclaim.

An insurance company has 200 policies on its books and wishes to takeout individual excess of loss reinsurance to cover all the policies in theportfolio. The reinsurer has quoted premiums for two levels ofreinsurance as follows (figures in £000s):

Retention Limit Premium

25 48.530 38.2

(a) Calculate the probability, under each reinsurance arrangement,that a claim arising will involve the reinsurer.

(b) By investigating the average amount of each claim ceded to thereinsurer, calculate which of the retention levels gives the bestvalue for money (ignoring the insurer�s attitude to risk).

(c) The following year, assuming all other things equal, the insurerbelieves that inflation will increase the mean and standarddeviation of the claims in its portfolio by 8%. If the reinsurercharges the same premiums as before, which of the retention levelswill be best value for money next year?

[18][Total 22]


EXAMINATIONS

April 2001


EXAMINERS� REPORT


Subject 106 (Actuarial Mathematics 2) � April 2001 � Examiners� Report

Page 2

Examiners Comments:

In the numerical questions, students were not unduly penalised for errors in earlier parts of eachquestion which affected their answers to the rest of the question. This mostly occurred in question 7for students who had not calculated the transition matrix correctly. Marks were deducted for errors inthe transition matrix but full credit was given for the later parts of the question where students usedthe correct method to determine the proportions at each NCD level and their final answer were wrongonly due to the original mistake.

In question 5, some students rightly concluded that ignoring fixed costs would not affect the finalanswer hence their figures calculated differed from the examiners' solution. In these cases, full markswere awarded where the student explicitly stated that fixed costs could be ignored.


Page 3

1 Cumulative claims:

2,541 3,570 3,7872,824 3,6141,981

λ3 = 3,7873,570

= 1.06078

λ2 = 7,1845,365

= 1.33905

Grossing up factors:

94.27%70.40%

2 (i) fY (y; θ, ϕ) = n

ny� ��

µny (1 − µ)n−ny

= exp log log(1 ) log1

nn y

ny

� �� µ� �+ − µ +� �� − µ � ��

This is of the form ( )

exp ( , )( )

y bc y

a� �θ − θ + ϕ� �ϕ� �

i.e. exponential family form

(ii) The natural parameter is

θ = log1� �µ� �− µ� �

so =1

ee

θ

θ

� �µ � �

+� �

ϕ = n, a(ϕ) = 1ϕ

b(θ) = −log(1 − µ) = 1

log1 eθ

� �− � �+� �

= log(1 + eθ)

c(y, ϕ) = logn

ny� ��


Page 4

Canonical link is g(µ) = θ(µ)

so ( ) = log1

g� �µµ � �− µ� �

V(µ) = ( )b′′ θ

( )b′ θ = 1

ee

θ

θ+

( )b′′ θ = 2(1 )

ee

θ

θ+= µ(1 − µ)

so V(µ) = µ(1 − µ) is the variance function

3 Let X be a typical claim. Then

E(X) = ¼ × 100 + ½ × 200 + ¼ × 250 = 187.5

E(X2) = ¼ × 1002 + ½ × 2002 + ¼ × 2502 = 38,125

Let S be the aggregate claims at time 3, then

E(S) = E(X) × 3λ = 11,250

Var(S) = E(X2) × 3λ = 2,287,500

standard deviation of S is 1,512.4483

So S is approximately N(11,250, 2,287,500)

Want 0.05 ≥ P(1,000 + (1 + θ) E(X) . 3λ − S < 0)

= P(S >1,000 + (1 + θ) × 11,250)

= + + θ × −� �>� �

� �

1,000 (1 ) 11,250 11,250(0,1)

1512.4483P N

= + θ� �>� �

� �

1,000 11,250(0,1)

1512.4483P N

So+ θ >1,000 11,250

1.6451512.4483

i.e. θ > 0.1323


Page 5

4 (i) The moment generating function of S is

MS(t) = E(eSt) = E[E(eStN)]

= E[{E(eXt)}N]

= ( ( ))N XG M t

(ii) (a) Let N be the number of claims.

P(N = n) = 2

0

( = ) 2P N n e d∞

− λλ λ�

= 2

0

2!

nee d

n

∞ −λ− λλ λ�

= ∞ + − λ

+λ λ�

1 3

10

2 3!3

n n

n

ed

n

= 1

23n+ n = 0, 1, 2, �

(b) N has pgf 1

0 0

2 2 2= =

3 3 33

nn

n

ss

s

∞ ∞

+� �� −� �

� �

For a gamma(α,λ), 2

mean = / = 2= 2 = 1

variance = / = 2α λ �

� α λ�α λ �

So MX(t) = 2

11 t� �� −� �

So MS(t) = 2 2

2 2

2

2 2(1 ) 2(1 )= =

1 3(1 ) 1 3 6 23(1 )

t tt t t

t

− −− − − +−

−

Note: candidates do not have to simplify the expression MS(t): anycorrect expression is satisfactory.


Page 6

5 (i) d1 100 policies premium £85 per annumd2 150 policies premium £81 per annumd3 200 policies premium £79 per annum

θ1 Intensity I1 claim costs £40 per policy per annumθ2 Intensity I2 claim costs £45 per policy per annumθ3 Intensity I3 claim costs £57 per policy per annumθ4 Intensity I4 claim costs £60 per policy per annum

Figures in £000sStrategy d1 d2 d3

Total premiumsFixed expensesPer policy expenses

8,5001,5001,800

12,1501,5002,700

15,8001,5003,600

Premium less expenses 5,200 7,950 10,700

Hence annual profits (£000s)

θ1 θ2 θ3 θ4

d1 1,200 700 −500 −800d2 1,950 1,200 −600 −1,050d3 2,700 1,700 −700 −1,300

Minimax = minimise maximum loss

d1 −800 ← choose d1, set premiums at £85 per annumd2 −1,050d3 −1,300

(ii) Bayes criterion

d1 = 0.1 × 1,200 + 0.4 × 700 − 0.3 × 500 − 0.2 × 800 = 90d2 = 0.1 × 1,950 + 0.4 × 1,200 − 0.3 × 600 − 0.2 × 1,050 = 285d3 = 0.1 × 2,700 + 0.4 × 1,700 − 0.3 × 700 − 0.2 × 1,300 = 480 ← choose d3

Choose d3 , premiums of £79 per annum


Page 7

6 (i) (a) Likelihood function

( )f xθ�

∝ 1

i

nx

i

e−θ

=

θ∏

∝ 1

n

ii

xne =

−θ

θ�

As a function of θ, this is in the form of a gamma distribution.

Hence the conjugate prior distribution is

f(θ) = 1

( )eα α− −λθλ θ

Γ α

(b) Find values for the parameters α and λ

E(θ): α / λ = 0.315

V(θ): α / λ2 = 0.2512

λ = 0.315 / 0.2512

= 5.000

α = 0.315λ

= 1.575

Posterior = Prior × Likelihood

∝ λα θα−1 e−λθ. θn ixe−θΣ

∝ θn+α−1 ( )ixe−θ λ+Σ

Hence Posterior

= gamma (n + α, λ + Σxi)

= gamma (11.575, 41.720)

Estimator under quadratic loss = mean of the Posterior

= 11.575 / 41.720 = 0.27744


Page 8

(ii) Hence Posterior

= gamma (n + α, λ + Σxi)

= gamma (81.575, 298.760)

Estimator under quadratic loss = mean of the Posterior

= 81.575 / 298.760 = 0.27305

(iii) The mean of the prior distribution is 0.315.

The maximum likelihood estimate of θ is 1

3.672 = 0.272. As more data is

collected, the Bayesian estimate approaches the maximum likelihoodestimate.

7 (i) 0% level P(0 claims) = e−0.29 = 0.7483P (≥1 claim) = 1 −0.7483 = 0.2517

25% level P(0 claims) = e−0.22 = 0.8025P(≥ 1 claim) = 1 − 0.8025 = 0.1975

50% level P(0 claims) = e−0.18 = 0.8353P(1 claim) = 0.18e−0.18 = 0.1503P(≥ 2 claims) = 1 − 0.8353 − 0.1503 = 0.0144

75% level P(0 claims) = e−0.10 = 0.9048P(1 claim) = 0.10e−0.10 = 0.0905P(≥ 2 claims) = 1 − 0.9048 −0.0905 = 0.0047

Therefore the transition matrix is:

0.2517 0.7483 0 00.1975 0 0.8025 00.0144 0.1503 0 0.8353

0.0047 0 0.0905 0.9048

� ��

(ii) In a steady state

0.2517 0.7483 0 00.1975 0 0.8025 0

=0.0144 0.1503 0 0.8353

0.0047 0 0.0905 0.9048

� �� Π Π� ��


Page 9

0.2517 π0 + 0.1975 π1 + 0.0144 π2 + 0.0047π3 = π0

0.7483 π0 + 0.1503 π2 = π1

0.8025 π1 + 0.0905 π3 = π2

0.8353 π2 + 0.9048 π3 = π3

and π0 + π1 + π2 + π3 = 1

Hence

π3 = 8.7742 π2

π2 = 3.8969 π1

π1 = 1.8062 π0

π0 + 1.8062 π0 + 7.0386 π0 + 61.7579 π0 = 1

π0 = 0.0140π1 = 0.0252π2 = 0.0983π3 = 0.8625

8 (i) (a) f(θ) ∝ θβ−1 (1 −θ)β−1

f(xθ) = 1

(1 )i i

nx m x

ii

m

x−

=

� �θ − θ� �

� �∏

∝ (1 )i ix nm xΣ −Σθ − θ

so f(θx) ∝ 1 1(1 )i ix nm xΣ +β− +β−Σ −θ − θ

so θx ~ Beta(Σxi + β, nm + β − Σxi)

(b) l = log like (θ) = (Σxi) log θ + (nm − Σxi) log (1 − θ) + constant

1i ix nm xl −∂ = −

∂θ θ − θ� � = ( )1

ixnmnm

� �� − θ� �θ − θ � �

�


Page 10

This is zero when Σxi (1 − θ) = (nm − Σxi) θ

i.e. �θ = ixnmΣ

0 formax

0

i

i

xlnmx

nm

Σ∂ �> θ < ��∂θ��Σ �< θ >

��

So mle = = ( )ixg x

nmΣ

(c) Need posterior mean.

E[θx] = 2

ixnmΣ + β

+ β

= 2 1

2 2 2ixnm

nm nm nmΣ β+

+ β + β

which is of the form Z g(x) + (1 − Z) µ

where g(x) = ixnmΣ

, µ = ½

Find µ: f(θ) = θβ−1 (1 − θ)β−1

symmetric about θ = ½

mean = ½ µ = prior mean

(d) Z = 2

nmnm + β

= 2

m

mnβ+

→ 1 as n → ∞

As more data obtained, put more relative weight on the estimatebased in data, rather than the prior estimate.


Page 11

(ii) Z = 2

nmnm + β

g(x) (Σxi = 12) Cred est

β = 1 0.9677 0.2 0.2097β = 4 0.8824 0.2 0.2353

Prior variance = 2

2(2 ) (2 1)β

β β +

= 1

4(2 1)β +

If β = 1 prior var is 0.0833β = 4 prior var is 0.0278

Credibility factor is lower when β = 4, and prior variance is lower,indicating greater precision in our prior estimate, so give relatively moreweight to prior estimate.

9 (i)log

( ) = ( )y

a a

xf x dx e f y dy∞ ∞

� �

= ( )2 2 2

2

2 2

22

log

1

2

y y y

a

e dy

− − µ+µ − σ∞σ

πσ�

=

2 2

2 2( )

½ 22

log

1

2

y

a

e e dy− −µ−σ∞

µ+ σ σ

πσ�

= 2

2½ log

1a

eµ+ σ � �� − µ − σ� �− Φ � �� σ� � ��

(ii)

(a) Find value of σ and µ

exp (µ + σ2/2) = 9.070exp 2 (µ + σ2 / 2) × (exp σ2 − 1) = 10.1322

→ exp σ2 = 1 + 2

2

10.1329.070

= 2.248

→ σ2 = 0.80999 σ = 0.90 and µ = 1.80


Page 12

P(X > 25) = 1 − Φ log 25 1.800.90

−� ��

= 1 − Φ (1.577)

= 1 − 0.9426

= 0.0574

P(X > 30) = 1 − Φ log30 1.800.90

−� ��

= 1 − Φ (1.779)

= 1 − 0.9624

= 0.0376

(b) Expected amount ceded to reinsurer:

( ) ( )M

xf x dx MP X M∞

− ≥�

= 2log 1.80 0.90

9.070 1 ( )0.90

MMP X M

� �� − −� �− Φ − >� ��

(1) = 9.070 {1 − Φ (0.677)} − 25 {1 − Φ (1.576)}

= 9.070 (1 − 0.7507) − 25 × 0.0574

= 0.82524 i.e. £825.24

(2) = 9.070 {1 − Φ (0.879)} − 30 {1 − Φ (1.779)}

= 9.070 (1 − 0.8103) − 30 × 0.0376

= 0.59258 i.e. £592.58

Expected number of claims = 200 × 20% = 40 claims

Reinsurancepolicy

Amount cededto reinsurer

Premium paid

Reduction ininsurer�s profit

(1)(2)

40 × 825.2440 × 592.58

= £33,010= £23,703

£48,500£38,200

£15,490£14,497 ←

Choose policy (2) for best value for money.


Page 13

(c) Mean = 9.796Standard Deviation = 10.943

σ = 0.900 as beforeµ = 1.877

(1) = 9.796 {1 − Φ(0.591)} − 25 {1 − Φ (1.491)}

= 9.796 (1 − 0.7227) − 25 × (1 − 0.9320)

= 1.01643 i.e. £1,016.43

(2) = 9.796 {1 − Φ (0.794)} −30 {1 − Φ(1.694)}

= 9.796 (1 − 0.7864) − 30 × (1 − 0.9549)

= 0.73943 i.e. £739.43

Hence:

Amount ceded Premium paid Reduction in profit

(1)(2)

£40,657£29,577

£48,500£38,200

£7,843 ←£8,623

Choose policy (1) for best value for money.


EXAMINATIONS





1. Write your surname in full, the initials of your other names and yourCandidate�s Number on the front of the answer booklet.







� Faculty of Actuaries106�S2001 (9.2.01) � Institute of Actuaries

106 S2001�2

1 The distribution of X, which represents the claim severity from a portfolio of non-life insurance policies, has a Pareto distribution with mean £350 and standarddeviation £452.

If the insurer arranges excess of loss reinsurance, with retention £1,200,calculate the probability that a claim will involve the reinsurer. [3]

2 On a portfolio of insurance policies, the claim size, Y is assumed to depend on theage of the policyholder, X. Suppose that the conditional mean and variance of Yare

E[YX = x] = 2x + 400

and V[YX = x] = 2

2x

.

The distribution of X over the portfolio is assumed to be normal with mean 50and standard deviation 14.

Calculate the unconditional mean and standard deviation of Y. [5]

3 A specialist motor insurer writes policies with individual excesses of £500 perclaim. The insurer has taken out a reinsurance policy whereby the insurer paysout a maximum of £4,500 in respect of each individual claim, the rest being paidby the reinsurer. The individual claims, gross of reinsurance and the excess, arebelieved to follow an exponential distribution with parameter λ.

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

• There were 5 claims which were not processed because the loss wasless than the excess.

• There were 11 claims where the insurer paid out £4,500 and thereinsurer the remainder.

• There were 26 other claims in respect of which the insurer paid out atotal of £76,457.

Derive the loglikelihood function of λ. [6]

4 An actuarial student enters a quiz evening. All the questions will be picked bythe organiser, from one of four subjects: History, Literature, Sport or GeneralKnowledge. Under the rules of the quiz, entrants choose one opponent to playagainst for a cash prize, determined in advance according to the level ofknowledge of each competitor.

The student can choose from three opponents, Person A, Person B or Person C.He knows that he is better than A at Literature, Sport and General Knowledge

106 S2001�3 PLEASE TURN OVER

but not History; he is better than B at Sport and General Knowledge only and heis better than C at History and Literature only.

If the student wins against A he will receive £100, if he wins against B hereceives £150 and if he wins against C he receives £200. It is assumed that thebetter student always wins.

From his experience of past quiz nights, the student believes that History has a15% chance of being the chosen subject, Literature a 25% chance, Sport a 20%chance and General Knowledge 40%.

(i) Determine which opponent the student should choose to play against tomaximise his expected winnings. [4]

(ii) If the student decides instead that Sport will definitely not be picked butthe other three subjects have the same chances of being picked as beforerelative to each other, determine which opponent he should choose. [2]

[Total 6]

5 Consider a portfolio of insurance policies, on which the number of claims has abinomial distribution, with parameters n and p. The claim size distribution is

assumed to be exponential with mean 1λ

. Claims are assumed to be independent

random variables and to be independent of the number of claims.

The insurer arranges excess of loss reinsurance, with retention M.

Calculate the moment generating function of SI , where SI is aggregate annualclaims paid by the insurer (net of reinsurance). [6]

6 In a No Claims Discount system for motor insurance, there are three discountlevels, 0%, 20% and 30%. The full annual premium is £190.

If no claims are made during a year, the policyholder moves up to the next higherlevel of discount, or stays at 30%. If one or more claims are made, thepolicyholder moves down to 0% discount, or stays at 0%. The probability that apolicyholder has one accident in a year is 0.1, and the probability of more thanone accident is negligible. In the event of an accident, the natural logorithm ofthe loss has a N(µ, σ2) distribution, with µ = 4.5 and σ2 = 0.84. The policyholderwill only make a claim when an accident occurs if the loss is greater than thetotal extra premiums that would have to be paid over the infinite time horizon,assuming no further accidents occur.

(i) Determine the transition matrix for the Markov chain X1 , X2 , � where Xn

is the discount level at the beginning of year n. [6]

(ii) For a policyholder who pays the full premium in the current year,calculate the expected premium at the beginning of the next year. [1]

[Total 7]

106 S2001�4

7 Claim amounts under a particular insurance portfolio are believed to follow aNormal distribution with standard deviation 2

1σ and an unknown mean θ. Theinsurer observes a sample of n policies which have given rise to a claim for whichthe mean amount is a . The prior distribution of θ is assumed to be Normal withmean µ and standard deviation 2

2σ .

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

(b) Show that the posterior mean of θ can be expressed as a weightedaverage of the prior mean and the sample mean, and derive anexpression for the weight placed on the sample mean. [3]

(ii) An insurer believes that individual claim amounts follow a Normaldistribution with an unknown mean and a standard deviation of £210.Prior information suggests that the mean should be assumed to follow aNormal distribution with mean = £155 and standard deviation = £84.Over the previous year, the insurer collected data from 15 claims wherethe 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 andexpenses, calculate the premium the insurer should charge. [2]

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

22σ increases and briefly describe how it is affected by the insurer�s

assumptions and the observed data. [3][Total 8]

8 The number of claims per month are independent Poisson random variables withmean λ, and the prior distribution for λ is exponential with mean 0.2.

(i) Determine the posterior distribution for λ given the observed valuesx1 , �, xn of the number of claims in each of n months. [2]

(ii) Determine the Bayesian estimate of λ

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

(iii) If n = 5 and 5

1ix� = 1, calculate to 2 significant figures the Bayesian

estimate of λ under absolute error loss. [4][Total 9]


9 Claims paid to date on a motor insurance account are as follows:

Figures in £000s

Development yearPolicy year 0 1 2 3

1996199719981999

1,7082,0792,4172,935

1,3661,6011,863

9421,195

461

Inflation for the 12-month period to the middle of each year was as follows:

1997 6.2%1998 4.9%1999 5.2%

You are given the following further information:

• Annual premiums written in 1999 were £7,731,000.

• Future inflation from mid-1999 is estimated to be 5.0% per annum.

• The ultimate loss ratio (based on mid-1999 prices) has been estimated at77%.

• Claims are assumed to be fully run-off by the end of development year 3.

Estimate the outstanding claims arising from policies written in 1999 onlytaking explicit account of the inflation statistics, using the Bornhuetter-Fergusonmethod. [15]

106 S2001�6

10 Claims for a particular risk arrive in a Poisson process rate λ. The claim sizesare independent and identically distributed with density f(x) and are independentof the claims arrival process. Assume there is a constant γ (0 < γ < ∞) such that

( )limr

M r→γ

= ∞ where M(r) is the moment generating function of a claim.

Premiums are received continuously at constant rate with premium loadingfactor θ > 0.

(i) (a) Define the adjustment coefficient, R.

(b) Define the surplus process and the probability ψ(u) of ruin withinitial surplus u > 0.

(c) Write down Lundberg�s inequality. [3]

(ii) Derive the adjustment coefficient if f(x) = 1

x

e−

µ

µ, x > 0, and θ = 0.25. [3]

(iii) Consider the case where f(x) = ½ (1 2 )x xe e− −+ , x > 0, and θ = 0.25.

(a) Calculate the expected claim size µ.

(b) Calculate the corresponding adjustment coefficient, and determinean upper bound for ψ(15).

(c) Compare your answers to (iii)(b) with those obtained if the claimsizes are mistakenly assumed to be exponentially distributed withmean µ, and comment briefly. [11]

[Total 17]

106 S2001�7

11 Observations Y1 , �, Yn are independent Poisson random variables with E(Yi) = µi

where

log µi = ,=1, ...

= 1, ..., .mi

i m n

α��α + β +�

(*)

(i) (a) Show that the loglikelihood can be written as

−meα − (n − m) eα+β + 1 1 1

log !nn n

i i ii i m i

y y y= = + =

� �α + β − � ��

� �� ∏

(b) Derive the maximum likelihood estimates of α and β.

(c) Derive an expression for the deviance for this model.

(d) Derive an expression for the deviance for the model where β = 0.[13]

(ii) Suppose n = 20, m = 10 and the observations are

8, 6, 4, 7, 5, 8, 2, 8, 2, 9, 10, 7, 6, 4, 5, 7, 9, 6, 8, 5.

(a) Complete the following table, and hence calculate the deviance forthe model in (i)(d).

yi Frequency Contribution to deviance

2 2 −9.17924 2 −7.26815 3 −6.93346 3 −1.75647 3 4.42518 4 15.28919 2 12.8403

10 ? ?

(b) Given that the deviance for the model in (i)(c) is 16.1499, commenton the fit of (*) and of the model with β = 0. [5]

[Total 18]


EXAMINATIONS

September 2001


EXAMINERS� REPORT


Subject 106 (Actuarial Mathematics 2) � September 2001 � Examiners� Report

Page 2

Examiners� Comments

In numerical questions, candidates were not unduly penalised for errors inearlier parts of each question which affected their answers to the rest of thequestion.

Question 7 contained two misprints. The question should say that the variancesare 2

1� and 22� , not the standard deviations. Many candidates corrected this error

and gave the solutions given here. Others interpreted the question as written,thus obtaining 4

1� and 42� (instead of 2

1� and 22� ) in their solutions. Both

approaches, provided done correctly, were awarded full marks.

Question 11 on generalised linear models was poorly done in general.Candidates should bear in mind that the examiners will continue to set questionson this part of the syllabus.


Page 3

1 If f(x) = 1( )x

�

��

��

� � x > 0

then1

�

� � = 350

2

2 2( 1)� �

� �� = 4522

so2

�

� �

= 2

2

452350

= 1.6678

and � = 2 1.6678

0.6678�

= 4.995

� = 350 � 3.995 = 1,398.2

so P(X > 1,200) = 1,200

�

� ��

= 0.0453

2 E[Y] = E[E[Y�X]]

= E[2X + 400]

= 2E[X] + 400

= 500

V[Y] = V[E[Y�X]] + E[V[Y�X]]

= V[2X + 400] + 2

2X

E� ��

= 4V[X] + ½E[X2]

= 4 � 142 + ½(V[X] + (E[X])2)

= 784 + 1,348 = 2,132

Hence standard deviation of Y is 46.17.

3 Let X = gross claim amount.

For 5 claims, X < 500For 26 claims, �x = 76,457 + 26 � 500 = 89,457For 11 claims, X > 5,000


Page 4

Now:

P(X < 500) = 1 � e�500�

P(X > 5,000) = e�5,000�

Likelihood is:

26

1

ix

i

e��

�

� ��

� �� [1 � e�500�]5 [e�5000�]11

log Likelihood is:

26 log � � �26

1i

i

x�

� + 5 log[1 � e�500�] + 11 [� 5,000�]

= 26 log � � 89,457� + 5 log[1 � e�500�] ��55,000�

= 26 log � � 144,457� + 5 log[1 � e�500�]

4 (i) The decision matrix is:

WinningsHistory Literature Sport General Knowledge

A 0 100 100 100B 0 0 150 150C 200 200 0 0

Expected winnings against:

Player A 0.15 � 0 + 0.25 � 100 + 0.20 � 100 + 0.40 � 100 = 85Player B 0.15 � 0 + 0.25 � 0 + 0.20 � 150 + 0.40 � 150 = 90 �Player C 0.15 � 200 + 0.25 � 200 + 0.20 � 0 + 0.40 � 0 = 80

Choose Player B

(ii) Expected winnings against:

Player A (0.15 � 0 + 0.25 � 100 + 0.40 � 100) / (1 ��0.20) = 81.25Player B (0.15 � 0 + 0.25 � 0 + 0.40 � 150) / (1 � 0.20) = 75Player C (0.15 � 200 + 0.25 � 200 + 0.40 � 0) / (1 � 0.20) = 100 �

Choose Player C


Page 5

5 ( )ISM t = (log ( ))

IN XM M t

( )IXM t =

0

M tx x tM Me e dx e e��

= ( ) ( )

0

Mt x t Me e

t� ��

� ��

= ( )1 t Met t

� ��

MN(t) ((1 � p) + pet)n

( )ISM t = ( )1 1

n

t Mp p et t

� ��

� � � ��

6 (i) Claims Doesn�t claim

0% 0 + 190 + 152 + 133+ � L + 152 + 133 + �20% 0 + 190 + 152 + 133 + � L +133 + 133 + �30% 0 + 190 + 152 + 133 + L + 133 + 133 +

0%: claims if L + 152 + 133 > 190 + 152 L > 5720%, 30%: claims if L + 133 + 133 > 190 + 152 L > 76

where L ~ lognormal (, �2) is the loss.

P(L > 57) = 1 � ln57 4.5

0.84

� ��

= 1 �� (�0.499) = 0.6911

P(L > 76) = 1 � ln76 4.5

0.84

� ��

= 1 �� (�0.18) = 0.5714

so P = 0.1 0.6911 1 0.1 0.6911 0

0.1 0.5714 0 1 0.1 0.57140.1 0.5714 0 1 0.1 0.5714

� � ��

(ii) E(premium at beginning of next year)

= 190 � 0.1 � 0.6911 + 152 � (1 ��0.1 � 0.6911)

= 154.63


Page 6

7 (i) (a) Posterior mean:

= 2 21 22 21 2

n an

��

� � �

(b) = 2 21 2

2 2 2 21 2 1 2

na

n n� �

� ��

Hence posterior mean can be expressed as:

(1 )Za Z� � �

Where: Z = 22

2 21 2

nn�

� � �

= 2122

n

n�

�

�

And: 1 � Z = 2 2 21 2 2

2 21 2

n nn

�

=

21

2 21 2n�

� � �

(ii) a = £163.73

And: Z = 2

2

15210

1584

�

= 0.70588

Hence premium is (0.29412 � £155 + 0.70588 � £163.73) � 1.30 = £209.51

(iii) As n �, Z 1 because more weight is placed on the sample data if wehave a greater volume of data.

As 21� �, Z 0 because less weight is placed on the sample data if we

believe the variability of the observed data to be higher.

As 22� �, Z � 1 because more weight is placed on the sample data if we

believe the variability of the prior estimate to be higher.


Page 7

8 (i) f(x��) � e�n� ix��

f(�) � e�5�

so f(��x) � 1

n

ix�

� e�(n+5)�

and1

~ gamma 1, 5n

ix x n� �

��

(ii) (a) Under quadratic loss, the Bayesian estimate is the posterior mean

1

1( ) =

5

n

ixE x

n

�

��

�

(b) Under all-or-nothing loss, the Bayesian estimate is the posteriormode.

f(��x) � ( 5)ix ne� � � ��

If �xi = 0, then mode is at � = 0

Otherwise, the posterior density has a maximum when

1 ( 5) ( 5)( ) = ( 5)i ix xn nix e n e� � ��

� � � �

i.e. =5ix

n�

��

(iii) Under absolute error loss, the Bayesian estimate is the posterior median

n = 5, �xi = 1 so ��x ~ gamma(2, 10)

Median given by g where 2 1010g

e d�

� �� = ½

i.e. 10 10[ 10 ] 10 = ½gg

e e d�

� � � � ��

i.e. (10g + 1) e�10g = ½

i.e. 2(10g + 1) = e10g


Page 8

g 2(1 + 10g) e10g root

0.10.20.150.160.170.165

4655.25.45.3

2.77.44.54.955.475.21

> 0.1< 0.2> 0.15> 0.16< 0.17> 0.165

So root between 0.165 and 0.17. To 2 sig. figs, g = 0.17

9 Adjust individual claim amounts to mid-1999 prices:

Figures in £000s

Development year

Policy year 0 1 2 3

1996199719981999

2,0022,2942,5432,935

1,5071,6841,863

9911,195

461

1

2

gmedian

e10g

2(1 + 10g)


Page 9

Cumulative claim amounts:

Figures in £000s

Development year

Policy year 0 1 2 3

1996199719981999

2,0022,2942,5432,935

3,5093,9784,406

4,5005,173

4,961

Column Sum 11,893 9,673 4,961Column sum minus last entry 6,839 7,487 4,500

Development factors for each year:

d3 = 4,961/4,500 = 1.10244

d2 = 9,673/7487 = 1.29197

d1 = 11,893/6,839 = 1.73900

Ultimate claim amount for 1999 policies = 0.77 � 7,731 = 5,953

Hence outstanding amounts arising from 1999 policies:

1999, 3 = 5,953 � 1

11.10244

� ��

� � = 553

1999, 2 = 5,953 � 1 1

1.10244 1.10244 1.29197� �

��

= 1,220

1999, 1 = 5,953 � 1 1

1.10244 1.29197 1.10244 1.29197 1.73900� �

��

= 1,776

Adjust for future inflation = 553 � 1.053 + 1,220 � 1.052 + 1,776 � 1.05

= £3,850,000


Page 10

10 (i) (a) Let be the mean claim size.

Then the adjustment coefficient R is the positive solution ofM(r) = 1 + (1 + �) r.

(b) The surplus at time t is U(t) = u + (1 + �) �t � S(t) where S(t) isthe aggregate claims at time t.

The surplus process is {U(t): t > 0}.

The probability of ruin with initial surplus u is �(u) = P(U(t) < 0 forsome t, 0 < t < �).

(c) Lundberg�s inequality says: �(u) � exp{�Ru}.

(ii) For exponential mean , the mgf is M(r) = 1

1 r� �, r <

1�

.

So R > 0 solves

11 r� �

= 1 + 54

r�

4 = 4(1 ��r) + 5r(1 � r)

52 r2 � r = 0

R > 0 so 1

=5

R�

(check: R < 1�

)

(iii) (a) Claims ~ ½ � (exponential mean 1) + ½ � (exponential mean ½) (*)

So = ½ � 1 + ½ � ½ = ¾

[or evaluate 0

x�

� ½e�x (1 + 2e�x) dx]

(b) From (*), M(r) = ½ � 1

1 r�+ ½ �

22 r�

, r < 1

[or evaluate 0

rxe�

� ½e�x(1 + 2e�x) dx]


Page 11

R > 0 solves

½1 2 5 3

= 1(1 ) 2(2 ) 4 4

rr r

� � �

� �

8(2 � r) + 16(1 � r) = 16(2 � 3r + r2) + 15r(2 ��3r + r2)

24r = �18r � 29r2 + 15r3

R � 0 15r2 � 29r + 6 = 0

Solution r = 229 29 4 6 1530

� � � � = 1.6977 or 0.2356

Need R < 1, i.e. R in domain of definition of M(r)

So R = 0.2356

Lundberg�s inequality gives �(15) � e�R�15 = 0.0292

(c) For exponential mean = ¾, we know from (ii) that the adjustment

coefficient R* is R* = 15�

= 4

15 = 0.2667 and so Lundberg�s

Inequality gives �(15) � e�R*�15 = 0.0183.

R* is larger than the true value, which means that the bound on�(15) is smaller than the true bound, and may lead to a false senseof security.

11 (i) (a)

f(y) = 1 !

i iyni

ii

ey

��

�

�

�

Log likelihood is

l(�, �) = 1 1

log log( !)n n

i i i iy y� � � � � ��

= ��me� ��(n � m) e�+� + 1 1

( ) log( !)m n

i i im

y y y�

� � � ��

= 1 1

( ) log( !)n n

i i im

me n m e y y y� ��

�

� � � ��

� ��


Page 12

(b)

l��

= 0: ��me� � (n � m) e�+� + 1

n

iy� = 0 (1)

l��

= 0: ��(n � m) e�+� + 1

n

im

y�

� = 0 (2)

Substitute (2) in (1): me� =

1

m

iy�

so 1� = log

mi

e

y

m

� ��

�

Substitute �� in (2): �� = 1log

nim

e

y

n m�

� ��

�

so 1 1� log log

n mi im

e e

y y

n m m�

� � � ��

� �

� �

(c)

Deviance = 2{log likelihood for full model � ��( , )}l � �

In the full (or saturated) model, fitted values are i�� = yi .

In model (*) the fitted values are � i� = 1

2

=1, ...,= 1,...,

t i m

t i m n��

��

where t1 = 21 1, = .

( )m ni i

m

y yt

m n m��

� �

So Deviance

= 1 1

2 logn n

i i ii

y y y�

� � ��

�� 1 1 2 2

1 1 1 1

log logm m n n

i ii m m

t y t t y t� � �

� � � � � �

��

= 1 21 21 1

2 log ( ) 2 log ( )m n

i ii i i i

m

y yy y t y y t

t t�

� � � ��

� � � ��

or equivalent expression


Page 13

(d)

When � = 0, i = e� for all i

maximum likelihood estimate of � is �� = log iyn��

� �

with fitted values i�� = =iyy

n�

for all i

The deviance is � � �2{ log log }i i iy y y y

= 2 log ii

yy

y� �

� � ��

(ii) (a) y = 6.3

iy freq Contribution todeviance

2 2 �9.17924 2 �7.26815 3 �6.93346 3 �1.75647 3 4.42518 4 15.28919 2 12.8403

10 1 9.240716.6581

So Deviance is 16.6581 on 19 df.

(b) Deviance for model (*) is 16.1499 on 18 df.

Both models fit well.

Drop in deviance resulting from including � is 0.5083 on 1 df.

This is not significant when referred to a 21� .

So no evidence to suggest we need � in the model.


EXAMINATIONS

12 April 2002 (am)




1. Enter all the candidate and examination details as requested on the front of your answerbooklet.

2. You must not start writing your answers in the booklet until instructed to do so by thesupervisor.


4. Attempt all 10 questions, beginning your answer to each question on a separate sheet.



Hand in BOTH your answer booklet, with any additional sheets firmly attached, and thisquestion paper.

In addition to this paper you should have available Actuarial Tables andyour own electronic calculator.


106 A2002�2

1 For each of m independent risks, there is probability 0.2 that a claim is made in a yearand probability 0.8 that no claim is made. Claim sizes are independent with mean 400and variance 110.

Determine the expected value and the variance of the total amount claimed in oneyear. [3]

2 The loss function under a decision problem is given by:

�1 �2 �3

d1 14 12 13d2 13 15 14d3 11 15 5

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

(ii) Given the probability distribution p(�1) = 0.25, p(�2) = 0.25, p(�3) = 0.5determine the Bayes criterion solution. [2]

[Total 4]

3 Claims arrive according to a Poisson process. Individual claim sizes are independentwith density:

f(x) = xe�x (x > 0)

and the insurer uses a premium loading factor of �.

(i) Derive the equation for the adjustment coefficient for this process. [2]

(ii) If � = 0.4, calculate the adjustment coefficient, and determine an upper boundfor the probability of ultimate ruin if the initial surplus is 50. [4]

[Total 6]


4 The loss amount, X, on a certain type of insurance policies, has a Pareto distributionwith density function f(x), where:

f(x) = 3

43 400

(400 )x�

�

(x > 0)

A policyholder deductible of £100 is applied to these policies.

(i) Calculate the expected claim size paid by the insurance company. [5]

(ii) Comment on the difference between your answer to (i) and the expected lossamount, E[X]. [2]

[Total 7]

5 The delay triangles given below relate to a certain portfolio of insurance policies, forwhich it may be assumed that the claims are fully run off by the end of developmentyear 2.

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

Development YearAccident Year 0 1 2

1999 2,317 1,437 5822000 3,287 1,7922001 4,816

The cumulative number of reported claims is shown in the table below:

Development YearAccident Year 0 1 2

1999 132 197 2072000 183 2582001 261

(i) Given that the total claims paid to date is 10,237 calculate the outstandingclaims reserve for this cohort using the average cost per claim method. [7]

(ii) State the assumptions that underlie your result. [2][Total 9]

106 A2002�4

6 The no claims discount system operated by an insurance company for their annualmotor insurance business has four levels of discount:

Level 1: 0%Level 2: 25%Level 3: 45%Level 4: 60%

If a policyholder does not make a claim under the policy in a particular year then heor she will go up one level (or stay at level 4), whereas if any claims are made he orshe will go down by two levels (or remain at, or move to, level 1). The full premiumpayable at the 0% discount level is 900.

The probability of an accident occurring is assumed to be 0.2 each year for allpolicyholders and losses are assumed to follow a lognormal distribution with mean1,188 and standard deviation 495. However, policyholders claim only if the loss isgreater than the total additional premiums that would have to be paid over the nextthree years, assuming that no further accidents occur.

(i) Calculate the smallest loss amount for which a claim will be made for apolicyholder at the 0% discount level. [2]

(ii) Complete the transition matrix below by calculating the missing values.

* * * *0.147 0 0.853 00.120 0 0 0.880

0 0.197 0 0.803

� ��

[5]

(iii) Calculate the proportion of policyholders at each discount level when thesystem reaches a stable state. [4]

[Total 11]


7 Claims occur on a portfolio of 100 general insurance policies according to a Poissonprocess. The expected number of claims per annum on each policy is �, and the claimsize distribution has density function f(x), where

f(x) = /100110,000

xxe� (x > 0)

The parameter � is not the same for all policies in the portfolio, but is modelled as arandom variable (independent of the claim sizes) with density function g(�), where:

g(�) = 100�e�10� (� > 0)

(i) Calculate the mean and variance of annual aggregate claims. [6]

(ii) Given an initial reserve of 2,000, use a normal approximation to thedistribution of aggregate claims to find the relative premium loading thatshould be used in order to be 95% sure that the reserve at the end of the firstyear is positive. [3]

(iii) If � were fixed at its mean value, describe the effect that this would have onthe value of the relative premium loading. (No further calculations arerequired.) [2]

[Total 11]

106 A2002�6

8 An insurance portfolio consists of m groups of individuals. In the ith group there areni (> 1) individuals aged xi. The number of claims from this group of ni individuals isa binomial random variable Yi with parameters ni and �i, with 0 � �i � 1 (i = 1, �, m).The random variables Y1, �, Ym are independent.

(i) Derive the maximum likelihood estimators 1� �, ..., m� � of �1, �, �m. [3]

(ii) (a) If

logit(�i) = log1

i

i

� ��

� �� = � + �xi,

show that the log-likelihood is

1 1 1log(1 exp( )) ,

m m m

i i i i ii i i

y x y n x c� � �

� � � � � � � � ��

where c does not depend on � or �.

(b) Derive, but do not attempt to solve, the equations satisfied by themaximum likelihood estimators �� and �� of � and �. [5]

(iii) Let ei = ni i�� where i�

� = � �� exp( ) /(1 exp( )).i ix x� � � � � � �

Derive the scaled deviance for the model in (ii). [3]

(iv) Data are available for six ages, 17, 18, 19, 20, 21 and 22.

Two models are fitted to the data with the following results:

Scaled Deviance Degrees of FreedomModel 1 logit(�i) = � 13.33 5Model 2 logit(�i) = � + �xi 1.67 *

(a) State the degrees of freedom for Model 2.

(b) Comment on the fit of these two models.

(c) In Model 2, the estimate of � is �0.2492, with standard error 0.07217.Interpret these results and explain how the odds �i+1/(1 � �i+1) arerelated to the odds �i/(1 � �i), when xi+1 = xi + 1. [5]

[Total 16]


9 The number of e-mail messages received each day by an actuarial student has aPoisson distribution with mean �, where from past experience, the prior distributionof � is exponential with mean �.

(i) The student has data x1, �, xn, where xi is the number of messages arriving onday i, i = 1, �, n.

(a) Derive the posterior distribution of �.

(b) Show that the Bayesian estimate of � under quadratic loss can bewritten in the form of a credibility estimate, and state the credibilityfactor.

(c) If � = 50 and the student receives a total of 550 messages over 10 days,calculate the Bayesian estimate of � under quadratic loss. [6]

(ii) 60% of messages require an answering time (in minutes) that is exponentiallydistributed with mean 1, and the remaining messages require an answeringtime that has a Pareto distribution with mean 2 and variance 12.

(a) Determine the probability that a randomly chosen message requiresmore than M minutes answering time.

(b) Using the value of � estimated in (i)(c), calculate the expected totalamount of time the student spends answering all the messages thatarrive on one day.

(c) The student decides to reduce the amount of time spent with e-mail,and adopts a strategy of imposing a maximum cut-off time of 1.5minutes for answering each message. Still using the estimated value of�, calculate the reduction in the expected amount of time spentanswering the e-mail messages from one day. [10]

[Total 16]

106 A2002�8

10 The last ten claims under a particular class of insurance policy were:

1,330 201 111 2,368 617309 35 4,685 442 843

(i) Assuming that the claims came from a lognormal distribution with parameters� and �, derive the formula for the maximum likelihood estimates of theseparameters and estimate the parameters based on the observed data. [5]

(ii) Assuming that the claims come from a Pareto distribution with parameters �and �, use the method of moments to estimate these parameters. [3]

(iii) Assuming that the claims come from a Weibull distribution with parameters cand , use the method of percentiles (based on the 25th and 75th percentiles) toestimate these parameters. [5]

(iv) If the insurance company takes out reinsurance cover with an individualexcess of loss of 3,000 estimate the percentage of claims that will involve thereinsurer under each of the three models above. [4]

[Total 17]


REPORT OF THE BOARD OF EXAMINERS ONTHE EXAMINATIONS HELD IN

April 2002


Introduction

The attached subject report has been written by the Principal Examiner with the aim ofhelping candidates. The questions and comments are based around Core Reading as theinterpretation of the syllabus to which the examiners are working. They have howevergiven credit for any alternative approach or interpretation which they consider to bereasonable.

K FormanChairman of the Board of Examiners

11 June 2002



Page 2

COMMENT

Question 1This question was generally well answered though some candidates struggled to determinethe variance.

Question 2This question was well answered though some candidates did not state the correct conclusionin part (ii) despite making the correct calculations.

Question 3This question was generally well answered though some candidates struggled with thealgebra.

Question 4Many candidates did not consider the conditional probability in part (i) giving an answer of£128 rather than £250. Candidates were only penalised once for this error.

Question 5Overall this question was answered well though many candidates were unable to state all theassumptions required in part (ii).

Question 6Many candidates answered this question well though some struggled with completing thetransition matrix in part (ii) but went on to give good answers for part (iii). Part (iii) used theanswer to part (ii) but candidates were given full credit if they had used the correct method.

Question 7This question had a wide variety of marks ranging from very good to very poor.

Question 8Some candidates found this question difficult with many omitting it altogether although manycandidates answered this question well. Candidates should note that the examiners willcontinue to set questions on this topic.

Question 9Overall this question was answered reasonably well though many candidates struggled withpart (ii)(c).

Question 10This question had a wide variety of answers ranging from very good to very poor. Candidateshad most trouble with parts (iii) and (iv) although many strong candidates scored well inthese parts.


Page 3

1 Let N be the number of claims in a year, so N ~ Bin(m, 0.2), and let X be a typicalclaim.

Let S be the total amount claimed in a year, so S has a compound Binomialdistribution.

The expected value of S is

E(S) = E(N) E(X) = m � 0.2 � 400 = 80m.

The variance of S is

V(S) = E(N)V(X) + V(N) (E(X))2

= m � 0.2 � 110 + m � 0.2 � 0.8 � 4002

= 25,622m.

2 25% 25% 50%�1 �2 �3 max Expected loss

d1 14 12 13 14 13d2 13 15 14 15 14d3 11 15 5 15 9

(i) The minimax solution is d1.

(ii) The Bayes criterion solution is d3.

3 (i)

Claim sizes have a Gamma(2,1) distribution and hence have mean 2 and momentgenerating function

M(r) = (1 � r)�2, r < 1.

The adjustment coefficient R > 0 satisfies

(1 � R)�2 �1 ��2(1 + �) R = 0,

(or equivalent). Hence we have

1 � (1 ��R)2 � 2(1 + �) R(1 � R)2 = 0,


Page 4

and, using R > 0, this simplifies to

2(1 + �)R2 � (3 + 4�) R + 2� = 0.

(ii)

When � = 0.4 this is

2.8R2 ��4.6R + 0.8 = 0,

and this has solutions 0.1977 and 1.4452.

Only the first of these is in the range of definition of the moment generating function,so the adjustment coefficient is R = 0.198.

The probability of ultimate ruin satisfies �(50) � e�50R by Lundberg�s inequality, i.e.�(50) � 5.09 � 10�5.

4 (i) E[X � 100�X > 100] = 100( 100) ( )

( 100)

x f x dx

P X

�

�

�

�

3

4100

3( )

x dxx

� �

� �� =

3 3

3100100

( )x dx

x x

�

��

� � � ��

= 3 3

2100

400100500 2( )x

�

� ��

� ��

= 100 � 3 3

24 4005 2 500

� ��

��

100100 ( )f x dx

�

� = 3400100

500� ��

� E[X � 100�X > 100] =

3

2

3

4002 500

45

�

� ��

= 312845

� ��

= 250


Page 5

(ii) E[X] = 4002

= 200

The expected claim size increases due to the heavy tail of the distribution.

5 (i) This solution has been completed using simple averages of the grossing-upfactors, but use of the basic chain ladder method to project average claimsamounts and numbers of reported claims would also be acceptable.

Cumulative cost of incurred claims:

Development yearAccident year 0 1 2

1999 2,317 3,754 4,3362000 3,287 5,0792001 4,816

Project number of reported claims:

Development yearAccident year 0 1 2 ultimate

1999 132 197 207 20763.77% 95.17% 100.00%

2000 183 258 27167.50% 95.17%

2001 261 39865.64%

Project average incurred cost per claim:

Development yearAccident year 0 1 2 ultimate

1999 17.553 19.056 20.947 20.94783.80% 90.97% 100.00%

2000 17.962 19.686 21.64083.01% 90.97%

2001 18.452 22.12383.41%


Page 6

Projected total claims:

1999: 207 � 20.947 = 4,3362000: 271 � 21.640 = 5,8642001: 398 � 22.123 = 8,805Total: 19,005

Less claims paid �10,237

O/s claims reserve 8,768

(ii) Assumptions are:

For each accident year, the number of claims reported in each developmentyear is a constant proportion of the total number of claims arising from thataccident year.For each accident year, the average claim amount in each development year isa constant proportion of the ultimate average claim amount for that accidentyear.Weighted average past inflation is appropriate estimate of future inflation.

6 (i) Discount If claim If don�t claim Difference0% 900, 675, 495 675, 495, 360 540

So the smallest loss for which a claim will be made at the 0% level is 540.

(ii) P(Claim) = P(Claim|Accident) . P(Accident)= P(X > x) * 0.2

where X is the loss, which has a lognormal distribution, and x is the minimumloss for which a claim will be made. Now we know that

E(X) = exp{ + ½2} = 1,188V(X) = exp{2( + ½2)}.(exp{2} � 1) = (495)2

hence

(exp{2} � 1) = (495)2 / (1,188)2

exp{2} = 1.17362 = 0.16

So

�= 0.4 and = 7.

P(X > x) = 1 � P(X < x) = 1 � ln x � �� = 1 � ln 7

0.4x ��

��


Page 7

P(Claim�Accident):

P(X > 540) = 1 � �(�1.771) = �(1.771) = 0.9617

P(Claim) = 0.9617 * 0.2 = 0.1923

So the transition matrix is:

0.192 0.808 0 00.147 0 0.853 00.120 0 0 0.8800 0.197 0 0.803

� ��

(iii) The steady state distribution is therefore the solution to:

0.192�0 + 0.147�1 0.120�2 = �00.808�0� + 0.197�3 = �1

0.853�1 = �2 0.880�2 + 0.803�3 = �3

�0 + �1 + �2 + �3 = 1

Expressing �0, �1 and �3 in terms of �2:

�3 = 0.8800.197

�2 = 4.4670 �2

�1 = 10.853

�2 = 1.1723 �2

�0 =

0.147 0.1200.853

0.808

� �2 = 0.3618 �2

And so:

0.3618 �2 + 1.1723 �2 + �2 + 4.4670 �2 = 1

which gives

�2 = 0.1428


Page 8

Proportions at each discount level in stable state:

0%: 5.2%25%: 16.7%45%: 14.3%60%: 63.8%

(Check: 0.052 + 0.167 + 0.143 + 0.638 = 1.000)

7 (i) Let iS be aggregate claims from the ith policy. Then

� � � � � �� ii SESESE and � � � � � �� ii SVSVSV

� � � �� XEEXEESEESE iiii �� |

� � � �� iiiii SVESEVSV �� || ��

� �� 2XEEXEV ii ��

� � � �� 22 XEEXEV ��

X ~ 12,100

� ��

� ~ �(2, 10)

E[X] = 200

E[�] = 0.2

� E[S] = 100 � 0.2 � 200 = 4,000

V[�] = 0.02

V[X] = 20,000

�E[X2] = 20,000 + 2002 = 60,000

V[S] � � � �� 22100 XEEXEV ��

=100 � 2002 � 0.02 + 100 � 0.2 � 60,000

= 1,280,000


Page 9

(ii) P(S < U + (1 + �) E[S]) = 0.95

[ ] (1 ) [ ] [ ] = 0.95[ ] [ ]

S E S U E S E SPV S V S

� ��

[ ][ ]

U E SV S� � = 1.645

� = 1.645 [ ][ ]V S U

E S� = 035.0

400020001861

��

�

The company does not need to add a premium loading since the reserve issufficient for the first year to provide the necessary security. It could even sellat (very slightly) less than expected cost (3.5% less than expected cost).

(iii) � fixed would reduce variability and therefore the value of � would decrease.This would provide further scope for competitive pricing in the first year.

8 (i) Writing y for (y1, �, ym), the likelihood is

f(y��1, �, � m) = 1

(1 ) ,i i im

yi n yi i

ii

ny

�

�

� ��

� ��

so the log-likelihood is

l(�1, �, �m) = 1 1 1

log ( ) log(1 ) log .m m m

ii i i i i

ii i i

ny n y

y� � �

� ��

� � � �

Differentiating with respect to �i and setting to zero gives

1i i i

i i

y n y��

� � � = 0,

and so

yi(1 ��i) = (ni � yi) �i ,

which gives �i� = yi/ni.

(ii) (a) First note that 1 � �i = 1(1 )ixe�� . Using the expression for the

log-likelihood derived in (i) above, the log-likelihood can be written


Page 10

l = 1 1

log ( )log(1 )m m

i i i i ii i

y n y c� �

� � � � � ��

= 1 1

log log(1 )1

im m

xii i

ii iy n e c��

� �

� ��

� �

= 1 1

( ) log(1 )im m

xi i i

i iy x n e c��

� �

� � � � � �� .

= 1 1 1

log(1 exp( ))m m m

i i i i ii i i

y x y n x c� � �

��

(b) Differentiating with respect to � and � in turn, and setting to zero givesthe equations satisfied by �� and �� , i.e.

��

��1 1 1

i

i

m m x

i i xi i

ey ne

��

��

�

�

� � = 0

��

��1 1 1

i

i

m m xi

i i i xi i

x ex y ne

��

��

�

�� = 0 .

(iii) The required deviance D is twice the difference between the maximised log-likelihoods for the models in (i) and (ii).

Substituting the maximum likelihood estimators into the log-likelihood in (i),and ei/ni for �i in the model in (ii), we obtain

1 1

1 1

2 log ( ) log 1

log ( ) log 1

m mi i

i i ii ii i

m mi i

i i ii ii i

y yD y n yn n

e ey n yn n

� �

� �

� � ��

� �

��

�

� �

� �

= 1 1

2 log ( ) logm m

i i ii i i

i i ii i

y n yy n ye n e

� �

� ��

� � .

(iv) (a) Model 2 has one extra parameter fitted, so the degrees of freedom is 4.

(b) The first model does not fit all that well (deviance = 13.33 compared to25� distribution), but the second model is better (deviance 1.67

compared to a 24� distribution).


Page 11

The drop in deviance resulting from including regression on age is13.33 � 1.67 = 11.66, which is significant when compared to a �2

distribution on 5 ��4 = 1 degree of freedom, implying that the age termshould be included in the model.

(c) In Model 2, �� /s.e. �( )� = �3.45 and is significant confirming that �should be in the model. The estimate of � is negative, so logit(�i)decreases as age increases.

Comparing odds,

1

11i

i

�

�

�

� �= 1ixe �

��

= ( 1)ixe��

= 1

i

ie��

� �.

Substituting the estimate of �, the odds �i/(1 � �i) are multiplied by�

e� = e�0.2492 = 0.78 to get the odds for age xi+1.

9 (i) (a) The prior density for � is f(�) = e��/�/, and, writing x for (x1, �, xn),

f(x��) = 1

1!

ni

ix

n

n

ii

e

x

��

�

��

�

so that the posterior density is

f(��x) � f(�) � f(x��)

� 1/n

ii xne e ��

�

= 1 1exp ,n

ii x n�

� ��

i.e. gamma(1 + 1 , (1/ )).nii x n

�

� ��


Page 12

(b) The Bayesian estimate under quadratic loss is the posterior mean,

1

11

= ,1 1 1

nii x n x

n n n�

� ��

� � ��

�

which is in the form of a credibility estimate with credibility factorn/(n + (1/)).

(c) With the given values, the estimate of � is

550 110 (1/50)

�

�

= 54.99.

(ii) (a) First note that the Pareto(�, �) distribution has �/(� �1) = 2 and��2/((� � 1)2(� � 2)) = 12. Solving these gives � = 3 and � = 4.

Let X be a randomly chosen claim, so X has density

f(x) = 0.6e�x + 0.4 3

43 4

(4 )x�

�

.

Then,

P(X > M) = 0.6e�M + 0.43

4 .4 M

� ��

��

(b) Let T be the total amount of time spent answering the messages fromone day.

E(T) = �(0.6 � 1 + 0.4 � 2) = 54.99 � 1.4 = 76.99.

(c) Let Y be the time to answer an e-mail message under the new strategy,so Y = min(X, 1.5) and

E(Y) = E(min(X, 1.5)) = 1.5

0( ) 1.5 ( 1.5)xf x dx P X� ��

= E(X) �0

( 1.5) .xf x dx�

��


Page 13

The reduction in expectation per message is

0( 1.5)xf x dx

�

�� = 3

1.540

3 40.6 0.4(4 1.5 )

xx e dxx

��

� ��

�

= 3 3

1.53 40 0

0.4 4 3 5.50.65.5 (5.5 )

xe xe dx x dxx

� ��

�

��

= 0.6e�1.5 + 3

30.4 4 5.5

25.5�

�

= 0.5570

Hence the total reduction for one day�s messages is54.99 � 0.5570 = 30.63 minutes.

(Or any correct calculation method.)

10 (i) The likelihood function is:

210

122

ln1exp2

( , ) =2

i

i

i

x

Lx

�

� ��

�

and l = log likelihood is:

l(,2) = 210 10

1 1

ln1 10ln 10ln 2 ln2

ii

i i

x x� �

� ��

So:

l��

= 10

1

ln1 i

i

x

�

� ��

�

and

l��

= 210

1

ln1 10i

i

x

�

� ��


Page 14

Equating both to zero gives:

10

1

�ln = 0�i

i

x

�

� ��

� which gives �� = 10

1

1 ln10 i

ix

�

�

and

210

1

�ln = 10�i

i

x

�

� ��

�

10 2 22

1

� �ln 2 ln� =10

i i

i

x x

�

� ��

�

10 102 2 2

2 21 1�ln 10 ln

� �= =10 10

i ii i

x x� �

� �

� � �

� �

Now from the data we have:

10

1ln i

ix

�

� = 61.9695 and 10

2

1ln i

ix

�

� = 403.1326 so

�� = 6.197 and 2�� = 1.911 i.e. �� = 1.382

(ii) Under a Pareto distribution we have:

E(X) = ( 1)

�

� � and

2

2( ) =( 1) ( 2)

V X ��

� � � �

Equating to sample moments, which are:

Mean = 10

1

110 i

ix

�

� = 1,094.1

and

Variance = 2

2 230,761,6791,094.1 = 1,094.1 = 1,879,11310 10

ix� �

�

gives:

21,879,1131,094.1

= �

� 2�

� �


Page 15

�� = 2

2

1,879,11321,094.1

1,879,113 11,094.1

� ��

� = 5.51013

and

�� = 1,094.1 � �( 1)� � = 4,934.5

[Some candidates divided by 9 rather than 10 in the variance and were notpenalised. This gives �� = 4.68745 and � = 4,034.44� . Following through topart (iv), the answer for the Pareto then becomes 0.07383.]

(iii) 25th percentile is ½ � (111 + 201) = 156 and75th percentile is ½ � (843 + 1,330) = 1,086.5

Then we need to solve:

1 � exp(�c � 156�) = 0.25 and 1 � exp(�c � 1,086.5�) = 0.75

Simplifying and taking logs gives:

��156c �� = ln(0.75) and ��1,086.5c �

� = ln(0.25)

Dividing out:

�1,086.5 ln 0.25=156 ln 0.75

�

� ��

Taking logs:

1,086.5 ln 0.25� ln = ln156 ln 0.75

� � � ��

So �� = 0.81022 and �c = 0.00481

(iv) P(X > 3,000) = 1 � F(3,000):

Lognormal: ln 3,000 6.19711.382

��

� � = 1 � �(1.309) = 0.09527


Page 16

Pareto: 5.510134,934.5

4,934.5 3,000� ��

= 0.073011

Weibull: exp{�0.00481 � 3,0000.81022} = 0.047542


EXAMINATIONS













� Faculty of Actuaries106�S2002 � Institute of Actuaries

106 S2002�2

1 Claims on a portfolio of insurance policies are exponentially distributed with mean1/�, where previous experience with similar portfolios suggests that the priordistribution of � is gamma with mean 1 and variance ½. Twenty claims are observedwith average value 1.2.

Determine the posterior distribution of �. [4]

2 A sample of claims from 50 individuals are independent normal random variables Yij,(i = 1, 2; j = 1, �, 25) with variance �2 and

E(Yij) = �i + �xij,

where group 1 consists of males (i = 1) and group 2 of females (i = 2) and xij is theage of the jth individual in group i.

(i) Show that this is a generalised linear model, by writing it in the form of anexponential family, and by stating the link function and the linear predictor.[3]

(ii) Sketch on the same diagram the dependence specified by the model ofexpected claim size on age for males and females. [1]

[Total 4]

3 (i) Derive the moment generating function of the total amount, T, claimed if thenumber of claims, N, has a Poisson distribution with mean � > 0 and the claimseverity distribution has moment generating function M(t). [2]

(ii) A portfolio consists of 210 risks each of which gives rise to claims as aPoisson process. The claim severity distribution is exponential. The portfoliois divided into 3 groups, as follows:

Number of Poisson rate Mean of claimGroup risks per risk severity distribution

1 40 1 4002 120 2 5003 50 2.5 600

(a) Derive the moment generating function of the total claim amount Sfrom all 210 independent risks in one time unit.

(b) Show that S has a compound Poisson distribution and determine thecorresponding Poisson parameter and the claim severity density. [6]

[Total 8]


4 A market trader has the option for one day of selling either ice-cream (d1), hot food(d2) or umbrellas (d3) at an outdoor festival. He believes that the weather is equallylikely to be fine (�1), overcast (�2) or wet (�3) and estimates his profits under eachpossible scenario to be:

�1 �2 �3

d1 25 19 7d2 10 30 8d3 0 2 34


(ii) The trader�s partner is very optimistic and believes that the criterion to adoptin deciding which product to sell should be to maximise the maximum profit.What decision would the trader�s partner make based on these predictedprofits? [1]

(iii) Determine the Bayes criterion solution to this problem. [2]

(iv) The trader�s partner agrees that it is equally likely to be either fine or wet butbelieves that there is more than an evens chance of it being overcast. Bysketching a graph of the Bayes risk for each of the three possible decisionsagainst the probability of it being overcast (p), or otherwise, determine therevised Bayes criterion solution. [4]

[Total 9]

5 (i) Claims arrive as a Poisson process with rate �. The claim sizes areindependent, identically distributed random variables X1, X2, � with

P(Xi = k) = pk , k = 1, �, M, 1

= 1M

kk

p�

� .

If the premium loading factor is �, show that the adjustment coefficient Rsatisfies:

1

2

1 2log(1 ) ,mRM m

��

where mi = 1( )iE X , i = 1, 2. [7]

[The inequality 1Rx RMx xe eM M

� � � (0 � x � M) may be used without proof.]

(ii) If � = 0.2 and Xi is equally likely to be 1 or 2, determine an upper and a lowerbound for R, and hence derive an upper bound on the probability of ruin whenthe initial surplus is u. [3]

[Total 10]

106 S2002�4

6 The no claims discount system for a particular class of annual insurance policy hasthree categories with discount levels of 0%, 30% and 50%. If a policyholder makesany claims during the year he or she moves down a single category (or stays at the 0%discount level). If no claims are made, then the policyholder moves to the next highercategory (or stays at the 50% discount level).

The probability that a policyholder will make at least one claim in any one year is:

p in the 0% discount category0.8p in the 30% discount category0.6p in the 50% discount category

The premium charged at the 0% discount category is c.

(i) Write down the transition matrix for this system in terms of p. [2]

(ii) Derive the steady state distribution of policyholders in each discount categoryin terms of p. [5]

(iii) Calculate the average premium paid, A(p, c), in the steady state in terms of pand c. [3]

(iv) Suppose that the insurance company calculates premiums for this class ofinsurance by grouping policyholders into two types. The first type is known tohave p = 0.1 and the second type has p = 0.15, while the premium charged forthe first type of policyholder at the 0% discount level is 1,000. Ignoringexpenses and profit loadings and assuming that all other characteristics ofthese risks are the same, calculate the value c, the premium in the 0% discountcategory for the second type of policyholder, in order that A(0.15, c)= 1.5A(0.1, 1,000). [2]

[Total 12]


7 The cumulative claims paid each year under a certain cohort of insurance policies arerecorded in the table below, for accident years 1998, 1999, 2000 and 2001.

Development yearAccident year 0 1 2 3

1998 2,457 4,196 4,969 5,0101999 2,648 4,715 5,5612000 3,084 5,3152001 3,341

(i) Calculate the development factors under the basic chain ladder technique andstate the assumptions underlying the use of this method. [4]

(ii) The rate of claims inflation over these years, measured over the 12 months tothe middle of each year, is given in the table below.

1999 2.1%2000 10.5%2001 3.2%

Calculate the development factors under the inflation-adjusted chain laddertechnique and state the assumptions underlying the use of this method. [6]

(iii) Based on the development factors calculated in parts (i) and (ii), calculate thefitted values under these two models and comment on how these compare withthe actual values. [7]

[Total 17]

8 (i) Show that

222

1 2 2(ln ) ½22

1 ln ln=2

xb

a

b ae dx e� ��

��

�[4]

(ii) Individual claim amounts on a certain type of general insurance policy have alog-normal distribution, with mean 264 and standard deviation 346. Apolicyholder excess of 100 is a standard condition on each policy, so that theinsurance company only covers the loss amount in excess of 100.

(a) Calculate the expected claim size payable by the insurance company.

(b) Next year, claims are expected to increase by 10%. Also, a newcondition will be introduced on all policies so that the maximumamount that the insurance company will pay on any claim will be1,000. The policyholder excess will remain unchanged at 100.

Calculate the expected claim size payable by the insurance company.[14]

[Total 18]

106 S2002�6

9 An insurance company has insured a fleet of cars for the last four years. For year j(j = 1, �, 4), let Yj and Pj be the total amount claimed and the number of cars in thefleet, respectively. Let Xj = Yj/Pj be the average amount claimed per car in year j.Assume that the distribution of Xj depends on a risk parameter � and that theconditions of Empirical Bayes Credibility Theory Model 2 are satisfied. Let

m(�) = E(Xj��),

s2(�) = PjV(Xj��),

m = E(m(�))

and c = V(m(�)) > 0.

(i) (a) Derive E(Xj).

(b) Derive E(XjXk), for j � k.

(c) Determine whether Xj and Xk are independent (j � k). [3]

(ii) The credibility premium per car is obtained by minimising

24

01

( ) j jj

E m a a X�

� ��

� .

Derive the credibility premium per car, and state the credibility factor. [10]

(iii) The company has insured 10 similar fleets over the last four years. Using thedata from these years, m, E(s2(�)) and V(m(�)) are estimated to be 62.8,106.32 and 5.8 respectively.

Calculate next year�s credibility premium for a fleet of cars with claims overthe last four years given below, if the fleet will have 16 cars next year.

Year1 2 3 4

Total amount claimed 1,000 1,200 1,500 1,400Number of cars 15 16 18 15

Explain how and why the credibility factor would be affected if the estimate ofV(m(�)) increases, and comment on the effect on the credibility premium. [5]

[Total 18]


EXAMINATIONS

September 2002


EXAMINERS� REPORT

Introduction


K G FormanChairman of the Board of Examiners12 November 2002



Page 2

EXAMINERS� COMMENTS

In numerical questions, candidates were not unduly penalised forerrors in earlier parts of each question which affected their answersto the rest of the question.

There are various alternative methods for Question 7(iii), all of whichgained full credit if done correctly. A common error in Question 7 was tomake inflation adjustments to cumulative (rather than incremental) figures.

Question 8 was poorly done. Many candidates omitted the conditioningin Question 8 (ii) and (iii).

The examiners were disappointed that many candidates found difficultyin reproducing the standard bookwork in Question 9 (ii).


Page 3

1 Writing x for (x1, �, x20), we have

f(x��) = �20 20

1exp .i

ix

�

� ��

� ��

The prior distribution is gamma(2,2).

The posterior density of � is

f(��x) � f(x��)f(�)� �20e�24� � �e�2�

= �21e�26�.

This means that the posterior distribution of � is a gamma (22,26).

2 (i) The normal density can be written

222

1 1exp ( )22

y� ��

= 2 2

22 2

/ 2exp ½ log(2 ) .y y� ��

This is of exponential family form

( )exp ( , )( )

y b c ya

� ��

,

with natural parameter � = � and scale parameter � = �2.

The link function is the canonical link function g(�) = � and the linear predictor isi + xij.

So this is a generalised linear model.

(ii) Males

Females

1

2

(if 1 > 2)

Expectedclaims


Page 4

3 (i) The total claim amount has moment generating function

MT(t) = E(etT) = E(E(etT�N)) = E(M(t)N) = exp(�(M(t) � 1)).

(ii)

(a) The total amount claimed from all 210 risks is the sum of three independentrandom variables S1, S2 and S3, where Si is the total amount claimed from risksof type i, i = 1, 2, 3. Using (i),

MS(t) = 1 2 3( ) ( ) ( )S S SM t M t M t

120 50401 1 1= exp 1 exp 2 1 exp 2.5 1

1 400 1 500 1 600t t t� � � ��

� � ��

1 1 1= exp 40 1 2 120 1 2.5 50 11 400 1 500 1 600t t t

� ��

� � ��

40 1 240 1 125 1= exp 405 1405 (1 400 ) 405 (1 500 ) 405 (1 600 )t t t

� ��

� � ��

(b) This can be recognised as the moment generating function of a compoundPoisson distribution with Poisson parameter � = 405. The claim severitydistribution has moment generating function

M(t) = 40 1 240 1 125 1 ,405 (1 400 ) 405 (1 500 ) 405 (1 600 )t t t

� �

� � �

and this corresponds to density function

f(x) = / 400 /500 / 60040 1 240 1 125 1 .405 400 405 500 405 600

x x xe e e� � �

� �

[Note: the fractions can be simplified, and this is acceptable.]


Page 5

4 1/3 1/3 1/3�1 �2 �3 min max Expected profit

d1 25 19 7 7 25 17d2 10 30 8 8 30 16d3 0 2 34 0 34 12

(i) minimax means minimise the maximum loss, which is the same as maximisethe minimum profit, so the minimax solution is d2 (sell hot food).

(ii) The trader�s partner would choose d3 (sell umbrellas).

(iii) The Bayes criterion solution is d1 (sell ice-cream).

(iv) If p(�2) = p then p(�1) = p(�3) = 1/2 � (1 � p) and the Bayes risk

for d1 is 25/2 � (1 � p) + 19p + 7/2 � (1 � p) = 16 + 3pfor d2 is 10/2 � (1 � p) + 30p + 8/2 � (1 � p) = 9 + 21pfor d3 is 0/2 � (1 � p) + 2p + 34/2 � (1 � p) = 17 � 15p

and the maximum values are:

d3 for 0 < p < 1/18d1 for 1/18 < p < 7/18d2 for 7/18 < p < 1

so for p > 1/2 the Bayes criterion solution is d2.

30

25

20

15

10

5

010.80.60.40.20

Series 1

Series 2

Series 3


Page 6

5 (i) Upper bound: The adjustment coefficient R > 0 satisfies

�M(R) � � � (1 + �) �m1R = 0,

i.e.

M(R) �� m1R = 0,

where M(r) is the moment generating function of X1. Using M(R) =

1M Rk

kk p e�

� and expanding the exponential gives

1 + (1 + �)m1R =1

MRk

kk

e p�

�

>2 2

11

2

M

kk

R kRk p�

� ��

� ��

= 1 + Rm1 + 2

22R m .

Then we have

((1 + �) �1)m1 R > 2

22R m

R < 1

2

2 .mm�

Lower bound: We use the inequality

eRx � 1 ,RMx xeM M

� � 0 � x � M.

Then R satisfies

1 + (1 + �)m1R = 1

MRk

kk

e p�

�

� 1

1M

RMk

k

k ke pM M

�

� ��

� ��

= 1 11 ,RMe m mM M

� �


Page 7

so

1 + � � 1 ( 1) .RM RMe eRM

� �

Then

R > 1M

log(1 + �).

(ii) We have M = 2 and p1 = p2 = 1/2, so m1 = 1.5 and m2= 0.5(1 + 4) = 2.5.

The upper bound for R is

1

2

2 mm� = 2 0.2 1.5

2.5� � = 0.24.

The lower bound is

1 log(1 ) = ½ log1.2 = 0.09116.M

� �

Using Lundberg�s inequality and the lower bound above, theprobability �(u) of ultimate ruin with initial reserve u satisfies

�(u) � e�Ru � e�0.09116u.


Page 8

6 (i) The transition matrix is:

1 01 0.80.8 01 0.60.60

pppppp

��

��

(ii) The steady state distribution is the solution to:

p�� + 0.8p�1 = �0(1 � p)�0 + 0.6p�2 = �1(1 ��0.8p)�1 + (1 � 0.6p)�2 = �2�0 + �1 + �2 = 1

Therefore, expressing �1 and �2 in terms of �0:

�1 = �0 (1 � p) / 0.8p�2 = �0 (1 � p) (1 � 0.8p) / 0.48p2

And so:

�0 + �0 (1 � p) / 0.8p + �0 (1 � p) (1 � 0.8p) / 0.48p2 = 1�0 {0.48p2 + 0.6p(1 � p) + (1 � p) (1 � 0.8p)} = 0.48p2

Which gives:

�0 = 0.48p2 / (1 � 1.2p + 0.68p2)�1 = 0.6p(1 � p) / (1 � 1.2p + 0.68p2)�2 = (1 � p) (1 � 0.8p) / (1 � 1.2p + 0.68p2)

(iii) A(p, c) = �0 c + �1 c (1 � 0.3) + �2 c (1 � 0.5)= c {0.48p2 + 0.6p(1 � p)0.7 + (1 � p) (1 � 0.8p)0.5}

/ (1 � 1.2p + 0.68p2)= c (0.5 � 0.48p + 0.46p2) / (1 � 1.2p + 0.68p2)

(iv) A(0.15, c) = 1.5 * A(0.1, 1,000)0.52478 c = 1.5 * 0.51488 * 1,000 c = 1,471.


Page 9

7 (i) Under the basic chain ladder method development factors are:

(4196 + 4715 + 5315) / (2457 + 2648 + 3084) = 14226 / 8189 = 1.7372(4969 + 5561) / (4196 + 4715) = 10530 / 8911 = 1.18175010 / 4969 = 1.0083

The central assumption underlying this method is that, for each accident year,the amount of claims paid in each development year is a constant proportionof the total claims paid from that accident year. Implicit in this is either thatthe rate of claims inflation is constant or that weighted average past claimsinflation will be repeated in the future.

(ii) First need to calculate incremental claims:

Development yearAccident year 0 1 2 31998 2457 1739 773 411999 2648 2067 8462000 3084 22312001 3341

Adjust for past inflation to convert to mid-2001 prices:


Calculate inflation-adjusted cumulative payments:


So development factors are 1.7002, 1.1644 and 1.0073.

The assumption underlying the use of this model is that, for each accidentyear, the amount of claims paid, in real terms, in each development year is aconstant proportion of the total claims, in real terms, from that accident year.Separate, explicit assumptions are needed for both past and future claimsinflation.


Page 10

(iii) Note that this solution follows the method given in the core reading forobtaining the fitted value. This is not necessarily the method that would beused in practice.

Under the basic chain ladder method �fitted� cumulative payments are:


Analysis of �fitted� versus �actual� incremental payments:

Development yearAccident year 0 1 2 31998 Actual 2457 1739 773 41

Fitted 2457 1811 776 42Error - +72 +3 +1

1999 Actual 2648 2067 846Fitted 2648 1952 836Error - �115 �10

2000 Actual 3084 2231Fitted 3084 2274Error - +43

2001 Actual 3341Fitted 3341Error -

Under the inflation-adjusted chain ladder method �fitted� cumulativepayments before allowing for past-inflation are:



Page 11

�Fitted� incremental claims before allowing for past inflation:


Analysis of �fitted� versus �actual� incremental payments after allowing forpast inflation:

Development yearAccident year 0 1 2 31998 Actual 2457 1739 773 41

Fitted 2457 1756 775 42Error - +17 +2 +1

1999 Actual 2648 2067 846Fitted 2648 2049 844Error - �18 �2

2000 Actual 3084 2231Fitted 3084 2229Error - �2

2001 Actual 3341Fitted 3341Error -

Clearly the inflation-adjusted chain ladder method with the given assumptionsabout past inflation gives a better fit to the observed data than the basic chainladder method. However, even under the basic chain ladder method none ofthe errors are large enough to suggest that this model is inaccurate. There isno guarantee that either model will prove to be a good guide to the future.


Page 12

8 (i) I =2

21 (ln )

22

1

2

xb

ae dx� ��

�

��

�

Put y = lnx, so dy = 1x

dx and dx = eydy

I =2

21 ( )ln 2

ln 2

1

2

yb ya

e e dy� ��

�

��

�

=ln 2 2 2

2ln 2

1 1exp ( 2 2 )22

b

ay y y dy� ��

� ��

=ln 2 2 2 4

2ln 2

1 1exp (( ) 2 )22

b

ay dy� ��

� ��

=2 2

2 21 ( )ln½ 2

ln 2

1

2

yb

ae e dy

� � ��

��

�

=2 2 2

½ ln lnb ae��

(ii)

(a)2 2 2½ ½ 2 2= 264 ( ) ( 1) = 346e e e��

�

2 2

23461 =264

e� �

�2 = ln(1 + 1.7177) = 1

� = ln264 ��0.5 = 5.076

XI = 1000

100 100X

X X��

��

Claims paid by the insurance company are X �100�X > 100.

E[X �100�X > 100] = 100( )

100( 100)

xf x dx

P X

�

�

�

�


Page 13

100( )xf x dx

�

� = 2 2

½ ln1001e��

� ��

= 264(1 � �(�1.471))

= 264 � 0.9294 = 245.35

P(X > 100) = ln1001 � �� = 1 � �(�0.471)

= 0.6812

E[X � 100�X > 100] = 245.35 1000.6812

� = 260.18

(b)

Let Y = 1.1X

E[Y] = 1.1E[X] = 290.4V[Y] = 1.12V[X] = (380.6)2

� 2

e� � 1 = 1.717�2 = 1

� = ln290.4 ��0.5 = 5.1714

YI = 1000

100 100 1,1001,1001,000

YY Y

Y

��

� � ��

Claim amount payable by the insurer is YI�Y > 100

E[YI�Y > 100] =

1,100

100( 100) ( ) 1,000 ( 1,100)

( 100)

y f y dy P Y

P Y

� � �

�

�

=

1,100

100( ) 100 (100 1,100) 1,000 ( 1,100)

( 100)

yf y dy P Y P Y

P Y

� � � � �

�

�


Page 14

1,100

100( )yf y dy� = ln1,100 5.1714 1 ln100 5.1714 1290.4

1 1� ��

= 290.4(�(0.832) � �(�1.566))

= 290.4(0.7973 ��0.0587)

= 214.49

P(Y < 1,100) = ln1,100 5.1714 = (1.832) = 0.96651��

� ��

P(Y < 100) = ln100 5.1714 = ( 0.566) = 0.28571��

�E[YI�Y > 100] = 214.49 100(0.9665 0.2857) 1,000(1 0.9665)1 0.2857

� � � �

�

= 251.87

9 (i) (a) E(Xj) = E(E(Xj��)) = E(m(�)) = m.

(b) E(XjXk) = E(E(XjXk��)) = E(m(�)2) for j � k

(c) E(XjXk) � E(Xj)E(Xk) for j � k

so Xj and Xk are not independent.

(ii) Differentiating with respect to a0 and setting to zero gives

4

01

( ) = 0,j jj

E m a a X�

� ��

�

and so

a0 = 4

11 .j

jm a

�

� ��

�

Note that

E(Xkm(�)) = E(E(Xkm(�)��)) = E(m(�)2),

E(XjXk) = E(m(�)2), j � k,


Page 15

2( )kE X = 2( ( ))kE E X ��

= E(V(Xk��) + (E(Xk��))2)

= 2 21 ( ( )) ( ( ) ).k

E s E mP

� � �

Differentiating with respect to ak (k � 0) and setting to zero gives

4

01

( ) = 0.k k j j kj

E X m a X a X X�

� ��

�

Using the above relationships this becomes

E(m(�)2) � a0m �4

1j

ja

�

� E(m(�)2) � 2( ( )) = 0,k

k

a E sP

�

which gives, using the expression above for a0,

ak = � �4

12

( ( )) 1

( ( ))

k jjP V m a

E s�

� �

�

�

Sum over k:

4 4 4

21 1 1

( ( ))= 1 ,( ( ))k k j

k k j

V ma P aE s

� � �

� � ��

� � �

and so

44

14 2

1 1

= .( ( )) / ( ( ))

jjj

j jj

Pa

P E s V m�

�

�

� � �

��

�

Substituting into the expressions for a0 and ak, we obtain

a0 = 2

4 21

( ( )) / ( ( ))

( ( )) / ( ( ))jj

mE s V m

P E s V m�

� �

� � ��

and

ak = 4 21 ( ( )) / ( ( ))

k

jj

P

P E s V m�

� � ��


Page 16

The credibility premium per car is

42

14 42 2

1 1

( ( )) / ( ( )) ,( ( )) / ( ( )) ( ( )) / ( ( ))

jj

j jj j

PE s V m m XP E s V m P E s V m

�

� �

� ��

� � � � � �

�

� �

where 4 41 1= / .j j jj jX P X P� �

� � The credibility factor is

Z = 4

14 2

1

.( ( )) / ( ( ))

jj

jj

P

P E s V m�

�

� � �

�

�

(iii) The credibility premium per car is (1 ) .ZX Z m� � We have

= 64jP�

E(s2(�))/V(m(�)) is estimated as 106.32/5.8 = 18.3310, so Zis 64/(64 + 18.3310) = 0.7773.

Further, X = 5,100/64 = 79.6875,so that credibility premium per car is 75.9275.

Hence for the fleet of 16 cars the premium is 16 � 75.9275 = 1,214.84.

If the estimate of V(m(�)) increases, then the estimate of Z increases andrelatively more weight is put on the data from this particular fleet. This happens because an increase V(m(�)) means an increase in the variabilitybetween fleets and so less emphasis on collateral information.

If Z increases, then Z � 79.69 + (1 ��Z) � 62.8 also increases. The credibilitypremium moves closer to ,X and, since this is greater than the estimatedvalue of m, this implies an increase in the premium.


EXAMINATIONS

4 April 2003 (am)













106 A2003�2

1 The lengths of time taken to deal with each of n reports are independent exponentiallydistributed random variables with mean 1/�.

Show that the gamma distribution is the conjugate prior for this exponentialdistribution. [3]

2 For the following main categories of insurance product:

LiabilityProperty

(i) Describe the cover provided. [2](ii) List two examples of insurance cover in each category. [2]

[Total 4]

3 (i) A random variable Y has density of exponential family form:

f(y) = ( )exp ( , ) .( )

y b c ya

� ��

State the mean and variance of Y in terms of b(�) and its derivatives and a(�).[1]

(ii) (a) Show that an exponentially distributed random variable with mean �has a density that can be written in the above form.

(b) Determine the natural parameter and the variance function.[3]

[Total 4]


Θ1 Θ2 Θ3

D1 23 34 16D2 30 19 18D3 23 27 20D4 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 P(Θ1) = 0.25, P(Θ2) = 0.15, P(Θ3) = 0.60,determine the Bayes criterion solution to the problem. [2]

[Total 6]


5 Claims arrive in a Poisson process rate �, and N(t) is the number of claims arriving bytime t. The claim sizes are independent random variables X1, X2, �, with mean �,independent of the arrivals process. The initial surplus is u and the premium loadingfactor is �.

(i) (a) Give an expression for the surplus U(t) at time t.

(b) Define the probability of ruin with initial surplus u, �(u), and sketch arealisation of the surplus process that shows a ruin event (graph papernot needed).

(c) State the value of �(u) when � = 0.[4]

(ii) The unit of currency is changed so that one unit of the old currency is worththe same as 2.5 units of the new currency.

Determine a relationship between �(u) in (i)(b) and the probability of ruin forthe new process. [2]

(iii) Suppose instead that the old unit of currency is used, but the claim arrival rateis doubled. Let N*(t) be the new number of claims by time t.

(a) Show that N*(t) has the same distribution as N(2t).

(b) Derive an expression for the surplus U*(t) for the new process in termsof the surplus process in (i)(a), and determine the probability of ruinfor the new process in terms of the probability of ruin in (i)(b).

[3][Total 9]

6 The triangle below shows incremental claims for a portfolio of property insurancepolicies.

Development Year

Accident Year 0 1 2 3

1998 4,652 3,079 931 2751999 6,067 4,555 1,1832000 5,822 4,2972001 7,934

(i) Calculate the outstanding claims reserve for this portfolio using the basicchain ladder method. [8]

(ii) State any assumptions used in determining your answer in (i). [2][Total 10]

106 A2003�4

7 Claims on a portfolio of general insurance policies have a Pareto distribution, withdensity function f(x), where:

f(x) = 1 ( )xx

�

��

��

Excess of loss reinsurance is arranged with retention M (M > �).

(i) (a) Show that P(X > x) = ( )xx

�

��

� �.

(b) Derive an expression for the expected amount paid by the reinsurer, ona claim which involves the reinsurer.

[7]

(ii) Last year 10 claims were received, of which 4 involved the reinsurer. Theclaims which involved the reinsurer are denoted by {xi: i = 1, �, 4 (xi > M)}.

Write down the likelihood for these data. [4][Total 11]

8 The number of claims on a particular risk in a fixed time period has a Poissondistribution with mean �. There were x1 and x2 claims during the first two timeperiods.

(i) Suppose that � has prior density f(�) = 2e�2� (� > 0).

(a) Determine the Bayesian estimate under quadratic loss for the expectednumber of claims during the third time period, and show that it is ofthe form of a credibility estimate.

(b) Show that the corresponding credibility estimate under Model 1 ofEmpirical Bayes Credibility Theory (EBCT Model 1) is the same asyour answer in (i)(a).

[8]

(ii) Now suppose that � has prior density f(�) = 1, 0 < � < 1.

Derive an expression in the form of the ratio of two integrals for the Bayesianestimate under quadratic loss for the expected number of claims during thethird time period (do not evaluate the integrals). [2]


(iii) Assume there were no claims at all during the first two time periods.

(a) Evaluate the expressions in (ii) and in (i)(a).

(b) Derive the corresponding EBCT Model 1 estimate with the prior as in(ii).

(c) Compare it with the Bayesian estimate in (ii), and with the estimate in(i)(a).

[7][Total 17]

9 (i) Let N be the number of claims on a risk in one year. Suppose claims[X1, X2, �] are independent, identically distributed random variables,independent of N. Let S be the total amount claimed in one year.

(a) Derive E(S) and V(S) in terms of the mean and variance of N and X1.

(b) Derive an expression for the moment generating function MS(t) of S interms of the moment generating functions MX(t) and MN(t) of X1 and Nrespectively.

(c) If N has a Poisson distribution with mean �, show that:

MS(t) = exp(�(MX(t) ��1)).

(d) If N has a binomial distribution with parameters m and q, determine themoment generating function of S in terms of m, q and MX(t).

[5]

(ii) A portfolio consists of 500 independent risks. For the ith risk, with probability1 � qi there are no claims in one year, and with probability qi there is exactlyone claim (0 < qi < 1). For all risks, if there is a claim, it has mean �, variance�2 and moment generating function M(t). Let T be the total amount claimedon the whole portfolio in one year.

(a) Determine the mean and variance of T. [4]

The amount claimed in one year on risk i is approximated by a compoundPoisson random variable with Poisson parameter qi and claims with the samemean �, the same variance �2, and the same moment generating function M(t)as above. Let T� denote the total amount claimed on the whole portfolio inone year in this approximate model.

(b) Determine the mean and variance of T� , and compare your answers tothose in (ii)(a). [4]

106 A2003�6

Assume that qi = 0.02 for all i, and if a claim occurs, it is of size � withprobability one.

(c) Derive the moment generating function of T, and show that T has acompound binomial distribution. [2]

(d) Determine the moment generating function of the approximating T� ,and show that T� has a compound Poisson distribution. [2]

[Total 17]

10 An insurance company charges an annual premium of £300 and operates a No ClaimsDiscount system as follows:

Level 1 0% discountLevel 2 30% discountLevel 3 60% discount

The rules for moving between levels are as follows:

If the policyholder does not make a claim during the year, they move up one level orare eligible to stay at level 3.

If the policyholder makes 1 claim during the year, they move down one level or stayat level 1.

If the policyholder makes 2 or more claims during the year, they move straight downto, or remain at, level 1.

The insurance company has recently introduced a �protection� system where onreaching, or remaining eligible to remain at, level 3, policyholders are immediatelyoffered the opportunity to �protect� their discount for an additional annual premiumof £50. If they make no claims or 1 claim during the year they can remain at level 3.However, if they make 2 or more claims during the year, they move straight down tolevel 1.

Out of the policyholders who had �protected� their discount level at the beginning ofthe year and are still eligible to stay at level 3 at the end of the year, 25% choose to�protect� their discount again the following year. From all the other policyholderseligible for level 3 at the end of the year, 10% choose to take the �protection� option.

Policyholders at different levels are found to experience different rates of claiming.The number of claims made per year follows a Poisson distribution with parameter λas follows:

Level: 1 and 2 3 and �protected�

λ: 0.40 0.25

106 A2003�7

(i) Derive the transition matrix. [7]

(ii) Calculate the proportions at each of the levels 1, 2, 3 and the �protected� level,when the system reaches a steady state. [10]

(iii) Determine the average premium per policy. [2][Total 19]


Page 1


REPORT OF THE BOARD OF EXAMINERS

April 2003


EXAMINERS� REPORT

Introduction


J CurtisChairman of the Board of Examiners

3 June 2003



Page 2

General comments

Most candidates attempted all of the questions although some were clearly short on timetowards the end of the paper.

There was a tendency by some candidates to write out definitions from the core readingrather than applying the specifics given in the questions.

1 Let x1, �, xn be the observed times taken. Then

f(x1, �, xn��) = �nexp1

n

ii

x�

� ��

� �� .

Suppose the prior is a gamma distribution with density

f(�) = v��e�v�, � > 0.

Then the posterior satisfies

f(��x1, �, xn) � �n 1

1exp

nv

ii

x e��

�

� ��

� ��

= �n+��1 � �1 ,n

ii x ve �

��

so the posterior distribution is also a gamma. Thus the gamma is the conjugate prior.

This question was generally well answered. A number of candidates lost marks for usingsloppy notation.

2 (i) Liability cover provides indemnity where the insured, owing to some form ofnegligence, is legally liable to pay compensation to a third party.

Property cover indemnifies the policyholder against loss of or damage tohis/her own material property.

(ii) Liability cover; any two of the following

employers� liabilitymotor third party liabilitypublic liabilityproduct liability


Page 3

professional indemnity

Property cover; any two of the following

residential buildingmoveable propertycommercial buildingland vehiclesmarine craftaircraft

Both parts of this question were poorly answered. In (i) very few mentioned negligence andcompensation to a third party. In (ii) many used examples such as 'house insurance' or 'carinsurance'.

3 (i) E(Y) = ( )b� � and V(Y) = ( ) ( ).a b��

(ii) (a) f(y) = 1 exp y� ��

= ( / ) logexp 0 ,1

y� � � ��

which is of the exponential family form.

(b) The natural parameter is � = ��1.

b(�) = �log(��) so ( )b� � = ��1 and ( )b�� = ��2 = �2.

so the variance function is V(�) = �2.

Nearly all candidates correctly answered (i). Part (ii) was poorly answered by manycandidates with sloppy algebra and answers being given in terms of theta rather than mu.

4 (i) Decision D4 can be discounted immediately.

D4 is dominated by D2 because the loss in each scenario for D2 is either thesame as or less than that for D4.

(ii) The maximum loss under each of the remaining decisions is:

D1 34D2 30D3 27 ←


Page 4

Choose D3 which minimises the maximum loss.

(iii) The expected loss under each decision is:

D1 23 � (0.25) + 34 � (0.15) + 16 � (0.60) = 20.45 ←D2 30 � (0.25) + 19 � (0.15) + 18 � (0.60) = 21.15D3 23 � (0.25) + 27 � (0.15) + 20 � (0.60) = 21.80

Therefore choose D1 which has the smallest expected loss.

Very well answered by nearly all candidates. A few omitted to show any working todemonstrate that they had correctly determined the answers to (ii) and (iii).


Page 5

5 (i) (a) The surplus at time t is

U(t) = u + (1 + �) ��t �( )

1.

N t

ii

X�

�

(b) The probability of ruin is

(u) = P(U(t) < 0 for some t, 0 < t < ).

(c) When � = 0, then (u) = 1 for all u > 0.

(ii) Let ( )U t� be the new surplus at t, then ( )U t� = 2.5U(t) and the initial capital is2.5u in the new units.

Then, writing ( )� �� for the new probability of ruin,

(2.5 )u�� = ( ( )P U t� < 0 for some t, 0 < t < )= P(2.5U(t) < 0 for some t, 0 < t < )= (u),

or an equivalent relationship.

(Could give a graphical explanation, or an explanation in words.)

ruin

U(t)

u

0 t


Page 6

(iii) (a) From properties of Poisson processes, N(t) has a Poisson distributionwith mean �t, and N*(t) has a Poisson distribution with mean 2�t. SoN*(t) has the same distribution as N(�t) for � = 2.

(b) For initial capital u, using ~ to denote �has the same distribution as,�

U*(t) = u + (1 + �) 2��t � *( )

1

N t

ii

X�

�

~ u + (1 + �)��(2t) � (2 )

1,

N t

ii

X�

�

so U*(t) ~ U(2t) (or a graphical explanation, or an explanation inwords). So

*(u) = P(U*(t) < 0 for some t, 0 < t < )= P(U(2t) < 0 for some t¸ 0 < t < )= (u).

(i) Many candidates used the expression S(t) as the sum of claims without definition ratherthan sigma Xi. Graphs were often disappointing with many illustrating a recovery from theruin event.

(ii) This was was well understood although a few candidates failed to explain adequatelywhy there was no change. Either a graphical or written explanation was acceptable.

(iii) Relatively few candidates managed to derive the required expression in (b).

6 (i) Cumulative claims:

4652 7731 8662 89376067 10622 118055822 101197934

Development factors:

R1 = (10119 + 10622 + 7731) / (5822 + 6067 + 4652)= 28472 / 16541= 1.7213

R2 = (11805 + 8662) / (10622 + 7731) = 20467 / 18353= 1.1152


Page 7

R3 = 8937 / 8662

= 1.0317

Forecast cumulative claims:

4652 7731 8662 89376067 10622 11805 121795822 10119 11285 116427934 13657 15230 15713

Total outstanding claims

= (12179 + 11642 + 15713) � (11805 + 10119 + 7934)= 39534 � 29858= 9676

(ii) For each accident year the total claim amount in each development year is aconstant proportion of the ultimate claim amount for that accident year.

Weighted average past inflation is appropriate estimate of future inflation.

Claims are fully run off by the end of development year 3.

This straightforward question was generally well answered. Many failed to obtain full marksfor stating the basic chain ladder assumptions.

7 (i) (a) P(X > x) = 1xdy

y

��

��

��

�

= =x

xy

��

�

� ��

� ��

(b) E[X � M�X > M] = 1( )

Mx M dx

x

M

��

��

�

��

��

�

1( )M

x M dxx

��

��

�� =

Mdx M

Mx

��

�

��

�

= 1( 1) M

MMx

��

��

� ��

� � � � � �


Page 8

= 11M

MM

��

��

� � ��

= 1 1( ( 1)) =( 1) ( 1)M M

� �

��

� ��

� � � �

� E[X � M�X > M] = 1 =1( 1)

M MM

� �

��

�

� ��

(ii) The likelihood is

46

1( ( )) ( )i

iP X M f x

�

� �

= 6 4

11

1iiM x

� �

��

�

� ��

�

= 6 4 4

14

1

1

ii

Mx

� �

��

�

� ��

� ��

A variety of successful approaches were adopted by candidates for (i). Some used the YellowTables whilst others carried out calculations from first principles. Both approaches wereacceptable. Many candidates failed to follow through the algebra in (ii).

Part (ii) was disappointing with many candidates misunderstanding the question. A commonerror was to express the likelihood as the probability of 4 claims exceeding M multiplied bythe product of x1 to x6.

8 (i) (a) Let k = x1 + x2, so

f(x1, x2��) � e�2��k.

The posterior density of � is

f(��x1, x2) � f(x1, x2��)f(�)

� �ke�4�, � > 0,


Page 9

i.e. the posterior distribution is a gamma distribution with parametersk + 1 and 4.

The Bayesian estimate under quadratic loss is the posterior mean

1 1 1 1= ,4 2 2 2

k x��

where x = k/2 is the mean of x1 and x2.

The prior mean of � is m = ½.The Bayesian estimate is of the form

(1 ) ,Zx Z m� �

which is of the form of a credibility estimate.

(b) m(�) = E(X��) = �, so m(�) has an exponential distribution with mean½. Hence V(m(�)) = ¼.

Further s2(�) = V(X��) = �, so that E(s2(�)) = E(�) = ½.

The EBCT Model 1 credibility factor is

Z = 2( ( ))( ( ))

nE snV m

��

�

= 2½2¼

�

= ½,

and the corresponding credibility estimate is

1 1 .2 2

x m�

This is the same as in (i)(a).

(ii) With the alternative prior, the posterior is

f(��x1, x2) � e�2��k, 0 < � < 1.

The Bayesian estimate is the posterior mean

1 1 201 20

k

k

e d

e d

� � �

� �

� � �

� � �

(iii) Now k = 0 (so x = 0).


Page 10

(a) The numerator in (ii) is then

1 20

e d� �� =

12 21

00

2 2e e d� � � � ��

� � � � ��

�

= 12 2

02 4

e e� � � ��

��

= 21 3 .

4e��

The denominator is

1 20

e d� � �� = 21

2e�� .

Then the expression in (ii) is evaluated as

2

21 3

2(1 )ee

�

�

�

�

= 0.3435.

The expression in (i)(a) is the same as that in (i)(b). When k = 0 theyare

14

k � = 0.25.

(b) The prior in (ii) is a uniform distribution on (0, 1), so we still haveE(s2(�)) = E(�) = ½. But now V(m(�)) = V(�) = 1/12.

Then the credibility factor for EBCT Model 1 is

Z = 2½2

1/12�

= 0.25,

and the EBCT Model 1 credbility estimate is

Zx + (1 � Z)m = 0.75 � 0.5 = 0.375.

(c) For the uniform prior in (ii), the EBCT Model 1 estimate is not thesame as the Bayesian estimate.


Page 11

Comparing the EBCT Model 1 credibility estimates for the two priors,they are not the same. Both priors have the same value for E(s2(�)),but the prior in (ii) has a smaller value for V(m(�)). As V(m(�))decreases, the credibility factor decreases. In our case, the credibilityestimate is (1 � Z)m, so this will increase as V(m(�)) decreases, andthis is what is observed.

(Or other sensible comments.)

This question was very poorly answered apart from (i)(a). Few candidates managed (ii) or(iii).

9 (i) (a) S = X1 + � XN, and so

E(S) = E(E(S�N)) = E(NE(X1)) = E(N)E(X1),

and

V(S) = E(V(S�N)) + V(E(S�N))= E(NV(X1)) + V(NE(X1))= E(N)V(X1) + V(N)(E(X1))2.

(b) The moment generating function of S is

MS(t) = E(eSt) = E(E(eSt�N)) = E((MX(t))N) = MN(log(MX(t))).

(c) Since MN(t) = exp(�(et �1)), the mgf of S is

MS(t) = exp(�(MX(t) � 1)).

(d) Here the mgf of N is (1 � q + qet)m, so that the mgf of S is

MS(t) = (1 � q + qMX(t))m.

(ii) (a) Let Yi be the amount claimed on risk i, so that

Yi = 0 with probability 1

with probability ,i

i i

qZ q

��

where E(Zi) = �, V(Zi) = �2 and the mgf of Zi is M(t).

Moments of Yi are


Page 12

E(Yi) = qi�,and

2( )iE Y = 2( )i iq E Z = qi(�2 + �2),

so that

V(Yi) = qi(�2 + �2) � 2 2iq � = qi�2 + �2qi(1 � qi).

The mean of T is

E(T) = 500 500

1 1( ) = .i i

i iE Y q

� �

��

The variance of T is

V(T) = 500

2 2

1( (1 )).i i i

iq q q

�

� ��

(b) Let iY� have a compound Poisson distribution with Poisson parameterqi and claim sizes with mean �, variance �2 and mgf M(t). Then from(i)(a)

( )iE Y� = qi�,

so

( )E T� = 500 500

1 1( ) = .i i

i iE Y q

� �

��

Also from (i)(a), the variance of iY� is

( )iV Y� = qi(�2 + �2),

so

( )V T� = 500

2 2

1( ) .i

iq

�

� ��

Comparing with the mean and variance of T,

( )E T� = E(T),


Page 13

and

( )V T� = 500

2 2

1( )i i

iq q

�

� � ��

= 500 500

2 2 2 2

1 1( (1 ) ) ,i i i i

i iq q q q

� �

� � � � ��

so that ( ) ( ).V T V T��

(c) Now q1 = � = q500 = 0.02 and if claims occur they are equal to �, soall Yi�s have the same distribution. Write MY(t) for the mgf of Yi. Then

MY(t) = E(exp(Y1t)) = 0.98 + 0.02e��t,

so that

MT(t) = (MY(t))500 = (0.98 + 0.02e�t)500.

From (i)(c) this is the mgf of a compound binomial distribution, withbinomial parameters 500 and 0.02 and with claims equal to � withprobability one.

(d) In the approximation for this case, the �siY� all have the same compoundPoisson distribution with mgf

( )YM t� = exp(0.02(e�t �1)),

so that

500( ) = ( ( )) = exp(10( 1)).tT YM t M t e� ��

From (i)(b) this is the mgf of a compound Poisson distribution, withPoisson parameter 10 and claims equal to � with probability one.

Part (i) was well answered although a number of candidates failed to derive E(S) and V(S) in(a).

Part (ii) was not well answered. Only a few candidates distinguished the difference betweena summation of qi and 500q. Some credit was given for this approach. A small number ofcandidates made meaningful comments on the comparison requested in (b). For (c) and (d)most candidates failed to apply the formulae from (i) correctly.


Page 14

10 (i) Level 1 & 2 P (0 claims) = e�0.40 = 0.6703P (> 0 claims) = 1 � 0.6703 = 0.3297

Level 3 & P P (0 claims) = e�0.25 = 0.7788P (1 claim) = 0.25e�0.25 = 0.1947P (> 1 claim) = 1 � 0.7788 � 0.1947 = 0.0265

Moving to level PFrom level 2 = 0.6703 � 0.10 = 0.0670From level 3 = 0.7788 � 0.10 = 0.0779From level P = (1 � 0.0265) � 0.25 = 0.2434

Moving to level 3From level 2 = 0.6703 � 0.90 = 0.6033From level 3 = 0.7788 � 0.90 = 0.7009From level P = (1 � 0.0265) � 0.75 = 0.7301

Therefore the transition matrix is:

0.3297 0.6703 0 00.3297 0 0.6033 0.06700.0265 0.1947 0.7009 0.07790.0265 0 0.7311 0.2434

(ii) In a steady state

0.3297 0.6703 0 0Π 0.3297 0 0.6033 0.0670 = Π

0.0265 0.1947 0.7009 0.07790.0265 0 0.7311 0.2434

0.3297π1 + 0.3297π2 + 0.0265π3 + 0.0265πP = π10.6703π1 + 0.1947π3 = π2

0.6033π2 + 0.7009π3 + 0.7311πP = π30.0670π2 + 0.0779π3 + 0.2434πP = πP

and π1 + π2 + π3 + πP = 1

Hence

πP = 0.0886π2 + 0.1030π3π3 = 2.0171π2 + 2.4443πP

πP = 0.0886π2 + 0.2078π2 + 0.2518πPπ3 = 2.0171π2 + 0.2166π2 + 0.2518π3


Page 15

πP = 0.3962π2π2 = 0.3350π3π3 = 4.7776π1

π1 + 4.7776π1 + 1.6005π1 + 0.6341π1 = 1

π1 = 0.1248π2 = 0.1998π3 = 0.5962πP = 0.0791

(iii) The premium at each level is:

Level 1 = £300Level 2 = £210Level 3 = £120Level P = £120 + £50 = £170

Therefore the average premium is:

0.1248 � £300 + 0.1998 � £210 + 0.5962 � £120 + 0.0791 � £170:

= £164.39

This was generally well done although a number of candidates presented a 3x3 matrix whichignored the Protected Level. Marks for the following part were reduced accordingly.

Some candidates struggled with the algebra in (ii), possibly as a result of time constraints.Credit was awarded even where the candidate had made errors in (i).

Part (iii) was well answered although a few candidates did not identify the correct premiumfor the Protected level (£300 * .4 + £50).


EXAMINATIONS













� Faculty of Actuaries106—S2003 � Institute of Actuaries

106 S2003—2

1 When rating general insurance business, such as motor insurance, state the mainissues that have to be considered to convert a pure risk premium into an officepremium. [4]

2 A portfolio consists of a total of 120 independent risks. On each risk, no more thanone event can occur each year, and the probability of an event occurring is 0.02.When such an event does occur, the number of claims, N, has the followingdistribution.

P(N = x) = 0.4 � 0.6x�1 (x = 1, 2, …)

Determine the mean and variance of the distribution of the number of claims whicharise from this portfolio in a year. [4]

3 Claims occur in a Poisson process rate 20. Individual claims are independent randomvariables with density

f(x) = 43 , 0,

(1 )x

x�

�

independent of the arrivals process.

(i) Calculate the mean and variance of the total amount claimed by time t = 2. [3]

(ii) Using a normal approximation, derive approximately the probability of ruin att = 2 if the premium loading factor is 30% and the initial surplus is u = 10. [2]

[Total 5]

4 The claims process is a compound Poisson process with rate �, and individual claimamounts have a gamma distribution with mean 2��1 and variance 2��2. A premiumloading factor of 50% is used.

(i) Determine the adjustment coefficient. [4]

(ii) (a) Derive an upper bound on the probability of ruin if the initial surplusis u.

(b) State how this upper bound depends on �, and explain this dependenceusing general reasoning. [3]

[Total 7]

106 S2003—3 PLEASE TURN OVER

5 (i) The distribution of claims on a portfolio of general insurance policies is aWeibull distribution, with density function f1(x) where

f1(x) = 2

2 cxcxe� (x > 0)

It is expected that one claim out of every 100 will exceed £1,000. Use thisinformation to estimate c. [2]

(ii) An alternative suggestion is that the density function is f2(x), where

f2(x) = �e��x (x > 0)

Use the same information as in part (i) to estimate �. [2]

(iii) (a) For each of f1(x) and f2(x) calculate the value of M such that

P(X > M) = 0.001

(b) Comment on these results. [3][Total 7]

6 Cumulative claims incurred on a motor insurance account are as follows:

Figures in £000’sDevelopment year

Policy year 0 1 2 3

1998 1,417 1,923 2,101 2,2091999 1,701 2,140 2,8402000 1,582 1,7402001 2,014

The data have already been adjusted for inflation. Annual premiums written in 2001were £3,912,000 and the ultimate loss ratio has been estimated as 92%. Claims paidto date for policy year 2001 are £561,000, and claims are assumed to be fully run-offby the end of development year 3.

Estimate the outstanding claims to be paid arising from policies written in 2001 only,using the Bornhuetter-Ferguson technique. [9]

106 S2003—4

7 In a portfolio of property insurance policies, let θ denote the proportion of policies onwhich claims are made in the year. The value of θ is unknown and is assumed to havea Beta prior distribution with parameters α and β. A claims analyst estimates that themean and standard deviation of θ are 0.20 and 0.25 respectively.

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

(i) Determine the values of the parameters, α and β, of the prior distribution. [3]

(ii) Determine the posterior distribution and hence the posterior mean of θ. [2]

(iii) For the general case where x is the number of claims arising from a samplesize n and µ is the mean of the Beta prior distribution, show that the posteriormean of θ can be expressed as:

Z.(x / n) + (1 � Z).µ

and express Z as a function of α, β and n. [3]

(iv) (a) Calculate the value of Z for the situation in part (ii) and explain what itrepresents.

(b) Without performing any further calculations, explain how you wouldexpect the value of Z to change if:

(1) The analyst now believes the standard deviation, σ, of the priordistribution to be 0.50.

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

(c) State the limiting value of Z as σ and n increase and explain what thismeans. [4]

[Total 12]

106 S2003—5 PLEASE TURN OVER

8 There are m male drivers in each of three age groups, and data on the number ofclaims made during the last year are available. Assume that the numbers of claims areindependent Poisson random variables. If Yij is the number of claims for the jth maledriver in group i (i = 1, 2, 3; j = 1, …, m), let E(Yij) = �ij, and suppose log(�ij) = �i.

(i) Show that this is a generalised linear model, identifying the link function andthe linear predictor. [3]

(ii) Determine the log-likelihood, and the maximum-likelihood estimators of �1,�2 and �3. [4]

For a particular data set with 20 observations in each group, several models are fitted,with deviances as shown:

Deviance

Model 1 log(�ij) = �i 60.40

Model 2 log(�ij) = if = 1,2if = 3

ii

��

61.64

Model 3 log(�ij) = � 72.53

(iii) Determine whether or not model 2 is a significant improvement over model 3,and whether or not model 1 is a significant improvement over model 2. [5]

(iv) Interpret these three models. [3][Total 15]

106 S2003—6

9 An insurance company, C, models the loss amount for each claim from its portfolio ofpolicies as having a mean and standard deviation of £500. A claims analyst assumesthat the loss amount follows an exponential distribution. There are 200 policies in theportfolio and C expects 30% of these policies to make a claim each year.

C is considering a reinsurance policy under which a reinsurer will pay the excess, ifany, over £2,500 for each individual claim made in the portfolio. A reinsurancecompany, R, has offered to provide cover for the next year for an annual premium of£300. C has asked you for your advice on whether or not to accept this offer.

(i) Calculate the expected total claim amount ceded to R over the whole portfolio[5]

(ii) Comment on your answer to (i) and indicate the main points you would makein advising C on whether to reinsure with R. [3]

A second claims analyst suggests that the claims follow a lognormal distribution.

(iii) Calculate the expected total claim amount ceded to R under the secondanalyst’s assumption.

(Note that for the lognormal distribution:

2 2½ ln( ) = 1

M

Mxf x dx e�

��

� ��

�

where Φ is the cumulative distribution function of the standard Normaldistribution.) [6]

(iv) Comment on your answer to (iii). [4][Total 18]

10 An insurance company has arranged a policy with a large organisation to cover aparticular type of disastrous event. Under the terms of the policy the insurer willcover the total cost of the claim for up to three disasters in any year. The insurer isconsidering taking out reinsurance for the policy whereby the reinsurer will cover theexcess, if any, of the total cost over £2 million per disaster. The reinsurer has quotedthe premiums for two different types of cover as follows:

Cover provided by reinsurer PremiumD1 claims over £2m on the first disaster only £500,000D2 claims over £2m on each of the first two

disasters£1,000,000

The insurer needs to decide whether to opt for cover levels D1, D2 or not to reinsure(D0).

106 S2003—7

The insurer models the total loss from a single disaster as having a Pareto distributionwith mean £1.5 million and standard deviation £2.5 million.

(i) Determine the expected total loss per individual disaster, with and withoutreinsurance. [7]

(ii) Complete the decision matrix below based on total outgoings (you may ignoreexpenses):

no disastersΘ0

1 disasterΘ1

2 disastersΘ2

3 or moredisasters

Θ3D0D1D2

[6]

(iii) (a) Determine the minimax solution.

(b) Explain your answer using general reasoning. [2]

(iv) The insurer believes that the number of disasters each year follows a Poissondistribution, parameter 0.9. Determine the Bayes Criterion solution. [4]

[Total 19]


REPORT OF THE BOARD OF EXAMINERS

September 2003


EXAMINERS’ REPORT

Introduction The attached subject report has been written by the Principal Examiner with the aim of helping candidates. The questions and comments are based around Core Reading as the interpretation of the syllabus to which the examiners are working. They have however given credit for any alternative approach or interpretation which they consider to be reasonable. J Curtis Chairman of the Board of Examiners 11 November 2003 © Faculty of Actuaries © Institute of Actuaries


Page 3

In numerical questions, candidates were not unduly penalised for errors in earlier parts of each question which affected their answers to the rest of the question. 1 Loadings: commission expenses profit other contingencies Adjust basic values for: unusual experience large claims trends in claims changes in risk, cover, cost of reinsurance 2 Given an event, distribution of number of claims has mean, E[N] = 2.5 variance, V[N] = 3.75 (using formulae in tables) The mean and variance of the number of claims annually are 120 × 0.02 × 2.5 = 0.05 × 120 = 6 and 120 × (0.02 × 3.75 + 0.02 × 0.98 × 2.52) = 120 × 0.1975 = 23.7. In Question 2, a common mistake was to find the mean and variance of the wrong type of negative binomial. 3 (i) Let X be a typical claim size, so X has a Pareto distribution with parameters

α = 3 and λ = 1, and has mean 0.5 and second moment 1. The number N of claims by t = 2 has a Poisson distribution with mean 40. Let S be the total amount claimed by t = 2. Then E(S) = E(X)E(N) = 20, and V(S) = E(N)E(X2) = 40.


Page 4

(ii) The premium income by t = 2 is (1 + θ) × 40 × 0.5 = 26, so probability of ruin

at t = 2 is

P(S > 10 + 26) = 20 36 2040 40

SP − −⎛ ⎞>⎜ ⎟

⎝ ⎠

= 20 2.5340

SP −⎛ ⎞>⎜ ⎟

⎝ ⎠

≈ 1 − Φ(2.53) = 0.0057 Question 3 was well answered.

4 (i) Let M(r) be the moment generating function of the gamma distribution, so

M(r) = (α/(α − r))2, for r < α. The adjustment coefficient R solves

M(r) = 1 + (1 + 0.5) 2 ,rα

which is

2 3= 1 .r

rα⎛ ⎞ +⎜ ⎟α − α⎝ ⎠

We solve α3 = α(α − r)2 + 3r(α − r)2, which simplifies to r(3r2 − 5αr + α2) = 0. Solving the quadratic, we find roots

r = 2 25 25 12

6α ± α − α

giving r = (5 13) / 6α ± = 1.434α, 0.232α.


Page 5

Since R must be in (0, α), we have R = 0.232α. (ii) (a) By Lundberg’s inequality, the probability of ruin ψ(u) with initial

capital u satisfies ψ(u) ≤ e−0.232αu for all u > 0. (b) This upper bound decreases as α increases. This makes intuitive sense since if α increases, the claim sizes are on

average smaller and we expect a smaller probability of ruin.

5 (i) P(X > M) = 2cMe−

21,000 ce− = 0.01

c = 2log 0.011,000

− = 0.000004605

(ii) P(X > M) = e−λM e−1,000λ = 0.01

λ = log 0.011,000

− = 0.004605

(iii) (a) f1(x):

2cMe− = 0.001

M2 = log 0.001 = 1,500,000c

−

∴ M = 1,225 f2(x): e−λM = 0.001

M = log 0.001 = 1,500−λ

(b) The Weibull distribution has a lower amount for the 1 in 1,000 claim.

It has a lighter tail.


Page 6

6 Column totals: 4,700 4,063 2,101 6,714 5,803 4,941 2,209 Chain ladder development factors: 1.235 1.216 1.051 For 2001, f = 1.57867. Initial estimate of ultimate is 3,599.04. Estimate of outstanding:

13,599.04 1 = 1,319.251.57867

⎛ ⎞× −⎜ ⎟⎝ ⎠

.

Bornhuetter-Ferguson estimate of ultimate is 3,333.25. Paid to date: 561. Estimate of outstanding is 2,772.25.

Question 6 was well answered. 7 (i) For a Beta distribution, µ = α / (α + β) = 0.2 and

σ2 = αβ / [(α + β)2.(α + β + 1)] = 0.252 From µ we can see that β = 5α – α = 4α

From µ and σ2: 0.252/0.202 = β / [α.(α + β + 1)]

hence: 1.5625 = 4α / (5α2 + α) = 4 / (5α + 1) Rearranging gives: α = 0.312 and β = 1.248

(ii) Likelihood is Binomial where n = 50 and x = 12


Page 7

Posterior ∝ prior × likelihood ∝ θα−1.(1 − θ)β−1. θx.(1 − θ)n−x ∝ θ−0.688.(1 − θ)0.248. θ12.(1 − θ)38

∝ θ11.312.(1 − θ)38.248

= Beta (12.312, 39.248) Posterior mean = 12.312/51.560 = 0.239

(iii) Posterior ∝ θα−1.(1 − θ)β−1. θx.(1 − θ)n−x

∝ θα+x−1.(1 − θ)β+n−x−1

= Beta (α + x, β + n − x) Posterior mean = (α + x) / (α + x + β + n − x)

= α / (α + β + n) + x / (α + β + n)

= [(α + β) / (α + β + n)].[α / (α + β)] + [n / (α + β + n)].[x / n]

= Z.(x / n) + (1 − Z).µ

where Z = n / (α + β + n)

(iv) (a) Z = 50/51.56 = 0.970 Z represents the weight we place on the sample data.

(b) (1) If the standard deviation of the prior increases, we place less weight on the collateral data and more on the sample data, therefore Z would increase.

(2) If the sample size increases, we place more weight on the sample data, therefore Z would increase.


Page 8

(c) The limiting value of Z as σ and/or n tend to infinity, is 1. This means that we place all weight on the sample data and ignore the

collateral information

Question 7(iv)(a), many candidates used the posterior values of α and β in the formulae for Z, instead of the prior values found in 7(i). 8 (i) If Y has a Poisson distribution with mean µ, then

f(y, µ) = e−µµy / y! = exp log log ! ,1

y yµ − µ⎛ ⎞−⎜ ⎟⎝ ⎠

which is of exponential family form. The link function is g(µ) = log(µ). The linear predictor is η = αi. So this is a generalised linear model. (ii) The likelihood is

3

1 1,

!

ijij ymij

iji j

ey

−µ

= =

µ∏∏

so the log-likelihood is

3

1 1( log( ) log( !))

m

ij ij ij iji j

y y= =

−µ + µ −∑∑

i.e., in terms of αi’s, writing yi+ for the sum of the observations in the ith

group, the log-likelihood is

l(α1, α2, α3) = 3 3

1 1constant.i

i ii i

me yα+

= =

− + α +∑ ∑

Differentiating,

i

l∂∂α

= ,iime yα+− +


Page 9

so the maximum likelihood estimator of αi is ˆ iα = log(yi+/m). (iii) Comparing models 2 and 3: There are 60 observations altogether. Model 3 has one parameter estimate, and so has degrees of freedom 59. Model 2 has degrees of freedom 58. The drop in deviance in going from model 3 to model 2 is 72.53 − 61.64 =

10.89. The corresponding drop in degrees of freedom is 59 −58 = 1. So to test for a significant improvement, compare 10.89 to a 2

1 .χ The upper 5% point of 2

1χ is 3.841, the upper 1% point is 6.635, this is a significant improvement. We prefer model 2 to model 3.

Comparing models 2 and 1: Model 1 has degrees of freedom 57. The drop in deviance is 61.64 − 60.40 = 1.24, and this should be compared to

21 .χ

It is not significant; do not prefer model 1 to model 2. (iv) Interpretation of models: Model 3 says that there is no difference in the average number of claims for

the three age groups. Model 2 says that there is no difference in the average number of claims

between age groups 1 and 2, but that the third age group may be different. Model 1 gives the possibility of different average number of claims for each

age group. In Question (i), showing the exponential family form was well done, but some candidates did not state the linear predictor correctly. Question 8(ii) was poorly done, with common mistakes being to omit the product over I for the likelihood, and algebraic mistakes in


Page 10

deriving the log likelihood in terms of iα ’s from the likelihood. In Question 8(iii), full marks were also given if candidates used the method given in Core Reading of comparing the drop in deviance with twice the change in the number of parameters. Question (iii) was well done. However, some candidates did not state how they arrived at their conclusions about whether models were significant improvements or not. 9 (i) mean and standard deviation = 500

therefore λ = 0.002

The expected amount per claim ceded to the reinsurer is:

2,500

( 2,500) .xx e dx∞ −λ− λ∫

if z = x – 2500

this gives ( )0

.z Mz e dz∞ −λ +λ∫

= 0

.Me z e dz∞−λ −λλ∫

the expression in the integral is the mean of an exponential distribution, parameter λ, therefore:

= e−λM(1/λ) = 500e−5 i.e. 3.3690 per claim For the whole portfolio: = 3.3690 × 0.30 × 200 = £202.14 (ii) The expected amount ceded implies that the reinsurer has applied a loading for

expenses and profit of: = 300/202.14 – 1 = 48.4%

The insurer would pay a premium that is significantly greater than the

expected value of the corresponding risk ceded. This may be acceptable depending on the insurer’s attitude to risk


Page 11

(iii) exp(µ + ½σ2) = 500 exp(2µ + σ2).[exp(σ2) − 1] = 5002 exp(σ2) − 1 = 5002/5002 σ2 = ln(1 + 1) = 0.69315 µ = ln500 – 0.34657 = 5.86803 The expected amount per claim ceded to the reinsurer is:

2,500

( ). . ( )xf x dx M P X M∞

− >∫

2 2

½2,500

ln 2,500( ). = 1xf x dx e∞ µ+ σ ⎛ ⎞− µ − σ

− Φ⎜ ⎟⎜ ⎟σ⎝ ⎠∫

= 500[1 − Φ(1.517)] = 32.32 and 2,500.P[X > 2,500] = 2,500.[1 − Φ((ln2,500 − µ)/σ)] = 2,500.[1 − Φ(2.349)] = 23.54 therefore expected amount ceded per claim: = 32.32 – 23.54 = 8.78 per claim For the whole portfolio:

= 8.78 × 0.30 × 200

= £527 (iv) The expected amount ceded is greater than the reinsurance premium by 527/300 = 175.6%. The reason for the increase is that the lognormal distribution has a much

heavier tail than the exponential distribution. The insurer’s profit is expected to increase if C agrees to take out the

reinsurance policy.


Page 12

The insurer should take out the policy irrespective of its attitude to risk. 10 (i) λ / (α − 1) = 1,500,000 αλ2 / [(α − 1)2(α − 2)] = (2,500,000)2 (2,500,000)2 / (1,500,000)2 = α / (α − 2) = 2.77778 therefore α = 3.125

so λ = 3,187,500

Expected loss without reinsurance = £1,500,000 Expected amount ceded to reinsurer is:

2,000,000

( 2,000,000) ( ).x f x dx∞

−∫

3.125

3.1254.1252,000,000 2,000,000

3.125. . = .( /( ))( )

x dx x xx

∞∞ λ ⎡ ⎤− λ λ +⎣ ⎦λ +∫

3.125

3.1252,000,000.

( )x dx

x

∞ λ+

λ +∫

= 2,000,000.(3,187,500/5,187,500)3.1253.125

2.1252,000,0002.125.( )x

∞⎡ ⎤λ

+ −⎢ ⎥λ +⎢ ⎥⎣ ⎦

= 2,000,000 × (51 / 83)3.125 + 3,187,5003.125 / (2.125 × 5,187,5002.125) = 436,584 + 532,889 = 969,473

2,000,000

2,000,000. ( ).f x dx∞

∫ = 2,000,000.(3,187,500/5,187,500)3.125

= 436,584

Therefore amount ceded to reinsurer is £532,889. Expected amount payable with reinsurance = 1,500,000 – 532,889 = 967,211


Page 13

(ii) Annual expected profits D0 Reinsurance premium £0

Net total claims: Θ0 = £0

Θ1 = £1,500,000

Θ2 = £1,500,000 × 2 = £3,000,000

Θ3 = £1,500,000 × 3 = £4,500,000 D1 Reinsurance premium £500,000


Θ1 = £967,111

Θ2 = £967,111+ £1,500,000 = 2,467,111

Θ3 = £967,111 + 2 × £1,500,000 = 3,967,111

D2 Reinsurance premium £1,000,000


Θ1 = £967,111

Θ2 = £967,111 × 2 = £1,934,222

Θ3 = £967,111 × 2 + £1,500,000 = £3,434,222

Hence decision matrix

Θ0

‘000 Θ1

‘000 Θ2

‘000 Θ3

‘000 D0 0 1,500 3,000 4,500 D1 500 1,467 2,967 4,467 D2 1,000 1,967 2,934 4,434

(iii) (a) Minimax = minimise maximum loss. D0 4,500

D1 4,467

D2 4,434 ← answer is D2

(b) The insurer would expect to minimise the maximum loss by taking out the policy offering the most reinsurance, i.e. D2.


Page 14

(iv) P(0 claims) = e−0.90 = 0.407 P(1 claim) = 0.90e−0.90 = 0.365 P(2 claims) = 0.902e−0.90/2 = 0.165 P(>2 claims) = 1 – 0.407 – 0.365 – 0.165 = 0.063

Expected loss D0 = (0.407×0) + (0.365×1,500) + (0.165×3,000) + (0.063×4,500) = 1,326

D1 = (0.407×500) + (0.365×1,467) + (0.165×2,967) + (0.063×4,467) = 1,509

D2 = (0.407×1,000) + (0.365×1,967) + (0.165×2,934) + (0.063×4,434) = 1,888

Therefore the answer is D0 (no reinsurance). In Question 10(i), many candidates made errors in evaluating the first integral.


EXAMINATIONS

29 April 2004 (am)

Subject 106 Actuarial Mathematics 2



1. Enter all the candidate and examination details as requested on the front of your answer booklet.

2. You must not start writing your answers in the booklet until instructed to do so by the supervisor.





Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this question paper.

In addition to this paper you should have available Actuarial Tables and your own electronic calculator.

Faculty of Actuaries 106 A2004 Institute of Actuaries

106 A2004 2

1 The loss severity distribution for a portfolio of household insurance policies is assumed to be Pareto with parameters = 3.5, = 1,000.

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]


1 2 3

d1 120 97 131 d2 132 74 89 d3 117 141 37


(ii) Given the probability distribution p( 1) = 0.3, p( 2) = 0.3, p( 3) = 0.4, determine the Bayes criterion solution. [2]

[Total 4]

3 A risk consists of 5 policies. On each policy in one month there is exactly one claim with probability and there is negligible probability of more than one claim in one month. The prior distribution for is uniform on (0, 1). There are a total of 10 claims on this risk over a 12 month period.

(i) Derive the posterior distribution for . [2]

(ii) Determine the Bayesian estimate of under:

(a) quadratic loss (b) all-or-nothing loss

[3] [Total 5]

106 A2004 3 PLEASE TURN OVER

4 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 S be the total amount claimed on the whole portfolio in one year. All policies are assumed to be independent.

(i) Determine the mean and variance of S. [2]

(ii) Derive the moment generating function of S and show that S has a compound Poisson distribution. [4]

[Total 6]

5 Claims arrive in a Poisson process rate and they are exponentially distributed with mean . The premium loading factor is .

(i) Derive the formula for the adjustment coefficient, and state an upper bound on the probability of ruin if the initial capital is u. [4]

(ii) Suppose = 1.

(a) Determine a value of such that the probability of ruin is at most 0.01 when the initial capital is 20.

(b) State how this value of changes if the initial capital is increased. [2] [Total 6]

106 A2004 4

6 The reserving department of a general insurance company has obtained the following incremental claims data (in £ 000s). You may assume that all claims are paid at the end of the year.

Development Year Accident Year 0 1 2 3

2000 210 95 40 10 2001 225 105 45 2002 215 95 2003 220

Underlying claims inflation rates over the twelve months to the middle of each year were as follows:

2001 3.0% 2002 2.5% 2003 2.5%

Claims inflation from the middle of 2003 onwards is assumed to be 3.0% per annum.

(i) Calculate the outstanding claims reserve at 31 December 2003 using the inflation adjusted chain ladder, without adjusting forecast claims for inflation.

[6]

(ii) State the assumptions made. [2] [Total 8]

7 The total amount claimed for a particular risk in a portfolio is observed for each of 5 consecutive years.

(i) From past knowledge of similar portfolios, an insurer believes that the claims are normally distributed with mean and variance 25, and that the prior distribution of is normal with mean 125 and variance 36.

(a) Derive the Bayesian estimate for under quadratic loss, and show that it can be written in the form of a credibility estimate combining the mean observed claim size for this risk with the prior mean for .

(b) State the credibility factor, and calculate the credibility premium if the mean claim size over the 5 years is 122.

(c) Comment on how the credibility factor and the credibility estimate change if the variance of 25 is increased.

[6]


(ii) A second insurer does not believe that this is an appropriate prior distribution for risks in this portfolio, and 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(m( )). Data from 3 risks in this portfolio over 5 years are available. Let Xij be the claim for risk i in year j. The table shows various summary statistics for the observed data.

iX

5 2

1( )ij ijX X

Risk 1 (i = 1) 122 2,848 Risk 2 (i = 2) 164 1,628 Risk 3 (i = 3) 106 1,887

(a) Calculate the estimated credibility factor, and calculate the credibility premium for risk 1.

(b) Compare your answer to that obtained in (i)(b). [6]

[Total 12]

106 A2004 6

8 (i) Show that

0( )

d mx f x dx = 2 2 2

½ logm m d me

where f(x) =

21 1 log

exp ½2

x

x

[4]

(ii) The loss amount, X, from a portfolio of non-life insurance policies is assumed to be independently distributed with mean £800 and standard deviation £1,200.

Calculate the values of the parameters of a lognormal distribution with this mean and standard deviation. [3]

(iii) The company is considering purchasing reinsurance cover, and has to decide whether to purchase excess-of-loss or proportional reinsurance.

The amounts paid by the direct insurer and reinsurer respectively, are given by

(Prop)IX = (1

k) X (Prop)RX = kX

and ( )XLIX = min{X, d} ( )XLRX = max{0, X

d}

where X denotes the loss amount.

Using the loss distribution from (ii), calculate the value of k such that

(Prop)[ ]IE X = 0.7E[X]

and show that if d = 1189.4, ( )[ ]XLIE X = 0.7E[X] [4]

(iv) Using the values of k and d from (iii), calculate the values of (Prop)Var[ ]IX and ( )Var[ ]XLIX . [4]

(v) Comment on the results in (iii) and (iv). [2] [Total 17]


9 Let Yij be the number of accidents on a particular motorway in the jth quarter of year

i, i = 1, 2, 3, j = 1, , 4. Suppose that Yij has a Poisson distribution with mean ij.

(i) (a) Derive the log-likelihood function as a function of ij and determine

the maximum likelihood estimate of ij.

(b) If log( ij) = , determine the maximum likelihood estimate of .

(c) Define the scaled deviance, and derive an expression for the scaled deviance for the model in (i)(b).

[8]

(ii) Three models are shown below.

Deviance Degrees of Freedom

Model 1 log( ij) =

266.35 11

Model 2 log( ij) = i 202.19 9

Model 3 log( ij) = i + j 10.68 6

(a) Interpret each of these models.

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

(iii) It is found that the model log( ij) =

i + j provides a reasonable fit to the

data, with the estimate of given as 0.34.

Interpret this model. [2] [Total 17]

106 A2004 8

10 A marine insurer offers policies to boat owners to protect against collision damage. Cover is provided to coincide with the calendar year. For each policyholder the probability of having an accident during the year is 0.2. The policy meets the cost of repairs which, regardless of the timing of the accident, are always carried out at the end of the year. The insurer operates a no claims discount system. In the event of a claim free year, the policyholder moves to the next higher level of discount. In the event of a claim during the year, the policyholder moves in the next year to the next lower level of discount, unless the claim was as a result of drunken behaviour. In this case, the policyholder moves in the next year to the lowest level of discount.

The discount levels are 0%, 20% and 50%. 25% of claims are due to drunken behaviour.

The insurer charges a premium of £2,500 for policyholders at the 0% discount level.

A policyholder makes a claim following an accident if the cost of repairs is greater than the additional premiums payable at the next two renewals, assuming no further claims are made.

(i) Calculate the cost of a repair below which the policyholder will not claim, for each level of discount. [7]

(ii) Assuming that the cost of each repair follows a lognormal distribution with parameters = 6.5 and 2 = 3.5, calculate the probability that a policyholder makes a claim in the event of an accident, at each level of discount. [6]

(iii) Write down the transition matrix. [5]

(iv) Derive the proportion of policyholders at each level of discount, once a steady state has been reached. [4]

[Total 22]

END OF PAPER


EXAMINATIONS

April 2004


EXAMINERS REPORT


Subject 106 (Actuarial Mathematics 2) April 2004

Examiners Report

Page 2

1 P(1.05X < 100) = 100

1.05P X

= 1 100

1.05

=

3.51000

1 1001000

1.05

= 0.2727

This question was well answered.

2 0.3 0.3 0.4

1 2 3 Max Expected Loss

d1 120 97 131 131 117.5 d2 132 74 89 132 97.4 d3 117 141 37 141 92.2

(i) The minimax solution is d1.

(ii) The Bayes criterion solution is d3.

Again, this question was well answered. However, some candidates did not show much working or did not explicitly identify the solution. Although credit was given where the answers were clearly correct, some easy marks were lost.

3 (i) Let Xi be the number of claims in month i, so that Xi has a binomial

distribution with parameters 5 and , i = 1, , 12.

The posterior density is, for 0 < < 1,

f( x)

f(x1, , x12 ) f( )

12 121 112 5(1 )i i i ix x

= 10(1

)50.


Examiners Report

Page 3

So the posterior distribution is a beta with parameters = 11 and = 51.

(ii) (a) Under quadratic loss, the Bayesian estimate is the posterior mean, i.e. /(

+ ) = 11/62 = 0.1774.

(b) Under all-or-nothing loss, the Bayesian estimate is the posterior mode. Differentiating the posterior density, the mode solves

10 9(1

)50 50 10(1

)49 = 0

10(1 ) = 50 ,

and so the Bayesian estimate is 10/60 = 0.1667.

Most of the candidates who passed the exam did well on this question. Conversely, most of those who did badly failed. Although most candidates understood how to determine the Bayesian estimates in part (ii) many were unable to correctly determine the posterior distribution.

4 (i) Let S1 and S2 be the total amount claimed on type 1 and type 2 policies respectively in one year. Then

E(S) = E(S1) + E(S2) = 1.5 5 + 2 4 = 15.5,

and

V(S) = V(S1) + V(S2)

= 1.5 2 52 + 2 2 42

= 75 + 64

= 139.

(ii) The moment generating function of S is

MS(t) = 1 2( ) ( )S SM t M t

= 1 1

exp 1.5 1 exp 2 11 5 1 4t t

= 1.5 1 2 1

exp 3.5 13.5 1 5 3.5 1 4t t

.


Examiners Report

Page 4

This is of the form exp( (M(t) 1)), i.e. the moment generating function of a compound Poisson distribution, with Poisson parameter 3.5 and claim size density

/ 5 / 43 1 4 1,

7 5 7 4x xe e

which is a probability density function.

Part (i) was well answered. Most candidates were also able to derive the moment generating function in part (ii) although the number of algebraic errors was relatively high. Very few demonstrated correctly that the amount claimed had a compound Poisson distribution.

5 (i) The adjustment coefficient R > 0 satisfies 1

1 r = 1 + (1 + ) r, so that

1 = 1

r + (1 + ) r (1

r)

0 = r (1 + ) 2r2

= r(

(1 + ) r).

Since R > 0, we have R = .(1 )

The Lundberg inequality gives that the

probability of ruin (u) satisfies (u)

e Ru for all u > 0.

(ii) (a) We need exp20

0.011

, so that

20log 100

1 e

log 100,

20 log 100e

e

and

0.299.

(b) This value will decrease if u increases, which makes sense since we expect the probability of ruin to decrease if we increase the initial capital.

Most candidates scored reasonably well on this question. Algebraic manipulation was poor in many cases leading to the wrong answer for (i). The second part was generally well answered although the use of equalities rather than inequalities was careless in most cases.


Examiners Report

Page 5

6 (i) To adjust claims in 2002, we need to inflate from end 2002 to mid 2003 at 2.5% p.a., and from mid 2003 to end 2003 at 3.0% p.a.

i.e. (1.025)½ (1.03)½

Similarly for other years. So inflation adjustment factors are:

1.025½ 1.03½ = 1.0275 1.025½ 1.025½ = 1.025 1.03½ 1.025½ = 1.0275

Inflation adjusted incrementals are:

227.25 100.05 41.1 10 236.97 107.89 45 220.91 95 220

Cumulatives:

227.25 327.30 368.40 378.40 236.97 344.85 389.85 220.91 315.91 220

988.07 758.26 378.40 685.13 672.16 368.40

1.4422 1.1281 1.0271

Forecast cumulatives:

400.44 356.38 366.05

317.28 357.92 367.63

Forecast total: 1512.52 Paid to date (allowing for inflation): 1304.16

Reserve: 208.36

(ii) Development pattern is the same each year.

Inflation has been removed from data, so future claims should be inflated.

None of the candidates identified correctly the twist of applying the mid-year interest adjustments to claims payments. A substantial majority correctly worked through the


Examiners Report

Page 6

remainder of the question and scored most of the marks. The question proved difficult to mark as the inflation adjustment factors were not sufficiently different from period to period.

7 (i) (a) Let x1, , x5 be the observed claims. The Bayesian estimate under quadratic loss is the posterior mean, so first find the posterior distribution. The posterior density for is

f( x)

52 2

1

1 1exp ( ) ( 125)

50 72ii

x

21 5 1 5 125exp 2

2 25 36 25 36

x

so that the posterior distribution is a normal with mean

5 12525 365 125 36

x

=

5 125 36 125.

5 1 5 125 36 25 36

x

This is of the form (1 )125,Zx Z where 125 is the prior mean for .

(b) So it is of the form of a credibility estimate with credibility factor

Z = 525

536

= 0.8780.

If x = 122, then the credibility premium is

122Z + (1

Z)125 = 122.37.

(c) If the variance of 25 is increased, then the value of Z would decrease, and the credibility estimate would move closer to the prior mean. This makes sense, since increasing this variance means that the claim amounts within each risk are more variable, and so we should put relatively less weight on past data.

(ii) (a) The estimate of E(m( )) is x = (122 + 164 + 106)/3 = 130.7, so that the credibility premium for risk i is (1 )130.7.iZx Z

The estimate for E(s2( )) is (2848 + 1628 + 1887)/12 = 6363/12 = 530.25.


Examiners Report

Page 7

The estimate of var(m( )) is

32

1

1 1( )

2 60ii

x x 6363 = 1794.67/2 106.05 = 791.29.

The estimated credibility factor is

5530.25

5791.29

= 0.8818,

so that the credibility premium for risk 1 is

0.8818 122 + (1 0.8818) 130.7 = 123.03.

(b) This is similar to the value obtained in (i), so the assumptions made in the prior appear not to be inappropriate.

Very few candidates derived the Bayesian estimate as required. Nonetheless most used information from tables to complete the remainder of the question. Marks for part (i) were generally good. Although most scored reasonably on part (ii) a disappointing number failed to calculate var(m(theta)) correctly. Most who made it to (ii)(b) made sensible conclusions.

8 (i) 0

( )d mx f x dx

= 2

log 1 1exp

22

d my ye dy

= 2 2 2log

2

1 1 2 2exp

22

d y y m ydy

= 2 2 2 2 4log

2

1 1 ( ) 2exp

22

d y m m mdy

= 2 2

22log½ 1 1 (exp

22

dm m y me dy

= 2 2 2

½ logm m d me

(ii) 2½e = 800


Examiners Report

Page 8

2 2½ 2( ) ( 1)e e = 12002

21e =

2

2

1200

800

2 = 1.178655

= 6.095284

(iii) (Prop)[ ]IE X = (1

k) E[X]

k = 0.3

( )[ ]XLIE X =

0( ) ( )

dxf x dx dP X d

= 2 2

½ log log1

d de d

= 2log log

800 1d d

d

Hence log 7.27394 log 6.095284

800 11.08566 1.08566

d dd

= 0.7 800

= 560

when d = 1189.4, the left hand side is

800 ( 0.1775) + 1189.4(1

(0.9081))

= 800 0.42956 + 1189.4 0.1819 = 560


Examiners Report

Page 9

(iv) (Prop)Var[ ]IX = Var[(1

k)X] = 0.72 Var[X]

= 705600

( )Var[ ]XLIX =

2( ) ( ) 2( ( ))XL XLI IE X E X

2( )XLIE X

= 2 2

0( ) ( )

dx f x dx d P X d

= 2 2

2 2 2log 20.1819

de d

= 2080000 ( 1.2632) + 1189.42 0.1819

= 2080000 0.10326 + 257328.9

= 472130

Hence ( )Var[ ]XLIX = 472130 5602 = 158530

(v) The mean is the same and hence the cost will be the same, if the loadings are the same for proportional and excess-of-loss reinsurance.

The variance is lower for excess-of-loss, which is preferable for the insurer. This is due to the heavy tail of the lognormal, which is not removed by the proportional reinsurance.

Most candidates scored well on part (i) although notation was rather sloppy. Very few identified that the lower bound should be -infinity once the appropriate substitution had been made. Most of the candidates also scored reasonably well on part (ii). However, the later parts were only done well by a small number of candidates with most scoring few marks.

9 (i) (a) The likelihood is

3 4

1 1

,!

ijij yij

iji j

e

y

so that the log-likelihood is

3 4

1 1

( log( ) log( !)).ij ij ij iji j

y y


Examiners Report

Page 10

Differentiating with respect to ij and setting to zero gives ij = yij.

(b) When log( ij) = , this becomes

l( ) = 3 4 3 4

1 1 1 1

12 log( !).ij iji j i j

e y y

Differentiating with respect to , we get

l/ = 12e + 3 4

1 1

,iji j

y

so that the maximum likelihood estimator of is

= 3 4

1 1

log /12iji j

y = log( ).y

(c) The scaled deviance is 2 (maximum loglikelihood for saturated model maximum loglikelihood for the current model). The saturated model has parameters ij with maximum likelihood estimators

ij

= yij. The scaled deviance is

, , ,

2 log( ) 12 log( )ij ij ij iji j i j i j

y y y y y y = ,

2 log( / ).ij iji j

y y y

(ii) (a) Model 1 says that there is no difference in the mean number of accidents over quarters and years.

Model 2 says that there is no seasonal difference in mean number of accidents over quarters, but there may be a difference over years.

Model 3 says that there is a difference over both years and quarters, and that there is no interaction between years and quarters.

The effects on the mean are multiplicative, with e.g. the same factor for the first quarter for all three years.

(b) The models 1, 2, 3 are nested. The drop in deviance from model 1 to model 2 is 266.35 202.19 = 64.16 on 11 9 = 2 degrees of freedom.

Since 64.16 is significant when compared to a 2 distribution on 2


Examiners Report

Page 11

degrees of freedom, model 2 is a significant improvement over model 1.

The drop in deviance from model 2 to model 3 is 191.51 on 3 df.

Again this is significant (compared to a 2 on 3df), so model 3 is a significant improvement over model 2.

Hence model 3, with dependence on both year and quarter, is the preferred model out of these three models.

(iii) This model says that there is dependence on both year and quarter. In terms of the log(mean) the dependence on quarter allows a different value for each quarter, but the dependence on year is given by a linear increase of 0.34 per year. This translates into multiplying the mean by e0.34 per year.

This was very poorly done. Many candidates identified the appropriate log-likelihood functions in (i) but few differentiated correctly to determine the maximum likelihood estimates. Similarly the scaled deviance in (i)(c) caused problems for most.

Part (ii) was answered better but a number of candidates failed to identify sufficiently clearly that they had interpreted the models correctly or how they arrived at the recommended model.

Only the best candidates were able to interpret the model correctly in (iii).

10 (i) Premiums are 2500, 2000, 1250

0% level

Claim: 2500 2000 No claim: 2000 1250

Difference: 1250

Policyholder will not claim if cost is less than £1,250

20% level

Claim: 2500 2000 No claim: 1250 1250

Difference: 2000

Policyholder will not claim if cost is less than £2,000

50% level


Examiners Report

Page 12

Not drunk

Claim: 2000 1250 No claim: 1250 1250

Difference: 750

Drunk

Claim: 2500 2000 No claim: 1250 1250

Difference: 2000

Policyholder will not claim if cost is less than £750 if it was not as a result of drunken behaviour, and £2,000 if it was.

(ii) 0% level

P(X > 1250) = log1250 6.5

13.5

= 1

(0.3372)

= 0.3680

20% level

P(X > 2000) = log 2000 6.5

13.5

= 1

(0.5885)

= 0.2781

50% level

P(X > 750) = log 750 6.5

13.5

= 1

(0.0642)

= 0.4744

P(Claim) = 0.75 0.4744 + 0.25 0.2781

= 0.42525


Examiners Report

Page 13

(iii) P(Accident) = 0.2

Transition matrix is

0.2 0.3680 1 0.2 0.3680 0

0.2 0.2781 0 1 0.2 0.2781

0.2 0.2781 0.25 0.2 0.4744 0.75 1 0.2 0.42525

i.e.

0.0736 0.9264 0

0.05562 0 0.94438

0.013905 0.07116 0.914935

(iv) Solve P =

0.0736 0 + 0.05562 1 + 0.013905 2 = 0

0.9264 0 + 0.07116 2 = 1

0.94438 1 + 0.914935 2 = 2

2 = 0.94438

0.085065 1 = 11.102 1

0 = (1 0.07116 11.102)

0.9264 1

= 0.2267 1

0 + 1 + 2 = 1

0.2267 1 + 1 + 11.102 1 = 1

12.3287 1 = 1

1 = 0.0811

0 = 0.0184

and 2 = 0.9005

(0.0184, 0.0811, 0.9005)

This question tended to split those who passed and those who failed. Stronger candidates identified the information required for (iii) and worked methodically through the whole question relatively easily. Many failed to calculate the probabilities correctly in (ii) or the


Examiners Report

Page 14

transition matrix in (iii). Credit was given to those who followed through sensible answers from earlier parts.

END OF EXAMINERS REPORT


EXAMINATIONS





1. Enter all the candidate and examination details as requested on the front of your answer booklet.

2. You must not start writing your answers in the booklet until instructed to do so by the supervisor.





Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this question paper.

In addition to this paper you should have available Actuarial Tables and your own electronic calculator.

Faculty of Actuaries 106 S2004 Institute of Actuaries

106 S2004 2

1 The preparation times for coffee in a high-street coffee shop have density

f(y) = 2 /2

4.yye

(i) Show that this can be written in exponential family form, and determine the

natural parameter. [2]

(ii) Interpret the two models

Model I: 1

= i, i = 1, 2, 3

Model II: 1

= 1

2,3

i

i

where i = 1, 2, 3 correspond to filter coffee, cappuccino and espresso respectively. [2]

[Total 4]

2 The number of claims arising from a hurricane in a particular region has a Poisson distribution with mean . 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]

106 S2004 3 PLEASE TURN OVER

3 A consumer has to decide whether to take out a travel insurance policy which covers him for all trips over the next year, or to purchase a separate policy each time he makes a trip. He is unsure how many trips he will make.

An annual policy would cost £95.

A separate policy for each trip costs £30.

He estimates that the probability distribution of the number of trips he will take over the next year, X, is as follows:

Number of trips x

P(X = x)

1 0.1 2 0.2 3 0.3 4 0.3 5 0.1

Determine the minimax and Bayes decisions. [5]

4 The number of claims from one group of drivers in a year has a Poisson distribution with mean , and the number of claims from a second group of drivers has a Poisson distribution with mean 2 . In one year, there are n1 claims from group 1 and n2

claims from group 2.

(i) Derive the maximum likelihood estimator, , of . [3]

(ii) Suppose that past experience shows that has an exponential distribution with mean 1/v.

(a) Derive the posterior distribution of .

(b) Show that the Bayesian estimate of under quadratic loss may be written in the form of a credibility estimate combining the prior mean

of

with the maximum likelihood estimate in (i). State the credibility factor.

[4] [Total 7]

106 S2004 4

5 A no claims discount system has two levels of discount, such that the premiums paid at each level are:

Discount Category Premium (£ s) 0 1,000 1 1,000d

where 0

d 1.

If a policyholder does not claim, they are in discount category 1 next year. Otherwise, they are in discount category 0.

The probability that a policyholder does not claim is 1

q (0

q 1).

(i) Write down the transition matrix and derive the steady state distribution, in terms of q. [4]

(ii) Calculate the value of d which should be used, so that policyholders for whom q = 0.2 pay an average premium (in the steady state distribution) which is 50% greater than that paid by policyholders for whom q = 0.1. [3]

(iii) Comment on the effectiveness of the no claims discount system, using your answer to (ii). [2]

[Total 9]

6 An insurance company has to estimate the risk premium for the coming year for a certain risk.

(i) Describe how the credibility approach to calculating the risk premium differs from a conventional approach. [2]

(ii) State the advantages and disadvantages of using Bayesian estimation and empirical Bayes credibility theory estimation. [3]

(iii) State the differences between the assumptions in empirical Bayes credibility theory Model 1 and Model 2, and state why Model 2 is more likely to be useful in practice. [4]

[Total 9]


7 The tables below show the cumulative incurred claims data and number of reported claims, by accident year and development year (in £ 000s).

Cumulative incurred claims

Development Year Accident Year 0 1 2 3 Ult 2000 3,417 4,291 4,581 4,714 4,900 2001 4,814 6,888 7,007 2002 5,844 8,000 2003 6,654

Cumulative number of reported claims

Development Year Accident Year 0 1 2 3 Ult 2000 414 460 482 488 500 2001 453 506 526 2002 496 558 2003 540

(i) Estimate the ultimate number of claims, for each accident year, using the chain-ladder technique. [4]

(ii) Estimate the ultimate average incurred cost per claim, for each accident year, using the grossing-up method. [4]

(iii) Using the results from (i) and (ii), calculate the total reserve required, assuming that claims paid to date are £19,212,000. [3]

[Total 11]

106 S2004 6

8 Claims arrive in a Poisson process rate , and the claim severity distribution has mean and moment generating function M(t). The premium income per unit time is c

where c > .

(i) (a) Write down an equation satisfied by the adjustment coefficient.

(b) Derive the adjustment coefficient in terms of , and c, when the claims are exponentially distributed with mean .

(c) Calculate the adjustment coefficient Rexp when = 100 and the premium loading factor is 25%.

(d) State Lundberg s inequality for the probability of ruin with initial capital u.

(e) Determine and comment on the effect on Rexp if the mean claim size is increased but the premium loading factor remains the same.

(f) Determine and comment on the effect on Rexp if instead the premium loading factor is increased but the mean claim size stays the same.

[7]

(ii) Now suppose that the premium loading factor is 25%.

(a) Determine to one significant figure the adjustment coefficient R* if each claim is exactly £100.

(b) Compare R* with Rexp and comment. [3]

(iii) For the exponential case in (i)(c) with = 100, proportional reinsurance is arranged with retention . The insurer s premium income per unit time before paying the reinsurance premium is as in (i)(c) with premium loading factor 25%, and the reinsurer uses a premium loading factor of 30%.

(a) Determine a condition on

that ensures that the insurer s premium income per unit time, net of reinsurance, exceeds the insurer s expected aggregate claims per unit time, net of reinsurance.

(b) Determine the insurer s new adjustment coefficient, taking reinsurance into account, when = 0.8. Compare this value with Rexp.

[5] [Total 15]


9 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 calculate the coefficient of skewness. [8]

(ii) Using a normal distribution to approximate the distribution of annual aggregate claims, calculate the premium loading factor necessary such that the probability that 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 17]

106 S2004 8

10 (i) The random variable X has a Pareto distribution with parameters and . Show that, for L, d > 0:

( )L d

dxf x dx =

( )

1 ( ) ( )

d L d

d L d

[6]

(ii) Claims on a certain class of insurance policy have a Pareto distribution with mean £3,000 and standard deviation £6,000. The insurance company arranges a layer of excess-of-loss reinsurance with a retention level of £8,000. The maximum amount the reinsurer will pay on any individual claim is £6,000.

(a) Calculate the mean claim amount paid by the reinsurer on claims which involve the reinsurer.

(b) Next year the claim amounts on these policies are expected to increase by 10% but the reinsurance treaty will remain unchanged. Calculate the mean claim amount to be paid next year by the reinsurer on claims which involve the reinsurer. [12]

[Total 18]

END OF PAPER


EXAMINATIONS

September 2004


EXAMINERS REPORT


Subject 106 (Actuarial Mathematics 2) September 2004

Examiners Report

Page 2

In numerical questions, candidates were not unduly penalised for errors in earlier parts of the question which affected their answers to the later parts of the question.

1 (i) Write

f(y) = exp 2 log log log 4 .y

y

This is of exponential family form

f(y) = ( )

exp ( , )( )

y bc y

a

with natural parameter = 1/ (accept 1/ ).

(ii) Model I says that there is a possibly different mean preparation time for each type of coffee.

Model II says that cappuccino and espresso have the same mean preparation time, but that the mean preparation time for filter coffee is possibly different.

In Question 1(ii), a common mistake was to say that Model I implies no difference in the mean preparation times for the three types of coffee.

2 (i) Let S be the total amount of claims arising from a single hurricane, and let X be a typical claim. Then

E(S) = E(X) = /2,

and

V(S) = E(X2) = (1 + 0.52) = 5 /4.

(ii) Let T be the total amount of claims arising from the hurricanes in one year. Then T = S1 + + SN, where S1, S2, are iid with the same distribution as S

in (i), and N has a Poisson distribution with mean . Then

E(T) = E(S) = /2,

and

V(T) = E(S2) = 25

4 4 =

(5 ).

4


Examiners Report

Page 3

3

Per Trip Annual

1 30 95

2 60 95

3 90 95

4 120 95

5 150 95

Max: 150 95

E[Cost]:

93 95

Minimax decision is to buy the annual policy.

Bayes decision is to buy policies per trip

Question 3 was done well.

4 (i) The likelihood is

f(n1, n2 ) = 1 22

1 2

(2 )

! !

n ne e

n n

1 23 ,n ne

and so the loglikelihood is

l( ) = 3 + (n1 + n2) log( ) + constant.

Differentiating with respect to and setting to zero gives

= 1 2 .3

n n

(ii) (a) The prior density is f( ) = ve v , so that the posterior density of given n1, n2 is

f( n1, n2) f(n1, n2 ) f( )

1 23 n n ve e

= 1 2 ( 3)n n ve

The posterior distribution is a gamma with parameters n1 + n2 + 1 and v + 3.


Examiners Report

Page 4

(b) Under quadratic loss, the Bayesian estimate is the posterior mean, i.e. it is

1 2 1

3

n n

v = 1 23 1

,3 3 3

n n v

v v v

which is of the form of a credibility estimate (1 )Z Z prior mean, where the credibility factor is Z = 3/(v + 3) and the prior mean is 1/v.

Many candidates misinterpreted Question 4(i) to mean that there are n1 independent identically distributed random variables having a Poisson distribution with mean , and, in addition, n2 independent identically distributed random variables having a Poisson distribution with mean 2 , and this error led to the wrong likelihood. However, later marks were awarded for correct use of this wrong likelihood.

5 (i) The transition matrix is

1

1

q q

q q

( 0 + 1) q = 0

0 = q since 0 + 1 = 1

and 1 = 1

q

The steady state distribution is (q, 1

q)

(ii) Average premium is 1000q + 1000d (1

q) = 1000 (q + d(1

q))

1000(0.2 + 0.8d) = 1.5 1000 (0.1 + 0.9d)

0.2 + 0.8d = 0.15 + 1.35d

0.05 = 0.55d

d = 0.05

0.55 =

1

11 = 0.091

(iii) The value of d is extremely small, making this unrealistic. This is when the premium is 50% higher but it should be even higher. Thus, the NCD system is not effective.

Question 5 was done well.


Examiners Report

Page 5

6 (i) Conventional approach uses data from risk itself only. Credibility approach combines this with information from other sources using a credibility premium.

(1 )ZX Z

(ii) Bayes

Advantage: Not an approximation.

Disadvantage: Have to assume full distribution is correct.

EBCT

Advantage: Can be used when distribution is not known.

Disadvantage: An approximation. May not take account of tail of distribution (e.g.)

(iii) Model 1:

Xij i iid

( i, Xij), ( k, Xkm) are iid (i

k)

Model 2:

Xij i are independent (not id)

( i, Xij), ( k, Xkm) are independent (i

k)

1, , N iid

Difference is that Xij i are not identically distributed to allow for different volumes of business.

In Question 6(ii), credit was given for other sensible and correct advantages and disadvantages.


Examiners Report

Page 6

7 (i) 414

460

482

488

500

453

506

526

496

558

540

1524

1008

488

500

1363

966

482

Chain-ladder development factors:

1.11812

1.04348

1.01245

1.02459

Forecasts:

2000

500

2001

532.5

545.6

2002

582.3

589.5

604.0

2003

603.8

630.0

637.9

653.6

(ii) Average cost per claim:

2000

8.254

9.328

9.504

9.660

9.800

2001

10.627

13.613

13.321

2002

11.782

14.337

2003

12.322

Grossing-up method:

84.22%

95.19%

96.98%

98.57%

77.37%

99.10%

13.736

79.83%

14.758

15.312

Average: 80.47% Average: 97.14%

(iii)

Number of Claims

Cost per Claim

Projected Loss

2000

500

9.800

4900

2001

545.6

13.736

7494.9

2002

604.0

14.758

8914.1

2003

653.6

15.312

10007.5

31316.6

Claims paid to date: 19212 Reserve = 12104.6

i.e. £12,104,600.

Question 7 was done well, although it is important to read the question carefully to make sure that the required methods are applied to the relevant figures.


Examiners Report

Page 7

8 (i) (a) The adjustment coefficient R satisfies

M(R) = + cR

(or equivalent).

(b) When claims are exponentially distributed with mean , the adjustment coefficient Rexp solves

1

1 r

= + cr

Hence

= (1

r) + cr(1

r)

0 = c r2

cr + r

= c r ,c

rc

and so

R = c

c=

1.

c

(c) When the premium loading factor is = 0.25 and = 100, then c = 1.25 = 125 and so Rexp = 0.002.

(d) The probability of ruin with initial capital u satisfies (u)

e Ru for all u > 0.

(e) If the premium loading factor is , then c = (1 + ) , so that Rexp = /((1 + ) ).

Hence if is increased then the adjustment coefficient is reduced, leading to an increased bound on (u), which makes sense since if claims are on average larger, then we expect a higher risk of ruin.

(f) If the premium loading factor is increased then the adjustment coefficient is increased, leading to a decreased bound on (u), which makes sense since increasing the premium loading factor means reduced risk.


Examiners Report

Page 8

(ii) (a) R* solves

e100r = 1 + 1.25 100r

Search for root of this equation. For example,

R e100r 1 + 125r 0.002 1.22 1.25 0.004 1.49 1.50 0.005 1.65 1.63 0.0045 1.57 1.56

Hence R* is between 0.004 and 0.0045, so that R* = 0.004 to one significant figure.

(b) Then R* > Rexp, which makes sense since we expect the ruin probability to be greater for the exponential distribution as it has heavier tails than the distribution concentrated on one point (or other sensible comment).

(iii) (a) The insurer s new premium income per unit time, net of reinsurance is

c = 100 (1.25 1.30(1

)).

The insurer s expected aggregate claim net of reinsurance per unit time is 100 so that we require

1.25 1.30(1

) > ,

so that > 0.05/0.30 = 0.17.

(b) The insurer pays a proportion of the original claim, so that the new claim size distribution is exponential with mean = 100 .

Hence we can use the result in (i)(b) for exponentially distributed claims to find that the new adjustment coefficient is

R = 1

100 100 (1.30 0.05) =

1 1

80 99 = 0.0024.

which is larger than Rexp resulting in a reduced Lundberg bound.


Examiners Report

Page 9

9 (i) Mean: nqm1

Variance: nqm2

nq2 2

1m

Skewness: nqm3 3nq2m2m1+ 2nq3 3

1m

where mk = E[Xk]

For the exponential m1 = , m2 = 2 2, m3 = 6 3

Variance = 2nq 2

nq2 2 = nq 2(2

q)

Skewness = 6nq 3 6nq2 3 + 2nq3 3

= 2nq 3(3 3q + q2)

Mean Var Skewness Small 56.507 1379.647 50531.554

Medium 169.691 9302.743 765038.299

Large 46.517 12019.316 4658528.856

272.715 22701.707 5474098.710

Standard deviation = 150.671

Coefficient of skewness = 3/ 2

Skewness

(Variance) = 1.6

(ii) P(S > (1 + ) E(S)) = 0.05

( ) ( )

Var( ) Var( )

S E S E SP

S S = 0.05

= Var( )

1.645( )

S

E S

= 1.645

22701.707

272.715

= 0.909


Examiners Report

Page 10

(iii) P(S > 1.25E(S) + Capital) = 0.05

0.25 ( ) Capital

Var( )

E S

S = 1.645

Capital = 1.645 Var( )S 0.25E(S)

= 179.675

i.e. £179,675

(iv) The assumption of independence will depend on the proximity of the risks. For example, if they are close and there is the danger of a catastrophic fire caused, for example, by a natural disaster, the assumption may not be valid.

However, it is probably likely to be reasonable.

The normal approximation is unsuitable when looking at the upper tail, as here, especially when the distribution is positively skewed. The consequence is that the capital required and the premium loading will be underestimated.

In Question 9(i), the first part (finding the mean) was done well. Many candidates used wrong formulae for the variances for the three types of buildings, and many candidates omitted the coefficient of skewness.

10 (i) ( )L d

dxf x dx

= 1( )

L d

dx dx

x

= ( ) ( )

L dL d

dd

x dxx x

= 1

1

1( ) ( )

L d L d

d d

xx x

= 1( ) ( 1)( )

L d

d

xx x

= ( ( 1) )

( 1)( )

L d

d

x x

x


Examiners Report

Page 11

= ( 1) ( )

L d

d

x

x

=

( )

1 ( ) ( )

d L d

d L d

(ii) (a) 1

= 3000 and 2

1 2

= 60002

2 =

2

2

6000

3000 = 4

= 4

8

= 8/3 and = 3000 5/3 = 5000

XR =

0 X d

X d d X d L

L X d L

E[XR X > d] = [ ]

( )RE X

P X d

E[XR] = ( ) ( ) ( )d L

dx d f x dx LP X d L

= ( ) ( ) ( )d L

dxf x dx dP d X d L LP X d L

8000 6000

8000( )xf x dx

= 8/3

8/3 8/ 3

5000 8 / 3 8000 5000 8 / 3 14000 5000

5 / 3 (5000 8000) (5000 14000)

= 513.78

P(8000 < X < 14000) = 8000 14000

= 0.0498


Examiners Report

Page 12

E[XR] = 513.78 8000 0.0498 + 6000

14000

= 513.78 398.37 + 170.63

= 286.04

P(X > 8000) = 0.07824

E[XR X > 8000] = 286.04

0.07824 = £3656.14.

(b) RX

=

0 1.1

1.1 1.1

1.1

X d

X d d X d L

L X d L

=

01.1

1.11.1 1.1

1.1

dX

d d LX d X

d LL X

[ ]RE X = 1.1

1.1

1.1 ( )1.1 1.1 1.1

d L

dd d L d L

xf x dx dP X LP X

140001.1

80001.1

( )xf x dx

= 8/ 3

8/3 8/3

8000 140008 / 3 5000 8 / 3 50005000 1.1 1.1

5 / 3 8000 140005000 5000

1.1 1.1

= 535.72

8000

1.1P X

= 80001.1

= 0.0912173

14000

1.1P X

= 0.0342143


Examiners Report

Page 13

[ ]RE X = 1.1 535.72 8000 0.057003 + 6000 0.0342143

= 338.55

[ 8000]R RE X X =

338.55

0.0912173 = £3,711.51

In Question 10(i), there were many slips in the integration, and not all candidates showed how their expression simplified to the one given in the question.

In Question 10(ii)(a), many candidates had an incorrect expression for the amount paid by the reinsurer. Another common source of error was to forget about the conditioning, or to condition on the wrong event (e.g. conditioning on the claim being between £8000 and £14000, instead of correctly conditioning on the claim being bigger than £8000).

END OF EXAMINERS REPORT

past exams subject 106 2000-2004 (1)

Documents