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CT6: CMP Upgrade 2008/09

The Actuarial Education Company © IFE: 2009 Examinations

Subject CT6CMP Upgrade 2008/09

CMP Upgrade

This CMP Upgrade lists all significant changes to the Core Reading and the ActEdmaterial since last year so that you can manually amend your 2008 study material tomake it suitable for study for the 2009 exams. It includes replacement pages andadditional pages where appropriate. Alternatively, you can buy a full replacement set ofup-to-date Course Notes at a significantly reduced price if you have previously boughtthe full price Course Notes in this subject. Please see our 2009 Student Brochure formore details.

This CMP Upgrade contains:

• All changes to the Syllabus objectives and Core Reading.

• Changes to the ActEd Course Notes, Series X Assignments and Question andAnswer Bank that will make them suitable for study for the 2009 exams.


© IFE: 2009 Examinations The Actuarial Education Company

1 Changes to the Syllabus objectives and Core Reading

1.1 Syllabus objectives

There have been no changes to the Syllabus objectives.

1.2 Core Reading

Chapter 10

Page 10

The Core Reading has been changed to read:

Each entry, Cij , in the run-off triangle represents the incremental claims and canbe expressed in general terms

Replacement pages are attached.

Chapter 13

Page 32

New Core Reading has been added to the end of Example 13.9:

The eigenvalues of matrix A are the values llll such that - =det( ) 0A Illll .

Eg for a 2-dimensional time series this equation reduces to:

- - - =11 22 12 21( )( ) 0a l a l a aa l a l a aa l a l a aa l a l a a

where =[ , ] ijA i j aaaa = =( 1,2, 1,2)i j .

Supporting ActEd text has added around this change. Replacement pages are attached.



2 Changes to the ActEd Course Notes

Chapter 2

Page 24

An table of posterior distributions has been expanded.

Replacement pages are attached.



3 Changes to the Q&A Bank

Part 1

This part has been completely overhauled. Questions 2, 3, 4, 5, 6, 11, 12, 15, 16, 19, 25,28, 30, 33 and 34 have been removed and 15 new questions have been added.

The additional questions and their solutions are attached.

Part 2

This part has been completely overhauled. Questions 8, 16, 17, 18, 19 and 21 have beenremoved and 14 new questions have been added.


Part 3

This part has been completely overhauled. Questions 4, 6, 7, 10, 12, 21 and 25 havebeen removed and 8 new questions have been added.




4 Changes to the X Assignments

Assignment X1 and X4

These assignments have been completely rewritten to reflect the changing diet of theexams and to ensure all areas of the syllabus are adequately covered.

Therefore if you are having the X Assignments marked during the 2009 session, youwill be supplied with the new 2009 versions of the X Assignments, as well as having theassignments marked, for the cost of the marking fee (provided you have purchased theCMP in an earlier session).

Alternatively, if you are not having your assignments marked, but still wish to have the2009 versions then you can purchase them at a substantial discount (provided you havepurchased the CMP in an earlier session) – please email [email protected] for moreinformation.

Assignment X2

Question X2.6 has been replaced by a new question.

Replacement pages are attached for the questions and the solutions.



5 Other tuition services

In addition to this CMP Upgrade you might find the following services helpful withyour study.

5.1 Study Material

We offer the following study material in Subject CT6:

• Series Y Assignments

• Mock Exam 2008 and Mock Exam 2009

• ASET (ActEd Solutions with Exam Technique) and Mini-ASET

• Flashcards.

For further details on ActEd’s study materials, please refer to the 2009 StudentBrochure, which is available from the ActEd website at www.ActEd.co.uk.

5.2 Tutorials

We offer the following tutorials in Subject CT6:

• a set of Regular Tutorials (lasting 4 half days, 2 full days or 3 full days)

• a Block Tutorial (lasting 2 or 3 full days)

• a Revision Day (lasting 1 full day).

For further details on ActEd’s tutorials, please refer to our latest Tuition Bulletin, whichis available from the ActEd website at www.ActEd.co.uk.

5.3 Marking

You can have your attempts at any of our assignments or mock exams marked byActEd. When marking your scripts, we aim to provide specific advice to improve yourchances of success in the exam and to return your scripts as quickly as possible.

For further details on ActEd’s marking services, please refer to the 2009 StudentBrochure, which is available from the ActEd website at www.ActEd.co.uk.



6 Feedback on the study material

ActEd is always pleased to get feedback from students about any aspect of our studyprogrammes. Please let us know if you have any specific comments (eg about certainsections of the notes or particular questions) or general suggestions about how we canimprove the study material. We will incorporate as many of your suggestions as we canwhen we update the course material each year.

If you have any comments on this course please send them by email to [email protected] by fax to 01235 550085.


All study material produced by ActEd is copyright and issold for the exclusive use of the purchaser. The copyright

is owned by Institute and Faculty Education Limited, asubsidiary of the Faculty and Institute of Actuaries.

You may not hire out, lend, give out, sell, store or transmitelectronically or photocopy any part of the study material.

You must take care of your study material to ensure that itis not used or copied by anybody else.

Legal action will be taken if these terms are infringed. Inaddition, we may seek to take disciplinary action through

the profession or through your employer.

These conditions remain in force after you have finishedusing the course.

CT6-02: Bayesian statistics Page 23

The Actuarial Education Company IFE: 2009 Examinations

3.5 Some Bayesian posterior distributions

In this section we give a table of situations in which the Bayesian approach may work well. The likelihood function is given, together with the distributions of the prior and the posterior. Do not attempt to learn all the results given in this table. The results are here for reference purposes only, and you will not be expected to be able to quote all these results in the examination. However, you may like to select one or two of the results given here and check that you can prove that the distribution of the posterior is as given in the table. Use the table as a way of generating extra questions on particular Bayesian results. The negative binomial distribution here is that described in the Tables as the Type 2 negative binomial distribution. The results for the geometric distribution are also based on the Type 2 geometric (ie the Type 2 negative binomial with 1k = ). You may like to work out the corresponding results for the Type 1 negative binomial and geometric distributions. Notice that despite the large number of examples given, the posterior distribution in all these cases turns out to be gamma, beta or normal. So in most Bayesian questions it is worth checking whether your posterior density takes the form of one of these three distributions before you start thinking about other possibilities.

CT6-02: Bayesian statistics

IFE: 2009 Examinations The Actuarial Education Company

Distribution of parameter Likelihood of IID Observations

X Xn1, ,…

Unknown Parameter Prior Posterior

(0, )U • ( 1, )Gamma x n+Â

Exp( )′λ ( 1, )Gamma x n l+ + ¢Â Poisson( )λ λ > 0 ( , )Gamma a l¢ ¢ ( , )Gamma x na l+ +¢ ¢Â

(0, )U • ( 1, )Gamma n x+ Â

( )Exp l¢ ( 1, )Gamma n x l+ + ¢Â Exp( )λ 0l > ( , )Gamma a l¢ ¢ ( , )Gamma n xa l+ +¢ ¢Â

(0, )U • ( 1, )Gamma n xa + Â

( )Exp l¢ ( 1, )Gamma n xa l+ + ¢Â Gamma( , )α λ 0l > ( , )Gamma a l¢ ¢ ( , )Gamma n xa a l+ +¢ ¢Â

(0, )U • ( 1, )Gamma n xg+ Â

( )Exp l¢ ( 1, )Gamma n xg l+ + ¢Â ( , )W c g 0c >

( , )Gamma a l¢ ¢ ( , )Gamma n xga l+ +¢ ¢Â

( , )U -• • 21 ,N x

n nsÊ ˆ

Á ˜Ë ¯Â

N ( , )µ σ 2 −∞ < < ∞µ

N ( , )′ ′µ σ 2 2 2

2 2 2 2

1,1 1

Ê ˆ¢+Á ˜¢Á ˜Á ˜+ +Á ˜¢ ¢Ë ¯

Â x

N n n

ms s

s s s s

LogN ( , )µ σ 2 −∞ < < ∞µ U ( , )−∞ ∞ 21 log ,

Ê ˆÁ ˜Ë ¯ÂN x

n ns

(0,1)U ( 1, 1)Beta x nm x+ - +Â Â ( , )Bin m p 0 1p< <

( , )Beta a b¢ ¢ ( , )Beta x nm xa b+ - +¢ ¢Â Â

U ( , )0 1 ( 1, 1)Beta n x+ +Â Geo p( ) 0 1< <p

( , )Beta a b¢ ¢ ( , )Beta n xa b+ +¢ ¢Â

U ( , )0 1 ( 1, 1)Beta nk x+ +Â NB k p( , ) 0 1p< <

( , )Beta a b¢ ¢ ( , )Beta nk xa b+ +¢ ¢Â

CT6-10: Run-off triangles Page 9


Question 10.5

Project the known figure for Accident Year 2005 across to Development Year 4 using: (i) the largest ratio for each development year (ii) the smallest ratio for each year. Comment on the difference between your two results.

However, some sort of average of the ratios would seem more appropriate. It is possible to use a simple arithmetic average:

1.794 1.742 1.823 1.756 1.7794

+ + + =

The disadvantage of this is that it does not take into account that the years in which more claims occur provide more information. Thus, the greater the amount of claims, the more confidence you can have in the ratio. Note that we’re assuming a large number of claims here, which would lead to a more predictable average, not a small number of very large claims, which would probably have the opposite effect. This suggests using a weighted average and the usual choice of weights are the cumulative claims values.

Accident Year Ratio Weight

2001 2002 2003 2004

1.794 1.742 1.823 1.756

786 904 995

1,220

1.794 786 1.742 904 1.823 995 1.756 1,220 1.777786 904 995 1,220

¥ + ¥ + ¥ + ¥=

+ + +

This method of estimating the ratios which describe the run-off pattern is called the chain-ladder method. The most efficient mode of calculating the ratios is given in section 1.3.

CT6-10: Run-off triangles


Question 10.6

Using this method, what estimate would you give for the ratio to be used for calculating figures for Development Year 2 from Development Year 1?

1.2 A statistical model for run-off triangles

It is helpful to have the following statistical model in the back of your mind when you are considering the various methods that may be used to complete the run-off triangle. The general form of a run-off triangle can be expressed as follows:

Each entry, Cij , in the run-off triangle represents the incremental claims and can be expressed in general terms Cij = rj . si . xi+j + eij where: rj is the development factor for year j, representing the

proportion of claim payments in year j. Each rj is independent of the origin year i.

si is a parameter varying by origin year, i, representing the

exposure, for example the number of claims incurred in the origin year i.

xi+j is a parameter varying by calendar year, for example

representing inflation. eij is an error term.

Development year Accident Year 0

1 ... j ..............................

... n

0

C0,0 C0,1 ... C0, j ................................

C0,n

1 C1,0 C1,1 ... C1, j ............ C1,n−1 ! ! ! i Ci,0 Ci,1 Ci,n−i

! ! !

! ! !

Cn−1 1

CT6-13: Time series 2 Page 31


Example 13.8 A vector autoregressive process of order p, denoted VAR(p), is a sequence of m-component random vectors {{{{ }}}}1 2, ,X X ………… satisfying:

(((( ))))1

p

jt t j tj

X A X em mm mm mm m----====

= + - += + - += + - += + - +ÂÂÂÂ (13.2)

where e is an m-dimensional white noise process and the jA are m m¥ matrices. Example 13.9 We might believe that interest rates, ti , and tendency to invest, tI , are related to one another by the equations:

( )

11 1( )

21 1 22 1

( )

( ) ( )

it i t i t

It I t i t I t

i i e

I i I e

m a mm a mm a mm a m

m a m a mm a m a mm a m a mm a m a m----

- -- -- -- -

ÏÏÏÏ - = - +- = - +- = - +- = - +ÔÔÔÔÌÌÌÌ

- = - + - +- = - + - +- = - + - +- = - + - +ÔÔÔÔÓÓÓÓ (13.3)

where (((( ))))ie and (((( ))))Ie are zero-mean (univariate) white noises. They may have different variances and are not necessarily uncorrelated; that is, we do not

require (((( )))) (((( ))))(((( ))))cov , 0i Ie e ==== , although we do require (((( ))))( ) ( )cov , 0i Iste e ==== for s tππππ .

This model can be expressed as a 2-dimensional VAR(1):

( )

111( )121 22

0 it i t i t

It I t I t

ei iI I e

m mm mm mm maaaam mm mm mm ma aa aa aa a

----

----

Ê ˆÊ ˆÊ ˆÊ ˆ- -- -- -- -Ê ˆ Ê ˆÊ ˆ Ê ˆÊ ˆ Ê ˆÊ ˆ Ê ˆÊ ˆÊ ˆÊ ˆÊ ˆ= += += += + Á ˜Á ˜Á ˜Á ˜Á ˜Á ˜Á ˜Á ˜Á ˜ Á ˜Á ˜ Á ˜Á ˜ Á ˜Á ˜ Á ˜- -- -- -- - Á ˜Á ˜Á ˜Á ˜Ë ¯Ë ¯Ë ¯Ë ¯Ë ¯ Ë ¯Ë ¯ Ë ¯Ë ¯ Ë ¯Ë ¯ Ë ¯ Ë ¯Ë ¯Ë ¯Ë ¯

The theory and analysis of a VAR(1) closely parallels that of a univariate AR(1). Iterating from Equation (13.2) in the case 1p ==== , it is clear that:

(((( ))))1

00

tj t

t t jj

X A e A Xm mm mm mm m----

----====

= + + -= + + -= + + -= + + -ÂÂÂÂ

In order that X should represent a stationary time series, the powers of A should converge to zero in some sense. The appropriate requirement is that all eigenvalues of the matrix A should be less than 1 in absolute value.

CT6-13: Time series 2


The eigenvalues of matrix A are the values llll such that - =det( ) 0A Illll where I is the identity matrix. Eg for a 2-dimensional time series this equation reduces to: - - - =11 22 12 21( )( ) 0a l a l a aa l a l a aa l a l a aa l a l a a where =[ , ] ijA i j aaaa = =( 1,2, 1,2)i j .

This is because we require:

11 12 11 12

21 22 21 22

1 0det( ) det det 0

0 1a a a l a

l la a a a l

Ê ˆ -Ê ˆ Ê ˆÊ ˆ- = - = =Á Á ˜˜Á ˜ Á ˜-Ë ¯Ë ¯ Ë ¯Ë ¯

A I

The appendix to this chapter briefly revises the definition of an eigenvalue.

Question 13.17

Is the following multivariate time series stationary?

1

1

0.3 0.50.2 0.2

Xt t t

Yt t t

X X eY Y e

-

-

Ê ˆÊ ˆ Ê ˆÊ ˆ= + Á ˜Á ˜Á ˜ Á ˜Ë ¯Ë ¯ Ë ¯ Ë ¯

Question 13.18

Derive a definition of the process tX in Question 13.17 that doesn’t involve tY . Show that tX is stationary if considered as a process in its own right.

Similar, though more complicated, requirements can be set out under which a more general VAR(p) process is stationary. Fitting a vector autoregression is very similar to the process of fitting a univariate autoregression. Parameter estimation can be carried out by least squares or by method of moments. Some elements of the univariate theory, such as the use of Akaike’s Information Criterion, do not translate unchanged into a multivariate setting, but other topics carry across relatively easily.



Example 13.10 The following simple dynamic Keynesian model provides an example of a multivariate autoregressive process. You may have already met a Keynesian model of the economy in Subject CT7. We are not so interested here in the economic theory that leads to this model. If you haven’t come across these ideas before, don’t worry. Just accept the given equations. The important point is to understand the vector time series equations. Denote by tY the national income over a certain period of time, and denote by tC and tI the total consumption and investment over the same period. It is assumed that the consumption, tC , depends on the income over the previous period:

(((( ))))11t t tC Y eaaaa ----= += += += +

where (((( ))))1e is a zero-mean white noise. The investment, tI , is determined by the “accelerator” mechanism:

(((( )))) (((( ))))21 2t t t tI C C ebbbb - -- -- -- -= - += - += - += - +

where (((( ))))2e is another zero-mean white noise. Finally, any part of the national income is either consumed or invested; therefore:

t t tY C I= += += += + Eliminating the national income we arrive at the following two-dimensional VAR(2):

(((( ))))11 1t t t tC C I ea aa aa aa a- -- -- -- -= + += + += + += + +

(((( )))) (((( ))))21 2t t t tI C C ebbbb - -- -- -- -= - += - += - += - +

Using matrix notation we can rewrite the above equation as:

(1)1 2

(2)1 2

0 00 0

t t t t

t t t t

eC C CI I I e

a aa aa aa ab bb bb bb b

- -- -- -- -

- -- -- -- -

Ê ˆÊ ˆÊ ˆÊ ˆÊ ˆ Ê ˆ Ê ˆÊ ˆ Ê ˆ Ê ˆÊ ˆ Ê ˆ Ê ˆÊ ˆ Ê ˆ Ê ˆÊ ˆ Ê ˆÊ ˆ Ê ˆÊ ˆ Ê ˆÊ ˆ Ê ˆ= + += + += + += + + Á ˜Á ˜Á ˜Á ˜Á ˜ Á ˜Á ˜ Á ˜Á ˜ Á ˜Á ˜ Á ˜Á ˜ Á ˜ Á ˜Á ˜ Á ˜ Á ˜Á ˜ Á ˜ Á ˜Á ˜ Á ˜ Á ˜----Ë ¯ Ë ¯Ë ¯ Ë ¯Ë ¯ Ë ¯Ë ¯ Ë ¯ Á ˜Á ˜Á ˜Á ˜Ë ¯ Ë ¯ Ë ¯Ë ¯ Ë ¯ Ë ¯Ë ¯ Ë ¯ Ë ¯Ë ¯ Ë ¯ Ë ¯ Ë ¯Ë ¯Ë ¯Ë ¯



5.2 Cointegrated time series

Recall that a time series process X is called integrated of order d, abbreviated as I(d), if the process dY X= —= —= —= — is stationary. Two time series processes X and Y are called cointegrated if: (i) X and Y are I(1) random processes, (ii) there exists a non-zero vector (((( )))),a ba ba ba b such that X Ya ba ba ba b++++ is stationary. In other words, two processes are themselves non-stationary (technically I(1)), but their movements are correlated in such a way that a certain weighted average of the two processes is stationary. The vector (((( )))),a ba ba ba b is called a cointegrating vector. There are a number of circumstances when it is reasonable to expect that two processes may be cointegrated: • if one of the processes is driving the other

• if both are being driven by the same underlying process.

Example 13.11 As above, don’t worry too much if you haven’t come across the economic theory referred to in this example. You will meet it in Subject CT7. However, make sure that you understand why the processes are cointegrated. The following simple model of evolution of the USDollar/GBPound exchange rate

tX provides an example of a cointegrated model. It is assumed that the exchange rate fluctuates around the purchasing power t tP Q , where tP and tQ are the consumer price indices for US and UK, respectively. This is described by the following model:

ln ln tt t

t

PX YQ

= += += += +

1 1( )t t t tY Y e em a m bm a m bm a m bm a m b- -- -- -- -= + - + += + - + += + - + += + - + +

where e is a zero-mean white noise.



The evolution of lnP and lnQ is described by ARIMA(1,1,0) models:

(((( )))) (((( )))) (1)1 1 1 11 ln 1 lnt t tB P B P em a mm a mm a mm a m----È ˘È ˘È ˘È ˘- = + - - +- = + - - +- = + - - +- = + - - +Î ˚Î ˚Î ˚Î ˚

(((( )))) (((( )))) (2)2 2 1 21 ln 1 lnt t tB Q B Q em a mm a mm a mm a m----È ˘È ˘È ˘È ˘- = + - - +- = + - - +- = + - - +- = + - - +Î ˚Î ˚Î ˚Î ˚

where (((( ))))1e and (((( ))))2e are zero-mean white noises, possibly correlated. The logarithm of the exchange rate is also non-stationary. However: ln ln lnX P Q- +- +- +- + is the ARMA(1,1) random process Y and, therefore, is a stationary random process. It follows that the sequence of random vectors: (((( )))){{{{ }}}}ln ,ln ,ln : 1,2,t t tX P Q t ==== ………… is described by a cointegrated model with the cointegrating vector (1, 1,1)- .

Question 13.19

Show that the two processes tX and tY defined by:

1 1

1 1

0.65 0.35

0.35 0.65

- -

- -

= + +

= + +

Xt t t t

Yt t t t

X X Y e

Y X Y e

are cointegrated, with cointegrating vector ( )1, 1- .



6 Some special non-stationary and non-linear time series models

Although the ARIMA class of processes is the most important for us, there are many other types of model. This section briefly discusses a few of them.

6.1 Bilinear models

The general class of bilinear models can be exemplified by its simplest representative, the random process X defined by the relation:

(((( )))) (((( ))))1 1 1 1n n n n n nX X e e b X ea m m b ma m m b ma m m b ma m m b m- - - -- - - -- - - -- - - -+ - = + + + -+ - = + + + -+ - = + + + -+ - = + + + - Considered only as a function of X, this relation is linear; it is also linear when considered as a function of e only. This is why it is called “bilinear”. The main qualitative difference between the bilinear model and models from the ARMA class is that many bilinear models exhibit “bursty” behaviour: when the process is far from its mean it tends to exhibit larger fluctuations. The difference between this model and an ARMA(1, 1) process may be seen to lie in the last term on the right-hand side: when 1nX ---- is far from µµµµ and 1ne ---- is far from 0 – events which are far from being independent – the final term assumes a much greater significance.

6.2 Threshold autoregressive models

A simple representative of the class of threshold autoregressive models is the random process X defined by the relation:

1 1 , 1

2 1 1

( ) if , =

( ) , if .n n n

nn n n

X e X dX

X e X da ma ma ma m

mmmma ma ma ma m

- -- -- -- -

- -- -- -- -

- + £- + £- + £- + £ÏÏÏÏ++++ ÌÌÌÌ - + >- + >- + >- + >ÓÓÓÓ

The distinctive feature of some models from the threshold autoregressive class is the limit cycle behaviour. This makes the threshold autoregressive models suitable for the description of “cyclic” phenomena. In an extreme case we might set 2 0a = for example. Then Xn follows an autoregressive process until it passes the threshold value d. At this point Xn returns to µ and the process effectively starts again. Thus we get cyclic behaviour as the process keeps resetting.

CT6: Q&A Bank Part 1 – Questions Page 17


Question 1.35

A market trader has the option for one day of selling either ice-cream ( 1d ), hot food ( 2d ) or umbrellas ( 3d ) at an outdoor festival. He believes that the weather is equally likely to be fine ( 1q ), overcast ( 2q ) or wet ( 3q ) and estimates his profits under each possible scenario to be:

1q 2q 3q

1d 25 19 7

2d 10 30 8

3d 0 2 34 (i) Determine the minimax solution to this problem. [2] (ii) The trader’s partner is very optimistic and believes that the criterion to adopt in

deciding which product to sell should be to maximise the maximum profit. What decision would the trader’s partner make based on these predicted profits?

[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 but

believes that there is more than an evens chance of it being overcast. By sketching a graph of the Bayes risk for each of the three possible decisions against the probability of it being overcast ( p ), or otherwise, determine the revised Bayes criterion solution. [4]

[Total 9]

CT6: Q&A Bank Part 1 – Questions


Question 1.36

For each of the following derive the posterior distribution:

Sample distribution 1( , , )nx x… Unknown parameter Prior distribution of

unknown parameter ( , )Gamma a l 0l > ~ ( , )Gammal g d

Type 2 ( , )NBin k p 0 1p< < ~ (0,1)p U ( )Poi m 0>m conjugate

2( , )N m s -• < < •m not known [13] Question 1.37

For the estimation of a population proportion p a sample of n is taken and yields x successes. A suitable prior distribution for p is beta with parameters 4 and 4. (i) Show that the posterior distribution of p given x is beta and specify its

parameters. [2] (ii) Given that 14 successes are observed in a sample of size 20, calculate the

Bayesian estimate under all-or-nothing (0/1) loss. [4] [Total 6] Question 1.38

A single observation, x , is drawn from a distribution with the probability density function:

1 0

( | )0 otherwise

xf x

q qq

-Ï < <Ô= ÌÔÓ

The prior distribution of q is given by: ( ) exp( ), 0g q q q q= - > Derive an expression in terms of x for the Bayes estimator of q with respect to the absolute error loss function. [4]



Question 1.39

The time, in hours, between claims is assumed to follow an exponential distribution with parameter l .

The average time between the last 20 claims was 1.89 hour. The prior distribution for l has mean 0.6 and standard deviation 0.2. (i) Show that the gamma distribution is the conjugate prior distribution forl and

determine its parameters. [5] (ii) Calculate the Bayesian estimator of l under quadratic loss. [2] [Total 7] Question 1.40

The number of claims in a week that arises from a certain group of insurance policies has a ( )Poi m distribution. In the last 2 weeks, the numbers of claims incurred were 7 and 11, respectively.

(i) Derive the posterior distribution given that the prior distribution for m is: (a) gamma with parameters 7a = and 0.5l = (b) uniform on the integers 8, 10 and 12. [5] (ii) Hence for each case in part (i) obtain the Bayesian estimate of m using a: (a) quadratic loss function (b) absolute loss function (c) zero-one loss function. [8] [Total 13]



Question 1.41

An insurer insures a risk for which individual claim sizes (in £000s) have a distribution, X , with mean 500 and standard deviation 250. The insurer arranges excess of loss reinsurance for this risk with a retention limit of £1,000,000. What is the proportion of claims from this risk for which the insurer expects to receive a payment from the reinsurer if X is: (a) gamma (b) lognormal [5] Question 1.42

The number of claims from a certain portfolio of motor policies has a Poisson distribution with parameter l . The parameter l varies over the portfolio in such a way that l has an exponential distribution with parameter q . Show that the unconditional claim number distribution is geometric, and write down its parameters. [4] Question 1.43

A student actuary wishes to fit a Burr distribution with parameters 200l = and 1.2g = to a sample of claims data. The value of the parameter a is to be determined using the method of percentiles. Given that the sample data has a median of 420, determine the value of a . [3]



Question 1.44

(i) The distribution of claims on a portfolio of general insurance policies is a Weibull distribution, with density function 1( )f x where:

2

1( ) 2 ( 0)cxf x cxe x-= > It is expected that one claim out of every 100 will exceed £1,000. Use this

information to estimate c. [2] (ii) An alternative suggestion is that the density function is 2( )f x , where: 2( ) ( 0)xf x e xll -= > Use the same information as in part (i) to estimate l . [2] (iii) (a) For each of 1( )f x and 2( )f x calculate the value of M such that: ( ) 0.001P X M> = (b) Comment on these results. [3] [Total 7]



Question 1.45

The random variable Y which has range (1, )• , has a loggamma distribution, so that logY has a gamma distribution with parameters a and l . (i) Show that the probability density function of Y is:

1 1(log )( )

y ya

a lla

- - -G

[3]

(ii) Let 1 2, ,..., ny y y be a sample of n values of Y and let a and l be the maximum

likelihood estimates of a and l , respectively, based on these data. Show that a satisfies the equation:

( )( ) { }

1

1

ˆˆ.log log log 0

ˆlog

n

ini

ii

nnn yy

aaa =

=

Ï ¸Ô Ô G¢Ô Ô - + =Ì ˝ GÔ ÔÔ ÔÓ ˛

ÂÂ

where ( ) ( )ˆ

ˆ dd a a

aa

a =

GG =¢ [5]

[Total 8]



Question 1.46

An insurer has an excess of loss reinsurance arrangement in place with a retention limit of 50. Claim amounts (in £000s) have a Pareto distribution with parameters 2.5a = and 350l = . (i) Find the mean claim amount paid by the insurer. [4] (ii) Find the mean claim amount paid by the reinsurer on claims in which the

reinsurer is involved. [2] (iii) The insurer decides to investigate a proportional reinsurance arrangement with

the reinsurer taking a proportion 0.3 of each claim. Find the percentage difference in mean claim amounts that the insurer will have to pay compared to using the excess of loss arrangement. [2]

[Total 8] Question 1.47

The loss severity distribution for a portfolio of household insurance policies is assumed to be Pareto with parameters 3.5a = , 1,000l = . Next year, losses are expected to increase by 5%, and the insurer has decided to introduce a policyholder excess of £100. Calculate the probability that a loss next year is borne entirely by the policyholder. [3]



Question 1.48

(i) Show that:

2

2 2221 (ln ) ln ln½2

2

1

2

b x b a

a

e dx em m s m sm ss

s sps

- - - - - -+ Ê ˆÊ ˆ Ê ˆ= F -FÁ ˜Ë ¯ Ë ¯Ë ¯Ú [4]

(ii) Individual claim amounts on a certain type of general insurance policy have a

lognormal distribution, with mean 264 and standard deviation 346. A policyholder excess of 100 is a standard condition on each policy, so that the insurance 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 new

condition will be introduced on all policies so that the maximum amount that the insurance company will pay on any claim will be 1,000. The policyholder excess will remain unchanged at 100.

Calculate the expected claim size payable by the insurance company. [14] [Total 18]



Question 1.49

(i) Show that:

2 2 2

½

0

log( )d

m m m d mx f x dx e m s m ss

+ Ê ˆ- -= FÁ ˜Ë ¯Ú

where 21 1 log( ) exp ½

2xf x

xm

ss p

Ï ¸-Ô ÔÊ ˆ= -Ì ˝Á ˜Ë ¯Ô ÔÓ ˛. [4]

(ii) The loss amounts, X, from a portfolio of non-life insurance policies are 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)

(Prop)

(1 )I

R

X k X

X kX

= -

= and:

( )

( )

min{ , }

max{0, }

XLI

XLR

X X d

X X d

=

= -

where X denotes the loss amount. Using the loss distribution from (ii), calculate the value of k such that: (Prop)[ ] 0.7 [ ]IE X E X= and show that if 1,189.4d = , ( )[ ] 0.7 [ ]XL

IE X E 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]



Question 1.50

(i) On one class of risks an insurance company has an excess of loss reinsurance contract with retention £3,000. Over the last year, the insurer paid the following claims:

£1,641 £462 £2,690 £1,160 £1,478 £828 £2,028 In addition, the insurer also paid the retention limit of £3,000 on a further three

claims, with the excess being paid by the reinsurer. The insurer believes that the distribution of gross claim amounts is exponential,

with parameter l . Calculate the maximum likelihood estimate of l based on these data. [3] (ii) Another class of risks is thought to have gross claim amounts that have a

Pareto ( , )a l distribution. The insurance company has an excess of loss reinsurance contract on these with retention £2,500.

(a) Show that the distribution of the claim amounts paid by the reinsurer

given that the reinsurer is involved in the claim is another Pareto. (b) Over the last year the reinsurer paid the following amounts: £3,243 £603 £571 £466 £395 Use your distribution from part (ii)(a) to find the method of moments

estimates of a and l . [7] [Total 10]

CT6: Q&A Bank Part 1 – Solutions Page 37


Part 1 – Solutions Solution 1.35

This is from Subject 106, September 2002, Question 4. (i) Minimax solution If he chooses 1d , the worst possible result is a profit of 7. For 2d the corresponding figure is 8, and for 3d it is 0. So the minimax solution is strategy 2d . [2] (ii) Maximax solution The maximum profit for each strategy is 25 for 1d , 30 for 2d and 34 for 3d . So the maximum possible profit is 34, and this may be achieved by choosing strategy 3d , which is the maximax strategy. [1] (iii) Bayes solution Since each type of weather is equally likely we have: 1d 1 1 1

3 3 3(profit) 25 19 7 17E = ¥ + ¥ + ¥ =

2d 1 1 13 3 3(profit) 10 30 8 16E = ¥ + ¥ + ¥ =

3d 1 1 13 3 3(profit) 0 2 34 12E = ¥ + ¥ + ¥ =

So the strategy selected by the Bayes criterion is Strategy 1d . [2] (iv) Bayes solution for differing probabilities The probabilities are now p for overcast weather, and ½ ½ p- for each of the other two weather types. So we have: 1d (profit) 25(½ ½ ) 19 7(½ ½ ) 16 3E p p p p= - + + - = + 2d (profit) 10(½ ½ ) 30 8(½ ½ ) 9 21E p p p p= - + + - = + 3d (profit) 0(½ ½ ) 2 34(½ ½ ) 17 15E p p p p= - + + - = -

CT6: Q&A Bank Part 1 – Solutions


If we draw a graph of these three straight lines for 0 1p£ £ , we find that initially the line for 3d (umbrellas) lies above the other two, then 1d (ice cream) and then 2d (hot food).

0

5

10

15

20

25

30

35

0 0.2 0.4 0.6 0.8 1

p

expe

cted

pro

fit

umbrellashot foodice cream

Finding first where the line for 3d cuts 1d : 1

1816 3 17 15p p p+ = - fi = Now we find where 1d cuts 2d : 7

1816 3 9 21p p p+ = + fi = So the revised Bayesian criterion solution is as follows: If 1

180 p£ £ , 3d is optimal.

If 7118 18p£ £ , 1d is optimal.

If 718p ≥ , 2d is optimal. [4]



Solution 1.36

(i) Gamma/gamma posterior

Likelihood 1( | ) in xxx nf x e e e llla a al l l l --- Âµ ¥ ¥ =! [1]

Prior 1( )f eg dll l - -µ [½]

Posterior ( )1 1( | ) i ix xn nf x e e el d la g dl a gl l l l- - +- - + -Â Âµ ¥ = [½] ( , )+ +Â igamma n xa g d [1] (ii) Negative binomial/uniform posterior

Likelihood 1( | ) (1 ) (1 ) (1 ) in xxxk k nkf x p p p p p p p Âµ - ¥ ¥ - = -! [1] Prior ( ) 1f p = [½]

Posterior ( | ) (1 ) 1 (1 )i ix xnk nkf p x p p p pÂ Âµ - ¥ = - [½] ( 1, 1)ibeta nk x+ +Â [1] (iii) Poisson/conjugate posterior

Likelihood 1( | ) in xxx nf x e e em m mm m m m- - -Âµ ¥ ¥ =! [1] Since this is of the form of a gamma distribution, the posterior will have a gamma distribution, Therefore , the conjugate prior is a gamma distribution. Prior ~ ( , )gammam a l [½]

1( ) - -µf ea lmm m [½]

Posterior 11 ( )( | ) - +- - - - +Â Âµ ¥ =i ix xn nf x e e eam a lm l mm m m m [½] ( , )+ +Â igamma x na l [½]



(iv) Normal/uninformative posterior

Likelihood 2 2 2

12 2 21 1 1( ) ( ) ( )

2 2 2( | )- - - - - -Â

µ ¥ ¥ =n ix x x

f x e e e…m m m

s s sm [1] We have no information about the prior – so we use a uniform distribution over the range of possible values ( , )U k k- where k Æ• . Essentially this is just a constant and so is absorbed into the posterior constant. [½]

Posterior ( )

( )

2 2 2 2 22 2 2

22

1 1 1( ) ( 2 ) 22 2 2

1 22

( | )- - - - + - - +

- - +

Â Â Â Âµ = =

Âµ

i i i i i

i

x x x x x n

x n

f x e e e

e

m m m m ms s s

m ms

m

[1] This is in the form of a normal distribution – but remember the posterior is a function of m . So we want it to look like:

2 2 2

* * *2 2* *

1 1( ) ( 2 )2 2( | )

- - - - +µ =f x e e

m m m mm ms sm

Taking a factor of n outside so we just have 2m inside:

( ) ( )2 22 * *2 22 * *

1 12 ( )2 2 22( | )Â - - - -- -

µ = µxi

nn

f x e e em mm m mm m s ssm [½]

where:

2

2* and *= =Â ix

n nsm s [1]



Solution 1.37

This is Subject C2, April 1996, Question 9 (i) Posterior distribution

Likelihood ( | ) (1 ) (1 )- -Ê ˆ= - µ -Á ˜Ë ¯

x n x x n xnf x p p p p p

x [½]

Prior 3 3( ) (1 )µ -f p p p [½] Posterior 3 3 3 3( | ) (1 ) (1 ) (1 )- + - +µ - ¥ - = -x n x x n xf p x p p p p p p [½] ( 4, 4)+ - +beta x n x [½] (ii) Bayesian estimate under all-or-nothing loss Given 14=x and 20=n , the posterior is (18,10)beta with PDF: 17 9( ) (1 )f p Cp p= - [1] The Bayesian estimate under all-or-nothing loss is the mode of the posterior, ie the maximum. Taking logs: ln ( ) ln 17 ln 9 ln(1 )f p C p p= + + - [1] Differentiating and setting the derivative equal to zero:

17 9 17ˆln ( ) 0 17(1 ) 91 26

d f p p p pdp p p

= - = fi - = fi =-

[1]

Checking we have a maximum:

2

2 2 217 9ln ( ) 0 max

(1 )d f pdp p p

= - - < fi-

[1]

Since this maximises the log it will also maximise the original PDF.



Solution 1.38

This is from Subject C2, September 1997, Question 7 We have to be careful here with the range of values for which the posterior is valid:

0( | ) prior likelihood0 otherwisee xf x

q qq-Ï < <Ôµ ¥ = Ì

ÔÓ [1]

We don’t have a standard distribution, so we’ll have to work from first principles to find the constant. ( | )f x ke xqq q-= > [½] Integrating over the full range of values that q can take we have:

1 1 1x xx

x

ke d ke ke k eq qq• •- - -È ˘= fi - = fi = fi =Î ˚Ú [½]

So the PDF of the posterior is: ( )( | ) xf x e xqq q- -= > [½] The Bayes estimate of q with respect to the absolute error loss function is the median of the posterior distribution. The median, m , is found from:

( ) ½x

m

e dq q•

- - =Ú [½]

ie ( ) ˆ½ log 2 log 2m xe m x m xq- - = fi - = fi = = + [1] So the Bayes estimator of q with respect to absolute error loss is log 2+x .



Solution 1.39

(i) Conjugate prior The distribution of the time between claims is exponential with parameterl , so: ( ) , 0tf t e tll -= ≥ The likelihood function is:

1 2( | ) in ttt t nf t e e e e lll ll l l l l --- - Â= ¥ ¥ ¥ =! [1] If we look at this expression as a function of l , we see that it has the functional form of a gamma distribution. So the posterior will be a gamma distribution. Hence the conjugate prior will also be a gamma distribution. Let the parameters of the prior gamma distribution be a and b , then:

1f ( ) ea bll l - -µ [1] Hence, the posterior

( )1( | ) itnf t e b lal l - ++ - Âµ [1] We see that this has the form of another gamma distribution – a

( , )iGamma n ta b+ +Â distribution, confirming that the gamma distribution is a conjugate prior. [1] Using the given prior mean and variance, we have:

220.6 and 0.2a a

b b= =

Solving these gives 9a = and 15b = . [1]



(ii) Bayesian estimate under quadratic loss Using the result from part (i), the parameters of our posterior distribution are:

29 and 52.8in ta b+ = + =Â [1] The Bayesian estimate under squared error loss is the mean of this posterior gamma distribution:

29 0.5492452.8i

nt

ab

+ = =+Â

[1]

Solution 1.40

Let jN denote the number of claims in the j th week. Assuming that 1 |N m and

2 |N m are independent, the likelihood function is:

( ) ( )7 11

2 181 27, 11

7! 11!e eL P N N e

m mmm mm m

- --= = = = ¥ µ [1]

(i)(a) Gamma prior The PDF of this prior distribution is: 6 0.5( )priorf e mm m -µ

So the posterior is: 6 0.5 2 18 24 2.5( )postf e e em m mm m m m- - -µ = [½]

We recognise this as the PDF of ( )25, 2.5Gamma . [½]



(i)(b) Uniform prior The prior distribution is now:

m 8 10 12 Prior probability 1

3 13 1

3

Using Bayes’ theorem, we obtain the following posterior probabilities:

( ) ( ) ( )( )

( )

( )

1 21 2

1 2

8 7 8 11

1 2

1 2

7, 11| 8 88 | 7, 11

7, 11

8 8 17! 11! 3

7, 11

0.0033597, 11

P N N PP N N

P N N

e e

P N N

P N N

m mm

- -

= = = == = = =

= =

¥ ¥=

= =

== =

[1]

( ) ( )

( )

10 7 10 11

1 21 2

1 2

10 10 17! 11! 310 | 7, 11

7, 11

0.0034157, 11

e e

P N NP N N

P N N

m

- -¥ ¥

= = = == =

== =

[½ ]

( ) ( )

( )

12 7 12 11

1 21 2

1 2

12 12 17! 11! 312 | 7, 11

7, 11

0.0016657, 11

e e

P N NP N N

P N N

m

- -¥ ¥

= = = == =

== =

[½ ]

The denominator in each of these expressions is: ( )1 27, 11 0.003359 0.003415 0.001665 0.008439P N N= = = + + = So the posterior distribution of m is:

m 8 10 12 Posterior probability 0.3980 0.4047 0.1973

[1]



(ii)(a) Bayesian estimate under quadratic loss The Bayesian estimate of m under quadratic loss is the mean of the posterior. For the gamma prior we have a ~ (25, 2.5)Gammam posterior, so:

( )1 225| 7, 11 102.5

E N Nm = = = = [1]

For the discrete prior, we have:

( ) ( ) ( ) ( )1 2| 7, 11 8 0.3980 10 0.4047 12 0.1973

9.599E N Nm = = = ¥ + ¥ + ¥

= [1]

(ii)(b) Bayesian estimate under absolute loss The Bayesian estimate of m under absolute loss is the median of the posterior. For the gamma prior we have a ~ (25, 2.5)Gammam posterior. To find the median, we need the value of m such that: ( ) 0.5P mm < = Multiplying through by 2l and using the 2

22 ~X al c relationship gives:

250

(2 2 ) 0.5

( 5 ) 0.5

P m

P m

lm l

c

< =

< = [1] From page 169 of the Tables we see that 2

50( 49.33) 0.5P c < = . Hence: 5 49.33 9.866m m= fi = [1] For the discrete prior we can see that: ( 8) 0.3980P m £ = ( 10) 0.3980 0.4047 0.8027P m £ = + = So the median must be 10. [1]



(ii)(c) Bayesian estimate under zero-one loss The Bayesian estimate of m under zero-one loss is the mode of the posterior. For the gamma prior we have a ~ (25, 2.5)Gammam posterior with PDF: 24 2.5( )f C e mm m -= To find is the mode of the posterior (ie the value that maximises the PDF), we first take logs: log ( ) ln 24log 2.5f Cm m m= + - [1] Differentiating this with respect to m and setting the derivative equal to zero:

24 24log ( ) 2.5 0 9.62.5

d fd

m mm m

= - = fi = = [1]

Since this maximises the log it will also maximise the original PDF. For the discrete prior the mode is 10m = as this has the biggest probability. [1] Solution 1.41

(a) Gamma distribution Using the formulae for the mean and variance of a gamma distribution:

22500 and 250a a

l l= =

This gives: 4 and 0.008a l= = [1] The reinsurer will make a payment if the claim size exceeds £1m. Multiplying through by 2l and using the 2

22 ~X al c relationship and the probabilities on page 165 of the Tables gives: 2

8( 1,000) (2 2,000 ) ( 16) 1 0.9576 0.0424P X P X Pl l c> = > = > = - = [1]



(b) Lognormal distribution Using the formulae for the mean and variance of a lognormal distribution:

2 2 2½ 2 2( ) 500 and var( ) ( 1) 250E X e X e em s m s s+ += = = - =

Squaring the first equation and dividing into the second gives:

2 2

22

2501 0.25 log1.25 0.22314500

es s- = = fi = = [1]

From the first equation: 2log500 ½ 6.1030m s= - = [1] We require:

log1,000 6.103036( 1,000) ( 1.70354)0.2231436

1 0.95576 0.04424

P X P Z P Z-Ê ˆ> = > = >Á ˜Ë ¯

= - = [1] Solution 1.42

Using the probability function for the Poisson distribution and the probability density function for the exponential distribution, we have:

|0 0

( ) ( ) ( )!

x

XeP X x f x f d e d

x

lql

l lll l q l

• • --= = =Ú Ú [1]

To evaluate this integral, we make it look like the probability density function of some distribution. We can make this look like the integral of the probability density function of a gamma distribution with parameters 1x + and 1q + by placing an appropriate constant both outside and inside the integral:

( )11

10 0

( 1)! ( 1)( 1)

x xx

xe e d e d

x x

lq lqll q qq l l l

q

• •- +- +-

++=

G ++Ú Ú [1]

since ! ( 1)x x= G + when x is an integer.



The value of the integral is equal to 1, since it is the area under the probability density function of the gamma distribution. So:

1( ) , 0, 1, 2,1 1

xP X x xq

q qÊ ˆ Ê ˆ= = =Á ˜ Á ˜Ë ¯ Ë ¯+ +

… [1]

This is the probability function for a geometric distribution, ie a negative binomial

type 2 distribution (using the title given in the Tables) with 1k = and 1

p qq

=+

,

11

qq

=+

. [1]

Solution 1.43

The distribution function for a Burr distribution with the given parameters is (from page 15 of the Tables):

1.2200( ) 1

200F x

x

aÊ ˆ= - Á ˜Ë ¯+ [1]

To use the method of percentiles, we solve the equation ( ) 0.5F m = where m is the sample median of 420. So we obtain here:

1.22001 0.5

200 420

aÊ ˆ- =Á ˜Ë ¯+ [1]

Rearranging this to find a , we find that:

log 0.5 0.33276log(0.1245546)

a = = [1]

So the method of percentiles estimate for a is ˆ 0.33a = .



Solution 1.44

This is from Subject 106, September 2003, Question 5 (i) Estimate c X has a Weibull distribution. Comparing the PDF given with the one on page 15 of the Tables, we can see that:

2 1

1( ) 2 2cx cxf x cxe c x eggg g- - -= = fi =

We are told that ( 1,000) 0.01P X > = . Using DF given on page 15 of the Tables gives:

21,000( 1,000) 1 (1,000) 0.01cP X F e- ¥> = - = = [1]

Hence:

62

ln 0.01ˆ 4.605 101,000

c -= - = ¥ [1]

(ii) Estimate λλλλ Here, X has an exponential distribution. Again, we are told that ( 1,000) 0.01P X > = . Using the DF given on page 11 of the Tables gives: 1,000( 1,000) 1 (1,000) 0.01P X F e l- ¥> = - = = [1] Hence:

3ln 0.01 4.605 101,000

l -= - = ¥ [1]



(iii)(a) Calculate M We require M such that: ( ) 1 ( ) 0.001P X M F M> = - = For the Weibull distribution with 64.605 10c -= ¥ , we get:

6 24.605 10 2

6ln 0.0010.001 1,500,000

4.605 10Me M

-- ¥ ¥-= fi = - =

¥

1,225Mfi # [1] For the Exponential distribution with 34.605 10l -= ¥ , we get:

34.605 10

3ln 0.0010.001 1,500

4.605 10Me M

-- ¥ ¥-= fi = - =

¥ [1]

(iii)(b) Comment Recall from Chapter 3 that loss distributions are positively skewed and long-tailed. We are looking at the tail probabilities here. A quick diagram will make clear exactly what these probabilities imply:

0.001

1,225

Weibull

Exponential

We can see that the Exponential has a longer tail than the Weibull (or we could say that the Exponential has a heavier tail – as we are more likely to have larger valued claims). [1]



Solution 1.45

This is from Subject C2, September 1998, Question 10. (i) PDF of Y This requires knowledge of functions of a random variable from Subject CT3 Chapter 3. Remember that CT3 is a prerequisite for studying CT6. Let logX Y= , so that X has a gamma distribution. We can write the expression for the area under the PDF of X as:

1

0

11( )

xx e dxa a lla

•- -=

GÚ [1]

Making the substitution logx y= :

1 log 1 1

1 1

1 1 11 (log ) (log )( ) ( )

yy e dy y y dyy

a a l a a ll la a

• •- - - - -= =

G GÚ Ú [1]

So the PDF of Y is:

1 11( ) (log ) , 1( )Yf y y y ya a lla

- - -= >G

[1]

(ii) MLE of α α α α The likelihood function based on the random sample 1 2, , , ny y y… is:

1 2

1 1 1 11 1

1 1

( , ) ( ) ( ) ( )

1 1(log ) (log )( ) ( )

1 (log )[ ( )]

n

n n

ni in

L f y f y f y

y y y y

y y

a a l a a l

a a l

a l

l la a

la

- - - - - -

- - -

=

= ¥ ¥G G

=G

’ ’

…

!

[1]

Taking logs, we get: log log log ( ) ( 1) log(log ) ( 1) logi iL n n y ya l a a l= - G + - - +Â Â [1]



Differentiating this expression with respect to a :

( )log log log(log )( ) iL n n y∂ al

∂a aG¢= - +G Â [1]

Differentiating the expression for log L with respect to l :

log log inL y∂ a

∂l l= -Â

Setting the second equation equal to zero and solving for l , we find that:

ˆlog i

ny

al =Â

[1]

Substituting this back into the first equation, which we also set to zero, we get:

ˆ ˆ( )log log(log ) 0

ˆ( )log ii

nn n yy

a aa

Ê ˆ G¢- + =Á ˜ GË ¯ÂÂ

[1]

Solution 1.46

(i) Mean amount paid by insurer If the amount paid by the insurer is Y, then:

50

50 50X X

YX

<= ≥

We require [ ]E Y :

501 1

0 50

[ ] ( ) 50 ( )E Y x x dx x dxα α α ααλ λ αλ λ∞

− − − −= + + +∫ ∫ [1]



The first integral can be evaluated by parts using u xααλ= :

5050

0 500

501

01 1

1

[ ] ( ) ( ) 50 ( )

50 ( ) 501( 50) ( 50)

( 50)1 1

( 50)1 1

E Y x x x dx x

x

α α α α α α

α α α α

α α

α α α α

α α

λ λ λ λ λ λ

λ λ λ λαλ λ

λ λ λ λα α

λ λ λα α

∞− − −

− +

− + − +

− +

= − + + + + − +

+= − + + − ++ +

+= −− + − ++= −

− + − +

∫

[2]

Substituting the parameter values into this:

2.5 1.5350 400 350[ ] 42.3529

1.5 1.5E Y

−×= − =− −

[1]

So the mean amount paid by the insurer is £42,350. (ii) Mean amount paid by reinsurer on claims involved Since ~ (2.5,350)X Pa , we have:

13

350( ) 2332.5 1

E X = =-

Hence, using the result from part (i): 1

3( ) ( ) ( ) 233 42.3529 190.9804E Z E X E Y= - = - = [1] This is the mean over all claims (including those not referred to the reinsurer). The mean of claims that are referred to the reinsurer is given by:

( )2.5350

400

( ) 190.9804 190.9804 266.67( ) 1 (50)E Z

P X M F= = =

> - [1]

Alternatively, we could show that the conditional distribution, |W Z M X M= - > , is a

( , )Pa Ma l + distribution. Hence the mean is 400( ) 266.671 1.5

ME W la+= = =-

.



(iii) Compare mean claim amounts under proportional and XOL Under the proportional reinsurance arrangement the insurer pays 0.7Y X= where X represents a claim amount before reinsurance. We require [ ]E Y :

350[ ] 0.7 [ ] 0.7 163.331.5

E Y E X= = × = [1]

The proportional reinsurance gives a higher mean by:

163.33 42.3529 2.8642.3529

− =

ie it is higher by 286%. [1] Solution 1.47

This is from Subject 106, April 2004, Question 1. Let the current loss be X so that: ~ (3.5,1000)X Pa We want (1.05 100)P X < : [1]

3.5

1001.05

100 100 1,000(1.05 100) 11.05 1.05 1,000

0.273

P X P X FÊ ˆÊ ˆ Ê ˆ< = < = = -Á ˜ Á ˜ Á ˜Ë ¯ Ë ¯ +Ë ¯

= [2]



Solution 1.48

This is from Subject 106, September 2002, Question 8. (i) Integral We want to find the value of the integral:

212

log

2

1

2

xb

a

e dxm

s

ps

-Ê ˆ- Á ˜Ë ¯Ú

We make the substitution log xu m ss-= - .

Using this substitution, we have:

21 and udu x e

dx xm s s

s+ += = [1]

So the integral becomes:

2 2

log

½( )

log

12

b

u u

ae e du

m ss

s m s s

m ss

ss p

- -

- + + +

- -

Ú [1]

Cancelling the s ’s, multiplying out the brackets in the exponent, taking out a common factor and simplifying, we have:

2 2

log

½ ½

log

12

b

u

ae e du

m ss

m s

m ss

p

- -

+ -

- -

Ú [1]

This integral is just a standard normal integral. So we have:

2

2

log½

2

½

1

2

log log

xb

a

e dx

b ae

ms

m s

psm ms s

s s

-Ê ˆ- Á ˜Ë ¯

+ È ˘- -Ê ˆ Ê ˆ= F - -F -Á ˜ Á ˜Í ˙Ë ¯ Ë ¯Î ˚

Ú

[1]



(ii)(a) Expected claim size We first need the parameter values m and 2s . Solving the equations for the mean and variance of the lognormal distribution:

2 2 2½ 2 2264 and ( 1) 346e e em s m s s+ += - = [1]

Squaring the first equation and substituting it into the second, we obtain:

22 2264 ( 1) 346es - =

Solving this for 2s , and substituting back for m , we find that 2 0.99978s = ,

5.07606m = . [1] The insurer pays an amount I , where: 100 | 100I X X= - > This is the mean of claims paid by the insurer given that they are referred to the insurer. This is the approach taken by the examiners in most of these questions (April 00 Q3, April 02 Q4 and this question). However there is some inconsistency in their solutions (see Sept 99 Q3 and April 07 Q3). We want the expected value of I :

100

( 100) ( )

( )( 100)

x f x dxE I

P X

•

-

=>

Ú [1]

where

2log½1( )2

x

f x ex

ms

s p

-Ê ˆ- Á ˜Ë ¯= is the probability density function of the

lognormal distribution.



Splitting the integral into two, and using the result proved in the first part of the question, the value of the numerator is:

2

100 100

2½ 2

( ) 100 ( )

log1001 100 log ( , ) 100

x f x dx f x dx

e P Nm s m s m ss

• •

+

-

È ˘Ê ˆ- - È ˘= -F - >Í ˙Á ˜ Î ˚Ë ¯Í ˙Î ˚

Ú Ú

[1]

Using the standard normal distribution tables, the first expression simplifies to:

[ ]log100 5.07606 0.99978264 1 264 1 ( 1.47083)0.99978

264 [1 0.07067 ] 245.34

È ˘- -Ê ˆ-F = -F -Í ˙Á ˜Ë ¯Î ˚= - = [1]

The second expression becomes:

log100 5.07606100 1 100[1 ( 0.47094)] 68.1160.99978

È ˘-Ê ˆ- F = -F - =Í ˙Á ˜Ë ¯Î ˚ [1]

The value of the denominator is 0.68116 (using the expression we have just found). So the total value of the integral is:

245.34 68.116 260.1840.68116

- = [1]

The expected amount paid by the insurer is about £260. (ii)(b) New expected claim size If claims in the next year rise by 10%, then a new claim can be written as 1.1Y X= , where X has the same lognormal distribution as before. The insurer pays an amount Z , where:

0 if 100 /1.11.1 100 if 100 /1.1 1,100 /1.11,000 if 1,100 /1.1

XZ X X

X

<ÏÔ= - £ <ÌÔ ≥Ó

[1]



We want to find ( | 100 /1.1)E Z X > . This will be given by:

1,100/1.1

100/1.1 1,100/1.1

(1.1 100) ( ) 1,000 ( )

( | 100 /1.1)( 100 /1.1)

x f x dx f x dx

E Z XP X

•

- +

> =>

Ú Ú [1]

where ( )f x is the PDF of the original lognormal distribution. So the numerator is equal to:

2

2 2

½

1,100 100log log1.1 1.11.1

1,100 100log log1.1 1.1100

1,100log1.11,000 1

em sm s m s

s s

m m

s s

+

È ˘Ê ˆ Ê ˆÊ ˆ Ê ˆ- - - -Í ˙Á ˜ Á ˜Á ˜ Á ˜Ë ¯ Ë ¯Í ˙F -FÁ ˜ Á ˜Í ˙Á ˜ Á ˜Á ˜ Á ˜Í ˙Ë ¯ Ë ¯Î ˚

È ˘Ê ˆ Ê ˆÊ ˆ Ê ˆ- -Í ˙Á ˜ Á ˜Á ˜ Á ˜Ë ¯ Ë ¯Í ˙- F -FÁ ˜ Á ˜Í ˙Á ˜ Á ˜Á ˜ Á ˜Í ˙Ë ¯ Ë ¯Î ˚

Ê ˆ -Á ˜Ë ¯+ - F

m

s

È ˘Ê ˆÍ ˙Á ˜Í ˙Á ˜Í ˙Á ˜Á ˜Í ˙Ë ¯Î ˚

[1]

Working out the relevant values, we have:

2 21,100 100log log1.1 1.10.83201 1.56615

m s m s

s s

Ê ˆ Ê ˆ- - - -Á ˜ Á ˜Ë ¯ Ë ¯= = - [1]

1,100 100log log1.1 1.11.83190 0.56626

m m

s s

Ê ˆ Ê ˆ- -Á ˜ Á ˜Ë ¯ Ë ¯= = - [1]

So the numerator is:

1.1 264 (0.79730 0.05866)

100 (0.96652 0.28561) 1,000 (1 0.96652) 179.889¥ -

- - + - = [1]



The denominator is: ( 100 /1.1) 1 0.28561 0.71439P X > = - = So the expected claim size payable by the insurance company is about 179.889 / 0.71439 £251.81= . [1] Alternatively, we calculate the parameters of the new lognormal distribution: 1.1Y X= We have:

2½( ) 1.1 ( ) 290.4E Y E X em s+= = =

2 22 2 2var( ) 1.1 var( ) (380.6) ( 1)Y X e em s s+= = = -

Solving this gives 2 0.99978s = , 5.17137m = . We then have:

0 if 100 /1.1

100 if 100 1,1001,000 if 1,100

YZ Y Y

Y

<ÏÔ= - £ <ÌÔ ≥Ó

Then we want:

( )( 100)

E ZP Y >



Solution 1.49

This is from Subject 106, April 2004, Question 8 (i) Proof The integral is:

2

12

0

1 1 lnexp2

dm xx dx

xm

ss p

Ï ¸-Ô ÔÊ ˆ-Ì ˝Á ˜Ë ¯Ô ÔÓ ˛Ú

We will use the substitution ln xu mm ss-= - . Define ln dU mm s

s-= - .

( ) ( )2

2 2 2

2 2 2 2 2

2 2 2

1 1exp([ ] ) exp22

1 1exp ( 2 ( ) 2 2 )22

1 1exp ( 2 2 2 2 )22

1 1 1exp ( 2 ) exp2 22

Um

U

U

U

u m u m du

m u m m u um m du

mu m m u um m du

m m u du

s s m sp

s s m s sp

s s m s sp

m sp

-•

-•

-•

-•

Ï ¸+ + - +Ì ˝Ó ˛

Ê ˆ= - - + - + + +Á ˜Ë ¯

Ê ˆ= - - - - + + +Á ˜Ë ¯

Ê ˆ Ê ˆ= - - - -Á ˜ Á ˜Ë ¯ Ë ¯

Ú

Ú

Ú

Ú [3]

The last integral expression is equivalent to integrating the PDF of the (0,1)N

distribution from -• to ln dU mm ss-= - . So we get:

2 21 lnexp2

dm m mmm s ss-Ê ˆ Ê ˆ+ F -Á ˜ Á ˜Ë ¯ Ë ¯

[1]

which is the required expression.



(ii) Parameter values We need to solve simultaneous equations to find the parameters:

2 2 2½ 2 2e 800 and e (e 1) 1, 200m s m s s+ += - = [1]

Substituting the square of the first equation into the second, we get: 2 1.1787 and 6.0953s m= = [2] (iii) Mean values (Prop)[ ] [(1 ) ] (1 ) [ ]IE X E k X k E X= - = - Comparing this to the expression given in the question, we can see that 0.3k = . [½]

( )

0 0

[ ] ( ) ( ) ( ) ( )d d

XLI

d

E X xf x dx df x dx xf x dx dP X d•

= + = + >Ú Ú Ú

since ( ) ( )d

f x dx P X d•

= >Ú .

Firstly, we will find this probability:

ln( ) (ln ln ) dP X d P X d P Z ms-Ê ˆ> = > = >Á ˜Ë ¯

Using the value of d from the question:

( )( 1,189.4) 0.908 1 ( 0.908)

1 0.81806 0.18194P X P Z P Z> = > = - <

= - = [1] To find the integral in the expression for the mean above, we will use part (i):

2

0

1 ln( ) exp 800 ( 0.178)2

800(1 0.57063) 343.496

d dxf x dx mm s ss-Ê ˆ Ê ˆ= + F - = F -Á ˜ Á ˜Ë ¯ Ë ¯

= - =

Ú [1]



Finally we get: ( )[ ] 343.496 1,189.4 0.18191 560XL

IE X = + ¥ = [1] Since 0.7 [ ] 0.7 800 560E X = ¥ = , we have the required result. [½] (iv) Variances Using 0.3k = , we get: (Prop) 2 2 2var[ ] var[(1 ) ] (1 ) var[ ] 0.7 1, 200 705,600IX k X k X= - = - = ¥ = [1]

( )2( ) ( ) ( )2

2 2 2

0

2 2 2

0

var[ ]

( ) ( ) 560

( ) ( ) 560

XL XL XLI I I

d

dd

X E X E X

x f x dx d f x dx

x f x dx d P X d

•

È ˘ È ˘= -Í ˙ Î ˚Î ˚

= + -

= + > -

Ú Ú

Ú [1]

We have already found the probability ( ) 0.18191P X d> = . To find the integral in the expression for the variance above, we will again use part (i):

( )( )

2 2

0

ln( ) exp 2 2 2

2,080,000 1.263 2,080,000(1 0.89671) 214,843

d dx f x dx mm s ss-Ê ˆ= + F -Á ˜Ë ¯

= F - = - =

Ú [1]

Finally we get: ( ) 2 2var[ ] 214,843 1,189.4 0.18191 560 159,000XL

IX = + ¥ - = [1] (v) Comments The means are the same under both arrangements, however the variance is much higher for the proportional arrangement. This is because under excess of loss you truncate the higher claims and therefore are left with a lower variance. Since an insurer wants a lower variance, it would usually go for excess of loss reinsurance. [2]



Solution 1.50

(i) MLE on censored data We have 7 known claims and 3 unknown claims. The likelihood is:

73

17 3

7 10,287 3,000 3

7 19,287

( ) ( ) [ ( 3,000)]

[1 (3,000)]

[ ]

i

ii

x

L f x P X

e F

e e

e

λ

λ λ

λ

λ

λ

λ

λ

=−

− −

−

= × >

∑= × −

= ×

=

∏

[2] So the log-likelihood is: ln ( ) 7 ln 19,287L λ λ λ= − Differentiating and setting it equal to zero:

7 ˆln ( ) 19,287 0 0.000363ˆd L

dλ λ

λ λ= − = ⇒ = [1]

Checking it’s a maximum:

2

2 27ln ( ) 0 maxd L

dλ

λ λ= − < ⇒

(ii)(a) Conditional distribution If X is the random variable for the gross claim amounts and W is the random variable for the claim amounts paid by the reinsurer given that the reinsurer is involved, then the PDF of W is given by:

( 2,500)( ) 0( 2,500)

XW

f wf w wP X

+= >>

[1]

The denominator is:

( 2,500) 1 (2,500)2,500

P X Fαλ

λ > = − = +

[½]



The numerator is:

1( 2,500)( 2,500)Xf w

w

α

ααλ

λ ++ =+ +

[½]

Hence, the PDF is:

1( 2,500)( )

( 2,500 )Wf ww

α

αα λ

λ ++=

+ + [½]

This is the PDF of a Pareto but with λ replaced by 2,500λ + . Hence the distribution is a Pareto ( , 2,500)α λ + . [½] (ii)(b) MLE on truncated (conditional) data Here we have

5 5

2

1 1

1 11,055.6 and 2,315,9765 5i i

i ix x

= == =Â Â [1]

Equating these to the population mean and variance, we get:

22

22,500 ( 2,500) 2,5001,055.6 and 2,315,976

1 1( 1) ( 2)λ α λ λ

α αα α+ + + = + = − −− −

[1]

ie 2

22

( 2,500) 2,315,976 1,055.6 1,201,685( 1) ( 2)α λα α

+ = − =− −

Combining this with the equation for the mean gives:

2 ˆ1,055.6 1,201,685 27.50( 2)

α αα

× = ⇒ =−

[1]

2,500 ˆ1,055.6 25, 473

27.50 1λ λ+

⇒ = ⇒ =−

[1]

Equating the sample mean and the unbiased sample variance (ie calculated with denominator 4) is also acceptable here. This leads to parameter values ˆ 7.75a = and ˆ 4,622l = .



Part 2 – Questions Question 2.23

One year ago a company insured a portfolio of 100 independent policies. Under each policy at most one claim can be made in any year and the probability of a claim being made in a year, denoted p , is the same for all policies. Let n denote the total number of claims occurring in the past year. (i) Show that the maximum likelihood estimate of p is 100n . [2] (ii) Now suppose that p , regarded as a random variable, has a beta prior distribution

with parameters a and b . (a) Show that the posterior distribution of p is also a beta and find the

parameters of this distribution in terms of a , b and n . (b) Show that the posterior mean of p can be written in the form of a

credibility estimate. [6] [Total 8] Question 2.24

The number of claims from one group of drivers in a year has a Poisson distribution with mean l , and the number of claims from a second group of drivers has a Poisson distribution with mean 2l . In one year, there are 1n claims from group 1 and 2n claims from group 2. (i) Derive the maximum likelihood estimator, l , of l . [3] (ii) Suppose that past experience shows that l has an exponential distribution with

mean 1n

.

(a) Derive the posterior distribution of l . (b) Show that the Bayesian estimate of l under quadratic loss may be

written in the form of a credibility estimate combining the prior mean of l with the maximum likelihood estimate l in (i). State the credibility factor. [4]

[Total 7]



Question 2.25

Claim amounts X are believed to follow a gamma distribution with parameters a and l , where a is a known constant. An insurer wishes to carry out a Bayesian analysis, using a gamma prior distribution for l with parameters b and d . (i) A random sample from the distribution gives the individual sample values

1 2, , , nx x x… . Show that the posterior distribution for l is also a gamma distribution, and write down its parameters in terms of ,a b ,d and the sample data. [4]

(ii) Write down the formula for the Bayesian estimate for l under squared error

loss. [1] (iii) Show that this formula can be written in the form of a credibility estimate:

ˆ (1 )Z Zl l m= + -" where l" is the maximum likelihood estimator for l calculated from the sample

data, and m is the mean of the prior distribution. Write down the formula for the credibility factor Z . [3]

(iv) Explain how the value of Z changes if the sample size increases, and explain

why it is reasonable to expect Z to change in this way. [1] [Total 9] Question 2.26

(i) The random variable X has a gamma distribution with parameters 3a = and 2l = . Y is a related variable with conditional mean and variance of:

( | ) 3 1E Y X x x= = + 2var( | ) 2 5Y X x x= = + Calculate the unconditional mean and standard deviation of Y . [3] (ii) The random variable V has a Poisson distribution with mean 5. For a given

value of V , the random variable U is distributed as follows: | ( ) ~ (0, )U V v U v= Obtain the mean and variance of the marginal distribution of U . [2] [Total 5]



Question 2.27

1S and 2S are independent random variables each with a compound Poisson distribution. The distribution of iS , 1,2i = , has Poisson parameter 1l and individual claim amount distribution ( )iF x . The distribution of 1 2( )S S+ : A has a compound Poisson distribution with Poisson parameter 1 2l l and

individual claim amount distribution 1 2( ) ( )F x F x+ B has a compound Poisson distribution with Poisson parameter 1 2( )l l+ and

individual claim amount distribution ( )1 2( ) ( ) 2F x F x+ C has a compound Poisson distribution with Poisson parameter 1 2( )l l+ and

individual claim amount distribution ( ) ( )1 1 2 2 1 2( ) ( )F x F xl l l l+ + D does not have a compound Poisson distribution. [2]



Question 2.28

The aggregate claim amount from a portfolio has a compound negative binomial distribution. (i) Show that if 1 NS X X= + +# , then: [ ]( ) log ( )S N XM t M M t= [3] (ii) If N is Type 2 negative binomial distribution with 2k = and 0.9p = , and X

has a gamma distribution with 10a = and 0.1l = , find ( )SM t . [2] (iii) Find the mean and variance of S . Using a suitable approximation, estimate the

aggregate amount which will be exceeded with probability 0.1%. [4] (iv) The insurer in fact has 100 identical independent portfolios of this type. If: 1 100T S S= + +# find the moment generating function for T , and, using a normal approximation,

estimate the total aggregate claim amount from the whole business which will be exceeded with probability 0.1%. Comment on your answers to parts (iii) and (iv). [4]

[Total 13] Question 2.29

Aggregate annual claims from a portfolio of general insurance policies have a compound Poisson distribution with Poisson parameter 20. Individual claim amounts have a uniform distribution over the range (0,200) . Excess of loss reinsurance is arranged so that the expected amount paid by the insurer on any claim is 50. Calculate the variance of the aggregate annual claims paid by the insurer. [6]



Question 2.30

Consider aggregate claims over a period of 1 year, S , on a portfolio of general insurance policies: 1 2 ... NS X X X= + + + The number of claims each year, N , has a Poisson distribution with mean 12.

1 2, ,...X X are assumed to be random variables, independent of each other and independent of N , with the following distribution: 0.01( ) 0.01 0 £200xf x e x-= < <

2( £200)P X e-= = The insurer effects excess of loss reinsurance with a retention of £100. Annual aggregate claims paid by the reinsurer are denoted by RS .

Calculate ( )RE S , var( )RS and 3( ( ))R RE S E SÈ ˘-Î ˚ . [12]

Question 2.31

0{ ( )}tS t ≥ and 0{ ( )}tS t*≥ are compound Poisson processes representing the aggregate

claims up to time t from two risks. Individual claim amounts have the same distribution for the two risks and premiums are calculated using the same premium loading factor for the two risks. The Poisson parameters for the two risks are l and l* , respectively. The probability of ruin in finite time and in infinite time for these two risks, given initial surplus U , are ( , )U ty and ( , )U ty * and ( )Uy and ( )Uy * ,

respectively. You are given that 2l l* = . Which of the following equations is/are true: I ( ) 2 ( )U Uy y *=

II ( ) ( )U Uy y *=

III ( , ) ( ,2 )U t U ty y *=

IV ( , ) ( , )U t U ty y *= [3]



Question 2.32

The aggregate claims process for a risk is a compound Poisson process. The expected number of claims each year is 0.5 and individual claim amounts have the following distribution:

[claim amount 1] 0.5



P

P

P

= =

= =

= =

Let ( )U t denote the insurer’s surplus at time t . The insurer’s surplus at time 0 is 0.5 and the insurer charges a premium of 1 each year to insure this risk. Calculate the probability: [ ( ) 0 for 1 or 2]P U t t< = [7] Question 2.33

An insurance company has a portfolio of policies, for which claims occur as a Poisson process at a rate of 25 claims per year. The claim amounts in pounds follow a generalised (three parameter) Pareto distribution with parameters 3k = , 500l = and

4a = . The insurer includes a premium loading of 15% in its premiums for this portfolio. You may assume that the aggregate claim amount for a year is approximately normally distributed. (i) Find u , the initial capital required in order to ensure that the probability of ruin

at the end of the first year is 2%. [4] (ii) If the insurer takes out proportional reinsurance, reinsuring 30% of the loss with

a reinsurer which loads its premiums by 45%, find the new level of initial capital required, and compare your answer with that in part (i). [5]

[Total 9]



Question 2.34

Aggregate annual claims from a portfolio of general insurance policies have a compound Poisson distribution with Poisson parameter l . Individual claim amounts have an exponential distribution with mean 1. The premium loading factor used to calculate the premium for these policies is 0.30. Given an initial surplus of 2, calculate the probability of ruin at the first claim. [6] Question 2.35

The aggregate claims produced by a risk have a compound Poisson distribution with Poisson parameter 100 and individual claim size density, ( )f x , where: 0.2( 5)( ) 0.2 5xf x e x- -= > The premium charged by the insurer to insure the risk is calculated using a premium loading factor of 0.15. The insurer is considering excess of loss reinsurance for this risk. The reinsurer’s premium would be calculated using a premium loading factor of 0.30. The table below shows, for various values of the retention limit M , the insurer’s expected profit in one year net of reinsurance, with some missing values indicated by asterisks.

Retention limit M Expected annual profit 7.5 59 10 * 15 * * 147 • *

Calculate the missing values of M and of the insurer’s expected profit in one year and set out the complete table. [12]



Part 2 – Solutions Solution 2.23

(i) MLE of p The likelihood is given by:

100100( ) (1 )n nL p p p

n-Ê ˆ

= -Á ˜Ë ¯ [½]

Taking logs gives:

100

ln ( ) ln ln (100 ) ln(1 )L p n p n pn

Ê ˆ= + + - -Á ˜Ë ¯

[½]

Differentiating and setting the derivate equal to zero gives:

100 ˆ ˆ ˆln ( ) 0 (1 ) (100 )ˆ ˆ1 100

d n n nL p n p n p pdp p p

-= - = fi - = - fi =-

[1]

(ii)(a) Posterior distribution The PDF of the prior is:

1 1 1 1( )( ) (1 ) (1 )( ) ( )

f p p p p pa b a ba ba b

- - - -G += - µ -G G

[1]

So the posterior is: 100 1 1 1 100 1( | ) (1 ) (1 ) (1 )n n n nf p x p p p p p pa b a b- - - + - - + -µ - ¥ - = - [1] This is a ( , )beta a b* * distribution with parameters na a* = + and 100 nb b* = - + . [1]



(ii)(b) Credibility estimate The Bayesian estimate under quadratic loss is the mean of the posterior:

100

na aa ba b

*

* *+=+ ++

[1]

We need to express this estimate as a weighted average of the MLE of p (usually the sample mean) and the prior mean:

100100 100 100 100

n na a b aa b a b a b a b+ += ++ + + + + + +

[2]

Solution 2.24

This is Subject 106, September 2004, Question 4. (i) Maximum likelihood estimate In our sample we have obtained a value of 1n from a Poisson distribution with parameter l , and a value of 2n from a Poisson distribution with parameter 2l . By definition, the likelihood is the probability of observing the sample obtained:

1 2

1 22

3

1 2

(2 )( )! !

n nn ne eL Ce

n n

l lll ll l

- -+-= ¥ = [1]

Taking logs gives: 1 2

1 2ln ( ) ln 3 ln ln 3 ( ) lnn nL C C n nl l l l l+= - + = - + + [½] Differentiating and setting the derivative equal to zero gives:

1 2 1 2( ) ˆln ( ) 3 0ˆ 3n n n nd L

dl l

l l+ += - + = fi = [1]

Checking whether we have a maximum

2

1 22 2

( )ln 0 maxn nd L

dl l+

= - < fi [½]



(ii)(a) Posterior distribution

Since the prior distribution for l is exponential with mean 1n

, the prior pdf is e nln - .

So we have: 1 2 1 23 (3 )n n n n

pdfpost e e enl l n ll l+ +- - - +µ ¥ = [½]

Comparing this to the standard distributions given in the Tables, we can see that this is the PDF of a gamma distribution with parameters 1 2 1n na = + + and 3d n= + . Hence the posterior distribution is 1 2( 1,3 )gamma n n n+ + + . [1] (ii)(b) Credibility estimate The mean of the posterior distribution is:

1 2 13

n nn

+ ++

[1]

This can be written in the form:

1 2 1 23 1 3 3 113 3 3 3 3 3

n n n nnn n n n n n

+ + Ê ˆ¥ + ¥ = ¥ + - ¥Á ˜Ë ¯+ + + + [1]

This is in the form of a credibility estimate with credibility factor 33

Zn

=+

. [½]



Solution 2.25

(i) Posterior distribution The likelihood of the sample 1 2, , , nx x x… is:

11 11( )

( ) ( )n

i

xxn

xn

L x e x e

e

a alla a

la

l lla a

l

--- -

-

= ¥ ¥G G

Âµ

"

[1] The PDF of the prior distribution for l is:

1 1( )( )

f e eb

b dl b dldl l lb

- - - -= µG

[1]

The posterior is proportional to the prior distribution multiplied by the likelihood function, so (ignoring all terms not involving l ):

1

( )1

( ) i

i

xn

xn

post e e

e

lb dl a

l da b

l l l

l

-- -

- ++ -

Âµ ¥

Â= [1] Looking at this, we see that it has the form of another gamma distribution, this time with parameters: * na a b= + and * ixl d= +Â So the posterior distribution is a gamma distribution with these parameters. [1] (ii) Bayesian estimate The Bayesian estimate for l under squared error loss is the mean of this posterior distribution:

*ˆ* i

nx

a a bll d

+= =+Â

[1]



(iii) Credibility estimate and factor The likelihood from earlier is:

11 11( )

( ) ( )in xxx n

nL x e x e C ea a

llla a al ll la a

---- - Â= ¥ ¥ =G G

"

Taking logs: ln ( ) ln ln iL C n xl a l l= + - Â Differentiating and setting the derivative equal to zero gives an MLE of:

ˆln ( ) 0ii

d n nL xd xx

a a al ll l

= - = fi = =Â Â [1]

We can now split the mean of the posterior from part (ii) into two terms:

(1 )

i

i i i i i i

xn n nx x x x x x

Z Zx

a b a b a d bdd d d d d

a bd

+ = + = ¥ + ¥+ + + + +

Ê ˆ Ê ˆ= + -Á ˜ Á ˜Ë ¯ Ë ¯

ÂÂ Â Â Â Â Â

where i

i

xZ

xd=

+ÂÂ

. [2]

(iv) Changes to sample size We can see that if the sample size increases, this will increase the value of ixÂ and hence of the credibility factor. We will put more weight on the sample data and less on the value of the prior mean, which seems sensible if we are using a larger sample. [1] [Total 9]



Solution 2.26

(i) Calculate the unconditional mean and variance The mean and variance of the gamma distribution are given by:

23 3( ) 1.5 var( ) 0.752 4

E X Xa al l

= = = = = = [½]

Using the result from page 16 of the Tables, ie ( ) [ ( | )]E Y E E Y X= , we get: ( ) [3 1] 3 [ ] 1 3 1.5 1 5.5E Y E X E X= + = + = ¥ + = [1] Using the result var( ) var[ ( | )] [var( | )]Y E Y X E Y X= + from page 16 of the Tables, we get:

2

2

var( ) [2 5] var[3 1]

2 [ ] 5 9var[ ]

Y E X X

E X X

= + + +

= + + [½] Using the fact that 2 2 2( ) var( ) ( ) 0.75 1.5 3E X X E X= + = + = , we get: [½] var( ) 2 3 5 9 0.75 17.75Y = ¥ + + ¥ = So the standard deviation is 17.75 4.21= . [½] (ii) Mean and variance of the marginal distribution We are given in the question that: | ( ) ~ (0, ) ~ (5)U V v U v V Poi= So: ( ) 5 var( ) 5E V V= = and: 21 1

2 12( | ) var( | )E U V V U V V= = [½]



Using the formulae on page 16 of the Tables, we have: [ ] [ ] [ ][ ] [ | ] var[ ] var [ | ] var[ | ]E U E E U V U E U V E U V= = + Therefore: [ ] 1 1 1

2 2 2[ ] ( | ) [ ] 2E U E E U V E V E VÈ ˘= = = =Î ˚ [½]

[ ] [ ]21 1

2 12

21 14 12

var[ ] var ( | ) var( | )

var

var[ ] [ ]

U E U V E U V

V E V

V E V

= +

È ˘È ˘= +Î ˚ Î ˚

= + [½] Since 2 2[ ] var[ ] [ ]E V V E V= + , we have: 2 31 1

4 12 4var[ ] 5 (5 5 ) 3U = ¥ + + = [½] Solution 2.27

This question is Subject C2, Specimen 1993, Question 3. C [2] Please note that the CT6 exam does not contain any multi-choice questions.



Solution 2.28

(i) Compound MGF The MGF of S is:

( ) ( )( )( ) |tS tSSM t E e E E e N= =

using the standard result for conditional means from page 16 of the Tables. Looking at the inner expression, we have:

( ) ( ) ( ) ( )1 1( )| n nt X X tXtXtSE e N n E e E e E e+ += = =" … [1]

Now each of these terms is just the MGF of the random variable X . So:

( ) [ ]| ( ) ntSXE e N n M t= = [1]

and:

{ } ( )[ ]

log ( )( ) ( )

log ( )

XN N M tS X

N X

M t E M t E e

M M t

È ˘= =Í ˙Î ˚

= [1] (ii) Compound negative binomial We first need the individual MGFs. Using results from the Tables, we have:

10

10( ) 1 (1 10 )0.1XtM t t

--Ê ˆ= - = -Á ˜Ë ¯

[½]

and: 20.9( )

1 0.1N tM te

Ê ˆ= Á ˜Ë ¯- [½]

Combining these, using the result from part (i):

[ ]2

100.9( ) log ( )

1 0.1(1 10 )S N XM t M M tt -

Ê ˆ= = Á ˜- -Ë ¯

[1]



(iii) Mean and variance of S We could differentiate this expression to find the mean and variance of S . However, it is much easier to use the standard compound distribution formulae:

( ) ( ) ( )E S E X E N= and: [ ]2var( ) ( ) var( ) var( ) ( )S E X N X E N= + Using the results from the Tables for the individual distributions:

10( ) 1000.1

E X al

= = = 2 210var( ) 1,000

0.1X a

l= = = [1]

2 0.1( ) 0.222220.9

kqE Np

¥= = = 2 22 0.1var( ) 0.246910.9

kqNp

¥= = = [1]

Using the formulae above: ( ) ( ) ( ) 22.222E S E X E N= = and:

[ ]2

2

var( ) ( ) var( ) var( ) ( )

100 0.24691 1,000 0.22222 2,691.358

S E X N X E N= +

= ¥ + ¥ = [1] We now assume that S has an approximate normal distribution with this mean and variance. So, standardising in the usual way, we have:

( ) ( )0.001 (0,1) 0.001var( )

k E SP S k P NS

Ê ˆ-> = fi > =Á ˜Ë ¯

Using the percentage points of the standard normal distribution, we find that:

( ) 3.0902 22.222 3.0902 2,691.358 182.54var( )

k E S kS

- = fi = + = [1]



(iv) Whole portfolio The moment generating function for T is:

[ ]1 100 1001 100( )

200

10

( ) ( ) [ ] ( ) ( ) ( )

0.91 0.1(1 10 )

t S S tStStTT SM t E e E E E e E e M t

t

+ +

-

= = = =

Ê ˆ= Á ˜- -Ë ¯

… …

[1]

The mean and variance of T will just be 100 times the corresponding results for S : ( ) 100 ( ) 2,222E T E S= = and: var( ) 100 var( ) 269,135.8T S= = [1] So the corresponding figure for the aggregate amount exceeded with probability 0.001 is: 2,222 3.0902 269,135.8 3,825.37+ = [1] This is substantially less than one hundred times the corresponding answer to part (iii). The central limit theorem tells us that as the number of portfolios increases, bad experience in some of the portfolios will be offset by better experience in others, leading to a situation where the overall variation is relatively smaller. Pooling of similar risks reduces the overall variance. We can see this happening here. [1] Solution 2.29

This is Subject C2, September 1995, Question 11. We have ~ (0, 200)X U and:

0X X M X M

Y ZM X M X M X M

< <Ï Ï= =Ì Ì> - >Ó Ó

The expected amount paid by the insurer on any claim is:

200

0

1 1( ) 50200 200

M

M

E Y x dx M dx= + =Ú Ú [1]



Solving this gives:

2002

0

50400 200

M

M

x MxÈ ˘ È ˘+ =Í ˙ Í ˙Î ˚Í ˙Î ˚

2 250

400 200M MMfi + - = [1]

2 400 20,000 0 58.579, 341.42M M Mfi - + = fi = [1] Since claims are a maximum of 200, it must be the first value of M . The variance of the aggregate annual claims, S , paid by the insurer is given by: 2 2var( ) ( ) 20 ( )S E Y E Yl= = where:

2002 2 2

0

2003 3 2

0

1 1( )200 200

376.44600 200 600 200

M

MM

M

E Y x dx M dx

x Mx M MM

= +

È ˘ È ˘= + = + - =Í ˙ Í ˙Î ˚Í ˙Î ˚

Ú Ú

[2]

Hence: var( ) 20 376.44 7,529S = ¥ = (4 SF) [1] Solution 2.30

This is Subject C2, September 1998, Question 12 (part) Let Z be the amount paid on each claim by the reinsurer. We have:

0 100

100 100 200100 200

XZ X X

X

<ÏÔ= - £ <ÌÔ =Ó

[½]



We first want ( )E Z , 2( )E Z and 3( )E Z . So we have:

200

0.01 2

100

( ) ( 100)0.01 100xE Y x e dx e- -= - +Ú [1]

We can integrate by parts to find the value of the integral:

200 2002000.01 0.01 0.01100

100 1002000.01 0.01100

1 2

( 100) 0.01 ( 100)

( 100) 100

100 200

x x x

x x

x e dx x e e dx

x e e

e e

- - -

- -

- -

È ˘- = - - +Î ˚

È ˘= - - -Î ˚

= -

Ú Ú

[2]

1 2 2 1 2( ) 100 200 100 100( ) 23.254416E Y e e e e e- - - - -fi = - + = - = We now find 2( )E Z :

200

2 2 0.01 2 2

100

( ) ( 100) 0.01 100xE Z x e dx e- -= - +Ú [1]

The integral is (integrating by parts again and using the result from the first integral):

200 2002 0.01 2 0.01100

100200

0.01

1002 2 1 2

1 2

( 100) 0.01 ( 100)

2( 100)

100 2 100 (100 200 )

20,000 50,000

x x

x

x e dx x e

x e dx

e e e

e e

- -

-

- - -

- -

È ˘- = - -Î ˚

+ -

= - + ¥ ¥ -

= -

Ú

Ú

[2] So adding in the final term, we get: 2 1 2( ) 20,000 40,000 1944.1775E Z e e- -= - = [1]



Finding 3( )E Z the same way:

200

3 3 0.01 3 2

100

( ) ( 100) 0.01 100xE Z x e dx e- -= - +Ú [1]

So the integral is:

200 2002003 0.01 3 0.01 2 0.01100

100 1003 2 1 2

( 100) 0.01 ( 100) 3( 100)

100 3 100 (20,000 50,000 )

x x xx e dx x e x e dx

e e e

- - -

- - -

È ˘- = - - + -Î ˚

= - + ¥ ¥ -

Ú Ú

[1] Adding in the final term, we get: 3 1 2( ) 6,000,000 15,000,000 177,247.4E Z e e- -= - = [1] Using the formulae for the mean, variance and skewness for a compound Poisson distribution, we get: ( ) ( ) 12 23.2544 279.053RE S E Zl= = ¥ = [½]

2var( ) ( ) 12 1944.1775 23,330.13RS E Zl= = ¥ = [½]

3skew( ) ( ) 12 177,247.4 2,126,969RS E Zl= = ¥ = [½] Solution 2.31

Equations II and III are true. Doubling l* means claims occur in half the time, but premium income comes in at double the rate. So graph of ( )U t is squashed compared to ( )U t* :

t

U*(t) U(t)

t

doubledλ

No change to ( )uy , but increases ( , )u ty . [3]



Solution 2.32

This is Subject C2, September 1996, Question 9 The surplus is:

(1) 0.5 1 (1) 1.5 (1)

(2) 0.5 2 1 (2) 2.5 (2)

U S S

U S S

= + - = -

= + ¥ - = -

Considering the probability of non-ruin we require: (1) 1.5 and (2) 2.5S S< < [1]

1st period 2nd period no. of claims

claim amount

no. of claims

claim amount

probability

0 claims 0 0.5 0.5( )( ) 0.36788e e- - =

1 0.5 0.5( )(0.5 ½) 0.09197e e- - ¥ = 1 claim

2 0.5 0.5( )(0.5 ¼) 0.04598e e- - ¥ = 0 claims 0

2 claims 1, 1 20.5 0.5 20.5

2( )( 0.5 ) 0.01150e e- - ¥ =

0 claims 0 0.5 0.5(0.5 ½)( ) 0.09197e e- -¥ = 1 claim 1

1 claim 1 0.5 0.5(0.5 ½)(0.5 ½) 0.02299e e- -¥ ¥ = [5] Hence [ ( ) 0 for 1 or 2] 0.6323 [ ( ) 0 for 1 or 2] 0.3677P U t t P U t t> = = fi < = = [1]



Solution 2.33

(i) Initial capital required We first need the moments of the generalised Pareto distribution. Using the formulae in the Tables, we have:

3 500( ) 5001 3

kE X la

¥= = =-

and: 2 2 2( 2) ( 2) (2) (5)( ) 500 500,000( ) ( ) (4) (3)

kE Xk

a la

G - G + G G= = =G G G G

[1]

( ) 25 500 12,500 and var( ) 25 500,000 12,500,000E S Sfi = ¥ = = ¥ =

where S is the aggregate claim amount in one year. [1] For ruin, S must exceed the initial capital u plus the premiums received, 1.15 ( )E S . So we want: ( )1.15 ( ) 0.02P S u E S> + = Standardising this in the usual way, we have:

( ) ( ) 0.15 ( )1.15 ( )var( ) var( )

0.15 ( )(0,1) 0.02var( )

S E S u E SP S u E S PS S

u E SP NS

Ê ˆ- +> + = >Á ˜Ë ¯

Ê ˆ+ª > =Á ˜Ë ¯ [1]

Using the tables of the standard normal distribution, this gives us the equation:

0.15 ( ) 2.0537var( )

u E SS

+ =

Using the values of the mean and variance found earlier to solve this for u , we find that: 2.0537 12,500,000 (0.15 12,500) 7, 260.93 1,875 5,385.93u = - ¥ = - = So the initial capital required is £5,386. [1]



(ii) Initial capital with reinsurance We now need to allow for the reinsurance. The reinsurer takes on 30% of the risks. So the reinsurance premium is: 1.45 0.3 ( ) 5.437.5P E S= ¥ =¢ [1] The insurer retains 70% of the risks, so the amount paid out on claims by the insurer now has moments: ( ) (0.7 ) 0.7 12,500 8,750E S E S= = ¥ =¢ 2var( ) var(0.7 ) 0.7 12,500,000 6,125,000S S= = ¥ =¢ [1] So the condition for ruin is now: ( )1.15 ( ) 5, 437.5 0.02P S u E S> + - =¢ which becomes:

( ) 8,937.5 8,750 0.02var( ) 6,125,000

S E S uPS

Ê ˆ- + -¢ ¢ > =Á ˜¢Ë ¯

Again using the normal distribution tables, this gives us the equation;

8,937.5 8,750 2.05376,125,000

u + - = [1]

Solving this equation, we find that 4,895.15u = . So the initial capital required is now £4,895. [1] The initial capital required has decreased. With the reinsurer taking on more of the risks the variance of the claim amounts paid by the insurer will have reduced and hence the amount of capital the insurer needs to write the business has decreased. This may make it worthwhile for the insurer to take out reinsurance (although the effect on profits and cashflows will also need to be considered). [1]



Solution 2.34

This is Subject C2, September 1999, Question 5. First, assume that the first claim occurs at time t . We want the probability that the first claim is big enough to cause ruin, ie it is bigger than the surplus accumulated at time t . Since the premium loading factor is 0.3 and the rate of premium income is 1.3l , the probability that the first claim will cause ruin if it occurs at time t is:

(2 1.3 )2 1.3

2 1.3

x x tt

t

e dx e e ll

l

• •- - - ++

+

È ˘= - =Î ˚Ú [3]

We now consider all the times at which the first claim could occur. Since we have a Poisson process, the time to the first claim has an exponential distribution with parameter l , and the unconditional probability of ruin at the first claim is:

2

(2 1.3 ) 2.3

0 0

2.32.3

t t tee e dt e dtl l ll l• •-

- - + -=Ú Ú [2]

But since this integral is just the density function of another exponential distribution, it integrates to 1. So the unconditional probability of ruin at the first claim is:

2

0.058842.3e- = . [1]

Solution 2.35

This is Subject C2, Specimen 1993, Question 16 (part). The insurer’s expected net annual profit will be: net net( ) 100 ( )Ic E S c E Y- = - [1] where: net 1.15 ( ) 1.30 ( ) 115 ( ) 130 ( )Rc E S E S E X E Z= - = - [1] Hence, the expected annual profit is: 115 ( ) 130 ( ) 100 ( )E X E Z E Y- -



Since ( ) ( ) ( )E X E Y E Z= + we can substitute for ( )E Y to simplify to: 15 ( ) 30 ( )E X E Z- [1] Now, integrating by parts:

0.2( 5) 0.2( 5) 0.2( 5)5

5 5

0.2( 5)5

( ) 0.2

5 5

5 5 10

x x x

x

E X x e dx xe e dx

e

• ••- - - - - -

•- -

È ˘= = - +Î ˚

È ˘= + -Î ˚= + =

Ú Ú

[2] And:

0 X M

ZX M X M

<Ï= Ì - >Ó

Hence, integrating parts:

0.2( 5) 0.2( 5) 0.2( 5)

0.2( 5)

0.2( 5)

( ) ( )0.2 ( )

5

5

x x xM

M M

xM

M

E Z x M e dx x M e e dx

e

e

• ••- - - - - -

•- -

- -

È ˘= - = - - +Î ˚

È ˘= -Î ˚

=

Ú Ú

[2] So the expected annual profit is:

( )0.2( 5) 0.2( 5)15 10 30 5 150 1M Me e- - - -¥ - ¥ = - [1]

The completed table is:

Retention limit M Expected annual profit 7.5 59 10 94.8 15 129.7

24.56 147 • 150

[4]



Question 3.27

Outline the steps used in setting premiums for general insurance business. [4] Question 3.28

The transition rules for moving between the three levels, 0%, 35% and 50%, of a No Claims Discount system are as follows: If no claim is made in a year, the policyholder moves to the next higher level of discount or remains at 50%. When at the zero or 35% level of discount, the policyholder moves to (or remains at) the zero level of discount when one or more claims are made in the year. When at the maximum level of discount (50%), the policyholder moves to the 35% level of discount if exactly one claim is made during the year, and moves to the zero level of discount if two or more claims are made during the year. It is assumed that the number of claims X made each year has a geometric distribution with parameter q such that: ( ) ( )1 0,1, 2,xP X x q q x= = − = … The full premium is £350. (i) (a) Write down the transition matrix. (b) Verify that the equilibrium distribution (in increasing order of discount),

is of the following form, for some constant k:

( ) ( ) ( )( )22 2 , 1 , 1kq q kq q k q− − −

and express k in terms of q . [8] (ii) The value of the expected premium in the stationary state, paid by “low risk”

policyholders (with 0.05q = ), is £178.51. (a) Calculate the corresponding figure paid by high risk policyholders (with

0.1q = ). (b) Comment on the effectiveness of the No Claims Discount system. [4] [Total 12]



Question 3.29

The following table shows incremental claims relating to the accident years 1997, 1998 and 1999. It is assumed that claims are fully run-off by the end of Development Year 2. Estimate total outstanding claims using the chain-ladder technique, ignoring inflation.

Development Year Incremental claims 0 1 2

1997 2,587 1,091 251 1998 2,053 1,298 Accident

Year 1999 3,190

[7] Question 3.30

The table below shows the payments, in £’000s, made in successive development years in respect of a particular class of general insurance business. It may be assumed that all claims are fully settled by the end of Development Year 3 and that all payments are made in the middle of a calendar year.

Development Year Year of origin 0 1 2 3 1992 342 82 68 37 1993 359 90 73 1994 481 120 1995 591

Use the inflation adjusted chain ladder method to estimate the amount of the reserve required at the end of 1995 to pay the outstanding amounts in respect of claims from 1993, 1994 and 1995. You should assume that past inflation has been at the rate of 5% per annum and that future inflation will also be at 5% per annum. [10]



Question 3.31

The following table shows cumulative incurred claims data, by year of accident and reporting development, for a portfolio of domestic household insurance policies:

Cumulative incurred

claims (£000) Development Year

0 1 2 3 2001 829 917 978 1,020 2002 926 987 1,053 2003 997 1,098

Accident Year

2004 1,021 The corresponding cumulative number of reported claims, by year of accident and reporting development, are as follows:

Cumulative number of reported claims

Development Year

0 1 2 3 2001 63 68 70 74 2002 65 69 72 2003 71 76

Accident Year

2004 70 Use the average cost per claim method with simple average grossing up factors to calculate an estimate of the outstanding claim amount for these policies for claims arising during these accident years. The claims paid to date are £3,640,000. State any assumptions used. [11]



Question 3.32

The table below shows the cumulative costs of incurred claims. The claims are assumed to be fully run-off by the end of Development Year 2.

£000s Development Year Accident Year 0 1 2

2001 2,670 3,290 4,310 2002 2,850 3,420 2003 3,030

Annual premiums written were:

Year Premiums (£000s)

2001 5,390 2002 5,600 2003 6,030

The ultimate loss ratio has been estimated at 80% and the total amount of claims paid to date is £5,720,000. Calculate the outstanding claims reserve using the Bornhuetter-Ferguson method. [6] Question 3.33

The random variable iZ has a binomial distribution with parameters n and iμ , where 0 1iμ< < . A second random variable, iY , is defined as /i iY Z n= . (i) Show that iY is a member of the exponential family, stating clearly the natural

and scale parameters and their functions ( )a ϕ , ( )b θ and ( , )c y ϕ . [4] (ii) Determine the variance function of iY . [2] [Total 6]



Question 3.34

Independent claim amounts 1 2, , , nY Y Y… are modelled as exponential distributions with mean iμ . The fitted values for a particular model are denoted by ˆiμ . Derive an expression for the scaled deviance. [5]



Solution 3.27

The steps used in the process for setting premiums for general insurance business are:

● Choose a base period of time over which to collect data. [½]

● Collect base data subdivided by claim frequency, cost per claim and exposureper policy. [½]

● If there is no internal data, use external data, eg market statistics. [½]

● Adjust the base data, eg for trends, exceptional claims or changes in the natureof the risk/cover over the base period. [½]

● Project the base data allowing for trends, inflation and anticipated changes inrisk/cover. [½]

● The projection period for the claims data is the period from the mean paymentdate of claims in the base period and the mean payment date of claims arisingduring the exposure period of the new premiums. [½]

● Calculate the projected risk premium. [½]

● Adjust the risk premium for factors such as investment income, reinsurance,commissions, expenses, profit and contingency margins. [½]

Solution 3.28

This is Subject C2, April 1999, Question 9.

(i)(a) Transition matrix

The probabilities of obtaining no claims, one claim and more than one claim are 1 q- ,

(1 )q q- and (by subtraction) 2q respectively. So the transition matrix is:

2

1 00 1

(1 ) 1

q qq q

q q q q

Ê ˆ-Á ˜-Á ˜Á ˜- -Ë ¯

[2]



(i)(b) Equilibrium distribution

If the equilibrium distribution is the one given in the question, then the following matrixequation must hold:

( )

( )

2 2

2

2 2

1 0(2 ) (1 ) (1 ) 0 1

(1 ) 1

(2 ) (1 ) (1 )

q qkq q kq q k q q q

q q q q

kq q kq q k q

Ê ˆ-Á ˜- - - -Á ˜Á ˜- -Ë ¯

= - - -

[1]

Verifying the first column of the calculation:

[ ]2 2 2

2 2 2 2

(2 ) (1 ) (1 )

2 1 1 2 (2 )

kq q q kq q q k q q

kq q q q q q kq q

È ˘ È ˘- + - + -Î ˚ Î ˚È ˘= - + - + - + = -Î ˚ [1]

Similarly for the second column:

2 2

2 2

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

(1 ) 2 1 2 (1 )

kq q q k q q q

kq q q q q q kq q

- ¥ - + - ¥ -

È ˘= - - + - + = -Î ˚ [1]

Finally the third column:

[ ]

2

2 2

(1 ) (1 ) (1 ) (1 )

(1 ) 1 (1 )

kq q q k q q

k q q q k q

- ¥ - + - ¥ -

= - + - = - [1]

So the given distribution is kept unchanged by multiplication by the transition matrix,and so it is the equilibrium distribution.

Note the word “verify” in the question, which suggests that you can tackle the questionthis way.

To express k in terms of q , we use the fact that the three probabilities in theequilibrium distribution must sum to one:

2 2

2 2 2

(2 ) (1 ) (1 ) 11 1

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

kq q kq q k q

kq q q q q q q q

- + - + - =

fi = =- + - + - - + -

[2]



(ii)(a) Expected premium for “high risk” policyholders

Using the formula above, we can find k when 0.1q = :

2 21 1 1.08814

(2 ) (1 ) 0.1 1.9 0.9k

q q q= = =

- + - ¥ +[1]

So the equilibrium proportions are:

2

2

(2 ) 0.020675(1 ) 0.097933

(1 ) 0.881393

kq qkq q

k q

- =- =

- =

[1]

So the expected premium for “high-risk” policyholders is:

350 0.020675 350(1 0.35) 0.097933350(1 0.5) 0.881393 £183.76

¥ + - ¥+ - ¥ = [1]

(ii)(b) Comment

The no claims discount system is not very effective at discriminating between differentgroups of policyholders. Although the high risk group have twice as many claims onaverage, they only pay an additional 3% in premiums, compared with the low riskgroup. An NCD system is not generally a good discriminator between different groups,even if high discounts are given. [1]



Solution 3.29

This is Subject 106, April 2000, Question 5 (and is very generous on marks).

First of all the claim data must be accumulated to form the table below:

Development yearCumulative claimsAccident Year 0 1 2

1997 2587 3678 39291998 2053 33511999 3190

[1]

Then the development factors should be calculated:

Development factor for DY1 = 3678 3351 1.5148712587 2053

+ =+

[1]

Development factor for DY2 = 3929 1.0682443678

= [1]

The lower half of the run-off triangle can now be completed:

Development yearCumulative claimsAccident Year 0 1 2

1997 2587 3678 39291998 2053 3351 3579.681999 3190 4832.44 5162.22

[3]

So the estimates amount of outstanding claims is:

(3579.7 3351) (5162.2 3190) 2201 (4 )SF- + - = [1]



Solution 3.30

This is Subject C2, April 1996, Question 13.

Adjust for past inflation (ie obtain figures in mid 1995 prices):

Development yearIncremental claimspaid mid 95 prices (£k) 0 1 2 3

1992 395.908 90.405 71.4 371993 395.798 94.5 731994 505.05 120

Accidentyear

1995 591[2]

Accumulate the data and project figures forward using the basic chain ladder:

Development yearCumulative claims paidmid 95 prices (£k) 0 1 2 3

1992 395.908 486.313 557.713 594.7131993 395.798 490.298 563.298 600.6681994 505.05 625.05 717.469 765.067

Accidentyear

1995 591 729.961 837.892 893.48[2]

The development factors are 1.235129, 1.147858 and 1.066342. [2]

Disaccumulate the data:

Development yearIncremental claimspaid mid 95 prices(£k) 0 1 2 3

19921993 37.37051994 92.4189 47.5986

Accidentyear

1995 138.961 107.931 55.5878[1]

Adjust for future inflation (ie obtain future amounts actually paid):

Development yearIncremental claimspaid – adjusted (£k) 0 1 2 3

19921993 39.2391994 97.040 52.477

Accidentyear

1995 145.91 118.99 64.345[2]

Totalling these up the reserve that needs to be held is 518. Since the figures are inthousands, the reserve is £518,000. [1]

× 1 .235129



Solution 3.31

Dividing each cell in the first table by the corresponding cell in the second table givesthe cumulative average incurred cost per claim, by year of accident and reportingdevelopment:

Cumulative averageincurred cost per

claimDevelopment year

0 1 2 32001 13.159 13.485 13.971 13.7842002 14.246 14.304 14.6252003 14.042 14.447

Accident year

2004 14.586[2]

Completing the ultimate cumulative number of claims reported table and the ultimatecumulative average incurred cost per claim table using the chain ladder technique withsimple grossing up factors gives:

Cumulative numberof reported claims Development year

0 1 2 32001 63 68 70 742002 65 69 72 76.1142003 71 76 83.267

Accident year

2004 70 82.095[1½]

The simple average grossing up factors used are:

Development year 2 to 3: 94.595%Development year 1 to 3: 91.273% (average of 91.892% and 90.653%)Development year 0 to 3: 85.267% (average of 85.135%, 85.398% and 85.268%)

[1½]

Cumulative averageincurred cost per

claimDevelopment year

0 1 2 32001 13.159 13.485 13.971 13.7842002 14.246 14.304 14.625 14.4292003 14.042 14.447 14.669

Accident year

2004 14.586 15.093[1½]



The simple average grossing up factors used are:

Development year 2 to 3: 101.361%Development year 1 to 3: 98.487% (average of 97.834% and 99.139%)Development year 0 to 3: 96.642% (average of 95.465%, 98.736% and 95.725%)

[1½]

Therefore, the ultimate claims incurred amount from accident years 2001 to 2004 is:

1,020 + (76.114 14.429) + (83.267 14.669) + (82.095 15.093) = 4,579¥ ¥ ¥ [1]

The claims paid to date (from accident years 2001 to 2004) amount to 3,640, resultingin a total outstanding claim amount of 939, ie £939,000. [½]

Assumptions:

● Claims incurred in the first accident year are fully run off. [½]

● For each accident year, the number of reported claims in each development year,is a constant proportion of the total for that accident year. [½]

● For each accident year, the average incurred cost per claim in monetary terms ineach development year, is a constant proportion of the total claims incurred inmonetary terns for that accident year. [½]

Solution 3.32

First we calculate the expected end of year figures (the initial ultimate liability):

2001: 0.8 5,390 4,3122002 : 0.8 5,600 4,4802003: 0.8 6,030 4,824

× =× =× =

[1]

Next we calculate the development factors:

3,290 3,420 6,710 1.2155802,670 2,850 5,5204,310 1.3100303, 290

+ = =+

=[1]



The emerging liabilities for each year are:

( )( )( )

11

11.310030

11.215580 1.310030

2001: 4,312 1 0

2002 : 4, 480 1 1,060.2

2003: 4,824 1 1,794.7×

× − =

× − =

× − =

[2]

So the ultimate liabilities for each year are:

2001: 4,310 0 4,3102002 : 3, 420 1,060.2 4,480.22003: 3,030 1,794.7 4,824.7

+ =+ =+ =

[1]

Thus the total ultimate liability is:

4,310 4,480.2 4824.7 13,614.9+ + = [½]

Therefore the outstanding claims reserve is 13,614.9 5,720 7,895- = (4 SF)

ie £7,895,000 (4 SF) [½]



Solution 3.33

(i) Show Yi is member of the exponential family

The PF of iZ is:

( ) (1 )i iz n zi i i

i

nf z

zm m -Ê ˆ

= -Á ˜Ë ¯

The PF of iY can be obtained by replacing iz with iny :

( ) (1 )i iny n nyi i i

i

nf y

nym m -Ê ˆ

= -Á ˜Ë ¯[½]

This can be written as:

( ) exp ln ln ln(1 ) ln(1 )

exp ln ln(1 ) ln1

ln ln(1 )1

exp ln1

i i i i i ii

ii i

ii

ii i

i

i

nf y ny n ny

ny

nny n

ny

yn

nyn

m m m

m mm

m mm

Ï ¸Ê ˆÔ Ô= + + - - -Ì ˝Á ˜Ë ¯Ô ÔÓ ˛

Ï ¸Ê ˆ Ê ˆÔ Ô= + - +Ì ˝Á ˜Á ˜- Ë ¯Ë ¯Ô ÔÓ ˛

Ï ¸Ê ˆ+ -Ô ÔÁ ˜- Ê ˆË ¯Ô Ô= +Ì ˝Á ˜Ë ¯Ô Ô

Ô ÔÓ ˛

[1]

Comparing this to the expression on page 27 of the Tables, we see that the naturalparameter is:

ln1

ii

i

mqm

Ê ˆ= Á ˜-Ë ¯

[½]

Rearranging this gives 1

i

iie

e

q

qm =+

, so the function ( )ib q is given by:

( )1( ) ln(1 ) ln 1 ln ln 11 1

ii

i ii ieb e

e e

qq

q qq mÊ ˆ Ê ˆ= - - = - - = - = +Á ˜ Á ˜Ë ¯+ +Ë ¯

[1]



The scale parameter and its functions are:

1, ( ) , ( , ) ln lnii i

nn a c y

ny yj

j j jjj

Ê ˆ Ê ˆ= = = =Á ˜ Á ˜Ë ¯ Ë ¯

[1]

(ii) The variance function

The variance function is given by var( ) ( )bm q= ¢¢ . Differentiating ( )b q gives:

( )1 i

ebe

q

qq m= =¢+

[½]

( )( ) ( )2 2

1( )

1 1

e e e e ebe e

q q q q q

q qq

+ -= =¢¢

+ +[½]

Substituting in ln1

i

i

mqm

Ê ˆ= Á ˜-Ë ¯

gives:

22

1( ) (1 ) (1 )1

11

i

iii i i

ii

i

b

mmmq m m mmm

m

-= = ¥ - = -¢¢-Ê ˆ

+Á ˜-Ë ¯

[1]



Solution 3.34

The scaled deviance is given by:

[ ]scaled deviance 2 ln lnS ML L= - [½]

where SL is the likelihood function for the saturated model, and ML is the likelihoodfunction for the fitted model.

First we need the log-likelihoods:

1 1111

1

1 11

1

1( ) ( ) ( ) n in in

y yyi n

nL f y f y e e em mm

m mmm m

- -- Â= ¥ ¥ = ¥ ¥ =¥ ¥

[1]

Taking logs:

1ln ( ) lnii i iL ymm m= - -Â Â [½]

So the log-likelihood of the fitted model, ln ML is given by:

1ˆˆln lniM i iL ymm= - -Â Â [1]

For the log-likelihood of the saturated model, ln SL , the fitted values, ˆim will just be theobserved values, iy . Hence:

1ln ln ln 1iS i i iyL y y y= - - = - -Â Â Â Â [1]

So the scaled deviance is:

( ) ( ){ }{ }

( ){ }

1ˆ

ˆ

ˆˆ

ˆscaled deviance 2 ln 1 ln

ˆ2 ln 1 ln

2 ln 1

i

i

i

i i

i i

i i i

yi i

yy

y y

y

m

m

mm

m

m

= - - - - -

= - - + +

= - +

Â Â Â Â

Â

Â [1]

CT6: Assignment X2 Questions Page 1


Question X2.1

Claims arrive in a Poisson process rate l , and ( )N t is the number of claims arriving by time t . The claim sizes are independent random variables 1 2, ,X X … with mean m , independent of the arrivals process. The initial surplus is u and the premium loading factor is q . (i)(a) Give an expression for the surplus ( )U t at time t. (i)(b) Define the probability of ruin with initial surplus u , ( )uY , and sketch a

realisation of the surplus process that shows a ruin event. [3] (ii) The unit of currency is changed so that one unit of the old currency is worth the

same as 2.5 units of the new currency. Determine a relationship between ( )uY in (i)(b) and the probability of ruin for

the new process. [2] [Total 5] Question X2.2

For the following main categories of insurance product: Liability Damage to property (i) Describe the cover provided. [2] (ii) List two examples of insurance cover in each category. [2] [Total 4] Question X2.3

S is a compound Poisson random variable with Poisson parameter 3 and individual claim size distribution P X P X P X( ) ( ) ( ) /= = = = = =1 2 3 1 3. T is a compound Poisson random variable with Poisson parameter 2 and individual claim size distribution P X P X( ) ( ) /= = = =1 2 1 2 . S and T are independent. If U S T= + , determine the individual claim size distribution of U . [3]

CT6: Assignment X2 Questions


Question X2.4

On a portfolio of insurance policies, the claim size, Y is assumed to depend on the age of the policyholder, X . Suppose that the conditional mean and variance of Y are: ( | ) 2 400= = +E Y X x x

and 2

var( | )2xY X x= =

The distribution of X over the portfolio is assumed to be normal with mean 50 and standard deviation 14. Calculate the unconditional mean and standard deviation of Y . [5] Question X2.5

Consider a portfolio of insurance policies, on which the number of claims has a binomial distribution, with parameters n and p . The claim size distribution is assumed to be exponential with mean 1 l . 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 IS , where IS is aggregate annual claims paid by the insurer (net of reinsurance). [6] Question X2.6

A portfolio consists of two types of policies. For type 1, the number of claims in a year Claims occur according to a compound Poisson process at a rate of ¼ claim per year. Individual claim amounts, X , have PF:

( 50) 0.8( 100) 0.2

P XP X

= == =

The insurer charges a premium at the beginning of each year using a 20% loading factor. The insurer’s surplus at time t is ( )U t . Find [ (2) 0]P U < if the insurer starts at time 0 with a surplus of 100. [4]

CT6: Assignment X2 Solutions Page 5


Conditioning on the number of claims N : ( ) ( ) [ ( | )]I I

It S t S

SM t E e E E e N= =

But:

1 2 1 2

1 2

( )( | ) [ ] ( ) ( ) ... ( )

( ) ( ) ... ( ) [ ( )]

n nI

n

t Y Y Y tYt S tY tY

nY Y Y Y

E e N n E e E e E e E e

M t M t M t M t

+ + += = =

= =

!

So: ( | ) [ ( )]ItS N

YE e N M t=

and: { }[ ( | )] [ ( )]It S NYE E e N E M t= [1]

This is the same as the PGF of N , but with t replaced by ( )YM t , ie

( ) [ ( )]IS N YM t G M t= . But ( ) [(1 ) ]n

NG t p pt= - + for the binomial distribution. So:

( )( ) ( )( ) (1 ) 1I

nt M t M

SM t p p e et

l lll

- - - -È ˘Ï ¸= - + + -Ì ˝Í ˙-Ó ˛Î ˚ [2]

This is the required MGF. Note to markers. Students will probably obtain the answer to this part using the formula on page 16 of the Tables. This should be given full credit.

CT6: Assignment X2 Solutions


Solution X2.6

The premium is: (1 ) ( ) 1.2 ( ) 1.2 ¼ 60 18c E S E Xq l= + = ¥ = ¥ ¥ = [½] Hence: (2) 100 2 18 (2) 136 (2)U S S= + ¥ - = - [½] So the probability of ruin is: [ (2) 0] [136 (2) 0] [ (2) 136]P U P S P S< = - < = > [½] Considering [ (2) 136]P S < remembering that (2) ~ (2 ¼)N Poi ¥ we have:

number of claims amount of claim(s) probability

0 claims 0 0.5( ) 0.60653e- =

50 0.5(0.5 ) 0.8 0.24261e- ¥ = 1 claim

100 0.5(0.5 ) 0.2 0.06065e- ¥ =

2 claims 50, 50 2 0.5 20.5

2( ) 0.8 0.04852e- ¥ =

[½ each probability = 2] Hence [ (2) 136] 0.9583 [ (2) 0] 0.0417P S P U< = fi < = [½]

subject ct6 - actuarial education company upgrades/ct… · · 2008-10-01subject ct6 cmp upgrade...

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