CT6 CMP Upgrade 09-10

CT6: CMP Upgrade 2009/10 Page 1

The Actuarial Education Company © IFE: 2010 Examinations

Subject CT6

CMP Upgrade 2009/2010

CMP Upgrade This CMP Upgrade lists all significant changes to the Core Reading and the ActEd material since last year so that you can manually amend your 2009 study material to make it suitable for study for the 2010 exams. It includes replacement pages and additional pages where appropriate. Alternatively, you can buy a full replacement set of up-to-date Course Notes at a significantly reduced price if you have previously bought the full price Course Notes in this subject. Please see our 2010 Student Brochure for more 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 and Answer Bank that will make them suitable for study for the 2010 exams.

Page 2 CT6: CMP Upgrade 2009/10

© IFE: 2010 Examinations The Actuarial Education Company

1 Changes to the Syllabus objectives and Core Reading

1.1 Syllabus objectives Chapter 6 (new) Page 1 Syllabus objectives (v) 8 to (v) 10 are new. These read: 8. Explain the empirical Bayes approach to credibility theory, in particular its

similarities with and its differences from the Bayesian approach. 9. State the assumptions underlying the two models in 8. 10. Calculate credibility premiums for the two models in 8. Chapter 6 (now Chapter 7) Page 1 Syllabus objective (iii) 1 has been removed from Subject CT6. Syllabus objectives (iii) 2 to (iii) 7 have been renumbered accordingly. Chapter 8 (now Chapter 9) Page 1 Syllabus objective (iv) has been enhanced. Syllabus objectives (iv) 2 to (iv) 4 are new. Syllabus objectives (iv) 2 has been renumbered to (iv) 5 accordingly. Syllabus objectives (iv) 6 to (iv) 8 are new. Overall, syllabus objective (iv) now reads: (iv) Explain the concept of ruin for a risk model. Calculate the adjustment

coefficient and state Lundberg’s inequality. Describe the effect on the probability of ruin of changing parameter values and of simple reinsurance arrangements.

1. Explain what is meant by the aggregate claim process and the cash-flow process for a risk.

2. Define a Poisson process, derive the distribution of the number of events

in a given time interval, derive the distribution of inter-event times, and apply these results.



3. Define a compound Poisson process and derive the moments and moment

generating function for such a process.

4. Define the adjustment coefficient for a compound Poisson process and for discrete time processes which are not compound Poisson, calculate it in simple cases and derive an approximation.

5. Define the probability of ruin in infinite/finite and continuous/discrete

time and state and explain relationships between the different probabilities of ruin.

6. State Lundberg's inequality and explain the significance of the

adjustment coefficient.

7. Describe the effect on the probability of ruin, in both finite and infinite time, of changing parameter values.

8. Analyse the effect on the adjustment coefficient and hence on the

probability of ruin of simple reinsurance arrangements. Chapter 9 (removed) Page 1 Syllabus objective (vi) has been removed from Subject CT6. Syllabus objectives (vii) onwards have been renumbered accordingly. Chapter 10 (now Chapter 11) Page 1 Syllabus objective (vii) has been renumbered to Syllabus objective (vi). Chapter 11 (now Chapter 10) Page 1 Syllabus objective (viii) has been renumbered to Syllabus objective (vii).



Chapter 12 Page 1 Syllabus objective (ix) has been renumbered to Syllabus objective (viii). Chapter 13 Page 1 Syllabus objective (ix) has been renumbered to Syllabus objective (viii). Chapter 14 Page 1 Syllabus objective (x) has been renumbered to Syllabus objective (ix).

1.2 Core Reading Chapter 5 A few references to Core Reading have been amended as a result of the new Core Reading in the new Chapter 6.

Page 3 The following Core Reading paragraph is no longer in Subject CT6: This material is covered in detail in Subject ST3 – General Insurance Specialist Technical.

Page 8 The following Core Reading paragraph is no longer in Subject CT6: In this course we will study Bayesian credibility. Empirical Bayes credibility theory is covered in Subject ST3, General Insurance Specialist Technical.



Page 20 After the first Core Reading paragraph in Section 2.5, a new Core Reading sentence has been added: The reason for doing this is that some of the observations will be helpful when empirical Bayes credibility theory is considered in Chapter 6.

Chapter 6 (new) All of the Core Reading in this chapter is new. A replacement chapter is provided.

Chapter 6 (now Chapter 7) Pages 2 to 4 The following Core Reading paragraphs are no longer in Subject CT6: This section gives an overview of the essential features that distinguish the main types of general insurance products and so helps to identify the crucial aspects that influence the nature and extent of the risk to be covered by the insurance.

The range of general insurance products is very wide and continually changing and therefore it is difficult to set out the features of all types of product.

Firstly, it is worth noting some overriding features of all types of insurance.

The sections below provide a general indication and examples of the knowledge that examiners would expect a candidate to have in relation to the features of the major types of general insurance product. These topics are covered in more detail in Subject CA1 – Core Applications Concepts. Pages 5 to 21 All the Core Reading in Section 1.2 and Section 2 has been moved from Subject CT6 to Subject CA1. Sections 3 onwards have been renumbered accordingly. A new Core Reading paragraph has been added to reflect this change: The range of general insurance products is very wide and continually changing. They are covered in Subject CA1 – Actuarial Risk Management.



Page 25 At the bottom of this page, a new Core Reading paragraph has been added. It reads: There are a number of additional elements included when setting the premium to be charged to policyholders, including the policyholders’ previous claims record and these are covered in Subject CA1 – Actuarial Risk Management. The allowance for policyholders’ claim experience could be based on claim frequency or total claim amounts. This is beyond the scope of CT6 but a specific example of how policyholder experience can be allowed form on a claim frequency basis, is covered in Subject CT4 – Models.

Chapter 8 (now Chapter 9) The Core Reading in this chapter has been enhanced considerably. A replacement chapter is provided.

Chapter 9 This chapter has been removed because the Core Reading in it is no longer covered in Subject CT6.



2 Changes to the ActEd Course Notes Chapter 5 A new exam-style question has been added to the end of the chapter. This is provided in the replacement pages.

Chapter 6 (new) This chapter is new and is provided in the replacement pages.

Chapter 6 (now Chapter 7) Section 1.2 and Section 2 has been moved from Subject CT6 to Subject CA1. Sections 3 onwards have been renumbered accordingly.

Page 2 The introduction has been trimmed to reflect the fact that the chapter has become smaller.

Chapter 8 (now Chapter 9) This chapter has been enhanced considerably due to additional Core Reading. A replacement chapter is provided.

Chapter 8 (now Chapter 9) A new exam-style question has been added to the end of the chapter. This is provided in any case as this chapter is in the replacement pages.

Chapter 9 This chapter has been removed because the Core Reading in it is no longer covered in Subject CT6.

Chapters 10 and 11 These two chapters have been swapped round. This is purely to reflect their relative levels of difficulty when it comes to running 3-Day CT6 tutorials.

Chapter 11 (now Chapter 10) A new exam-style question has been added to the end of the chapter. This is provided in the replacement pages.



Chapter 14 A new exam-style question has been added to the end of the chapter. This is provided in the replacement pages.



3 Changes to the Q&A Bank The Questions in the Question and Answer Bank have been annotated according to whether they are: bookwork questions

developmental questions or

exam-style questions.

This change is not sufficient to provide replacement pages but here a summary of the questions and what type they each are is provided with the replacement pages. Q&A Bank Part 1 Solution 4.18 (i) There was a minor typo in the calculations here. The first two paragraphs should now read:

First we need to obtain random variates from ~ ( 1,1)V U - . Since the distribution

function is 12( ) ( 1)F v v= + , we have 2 1v u= - using the inverse transform method.

Hence, 1 0.1834v = , 2 0.5872v = and 2 21 2 0.37844s v v= + = .

Solution 5.12 (ii)(a) There was a minor typo in the calculations here. The second covariance term should be 7

12 and not 13

12. The first two paragraphs should now read:

First note that:

( )cov , var( , ) 1t t t tZ X Z Z= =

( )1 1 1 15 1 7cov , cov( , ) var( )6 4 12t t t t tZ X X Z Z- - - -= - =



Q&A Bank Part 2 Questions 2.7 to 2.9 have been deleted.

Questions 2.6 to 2.10 are new and attached as replacement pages at the end.

Questions 2.6 is now Question 2.11. Questions 2.10 to 2.24 are now Questions 2.11 to 2.26.

Question 2.25 has been removed. Questions 2.26 to 2.32 have been moved to Q&A Bank Part 3. They are now Questions 3.1 to 3.7

Q&A Bank Part 3 Questions 3.1 to 3.7 have been replaced as they related to material that is no longer tested in Subject CT6. They have been replaced by Questions 2.26 to 2.32 from Q&A Bank Part 2.

Questions 3.8 to 3.18 are now Questions 3.19 to 3.29.

Questions 3.19 to 3.24 are now Questions 3.13 to 3.18.

Questions 3.8 to 3.12 are new and attached as replacement pages at the end.



4 Changes to the X Assignments

Assignment X1 Solution X1.5 This solution has been updated to be more efficient, since we observe strategy 3d is dominated. Replacement pages have been provided. Solution X1.10 (iii) & (iv) The notation here has changed. The confusing Ad , Bd and Cd have been replaced with 1d , 2d and 3d .

Solution X1.10 (iii) The minimax solution here should be 2d and not 3d , Bd or Cd . This solution now reads: The minimax criterion will select the risk function with the smallest maximum loss. This is 2d . Assignment X2 Question X2.1 This has been replaced with a new question. The new question is attached with the replacement pages. Question X2.2 This has been replaced with Question X2.6. Questions X2.7 to X2.10 have been renumbered X2.6 to X2.9



Solution X2.8 (now Solution X2.7) More explanation has been added to this solution after the first paragraph. The new paragraph reads: It is also important to note here that ES and CS are NOT independent because the

expenses depend on the claims. Any approach using [ ] [ ] [ ]var var varC ES S S= + is doomed to fail.

Assignment X3 Question X3.1 This has been removed. Questions X3.2 and X3.3 are now Questions X3.1 and X3.2. Question X3.4 This has been removed. Questions X3.5 to X3.8 are now Questions X3.3 to X3.6. Question X3.9 This has been removed. Questions X3.7 and X3.8 are new and replacement pages have been provided. Assignment X4 Question X4.1 This question has been rephrased to be more specific as to what is required. It now reads: Consider the following time series:

215 1 ( 5)t t tX Z X -= + + -

where tZ is a white noise process with mean 0 and standard deviation 1. Describe how tX fluctuates and state the long-term mean of the process.



5 Other tuition services In addition to this CMP Upgrade you might find the following services helpful with your study.

5.1 Study Material We offer the following study material in Subject CT6: • Series Y Assignments • Mock Exam 2009 and Mock Exam 2010 • ASET (ActEd Solutions with Exam Technique) and mini-ASET • Revision Notes • Sound Revision • Flashcards • Smart Revise. For further details on ActEd’s study materials, please refer to the 2010 Student Brochure, 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 two or three full days) • a Block Tutorial (lasting two or three full days) • a Revision Day (lasting one full day). For further details on ActEd’s tutorials, please refer to our latest Tuition Bulletin, which is 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 by ActEd. When marking your scripts, we aim to provide specific advice to improve your chances 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 2010 Student Brochure, 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 study programmes. Please let us know if you have any specific comments (eg about certain sections of the notes or particular questions) or general suggestions about how we can improve the study material. We will incorporate as many of your suggestions as we can when we update the course material each year. If you have any comments on this course please send them by email to [email protected] or by fax to 01235 550085.



Q&A Bank Question Classification Summary The first 4 parts of the Q&A Bank have been split into bookwork questions, developmental questions and exam-style questions. Part 5 consists only of exam-style questions. Please note that the following grid refers to the numbering in the 2009 Q&A Bank.

Question Part 1 Part 2 Part 3 Part 4 1 Developmental Developmental Deleted Developmental 2 Developmental Exam-style Deleted Developmental 3 Exam-style Exam-style Deleted Developmental 4 Developmental Exam-style Deleted Exam-style 5 Exam-style Exam-style Deleted Developmental 6 Developmental Bookwork Deleted Developmental 7 Deleted Bookwork Deleted Developmental 8 Deleted Bookwork Bookwork Developmental 9 Deleted Bookwork Developmental Developmental 10 Developmental Developmental Developmental Developmental 11 Exam-style Developmental Developmental Developmental 12 Exam-style Developmental Exam-style Developmental 13 Exam-style Developmental Exam-style Developmental 14 Exam-style Developmental Exam-style Developmental 15 Exam-style Exam-style Exam-style Exam-style 16 Developmental Exam-style Exam-style Developmental 17 Developmental Developmental Exam-style Developmental 18 Exam-style Exam-style Exam-style Developmental 19 Exam-style Exam-style Exam-style Exam-style 20 Exam-style Exam-style Exam-style Developmental 21 Exam-style Exam-style Developmental Developmental 22 Developmental Exam-style Exam-style Bookwork 23 Exam-style Developmental Developmental Exam-style 24 Exam-style Developmental Exam-style Developmental 25 Exam-style Deleted Exam-style 26 Developmental Developmental 27 Developmental Developmental 28 Developmental Exam-style 29 Exam-style Exam-style 30 Exam-style Exam-style 31 Developmental Exam-style 32 Exam-style Exam-style 33 Exam-style 34 Exam-style


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

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CT6-05: Credibility theory Page 25


3 Exam-style questions

Exam-style question 1 A statistician wishes to find a Bayesian estimate of the mean of an exponential

distribution with density function /1( ) xf x e mm

-= . He is proposing to use a prior

distribution of the form:

/

1( )( )

epriora q m

aqmm a

-

+=G

, 0m >

You are given that the mean of this distribution is 1

qa -

.

(i) Write down the likelihood function for m , based on a random sample of values

1, , nx x… from an exponential distribution. [1] (ii) Find the form of the posterior distribution for m , and hence show that an

expression for the Bayesian estimate for m under squared error loss is:

ˆ1ix

nq

ma

+=

+ -Â [4]

(iii) Show that the Bayesian estimate for m can be written in the form of a credibility

estimate, and write down a formula for the credibility factor. [3] (iv) The statistician now decides that he will use a prior distribution of this form with

parameters 40q = and 1.5a = . His sample data have statistics 100n = , 9,826ix =Â , and 2 1,200,000ix =Â . Find the posterior estimate for m , and

the value of the credibility factor in this case. [3] (v) Comment on the results obtained in part (iv). [2] [Total 13]

Page 26 CT6-05: Credibility theory


Exam-style question 2 The annual number of claims from a particular risk has a Poisson distribution with mean m . The prior distribution for m has a gamma distribution with 2a = and 5l = . Claim numbers 1, , nx x… over the last n years have been recorded. (i) Show that the posterior distribution is gamma and determine its parameters. [3]

(ii) Given that 8n = and 8

15i

ix

==Â determine the Bayesian estimate for m under:

(a) squared-error loss (b) “all-or-nothing” loss (c) absolute error loss. [5] [Total 8]



Solution to exam-style question 1 (i) Likelihood function The likelihood function is:

1/

//1 1( )i

nx

xxn

eL e em

mmmm m m

---

Â= ¥ ¥ =

(ii) Posterior distribution The posterior distribution for m is proportional to the prior distribution multiplied by the likelihood function. So we have (ignoring any terms not involving m ):

( ) ///

1 1( )ii xx

n ne e ePost

q mmq m

a amm m m

- +--

+ + +

ÂÂμ ¥ =

We see that this has the same form as the prior distribution, but with different parameters. So we have the same distribution as before, but with parameters: * ixq q= +Â and: * na a= + We can now use the formula for the mean of the distribution given in the question:

*( )* 1 1

ixE

nqqm

a a+

= =- + -

Â

This is the Bayesian estimate for m under squared error loss.



(iii) Credibility estimate We now have to show that this can be written in the form of a credibility estimate. Splitting the Bayesian estimate into two parts:

ˆ1 1 1

11 1 1

i i

i

x xn n n

x nn n n

q qma a a

q aa a a

+= = +

+ - + - + --= ¥ + ¥

- + - + -

Â Â

Â

We can see from this that m can be written as a weighted average of the prior mean

(1

qa -

) and the maximum likelihood estimator for m (the sample mean ixnÂ ). So the

Bayesian estimate m can be written in the form of a credibility estimate, and the credibility factor is:

1

nZn a

=+ -

(iv) Posterior estimate We now have:

40 9,826ˆ 98.16921 100 1.5 1ix

nq

ma

+ += = =+ - + -Â

The value of the credibility factor is:

100 0.99501 100 1.5 1

nZn a

= = =+ - + -

(v) Comments We see that the value of Z is very close to 1. So our credibility estimate comes out to be very close to the sample mean (98.26), and takes little account of the prior mean (80). This is because n is nuch bigger than a . We would use a prior like this if we are not very sure initially about the true value of m . We have chosen a prior distribution with a large variance, to reflect our high initial degree of uncertainty about m .



Solution to exam-style question 2 (i) Poisson/gamma posterior The likelihood is given by:

1

1( )

! !

ni

xxx n

nL e e e

x xm m mm mm m- - -Â= ¥ ¥ μ…

The prior distribution of ~ (2,5)gammam is given by:

2

5 55( )(2)

f e em mm m m- -= μG

So the posterior is given by:

5

1 (2 ) 1( 5) ( 5)

i

i i

x n

x xn n

posterior prior likelihood

e e

e e

m m

m m

m m

m m

- -

+ + -- + - +

μ ¥

Â= ¥

Â Â= = This is the form of a (2 , 5)igamma x n+ +Â distribution. (ii)(a) Squared-error loss We have 8n = and 5ix =Â , so the posterior is (7,13)gamma . The Bayesian estimate for m under squared-error loss is the mean of the posterior:

7ˆ 0.53813

m =

(ii)(b) “All-or-nothing” loss The Bayesian estimate under “all-or-nothing” loss is given by the mode of the posterior. To find the mode, we need to differentiate the PDF (or equivalently differentiate the log of the PDF) and equate it to zero. The PDF of a (7,13)gamma is given by: 6 13( ) constant ×f e mm m -=



Taking logs, differentiating and equating the derivative to zero gives: ln ( ) ln 6ln 13f km m m= + -

6ln ( ) 13 0d fd

mm m

fi = - =

6ˆ 0.46213

mfi =

Alternatively, students may differentiate the original PDF to give:

5 13 6 13

5 13

( ) 6 13

(6 13 )

d f e ed

e

m m

m

m m mm

m m

- -

-

= -

= -

Setting this equal to zero gives 6

130,m = . The 0m = solution can be discarded as

~ (7,13)gammam means that 0m > .

In general the mode of the gamma distribution is 1al- (provided that 1a > ).

(ii)(c) Absolute error loss The Bayesian estimate under absolute error loss is given by the median, M , of the posterior. Hence ( ) ½P Mm < = . Using the gamma-chi squared relationship on page 12 of the Tables, we get: 2

14( ) (2 13 2 13 ) ( 26 ) 0.5P M P M P Mm m c< = ¥ < ¥ = < = Using the chi-square percentage points on Page 169 of the Tables, we get:

13.3426 13.34 0.51326

M M= fi =

CT6-06: Emperical Bayes Credibility theory Page 1


Chapter 6

Emperical Bayes Credibility theory

Syllabus objectives (v) Explain the fundamental concepts of Bayesian statistics and use these concepts

to calculate Bayesian estimators. 8. Explain the empirical Bayes approach to credibility theory, in particular

its similarities with and its differences from the Bayesian approach. 9. State the assumptions underlying the two models in 8. 10. Calculate credibility premiums for the two models in 8.

0 Introduction

In this chapter, we will study Empirical Bayes Credibility Theory (EBCT) Models 1 and 2. The algebra for the EBCT models that we will study in this unit is fairly heavy going. Try not to get too bogged down in the mathematics and don’t learn the formulae. (Most of them are in the Tables.) Concentrate on appreciating the differences between the models and being able to follow the numerical examples.

Page 2 CT6-06: Emperical Bayes Credibility theory


1 Empirical Bayes Credibility Theory: Model 1

1.1 Introduction

In this chapter we will discuss two Empirical Bayes Credibility Theory (EBCT) models. As with the Poisson/gamma and normal/normal models discussed in the previous chapter, these models are used to estimate the “true” claim frequency or risk premium based on the total claim amounts in successive periods. Model 1 gives equal weight to each risk in each year. Model 2 is more sophisticated and takes into account the volume of business written under each risk in each year. The main similarities and differences between the Bayesian models and the EBCT models are outlined below. Risk parameter Both approaches use an auxiliary risk parameter θ . With the Bayesian approach, the quantity we wish to estimate is θ , whereas, for the EBCT models, it is m( )θ , ie a function of θ . Unlike the Bayesian approach, the EBCT models don’t assume any specific statistical distribution for θ . Conditional claim distribution Both approaches assume that the conditional variables X j θ s are independent and identically distributed. Unlike the pure Bayesian approach, the EBCT models don’t assume any specific statistical distribution for X j θ . Instead, formulae are developed

just using the assumption that the mean m( )θ and variance s2 ( )θ of X j θ can be

expressed as functions of θ . The values of m( )θ and s2 ( )θ are then estimated by the models. Credibility formula With both approaches, the resulting formula for estimating the claim frequency or risk premium for a given risk can be expressed using a credibility formula, ie a linear combination (= weighted average) of the average derived from past claims and an overall average.



Empirical Bayes Credibility Theory is the name given to a particular approach to the problems in Section 1 of the previous chapter. This approach has led to the development of a vast number of different models of varying degrees of complexity. In this chapter, two of these models will be studied. In this section the simplest possible model will be studied. Although this model, which will be referred to as Model 1, is not very useful in practice, it does provide a good introduction to the principles underlying Empirical Bayes Credibility Theory (EBCT). In particular, it shows the similarities and the differences between the Empirical Bayes and the pure Bayesian approaches to credibility theory. In Section 2 an extension of Model 1 will be studied that is far more useful in practice.

1.2 Model 1: specification

In this section the assumptions for Model 1 of EBCT will be set out. EBCT Model 1 can be regarded as a generalisation of the normal/normal model. This point will be considered in more detail later in this section. The problem of interest is the estimation of the pure premium, or possibly the claim frequency, for a risk. Let X X1 2, ,… denote the aggregate claims, or the number of claims, in successive periods for this risk. A more precise statement of the problem is that, having observed the values of X X Xn1 2, , ,… , the expected value of Xn+1 needs to be estimated. From now on X X Xn1 2, , ,… will be denoted by X . The following assumptions will be made for EBCT Model 1.

Assumptions for EBCT Model 1 The distribution of each X j depends on a parameter, denoted θ , whose value is

fixed (and the same for all the X j s) but is unknown. (1.1) Given θ , the X j ’s are independent and identically distributed. (1.2)

The parameter θ is known as the risk parameter. It could, as in Section 2 of the previous chapter, be a real number or it could be a more general quantity such as a set of real numbers. The risk parameter in the Poisson/gamma model is (the Poisson parameter) λ . The risk parameter for the normal/normal model is θ .



A consequence of these two assumptions is that: the random variables X jn s are identically distributed. (1.3)

An important point to note is that: the jX ’s are not (necessarily) unconditionally independent. (1.4) The above assumptions and consequences were all either made or noted for the normal/normal model of Bayesian credibility in Section 2.5 of the previous chapter. See (2.17), (2.18), (2.19), (2.23) and (2.24) in the previous chapter. Note that (1.1) corresponds to (2.17), (1.2) corresponds to (2.18) and (2.19), (1.3) corresponds to (2.23), and (1.4) corresponds to (2.24). Next some notation is introduced. Define ( )m q and 2 ( )s q as follows: =( ) ( | )jm E Xq q =2 ( ) var ( | )js Xq q Two things should be noticed about ( )m q and 2 ( )s q . The first is that since,

given q , the jX ’s are identically distributed, neither ( )m q nor 2 ( )s q depends on j , as their notation suggests. The second is that since q is regarded as a

random variable, both ( )m q and 2 ( )s q are random variables. Note also that they are functions of the unknown parameter θ . If the value of q and the distribution of jX given q were known, the obvious estimate of the expected aggregate claims, or the expected number of claims, in any future year would be ( )m q . Since it is assumed that q is not known, the problem is to estimate: ( )m q given X (1.5)



Example A specialist insurer concentrates on providing third party motor insurance to drivers of a certain type of car who are based in London. Explain what each of the variables and functions in EBCT Model 1 would represent if this model were used to describe the numbers of claims made by different drivers in different years. The only risk factor considered relevant is the safety standard adopted by each individual driver. Solution Here, the risk parameter q represents the “safety coefficient” of a particular driver. It is assumed that each driver has a constant inherent level of safety, which could (in theory) be measured by some means. The value of q for a particular driver influences the likely number of claims that the driver will make. For example, a value of 1q = may correspond to a driver who is “100% safe” (ie never makes any claims), while a value of

0q = may correspond to a driver who is “0% safe” (ie makes a claim every time the car is used). The distribution of the values of q , which will vary from one driver to the next, is assumed to be determined by a definite (but unknown) probability distribution.

jX is a random variable, representing the number of claims made by a particular driver in year j .

jX q represents the number of claims made by a particular driver whose safety coefficient is q in year j .

( ) ( )jm E Xq q= is the (theoretical) average number of claims made by a driver whose safety coefficient is q . In this example, ( )m q would be a decreasing function of q , since a higher safety coefficient would reduce the average number of claims.

2 ( ) var( )js Xq q= is the (theoretical) variance of the number of claims made by a driver

whose safety coefficient is q in different years. 2 ( )s q will take lower values for drivers with a high safety coefficient, since they are likely to make no claims, or possibly one claim, each year, whereas the numbers of claims made each year by drivers with low safety coefficients may range from none to five or more.



Question 6.1

Describe what each of the variables and functions in EBCT Model 1 would represent if this model was used to study the scores of different golf players on different days, if the scores for the players are considered to be dependent on their “handicaps”. (For those of you who are not golfers, a player’s handicap is a measure of his or her golfing ability. A golfer with a low handicap is a better player than a golfer with a high handicap. A good golfer will achieve a lower score on a round of golf than a bad golfer.)

The similarities between EBCT Model 1 and the normal/normal model can be summarised as follows. (i) The role of θ is the same for both models: it characterises the underlying

distributions of the processes being modelled, eg the aggregate claim distribution for each year of business. See (2.17) from the previous chapter and (1.1).

(ii) Assumptions concerning the unconditional distribution of the X j ’s are

the same: they are identically distributed in each case. See (2.23) and (2.24) from the previous chapter, and (1.3) and (1.4).

(iii) Assumptions concerning the (conditional) distribution of the X j ’s given

θ are the same: they are conditionally independent in each case. See (2.19) and (2.22) from the previous chapter, and (1.2).

EBCT Model 1 can be regarded as a generalisation of the normal/normal model. The particular points where it differs from, ie generalises, the normal/normal model are: (i) ( | )jE X q is some function of q , ( )m q , for EBCT Model 1 but is simply q

for the normal/normal model. Hence: [ ]var ( )m q EBCT corresponds to 2

2s ( = var ( )q ) normal/normal. (ii) var ( | )jX q is a function of q , 2 ( )s q , for EBCT Model 1 but is a

constant, 21s , for the normal/normal model. Hence:

È ˘

Î ˚2 ( )E s q corresponds to 2

1s ( ( )= var |jX q ).



(iii) The normal/normal model makes very precise distributional assumptions about both jX given q , which is 2

1( , )N q s , and q , which is 22( , )N m s .

EBCT Model 1 makes no such distributional assumptions. (iv) The risk parameter, q , is a real number for the normal/normal model but

could be a more general quantity for EBCT Model 1.

1.3 Model 1: the credibility premium

In the previous section the assumptions for Model 1 of EBCT were studied and the similarities between this model and the normal/normal model were emphasised. In this section a solution to the problem summarised in (1.5) will be considered, that is, the estimation of ( )m q given the data X . The derivation of the credibility premium under this model is beyond the scope of the CT6 syllabus and is not covered here.

Question 6.2

Without looking back to the previous chapter, write down the general credibility formula.

Result of EBCT Model 1 The estimate of ( )m q given X given by EBCT Model 1 is: ( ) [ ]- +1 ( )Z E m Z Xq where:

X X njj

n=

=∑

1/

and:

( )

=È ˘ È ˘+ Î ˚Î ˚

2 ( ) varnZ

n E s mq q (1.6)



The first, and most important, point to note about the solution is that it is in the form of a credibility estimate. In other words, it is in the form of formula (1.1) from the previous chapter with: [ ]( )E m q playing the role of m and

=Â

1/

n

jj

X n playing the role of X .

The second point to note is the similarity between the solution above and the solution in the normal/normal model, and, in particular, the similarity between the formulae for the credibility factors, ie (2.16) from the previous chapter and (1.6). Formula (1.6) is a straight generalisation of formula (2.16) from the previous chapter, since È ˘

Î ˚2 ( )E s q and [ ]var ( )m q can be regarded as

generalisations of 21s and 2

2s , respectively. We saw this earlier. The similarities between EBCT Model 1 and the normal/normal model lead to the similarity in the resulting answers. The final point to note is that the formula for the credibility estimate given by (1.6) involves three parameters, [ ]q( )E m , qÈ ˘

Î ˚2 ( )E s and [ ]qvar ( )m , which have

been treated so far as though they were known. EBCT makes no distributional assumptions about the risk parameter θ , unlike the Bayesian approach to credibility, but the values of these three parameters need to be known. It can be noted that these three parameters relate to first and second order moments as opposed to, say, higher order moments or some other quantities. This is due to the derivation of the credibility premium and the detail behind this is beyond the scope of this course. The way in which these parameters would be estimated is discussed in the next section.



1.4 Model 1: parameter estimation

In this section the solution to the problem of estimating ( )m q given X will be

completed by showing how to estimate [ ]( )E m q , [ ]var ( )m q and È ˘Î ˚

2 ( )E s q .

To do this a further assumption is needed: that there are available data from some other risks similar, but not identical, to the original risk. This requires the problem to be set up a little more generally, to make some extra assumptions and to change the notation slightly. An important distinction between a pure Bayes approach to credibility and EBCT is that no data are required to estimate parameters in the former case but data are required to estimate the parameters in the latter. What is of interest is estimating the pure premium, or the expected number of claims, for a particular risk, just as in Section 1.2 and Section 1.3, and that this risk is one of N risks in a collective. By a collective is meant a collection of different risks that are related in a way to be made clearer later in this section. For simplicity, suppose the particular risk that is of interest is Risk Number 1 in this collective. It is assumed that the aggregate claims, or claim frequencies, for each of these N risks for each of the past n years have been observed. Let Xij denote the aggregate claims, or number of claims, for Risk Number i , i N= 1 2, , ,… , in year j , j n= 1 2, , ,… . These values are summarised in the following table:

Table 1

Year 1 2 ... n Risk 1 X11 X12 ... X n1 Number 2 X21 X22 ... X n2 .

.

. .

. .

... ...

. .

N XN1 XN2 ... XNn The risk numbers could refer to risks from different companies, insurers, shops etc. It is helpful to remember that small n represents the number of years while big N represents the number of risks.



Each row of Table 1 represents observations for a different risk; the first row, relating to Risk Number 1, is the set of observed values denoted X in Section 1.2 and Section 1.3. (Notice that X X Xn1 2, , ,… in Section 1.2 and Section 1.3 have now become X X X n11 12 1, , ,… in this section.) In Section 1.2 two assumptions, (1.1) and (1.2), were made about the connection between the observed values for the single risk then being considered. In this section exactly the same assumptions for each of the N risks in the collective are made. These assumptions are given as (1.7) and (1.8) below. For each risk i , = 1, 2, ,i N… : The distribution of each ijX , = 1, 2, ,j n… , depends on the value of a parameter,

denoted iq , whose value is fixed (and the same for each value of j ) but is unknown. (1.7) Given θ i , the Xij ’s, j n= 12, , ,… , are independent and identically distributed. (1.8) Notice that the risk parameter, which was denoted θ in Section 1.2 and Section 1.3, is now denoted θ i , and that an implication of these two assumptions is that the risk parameters for different risks have different values. (However, as in Section 1.2 and Section 1.3, the risk parameter for a given risk does not change value from year to year.) These two assumptions show something about the relationships within each row of Table 1 but they do not show anything about the relationship between different rows, ie between different risks in the collective. The assumption that shows something about the relationship between different risks in the collective is the following. For πi k , the pairs ( , )i ijXq and ( , )k kmXq are independent and identically distributed. (1.9) This assumption shows that the rows of Table 1 are independent of each other. Two immediate consequences of this assumption are: For πi k , ijX and kmX are independent and identically distributed. (1.10) The risk parameters θ θ θ1 2, , ,… N are independent and identically distributed. (1.11)



The connection between the different risks, ie rows of the table, is a result of the assumption that the risk parameters, θ θ θ1 2, , ,… N , are identically distributed. Intuitively, this means that if, by some means, the values of θ θ θ2 3, , ,… N were known, then something about the common distribution of the θ i ’s would be known and hence something about θ1, or, at least, about the distribution it comes from, would be known. The functions ( )m and 2 ( )s were introduced in Section 1.2. Keeping the same definitions for these functions and applying them to all the risks in the collective: ( )=( ) |i ij im E Xq q

( ) ( )=2 var |i ij is Xq q

Notice that, as in Section 1.2, neither ( )|ij iE X q nor ( )var |ij iX q depends on j

since, given iq , the random variables 1 2, ,i i inX X X… are identically distributed. Notice also that, since 1 2, , , Nq q q… are identically distributed, [ ]( )iE m q ,

È ˘Î ˚

2 ( )iE s q and [ ]var ( )im q do not depend on i . These are precisely the

parameters denoted [ ]( )E m q , È ˘Î ˚

2 ( )E s q and [ ]var ( )m q in Section 1.2 and

Section 1.3 and which the collective will be used to estimate. More notation is needed. Denote:

=Â

1/

n

ijj

X n by iX

and:

= = =

Ê ˆÁ ˜=Á ˜Ë ¯

Â Â Â1 1 1

/ /( )N N n

i iji i j

X N X Nn by X .

Notice that what is now denoted Xi was denoted X in Section 1.2 and Section 1.3. It is important to recognise the difference between the meanings of the symbol X in this section and in earlier sections.

iX denotes the sample mean of the data from Risk Number i and 1 NX XXN

+=

denotes the sample mean of the data from all of the risks.



This new notation will be used to reformulate the credibility estimate of the pure premium, or the number of claims, for the coming year for Risk Number 1 in the collective as: ( ) [ ]- + 11 ( )Z E m Z Xq where:

=

= Â1 11

/n

jj

X X n

and:

[ ]

=È ˘+ Î ˚

2 ( ) var ( )nZ

n E s mq q (1.12)

It is important to realise that formula (1.12) is exactly the same as formula (1.6) but in this case the notation used for the data from the risk itself is X1 and X j1

rather than X and X j . Estimators for [ ]( )E m q , È ˘

Î ˚2 ( )E s q and [ ]var ( )m q can now be produced.

These estimators will be functions of = =1 1

Nnij j i

X , whose values will be known

when the credibility estimate of 1( )m q is computed. Each row of Table 1 corresponds to a fixed value of q . Bearing this and the definitions of ( )im q and 2 ( )is q in mind, obvious estimators for ( )im q and

2 ( )is q are:

iX and -

=- -Â1 2

1( 1) ( )

n

ij ij

n X X

respectively. So we estimate ( )qim by the sample mean of the values for Risk Number i and we

estimate ( )2 qis by the sample variance of the values for Risk Number i .



Now, [ ]( )E m q is the “average” (over the distribution of q ) of the values of

( )m q for different values of q . The obvious estimator for [ ]( )E m q is the average of the estimates of ( )im q for = 1, 2, ,i N… . In other words, the

estimator for [ ]( )E m q is X . So we estimate [ ]( )E m q by the average of the sample means 1,..., NX X . This is equivalent to the sample mean of the full set of (both direct and collateral) data. Similarly, È ˘

Î ˚2 ( )E s q is the “average” value of 2 ( )s q and so an obvious

estimator is the average of the estimates of 2 ( )is q , which is:

( ) ( )--

= =- -Â Â 211

1 11

N n

ij ii j

N n X X

Note that this is a “nested” sum, ie it is ( ) ( )211

1 11 --

= =

Ï ¸Ô Ô- -Ì ˝Ô ÔÓ ˛

Â ÂN n

ij ii j

N n X X . It is not two

sums multiplied together. For each row of Table 1, iX is an estimate of ( )im q , = 1,2, ,i N… . So it might be thought that the observed variance of these values, ie:

( ) ( )-

=- -Â 21

11

N

ii

N X X

would be an obvious estimator for [ ]var ( )m q . Unfortunately, this can be shown

to be a biased estimate of [ ]var ( )m q . In fact it is positively biased, ie it tends to overestimate slightly. It can be shown that an unbiased estimate of [ ]var ( )m q can be produced by subtracting a correction term from the above formula. An unbiased estimator for

[ ]var ( )m q is:

( ) ( ) ( ) ( )- --

= = =- - - - -Â Â Â 221 11

1 1 11 ( ) 1

N N n

i ij ii i j

N X X Nn n X X

The correction term is -1n times the estimator for È ˘

Î ˚2 ( )E s q .



This last observation can simplify the working in numerical calculations. Just make sure that you use 1/ n , where n is the number of data values for each risk, rather than 1/ N in your calculation. These estimators can be summarised as follows: Parameter Estimator [ ]E ( )m q X (1.13)

È ˘Î ˚

2 ( )E s q ( ) ( )--

= =- -Â Â 211

1 11

N n

ij ii j

N n X X (1.14)

[ ]var ( )m q ( ) ( ) ( ) ( ) ( )- - -

= = =- - - - -Â Â Â 221 1 1

1 1 11 1

N N n

i ij ii i j

N X X Nn n X X

(1.15) These formulae are given in the Tables. They are on Page 29. We will do some numerical examples shortly to help you see what is going on. An important point to note is that although [ ]var ( )m q is a non-negative parameter since it is a variance, the estimator given by (1.15) could be negative. Formula (1.15) is the difference between two terms. Each of these terms is non-negative but their difference need not be. It could be that the second term works out to be bigger than the first. If, in practice, (1.15) gives a negative value, the accepted procedure is to estimate [ ]var ( )m q as 0. Strictly speaking, this means that the estimator for

[ ]var ( )m q is the maximum of 0 and the value given by (1.15). Although (1.15)

gives an unbiased estimate of [ ]var ( )m q , taking the maximum of 0 and (1.15) does not give an unbiased estimate. However, this pragmatic approach avoids a nonsensical estimate for [ ]var ( )m q . The parameter È ˘

Î ˚2 ( )E s q must also be non-negative, but its estimator, given by

formula (1.14), will always be non-negative and so no adjustment to (1.14) is required. It can be shown that the estimators for [ ]( )E m q , È ˘

Î ˚2 ( )E s q and [ ]var ( )m q are

unbiased. The proof is beyond the scope of the syllabus.



Now consider formula (1.6) for the credibility factor for EBCT Model 1. The different “ingredients” of this formula have the following definitions or interpretations: n is the number of data values in respect of the risk. Usually this will correspond to the number of years of data available. È ˘Î ˚

2 ( )E s q is the average variability of data values from year to year for a

single risk, ie the average variability within the rows of Table 1

[ ]var ( )m q is the variability of the average data values for different risks, ie the variability of the row means in Table 1.

Looking at formula (1.6) for Z , the following observations can be made: (i) Z is always between zero and one. (ii) Z is an increasing function of n . This is to be expected – the more data

there are from the risk itself, the more it will be relied on when the credibility estimate of the pure premium or number of claims is calculated.

(iii) Z is a decreasing function of È ˘

Î ˚2 ( )E s q . This is to be expected – the

higher the value of È ˘Î ˚

2 ( )E s q , relative to [ ]var ( )m q , the more variable,

and hence less reliable, are the data from the risk itself relative to the data from the other risks in the collective.

(iv) Z is an increasing function of [ ]var ( )m q . This is to be expected – the

higher the value of [ ]var ( )m q , relative to È ˘Î ˚

2 ( )E s q , the more variability

there is between the different risks in the collective and hence the less likely it is that the other risks in the collective will resemble the risk that is of interest, and the less reliance should be placed on the data from these other risks.



There appears to be a contradiction in this section. In Section 1.2 of the previous chapter it was stated that the credibility factor should not depend on the data from the risk being rated. However, these data have been used to estimate [ ]var ( )m q and È ˘

Î ˚2 ( )E s q , whose values are then used to calculate Z .

The explanation is that in principle the credibility factor, Z , as given by formula (1.12), does not depend on the actual data from the risk being rated. Unfortunately, the formula for Z involves two parameters, [ ]var ( )m q and

È ˘Î ˚

2 ( )E s q , whose values are unknown but which can, in practice, be estimated

from data from the risk itself and from the other risks in the collective. In other words, we include the data for the risk we’re interested in in the calculation of

[ ]var ( )qm and 2 ( )qÈ ˘Î ˚E s .

There is one further comment to be made about the assumptions for this model. Assumptions have been made concerning the identical distribution of the ijX ’s

both from different risks and from the same risk. If the ijX ’s come from different risks it has been assumed that they are (unconditionally) identically distributed (see (1.10)). If the ijX ’s come from the same risk, i , it has been assumed they

are unconditionally identically distributed (see (1.3)) and conditionally, given iq , identically distributed (see (1.2) and (1.8)). A consequence of these assumptions is that neither ( )|ij iE X q nor ( )var |ij iX q depends on j . This consequence

was the only step in the derivation of the credibility estimate (1.12) that required assumptions about the identical distribution of the ijX ’s. In fact, assumptions (1.8) and (1.9) (and the corresponding assumption in Section 1.2, (1.2)) could have been replaced by the following assumptions and it would still have been possible to derive the same credibility estimate. Given iq , the ijX ’s, = 1, 2, ,j n… , are independent for each risk i , = 1, 2, ,i N… . (1.16) For πi k , the pairs ( , )i ijXq and ( , )k klXq are independent and the risk

parameters 1 2, , , Nq q q… , are also identically distributed. (1.17) For each risk, i , = 1, 2, ,i N… , neither ( )|ij iE X q nor ( )var |ij iX q depends on

j . (1.18) Assumptions (1.16) and (1.17) are weakened versions of (1.8) and (1.9). Assumption (1.18) now has to be included as a separate assumption since it is not a consequence of (1.16) and (1.17).



None of the results or formulae in Section 1 would be altered in any way by making these weaker assumptions. The reason for making the slightly stronger assumptions, as was done in Section 1.2, is that they make the presentation a little easier. The reason for pointing out now that weaker assumptions could have been made is that this will help to link EBCT Model 1 with EBCT Model 2 in Section 2.

Example The table below shows the aggregate claim amounts (in £m) for an international insurer’s fire portfolio for a 5-year period, together with some summary statistics. Fill in the

missing entries and calculate [ ]( )qE m and 2 ( )qÈ ˘Î ˚E s using EBCT Model 1.

Total claim amount

Year (j) Xi14 1

52( )X Xij i

j−

=∑

Country (i) 1 2 3 4 5

1 48 53 42 50 59 50.4 39.3

2 64 71 64 73 70 68.4 17.3

3 85 54 76 65 90 74.0 215.5

4 44 52 69 55 71 ? ?

Solution In this example, there are n = 5 years and N = 4 risks (= countries). Xi is the average claims for risk i for the 5-year period. So the missing entry is:

X4 44 52 69 55 71 5 58 2= + + + + =( ) / .

We can then calculate the other missing entry:

( )5 2 2 2

1

1 1 (44 58.2) (71 58.2) 132.74 4ij i

jX X

=

⎡ ⎤− = − + + − =⎣ ⎦∑



To find the estimates we need X , which is:

X Xii

= = + + + ==∑1

450 4 68 4 74 0 58 2 4 62 75

1

4( . . . . ) / .

We can then the evaluate the estimates directly:

[ ]( ) 62.75q ª =E m X

Also:

( )4 5 22

1 1

1 1( )4 4

(39.3 17.3 215.5 132.7) / 4 101.2

ij ii j

E s X Xq= =

È ˘ ª -Î ˚

= + + + =

Â Â

Question 6.3

Estimate the value of [ ]var ( )qm .



Example Find the credibility factor for the example above and hence calculate the EBCT premium for each of the countries for the coming year. Solution We can use the estimates we have calculated to find the credibility factor:

[ ]25 0.8169101.2( ) var ( ) 590.33

q q= = =

È ˘+ +Î ˚

nZn E s m

Because we have the same number of time periods for each country, the credibility factor is the same for each. We can then use the basic credibility formula [ ](1 ) ( )q= + -iP Z X Z E m to find the EBCT premiums for each country:

Country PCountry PCountry PCountry P

1 08169 50 4 1 08169 62 75 52 662 08169 68 4 1 08169 62 75 67 373 08169 74 0 1 08169 62 75 71944 08169 58 2 1 08169 62 75 59 03

: . . ( . ) . .: . . ( . ) . .: . . ( . ) . .: . . ( . ) . .

= × + − × == × + − × == × + − × == × + − × =

Question 6.4

Are the following statements true or false for EBCT Model 1? (a) θ represents the “true” risk premium for a given risk. (b) The variance of X j θ doesn’t depend on θ . (c) None of the random variables or parameters in the model are assumed to have a

normal distribution.



2 Empirical Bayes Credibility Theory: Model 2

2.1 Introduction

Model 2 is a generalisation of Model 1. This section is therefore very similar to Section 1. The important thing is to note the differences. In this section the techniques of Empirical Bayes Credibility will be applied to a second, and slightly more complicated, model. The format will be exactly the same as in Section 1. First of all, in Section 2.2, the problem will be stated and the assumptions set out. The problem will be the same as in Section 1, ie to estimate the pure premium, or the expected number of claims, in the coming year for a risk. The assumptions will be slightly different from those in Section 1. Next, in Section 2.3, the credibility estimate for the pure premium or expected number of claims will be considered. Finally, in Section 2.4, the method of estimating the values of the parameters that are part of the credibility estimate will be discussed. In keeping with the terminology in Section 1, the model in this section will be referred to as EBCT Model 2. The examiners will also refer to the model in this way.

2.2 Model 2: specification

The problem is to estimate the expected aggregate claims, or the expected number of claims, in the coming year for a given risk. Let Y Y Yn1 2, , ,… be random variables representing the aggregate claims or numbers of claims in successive years from this risk. In Model 1 these quantities were called X j . You will see shortly why a different letter has been used. It is assumed that the values of 1 2, , , nY Y Y… have already been observed and the expected value of +1nY needs to be estimated.



So far the problem looks exactly like the problem in Section 1. The important difference between EBCT Model 1 and EBCT Model 2 is that Model 2 involves an extra parameter known as the risk volume, jP . Intuitively, the value of jP measures the “amount of business” in year j . For example, jP might represent the premium income for the risk in year j or the number of separate policies comprising the risk in year j . An important point to note is that the value of

+1nP at the start of year + 1n is assumed to be known. Next a new sequence of random variables, 1 2, , ,X X … is defined as follows: = /j j jX Y P = 1, 2,j … These X j ’s, which have been adjusted for the risk volumes, correspond more closely to the X j ’s in Model 1. In Model 1, we effectively assumed that Pj was always equal to 1, ie the volume of business was the same for each risk group. The random variable jX represents the aggregate claims, or the number of claims, in year j standardised to remove the effect of different levels of business in different years. The assumptions that specify EBCT Model 2 are as follows.

Assumptions for EBCT Model 2 The distribution of each jX depends on the value of a parameter, q , whose value is the same for each j but is unknown. (2.1) Given q , the jX ’s are independent (but not necessarily identically distributed). (2.2)

( )|jE X q does not depend on j . (2.3)

( )var |j jP X q does not depend on j . (2.4)



As in previous sections, q is known as the risk parameter for the risk, and, as for EBCT Model 1, it could be just a single real valued number or a more general quantity such as a vector of real valued numbers. Assumption (2.1) is the standard assumption for all credibility models considered here. See assumptions (2.2) and (2.9) from the previous chapter, and assumption (1.1). Assumption (2.2) corresponds to Assumption (1.2) for EBCT Model 1, but notice that (2.2) is slightly weaker than (1.2). Assumption (2.2) does not require the X j ’s to be conditionally (given θ ) identically distributed, but only to be conditionally independent. There is no assumption in EBCT Model 2 that the X j ’s are unconditionally, or conditionally given θ , identically distributed. Comparing the above assumptions with assumptions (1.16), (1.17) and (1.18) for EBCT Model 1, if all the jP ’s are equal to 1, then EBCT Model 2 is exactly the same as EBCT Model 1. Having made assumptions (2.3) and (2.4), ( )m q and 2 ( )s q can be defined as follows:

( )=( ) |jm E Xq q

( )=2 ( ) var |j js P Xq q

The definition of ( )m q corresponds exactly to the definition for EBCT Model 1 in

Section 1 but the definition of 2 ( )s q is slightly different. Remember to include the Pj factor in the definition of s2 ( )θ . To gain a little more insight into assumptions (2.3) and (2.4), consider the following example. Suppose the risk being considered is made up of a different number of independent policies each year and that the number of policies in year j is jP . Suppose also that the aggregate claims in a single year from a single

policy have mean ( )m q and variance 2 ( )s q , where ( )m and 2 ( )s are functions of q and q is the fixed, but unknown, risk parameter for all these policies. Now let jY denote the aggregate claims from all the policies in force in year j .



Then: ( ) = ( )j jE Y P m q

( ) = 2var ( )j jY P s q

( ) = ( )jE X m q

( ) = 2var ( )j jP X s q

This example satisfies assumptions (2.3) and (2.4).

2.3 Model 2: derivation of the credibility premium

The problem was stated rather loosely in the previous section as the estimation of the expected value of +1nY , given the values of 1 2, , , nY Y Y… . We can now be rather more precise about this. The quantity to be estimated is the mean (given q ) of +1nY . This is given by +1 ( )nP m q . Since in the model +1nP is known at the start of year + 1n , the problem is to estimate ( )m q . The data available are the values of each jY and its corresponding jP for = 1, 2, ,j n… . As for Model 2 the full derivation of the credibility premium under this model is beyond the scope of the CT6 syllabus and is not covered here. The solution to the problem, ie the “best” linear estimate of ( )m q given X is given by:

[ ] [ ]

[ ]

=

=

È ˘ +Î ˚

È ˘+ Î ˚

Â

Â

2

1

2

1

( ) ( ) var ( )

( ) var ( )

n

jj

n

jj

E m E s m Y

P E s m

q q q

q q

Remember that Yj is just P Xj j .



This can be written more attractively as follows:

Result of EBCT Model 2 The estimate of ( )m q given X given by EBCT Model 2 is: ( ) [ ]+ -1 ( )Z X Z E m q (2.5) where:

= = = =

= =Â Â Â Â1 1 1 1

n n n n

j j j j jj j j j

X P X P Y P

and:

[ ]

=

=

=È ˘+ Î ˚

Â

Â1

2

1( ) var ( )

n

jj

n

jj

P

ZP E s mq q

This demonstrates the similarities to and the differences from the Model 1 result.

Question 6.5

Check that the “attractive” version really is the same.

Note carefully that the above result tells us how to estimate the value of X for the coming year. If we want to estimate 1nY + , the aggregate claim amount for the coming year, we have to multiply our estimate of X by 1nP + , the risk volume for the corresponding year.



Here are some additional points of note about this solution: (i) If all the kP ’s were equal to 1, the solution given by (2.5) is exactly the

same as the solution given by (1.6). This is as it should be since if all the kP ’s are equal to 1 then EBCT Model 2 is exactly the same as EBCT Model

1. (ii) As for EBCT Model 1, the solution given by (2.5) involves three

parameters, [ ]( )E m q , [ ]var ( )m q and È ˘Î ˚

2 ( )E s q . The way in which these

parameters are estimated is explained in the next section.

2.4 Model 2: parameter estimation

The procedure for estimating the parameters [ ]( )E m q , [ ]var ( )m q and

È ˘Î ˚

2 ( )E s q for EBCT Model 2 follows exactly the same steps as the procedure for

EBCT Model 1, which was studied in Section 1.4. As before, we are now moving to a two-dimensional data set with rows representing different risks. It is now assumed that the risk that is of interest is one of a collective of N risks and that there exist data in the form given in Section 2.2 for each of these N risks for each of the past n years. These data consist of values for the aggregate claims, or the number of claims, and the corresponding risk volumes. Let Yij be a random variable denoting the aggregate claims, or the number of

claims, for risk number i in year j , j n= 1 2, , ,… , i N= 1 2, , ,… , and let Pij be the corresponding risk volume. For each i and j define: X Y Pij ij ij= /



The data are summarised in the following table, which corresponds to Table 1 for EBCT Model 1:

Table 2

Year

1 2 … n

Risk Number

1 11 11,Y P 12 12,Y P … 1 1,n nY P

2 21 21,Y P 22 22,Y P … 2 2,n nY P

. . . … .

. . . … .

N 1 1,N NY P 2 2,N NY P … ,Nn NnY P

For simplicity it is assumed, as was done in Section 1.4, that the risk that is of particular interest is Risk Number 1 in this collective. This means that what were denoted jY , jP and jX in Section 2.2 and Section 2.3 are now denoted 1jY , 1jP and 1jX respectively in this section. The problem is to estimate the expected

value of +1, 1nX and the solution to this problem has already been given by

formulae (2.5), remembering that jX and jP in (2.5) are now denoted 1jX and

1jP . The purpose of the data from the other risks in the collective is purely to

help to estimate the parameters [ ]( )E m q , [ ]var ( )m q and È ˘Î ˚

2 ( )E s q that appear

in (2.5). These other risks in the collective satisfy assumptions exactly the same as assumptions (2.1), (2.2), (2.3) and (2.4) for Risk Number 1. These assumptions are as follows. For each risk i , = 1, 2, ,i N… : the distribution of each ijX , = 1, 2, ,j n… , depends on a parameter, iq , whose value is the same for each j but is unknown (2.6) given θ i , the Xij ’s are independent (but not necessarily identically distributed) (2.7) there exists a function ( )m such that ( )=( ) |i ij im E Xq q (2.8)

there exists a function ( )s such that ( )=2 ( ) var |i ij ij is P Xq q . (2.9)



These four assumptions show that each risk in the collective satisfies the same assumptions as the particular risk that is of interest. The following two assumptions, which correspond to (1.9), (1.10) and (1.11) in Section 1.4, show the connection between different risks in the collective. The risk parameters 1 2, , , Nq q q… , regarded as random variables, are independent and identically distributed. (2.10) For πi k , the pairs ( , )i ijXq and ( , )k kmXq are independent. (2.11) Notice that, since the iq ’s are identically distributed, the values of [ ]( )iE m q ,

[ ]var ( )im q and È ˘Î ˚

2 ( )iE s q do not depend on i so that they can be denoted by

[ ]( )E m q , [ ]var ( )m q and È ˘Î ˚

2 ( )E s q , respectively, as in Section 2.2 and

Section 2.3. As in Section 1.4, more notation is needed. Denote:

=Â

1

n

ijj

P by iP (2.12)

=Â

1

N

ii

P by P (2.13)

( ) ( )-

=- -Â1

11 1 /

N

i ii

Nn P P P by *P (2.14)

=Â

1/

n

ij ij ij

P X P by iX (2.15)

= = =

=Â Â Â1 1 1

/ /N N n

i i ij iji i j

P X P P X P by X (2.16)

This notation is all given in the Tables (Page 30). Notice that what was denoted X in formula (2.5) is now denoted X1 and that X now has a different definition.

(This was exactly what happened in Section 1.4.) Notice also that Xi and X are weighted averages of the Xij ’s, the weights being the risk volumes Pij .



With this new notation the credibility estimate of the pure premium, or number of claims, per unit of risk volume for the coming year for Risk Number 1 in the collective originally given by formula (2.5) can be reformulated as: ( ) [ ]+ -1 1 1 ( )Z X Z E m q (2.17) where:

= =

= Â Â1 1 1 11 1

n n

j j jj j

X P X P

and:

[ ]

=

=

=È ˘+ Î ˚

Â

Â

11

12

11

( ) var ( )

n

jj

n

jj

P

ZP E s mq q

Note that this credibility factor is specific to Risk 1, so is denoted by 1Z . For the other risks there will be different values of Z . It is important to realise that this is exactly the same as formula (2.5) but is written in the notation of this section rather than that of the previous section. Unbiased estimators for [ ]q( )E m , [ ]qvar ( )m and qÈ ˘

Î ˚2 ( )E s can be proposed

based on the observed values Y Pij ij jn

i

N,n s = =

RSTUVW1 1

.



These estimators are: Parameter Estimator [ ]( )E m q X (2.18)

È ˘Î ˚

2 ( )E s q ( ) ( )--

= =- -Â Â 211

1 11

N n

ij ij ii j

N n P X X (2.19)

[ ]var ( )m q ( ) ( )

( ) ( )

--

= =

--

= =

ÈÍ - -ÍÎ

˘˙- - -˙˚

Â Â

Â Â

211

1 1

211

1 1

* 1

1

N n

ij iji j

N n

ij ij ii j

P Nn P X X

N n P X X

(2.20)

The points of note about these estimators are: (i) They are in exactly the same form as the estimators for EBCT Model 1.

Look back at formulae (1.13), (1.14) and (1.15). In particular, if all the ijP ’s were equal to 1, then the two sets of estimators would be identical.

(ii) It can happen in practice that formula (2.20) gives a negative value even

though [ ]var ( )m q must be non-negative. In such a situation the estimate

of [ ]var ( )m q is taken to be zero. A similar point about the variance estimates of EBCT Model 1 was discussed in Section 1.4.

(iii) Formulae (2.18), (2.19) and (2.20) are all given in the Tables. And you’ll be pleased to know that … (iv) The proofs that (2.18), (2.19) and (2.20) are unbiased are beyond the scope

of the syllabus.



Example The table belows show the volumes of business for each country for the insurer in the example on Page Error! Reference source not found.17.

Volume of business (Pij)

Country (i) Year (j)

1 2 3 4 5 Current year

1 12 15 13 16 10 20

2 20 14 22 15 30 25

3 5 8 6 12 4 10

4 22 35 30 16 10 12

Calculate the EBCT premium for Country 1 using Model 2. Solution The original data for the total claims is Yij and the table above gives values of Pij . The claims per unit volume X Y Pij ij ij= / are shown in the table below.

Total claims per unit volume (Xij)

Country (i) Year (j)

1 2 3 4 5

1 4.000 3.533 3.231 3.125 5.900

2 3.200 5.071 2.909 4.867 2.333

3 17.000 6.750 12.667 5.417 22.500

4 2.000 1.486 2.300 3.438 7.100

We can then calculate Pi , P and P* .



The figures are given in the table below.

Country(i) Pi P P Pi i( / )1−

1 66 52.17

2 101 68.62

3 35 31.11

4 113 72.46

P = 315 P* .= 1181

Furthermore, Xi and X can be calculated as: X Y Pi ij ij

n=

=∑ /

1 and X Y Pij

j

n

i

N=

==∑∑ /

11.

Country(i) Xi P X Xij ij i( )−∑ 2 P X Xij ij( )−∑ 2

1 3.818 57.13 58.94

2 3.386 111.59 147.71

3 10.571 1,237.82 2,756.56

4 2.575 267.73 492.04

X = 3984.

So this gives: [ ]( ) 3.984q ª =E m X From the other columns in the table, we get:

( )4 5 22

1 1

1 1( )4 4

(57.13 111.59 1, 237.82 267.73) /16

104.64

q= =

È ˘ ª -Î ˚

= + + +

=

Â Â ij ij ii j

E s P X X



Also:

[ ] ( )4 5 2

*1 1

1 1var ( ) 104.644 5 1

1 [(58.94 147.71 2,756.56 492.04) /19 104.64]11.81

6.539

ij iji j

m P X XP

q= =

È ˘Í ˙ª - -

¥ -Í ˙Î ˚

= + + + -

=

ÂÂ

The credibility factor for Country 1 is:

[ ]

11

1 2

11

66 0.8048104.64( ) 666.539

var ( )

q

q

=

=

= = =È ˘ +Î ˚+

Â

Â

n

jj

n

jj

P

ZE s

Pm

The risk premium per unit volume is:

( ) [ ]1 1 11 ( ) 0.8048 3.818 (1 0.8048) 3.984 3.851q+ - = ¥ + - ¥ =Z X Z E m Since the volume for Country 1 for the coming year is 20 units, the EBCT premium is:

20 3851 77 01× =. .

Question 6.6

Calculate the EBCT premiums for Countries 2, 3 and 4.



Question 6.7

Explain the effect each of the following changes (acting in isolation) would have on the value of the credibility factor for EBCT Model 2.

(a) 2 ( )qÈ ˘Î ˚E s is increased.

(b) [ ]var ( )qm is reduced. (c) The unit of currency is changed from £ to $. (d) All the Pj ’s are increased by the same factor. Does the credibility factor “behave” as you would expect?

We now finish this chapter with two exam-style questions.

Past Exam Question (Subject 106, September 2000, Question 3) The table below shows annual aggregate claim statistics for 3 risks over 4 years. Annual aggregate claims for risk i , in year j , are denoted by ijX .

Risk, i 4

1

14i ij

jX X

== Â ( )

4 22

1

13i ij i

js X X

== -Â

1 2,517 4,121,280 2 7,814 7,299,175 3 2,920 3,814,001 (i) Calculate the value of the credibility factor for Empirical Bayes Model 1. [3] (ii) Using the numbers calculated in (i) to illustrate your answer, describe the way in

which the data affect the value of the credibility factor. [4] [Total 7]



Exam-style question An actuary wishes to analyse the amounts paid by a group of insurers on their respective portfolios of commercial property insurance policies using the models of Empirical Bayes Credibility Theory. The actuary obtains the following information about the amounts of claim payments made and the number of policies sold for each of three different insurers. The data obtained are as follows.

Year 1 Year 2 Year 3 Year 4

£14.2m £15.8m £22.7m £19.0mInsurer A

163 189 252 199

£58.6m £63.1m £81.0m £64.2mInsurer B

4,435 4,761 5,576 4,581

£123m £132m £161m £133mInsurer C

16,184 17,443 20,102 18,000

(i) Analyse the data using EBCT Model 1, and calculate the expected total claim

payment to be made by Insurer B in the coming year. [6] (ii) Analyse the data using EBCT Model 2, and again calculate the expected payout

amount for Insurer B in the coming year, assuming that the expected number of policies sold for the coming year for Insurer B is 4,800. You may use the summary statistics given below, which have been calculated using the formulae and notation given in the Tables, again working in millions of pounds. Subscripts 1, 2 and 3 refer to Insurers A, B and C respectively. [8]

P X Xj j1 1 12 0 014667( ) .− =∑ P X Xj j1 1

2 5106461( ) .− =∑

P X Xj j2 2 22 0 006103( ) .− =∑ P X Xj j2 2

2 0 336408( ) .− =∑

P X Xj j3 3 32 0 003979( ) .− =∑ P X Xj j3 3

2 0 292641( ) .− =∑ (iii) Comment on your results. [2] [Total 16]



Chapter 6 Summary Empirical Bayes credibility This approach to credibility theory assumes that the claims for each risk are dependent on an underlying risk parameter θ . However, no assumptions are made about the form of the distribution of the claim amounts. The credibility premium can be expressed in terms of a credibility factor, which depends on the mean and variance of the conditional claim distribution. These quantities can be estimated based on data derived from a number of different risks. The table below summarises the models.

Model Quantityestimated

Assumptions about theunderlying distributionof the risk parameter θ

Assumptions about theconditional claimdistribution X|θ

EBCT Model 1

EBCT Model 2

risk premium

or

claimfrequency

a definite but unknowndistribution

a distribution withmean m(θ ) andvariance s2(θ )

Model 2 adjusts forvolume of business

Empirical Bayes Credibility Theory model 1 Definitions: X j represents the amount (or number) of claims ( ) ( )2( ) | ( ) var |q q q q= =j jm E X s X

Credibility factor: [ ]

2 ( )

var ( )

q

q

Ê ˆÈ ˘Î ˚Á ˜= +

Á ˜Ë ¯

E sZ n n

m

Credibility premium: [ ] ( ) [ ]( )| 1 ( )q q= + -iE m X Z X Z E m



Empirical Bayes Credibility Theory model 2 Definitions: Yj represents the amount (or number) of claims X Y Pj j j= / ( Pj constant) ( ) ( )2( ) | ( ) var |q q q q= =j j jm E X s P X

Credibility factor: [ ]

2

1 1

( )

var ( )

q

q= =

Ê ˆÈ ˘Î ˚Á ˜= +

Á ˜Ë ¯Â Ân n

j jj j

E sZ P P

m

Credibility premium: ( ) ( ) [ ]( )| 1 ( )q q= + -iE m X Z X Z E m

The formulae for estimating [ ]( )qE m , 2 ( )qÈ ˘Î ˚E s and [ ]var ( )qm for the EBCT models

are given in the Tables (Pages 29 & 30).



Chapter 6 Solutions Solution 6.1 Here, the risk parameter θ represents the handicap for a particular player. The distribution of the values of θ , which will vary from one player to the next, are assumed to be determined by a definite (but unknown) probability distribution. The shape of this distribution could be estimated by finding out the handicaps of a random sample of players. (In order for the EBCT model to apply here, we’ll have to assume that players are secretive about their handicaps, so that the actual handicap for a given player isn’t known with certainty.) X j is a random variable, representing the score obtained by a particular golfer on the j th day.

X j θ represents the score obtained by a particular golfer whose handicap is θ on the j th

day.

( ) ( )jm E Xθ θ= is the (theoretical) average score for a golfer whose handicap is θ . Note that, although the average score and the handicap are related, m( )θ will not be equal to θ . In this example, m( )θ will be an increasing function, since golfers with high handicaps (ie not very good players) will have high average scores (ie they will take lots of shots to get round).

2 ( ) var( )js Xθ θ= is the (theoretical) variance in the scores obtained by a golfer whose

handicap is θ . s2 ( )θ will take lower values for players with a high standard of play (a low θ ), since their scores will be fairly consistent, whereas the scores for novices will tend to fluctuate over a much wider range. So again, s2 ( )θ will be an increasing function. You may find it helpful to keep this example in the back of your mind to help you understand how the EBCT models work. Solution 6.2 The general credibility formula is: ( )1Z X Z m+ -



Solution 6.3 We need to find:

[ ] ( )4 4 5 22

1 1 1

1 1 1var ( ) ( )3 4 5 4

q= = =

ª - - -¥Â Â Âi ij i

i i jm X X X X

We’ve already worked out the second term (more or less) when we found 2 ( )qÈ ˘Î ˚E s .

So, putting in the numbers gives:

[ ] ( ) ( )4 4 5 22

1 1 1

2 2

1 1 1 1var ( )3 5 4 4

1 1[(50.4 62.75) (58.2 62.75) ] 101.23 5

90.33

i ij ii i j

m X X X Xθ= = =

≈ − − × −

= − + + − − ×

=

∑ ∑ ∑

Solution 6.4 (a) False. q is just a risk parameter that reflects the likelihood of claims. The true

risk premium for a given risk is ( )qÈ ˘Î ˚E m X . (b) False. The variance of qjX is 2 ( )qs , which is a function of q . (c) True. In fact, none of the quantities in the model are assumed to have any

specific type of distribution.



Solution 6.5

We start with ( ) [ ]1 ( )q+ -Z X Z E m . This is equal to:

[ ]

[ ]

[ ]

[ ]

2

1 12 2

11 1

( )

var ( )( )

( ) ( )

var ( ) var ( )

q

qq

q q

q q

= =

== =

È ˘Î ˚

¥ + ¥È ˘ È ˘Î ˚ Î ˚+ +

Â Â

ÂÂ Â

n n

j j jj j

nn njj jjj j

E sP P X

mE m

E s E sPP Pm m

We can see that the second term here is equal to the first part of the original expression

for the best linear estimator. If we cancel a factor of Pjj

n

=∑

1 in the first term here, and

remember that P X Yj j jj

n

j

n=

==∑∑

11, we see that the first term here is the same as the

second part of the expression for the best linear estimator. So the alternative form is equivalent to what we got before. Solution 6.6 The table below shows the figures for all four countries:

Country Credibility factor Risk premium per unit volume

EBCT premium

1 0.8048 3.851 77.0

2 0.8632 3.468 86.7

3 0.6862 8.504 85.0

4 0.8759 2.750 33.0



Solution 6.7 The formula for the credibility factor for EBCT Model 2 is:

[ ]

2

1 1

( )

var ( )

q

q= =

Ê ˆÈ ˘Î ˚Á ˜= +

Á ˜Ë ¯Â Ân n

j jj j

E sZ P P

m

(a) 2 ( )qÈ ˘Î ˚E s represents the average deviation from the expected value of the claims

for different risks. Increasing this would mean that claims tended to vary more from their “true” values. So we should have less confidence in the accuracy of an estimate based on past claims.

Since increasing 2 ( )E s θ⎡ ⎤⎣ ⎦ increases the denominator, Z decreases, as expected.

(b) [ ]var ( )qm represents the variation in the average claim amount for different risks.

Reducing this would mean that the actual claims were less strongly influenced by the true value of the risk parameter. So we would be less concerned about the actual value of θ indicated by the past claims, and we should put more emphasis on the overall mean.

Since reducing [ ]var ( )qm increases the denominator, Z decreases, as expected.

(c) Changing the unit of currency should not affect the credibility factor.

Since the quantities 2 ( )qÈ ˘Î ˚E s and [ ]var ( )qm are measured in units of £2 , their

ratio is dimensionless. So changing the unit of currency to $ would not affect Z . (d) The Pj ’s specify the relative weightings to be put on the claims for each year.

We would not expect a uniform increase applied to all the weightings to affect the credibility factor.

Since the definition ( )2 ( ) varq qÈ ˘ =Î ˚ j jE s P X includes a Pj factor, but

[ ]var ( )qm doesn’t, the ratio [ ]2 ( ) / var ( )q qÈ ˘Î ˚E s m varies in proportion to the

Pj ’s. So any extra factor incorporated in the Pj ’s would cancel out, leaving Z unchanged.



Solution to Past Exam Question (Subject 106, September 2000, Question 3) (i) Credibility factor

( )2E s qÈ ˘Î ˚ is estimated by the average of the sample variances, ie:

4,121,280 7,299,175 3,814,001 5,078,1523

+ + =

The sample mean of the iX ’s is:

2,517 7,814 2,920 4,4173

+ + =

and the sample variance of the iX ’s is:

( ) ( ) ( )2 2 22,517 4,417 7,814 4,417 2,920 4,417

8,695,3092

- + - + -=

So ( )var m qÈ ˘Î ˚ is estimated by:

18,695,309 5,078,152 7,425,7714

- ¥ =

The credibility factor ( ) ( )2 var

nZn E s mq q

=È ˘ È ˘+ Î ˚Î ˚

is then estimated by:

4 0.8539974 5,078,152 / 7,425,771

=+



(ii) How the data affect the value of the credibility factor Z is an increasing function of n , the number of years of past data. If we have more than 4 years of past data, the credibility factor will increase.

Z is a decreasing function of ( )2E s qÈ ˘Î ˚ . If ( )2E s qÈ ˘

Î ˚ increases, eg if the variance of

the claim amounts from one or more of the risks were to increase, then the value of the credibility factor would fall.

Z is an increasing function of ( )var m qÈ ˘Î ˚ . If ( )var m qÈ ˘Î ˚ increases, eg if there was greater variation between the individual sample means, then Z would increase. Solution to Exam-style Question (i) Analysis using EBCT Model 1 Using EBCT Model 1, we obtain the following numerical values:

Year 1 Year 2 Year 3 Year 4 Xi 13

2( )X Xij i−∑

Insurer A 14.2 15.8 22.7 19.0 17.925 14.1158

Insurer B 58.6 63.1 81.0 64.2 66.725 96.4358

Insurer C 123 132 161 133 137.25 270.9167

Using the standard formulae for the EBCT estimates, we get: X = + + =1

3 17 925 66 725 137 25 73 967( . . . ) .

2 13( ) (14.1158 96.4358 270.9167) 127.1561E s qÈ ˘ = + + =Î ˚

[ ] 2 212

2 14

var ( ) (17.925 73.967) (66.725 73.967)

(137.25 73.967) 127.1561

3,567.1562

m q È= - + -Î

˘+ - - ¥˚

=



So the credibility factor for the model is:

[ ]

24 0.9912

127.1561( ) 43,567.1562

var ( )

nZE s

nm

q

q

= = =È ˘ +Î ˚+

So the EBCT premium for the coming year for Insurer B is: 66 725 0 9912 73967 0 0088 79. . . . £66.× + × = m (ii) Analysis using EBCT Model 2 We first need to use the data to calculate the statistics we need. Using the notation given in the Tables, we have: P1 163 189 252 199 803= + + + = The same approach for the other insurers gives the values: P2 19 353= , P3 71 729= , For the whole portfolio we have: P = + + =803 19 353 71 729 91885, , ,

and:

* 1 803 19,353 71,729803 1 19,353 1 71,729 111 91,885 91,885 91,885

2,891.5793

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

= Calculating Xi for the first insurer:

X114 2 158 22 7 19

8030 089290=

+ + +=

. . . .



Similar calculations for the other insurers give: X2 0 013791= . X3 0 007654= . For the whole portfolio we have:

X =+ + + +

=14 2 158 161 133

918850 009660. .

,.

Now using the summary statistics given in the question, we can find our estimates for

[ ]( )E m q , [ ]var ( )m q and 2 ( )E s qÈ ˘Î ˚ :

[ ]( )E m q 0.009660

[ ]var ( )m q 1 1 (5.106461 0.336408 0.292641)2,891.5793 11

1 (0.014667 0.006103 0.003979)3 3

0.0001794

È ¥ + +ÍÎ

˘- + + ˙¥ ˚

=

2 ( )E s qÈ ˘Î ˚ 1

3 30 014667 0 006103 0 003979 0 002750

×+ + =( . . . ) .

So for Insurer B, the credibility factor is:

19,353 0.9992080.00275019,3530.0001794

BZ = =+

So the credibility premium per unit of risk volume for Insurer B is: 0 999208 0 013791 0 000792 0 009660 0 013788. . . . .× + × = So assuming a risk volume in the coming year of 4,800, the risk premium for Insurer B is £66.18m.



(iii) Comment The two models give fairly similar results. The estimate in Model 2 will depend on the prediction of risk volume for the coming year. In both cases we have used a very high value for the credibility factor. So we are effectively ignoring the data from the other insurers, and are basing our estimate almost entirely on the data from Insurer B. This seems pretty sensible, given that both the volume figures and the average claim amounts appear to be pretty variable between the three different insurers. So we are tempted to ignore the data from Insurers A and C, and focus on the information that we have for Insurer B.









CT6-09: Ruin theory Page 1


Chapter 9

Ruin theory

Syllabus objectives (iv) Explain the concept of ruin for a risk model. Calculate the adjustment coefficient

and state Lundberg’s inequality. Describe the effect on the probability of ruin of changing parameter values and of simple reinsurance arrangements.

1. Explain what is meant by the aggregate claim process and the cash-flow

process for a risk. 2. Define a Poisson process, derive the distribution of the number of events

in a given time interval, derive the distribution of inter-event times, and apply these results.

3. Define a compound Poisson process and derive the moments and moment

generating function for such a process. 4. Define the adjustment coefficient for a compound Poisson process and

for discrete time processes which are not compound Poisson, calculate it in simple cases and derive an approximation.

5. Define the probability of ruin in infinite/finite and continuous/discrete

time and state and explain relationships between the different probabilities of ruin.

6. State Lundberg's inequality and explain the significance of the

adjustment coefficient. 7. Describe the effect on the probability of ruin, in both finite and infinite

time, of changing parameter values. 8. Analyse the effect on the adjustment coefficient and hence on the

probability of ruin of simple reinsurance arrangements.

Page 2 CT6-09: Ruin theory


0 Introduction

In the previous two chapters we used the collective risk model to look at the aggregate claims S arising during a fixed period of time. S was given by the equation

1 2 NS X X X= + + + , where N denotes the number of claims arising during the period. In this chapter we will extend this model by treating ( )S t as a function of time. This gives us the equation 1 2 ( )( ) N tS t X X X= + + + , where ( )N t denotes the number of

claims occurring before time t. ( )N t is called a Poisson process and ( )S t is called a

compound Poisson process. We can use ( )S t to model claims received by an insurance company and hence consider the probability that this insurance company is ruined. The notation and the other basic concepts are covered in Section 1. In Section 2 we will give formal definitions of both the Poisson process and the compound Poisson process. You may have already met these in Subject CT4. We will also introduce the concept of a premium security loading. Briefly, this is an additional amount charged on an insurance premium to reduce the likelihood of an insurance company becoming ruined. In Section 3 we will introduce the adjustment coefficient, a parameter associated with risk, and Lundberg’s inequality. Section 4 considers the effect of changing parameter values on the probability of ruin for an insurance company. Finally, in Section 5 we will consider the impact on the probability of ruin for an insurance company when reinsurance is introduced into the equation.



1 Basic concepts

1.1 Notation

One technical point needed later in this chapter is that a function ( )f x is described as being ( )o x as x goes to zero, if:

0

( )lim 0x

f xxÆ

=

You can use this notation to simplify your working. For example, the function

2 3( ) 3 0.5 0.004= + +g x x x x can be rewritten as ( ) 3 ( )= +g x x o x , since 2 30.5 0.004 0+ Æx x

x as 0Æx . Note that ( )o x does not represent an actual number so

that ( )co x ( c is a constant), ( )-o x and ( )o x are all equivalent.

Question 9.1

Which of the following functions are ( )o x as 0Æx ? (i) 2x (ii) xe (iii) 1- - +xe x

For the purposes of this course, you just need to be able to understand the notation. However, if you wish to know more, then it’s covered in more detail in the Foundation ActEd Course (FAC). In Chapters 7 and 8 the aggregate claims generated by a portfolio of policies over a single time period were studied. In the actuarial literature, the word “risk” is often used instead of the phrase “portfolio of policies”. In this chapter both terms will be used, so that by a “risk” will be meant either a single policy or a collection of policies. In this chapter this study will be taken a stage further by considering the claims generated by a portfolio over successive time periods. Some notation is needed.

( )N t the number of claims generated by the portfolio in the time interval [0, t], for all t ≥ 0

iX the amount of the i-th claim, i = 1, 2, 3, ... ( )S t the aggregate claims in the time interval [0, t], for all t ≥ 0.



•=1i iX is a sequence of random variables. ≥0( ) tN t and ≥0( ) tS t are both

families of random variables, one for each time t ≥ 0; in other words ≥0( ) tN t

and ≥0( ) tS t are stochastic processes. You can think of a stochastic process as being a whole family of different random variables. Consider a time line. On the line there are an infinite number of different time intervals. For each interval of time, there is a random variable that corresponds to the aggregate claim amount arising in that time interval. This is what we mean by a stochastic process. Of course, if you have previously studied Subject CT4, you should be familiar with these ideas already. It can be seen that:

=

= Â( )

1( )

N t

ii

S t X

with the understanding that S(t) is zero if N(t) is zero. The stochastic process ≥0( ) tS t as defined above is known as the aggregate claims process for the risk. The random variables N(1) and S(1) represent the number of claims and the aggregate claims respectively from the portfolio in the first unit of time. These two random variables correspond to the random variables N and S, respectively, introduced in Chapter 7. So we have just taken the idea of a compound distribution and generalised it to cover different time periods. The insurer of this portfolio will receive premiums from the policyholders. It is convenient at this stage to assume, as will be assumed throughout this chapter, that the premium income is received continuously and at a constant rate. Here is some more notation: c the rate of premium income per unit time so that the total premium income received in the time interval [0, t] is ct. It will also be assumed that c is strictly positive.



1.2 The surplus process

Suppose that at time 0 the insurer has an amount of money set aside for this portfolio. This amount of money is called the initial surplus and is denoted by U. It will always be assumed that U ≥ 0. The insurer needs this initial surplus because the future premium income on its own may not be sufficient to cover the future claims. Here we are ignoring expenses. The insurer’s surplus at any future time t (> 0) is a random variable since its value depends on the claims experience up to time t. The insurer’s surplus at time t is denoted by U(t). The following formula for U(t) can be written: = + -( ) ( )U t U ct S t In words this formula says that the insurer’s surplus at time t is the initial surplus plus the premium income up to time t minus the aggregate claims up to time t. Notice that the initial surplus and the premium income are not random variables since they are determined before the risk process starts. The above formula is valid for t ≥ 0 with the understanding that U(0) is equal to U. For a given value of t, U(t) is a random variable because S(t) is a random variable. Hence ≥0( ) tU t is a stochastic process, which is known as the cash flow process or surplus process.

Figure 1



Figure 1 shows one possible outcome of the surplus process. Claims occur at times T1, T2, T3, T4 and T5 and at these times the surplus immediately falls by the amount of the claim. Between claims the surplus increases at constant rate c per unit time. The model being used for the insurer’s surplus incorporates many simplifications, as will any model of a complex real-life operation. Some important simplifications are that it is assumed that claims are settled as soon as they occur and that no interest is earned on the insurer’s surplus. Despite its simplicity this model can give an interesting insight into the mathematics of an insurance operation. We are also assuming that there are no expenses associated with the process (or, equivalently, that S t( ) makes allowance for expense amounts as well as claim amounts), and that the insurer cannot vary the premium rate c . We are also ignoring the possibility of reinsurance. Simple forms of reinsurance will be incorporated into the model later in this chapter.

1.3 The probability of ruin in continuous time

It can be seen from Figure 1 that the insurer’s surplus falls below zero as a result of the claim at time T3. Speaking loosely for the moment, when the surplus falls below zero the insurer has run out of money and it is said that ruin has occurred. In this simplified model, the insurer will want to keep the probability of this event, that is, the probability of ruin, as small as possible, or at least below a predetermined bound. Still speaking loosely, ruin can be thought of as meaning insolvency, although determining whether or not an insurance company is insolvent is, in practice, a very complex problem. Another way of looking at the probability of ruin is to think of it as the probability that, at some future time, the insurance company will need to provide more capital to finance this particular portfolio. Now to be more precise. The following two probabilities are defined: = < < < •( ) [ ( ) 0, for some , 0 ]U P U t t ty = < < £( , ) [ ( ) 0, for some , 0 ]U t P U ty t t t



ψ ( )U is the probability of ultimate ruin (given initial surplus U) and ψ ( , )U t is the probability of ruin within time t (given initial surplus U). These probabilities are sometimes referred to as the probability of ruin in infinite time and the probability of ruin in finite time. Here are some important logical relationships between these probabilities for 0 < t1 ≤ t2 < ∞ and for 0 ≤ U1 ≤ U2: ψ ψ( , ) ( , )U t U t2 1≤ (1.1) ψ ψ( ) ( )U U2 1≤ (1.2) ψ ψ ψ( , ) ( , ) ( )U t U t U1 2≤ ≤ (1.3)

Æ•=lim ( , ) ( )

tU t Uy y (1.4)

The intuitive explanations for these relationships are as follows: The larger the initial surplus, the less likely it is that ruin will occur either in a finite time period, hence (1.1), or an unlimited time period, hence (1.2). For a given initial surplus U, the longer the period considered when checking for ruin, the more likely it is that ruin will occur, hence (1.3). Finally, the probability of ultimate ruin can be approximated by the probability of ruin within finite time t provided t is sufficiently large, hence (1.4).

Question 9.2

What is lim ( , )u

u t→∞

ψ ?

You may be wondering whether it is possible to find numerical values for these ruin probabilities. In some very simple cases it is. However, for most practical situations, finding an exact value for the probability of ruin is impossible. In some cases there are useful approximations to ψ ( )u , even if calculation of an exact value is not possible.

1.4 The probability of ruin in discrete time

The two probabilities of ruin considered so far have been continuous time probabilities of ruin, so-called because they check for ruin in continuous time. In practice it may be possible (or even desirable) to check for ruin only at discrete intervals of time.



For a given interval of time, denoted h, the following two discrete time probabilities of ruin are defined: ψ h U P U t t t h h h( ) = [ ( ) < , for some , = , , , ]0 2 3 … -( , )= [ ( )<0, for some , = , 2 , , , ]h U t P U h h t h ty t t t … Note that it is assumed for convenience in the definition of ψ h U t( , ) that t is an integer multiple of h. Figure 2 shows the same realisation of the surplus process as given in Figure 1 but assuming now that the process is checked only at discrete time intervals. The black markers show the values of the surplus process at integer time intervals (ie h = 1); the black markers together with the white ones show the values of the surplus process at time intervals of length ½.

Figure 2

It can be seen from Figure 2 that in discrete time with h = 1, ruin does not occur for this realisation of the surplus process before time 5, but ruin does occur (at time 2½) in discrete time with h = ½.



Listed below are five relationships between different discrete time probabilities of ruin for 0 ≤ U1 ≤ U2 and for 0 ≤ t1 ≤ t2 < ∞. Formulae (1.5), (1.6), (1.7) and (1.8) are the discrete time versions of formulae (1.1), (1.2), (1.3) and (1.4) above and their intuitive explanations are similar. The intuitive explanation of (1.9) comes from Figure 2. ψ ψh hU t U t( , ) ( , )2 1≤ (1.5) ψ ψh hU U( ) ( )2 1≤ (1.6) ψ ψ ψh h hU t U t U( , ) ( , ) ( )1 2≤ ≤ (1.7)

Æ•=lim ( , ) ( )h ht

U t Uy y (1.8)

ψ ψh U t U t( , ) ( , )≤ (1.9)

Question 9.3

Explain why Equation 1.9 is true.

Intuitively, it is expected that the following two relationships are true since the probability of ruin in continuous time could be approximated by the probability of ruin in discrete time, with the same initial surplus, U, and time horizon, t, provided ruin is checked for sufficiently often, ie provided h is sufficiently small.

Æ +=

0lim ( , ) ( , )hh

U t U ty y (1.10)

Æ +=

0lim ( ) ( )hh

U Uy y (1.11)

Formulae (1.10) and (1.11) are true but the proofs are rather messy and will not be given here.



2 The Poisson and compound Poisson processes

2.1 Introduction

In this section some assumptions will be made about the claim number process,

N t t( )l q ≥ 0 , and the claim amounts, X i il q =∞

1 . The claim number process will be

assumed to be a Poisson process, leading to a compound Poisson process S t t( )l q ≥ 0 for aggregate claims. The assumptions made in this section will hold

for the remainder of this chapter.

2.2 The Poisson process

The Poisson process is an example of a counting process. Here the number of claims arising from a risk is of interest. Since the number of claims is being counted over time, the claim number process N t t( )l q ≥ 0 must satisfy the

following conditions: (i) (0) 0N = , ie there are no claims at time 0 (ii) for any 0t > , ( )N t must be integer valued (iii) when s t< , ( ) ( )N s N t£ , ie the number of claims over time is non-

decreasing (iv) when s t< , ( ) ( )N t N s- represents the number of claims occurring in the

time interval ( ],s t . The claim number process N t t( )l q ≥ 0 is defined to be a Poisson process with

parameter l if the following conditions are satisfied: (i) (0) 0N = , and ( ) ( )N s N t£ when s t< (ii) ( )( ) | ( ) 1 ( )P N t h r N t r h o hl+ = = = - + ( )( ) 1| ( ) ( )P N t h r N t r h o hl+ = + = = + (2.1) ( )( ) 1| ( ) ( )P N t h r N t r o h+ > + = = (iii) when s t< , the number of claims in the time interval ( , ]s t is independent

of the number of claims up to time s . (2.2)



Condition (ii) states that in a very short time interval of length h , the only possible numbers of claims are zero or one. Note that condition (ii) also implies that the number of claims in a time interval of length h does not depend on when that time interval starts.

Example Explain how motor insurance claims could be represented by a Poisson process. Solution The events in this case are occurrences of claim events (ie accidents, fires, thefts etc) or claims reported to the insurer. The parameter λ represents the average rate of occurrence of claims (eg 50 per day), which we are assuming remains constant throughout the year and at different times of day. The assumption that, in a sufficiently short time interval, there can be at most one claim is satisfied if we assume that claim events cannot lead to multiple claims (ie no motorway pile-ups etc).

The reason why a process satisfying conditions (i) to (iii) is called a Poisson process is that for a fixed value of t , the random variable ( )N t has a Poisson distribution with parameter tl . This is proved as follows: Let p t P N t nn ( ) ( ( ) )= = . We want to show that:

p t t tnn

n( ) exp ( )

!= −λ λ (2.3)

This is the Poisson probability function taken from page 7 of the Tables, where the parameter is tl . This will be proved by deriving and solving a “differential-difference” equation. For a fixed value of 0t > and a small positive value of h , condition on the number of claims at time t and write:

1

1

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

( ) [1 ] ( ) ( )

n n n

n n

p t h p t h o h p t h o h o h

hp t h p t o h

l l

l l

-

-

+ = + + - + +

= + - +



What we are saying here is that the event that there are n claims in the interval up to time t h+ can occur in one of two different ways. Either there were n − 1 claims in total up to time t , followed by an additional claim in the interval ( , )t t h+ , or there were n claims already in the interval up to time t , and no additional claim in the subsequent time interval. The terms ( )+h o hl and 1 ( )- +h o hl come from condition (ii) of the Poisson process. Thus: p t h p t h p t p t o hn n n n( ) ( ) [ ( ) ( )] ( )+ − = − +−λ 1 (2.4) and this identity holds for 1,2,3,n = … . Now divide (2.4) by h , and let h go to zero from above to get the differential-difference equation:

1( ) [ ( ) ( )]n n nd p t p t p tdt

l -= - (2.5)

Remember that the definition of a derivative is 0

( ) ( )limÆ

+ -=h

df f x h f xdx h

. Note also

from the definition given at the start of the chapter that o hh( ) tends to zero.

When 0n = , an identical analysis yields:

0 0( ) ( )d p t p tdt

l= - (2.6)

Solve for ( )np t by introducing the probability generating function ( , )G s t defined by:

G s t s p tnn

n( , ) ( )=

=

∞

∑0

In Subject CT3 we would have expressed this as ( )

( ) ( ) [ ]= N tN tG s E s . Note that we use

s as the dummy variable for the generating function in order to avoid confusion with t denoting time. We are assuming here that s is a fixed number. By definition the expected value of a function of a discrete random variable, say ( )g X , is

[ ( )] ( ) ( )x

E g X g x P X x= =Â . Applying this formula with ( ) ( )( ) = N tg N t s gives us the

expression used in Core Reading.



So, differentiating with respect to t gives:

ddt

G s t s ddt

p tnn

n( , ) ( )=

=

∞

∑0

Now multiply (2.5) by sn and sum over all values of n to get:

11 1 1

( ) ( ) ( )n n nn n n

n n n

ds p t s p t s p tdt

l l• • •

-= = =

= -Â Â Â

Now add (2.6) to the above identity to get:

10 1 0

( ) ( ) ( )n n nn n n

n n n

ds p t s p t s p tdt

l l• • •

-= = =

= -Â Â Â

which can be written as:

ddt

G s t sG s t G s t( , ) ( , ) ( , )= −λ λ

Here we have used the definition of G s t( , ) directly in the first and last sums. If we

take a factor of s out of the middle sum it leaves 11

1( )

•-

-=Â n

nn

s p t , which is an equivalent

way of writing G s t( , ) . Or, equivalently:

( )1 ( , ) ( 1)d G s t s

G s, t dtλ= − (2.7)

Recall that ( )ln ( )( )¢=d f xf x

dx f x.

Since the left hand side of (2.7) is the same as the derivative with respect to t of logG(s, t), (2.7) can be integrated to find that: log ( ) ( )G s t t s c s( , ) = λ − +1 where c(s) is some function of s. As we are integrating with respect to t , the “constant” we add when integrating can be a function of the other variable s (but not of t ).



c(s) can be identified by noting that when 0t = , 0( ) 1p t = and ( ) 0np t = for 1,2,3,n = … , since there are no claims at time zero.

Hence ( ,0) 1G s = and log ( ,0) 0 ( )G s c s= = . Thus: G s t t s( , ) = (exp )λ − 1 which is the probability generating function for the Poisson distribution with parameter λt . This generating function is given on page 7 of the Tables. Since there is a one-to-one relationship between probability generating functions and distribution functions (you should recall this from Subject CT3), the distribution of N(t) is Poisson with parameter λt . This is what we were trying to prove. This study of the Poisson process concludes by considering the distribution of the time to the first claim and the times between claims. Time to the first claim This section will show that the time to the first claim has an exponential distribution with parameter l . Let the random variable 1T denote the time of the first claim. Then, for a fixed value of t , if no claims have occurred by time t , 1T t> . Hence: 1( ) ( ( ) 0) exp P T t P N t tl> = = = - This last step follows from equation (2.3). And: P T t t( ) = 1 1≤ − −exp λ so that 1T has an exponential distribution with parameter λ . This is because the RHS matches the formula for the distribution function of an exponential distribution.



Important information The time to the first claim in a Poisson process has an exponential distribution with parameter l .

Time between claims This section will show that the time between claims has an exponential distribution with parameter l . For 2,3,i = … , let the random variable iT denote the time between the ( 1)i - th and the i th claims. Then:

1

11 1 1

( | ) ( | )

( ( ) | ( ) )

( ( ) ( ) 0 | ( ) )

n n n

n i i ii i i

P T t T r P T t r T r

P N t r n N r n

P N t r N r N r n

+

+= = =

> = = > + =

= + = =

= + - = =

Â Â Â

By condition (2.2) (ie using the independence of claim numbers in different time periods):

( ( ) ( ) 0 ( ) ) ( ( ) ( ) 0)P N t r N r N r n P N t r N r+ - = Ω = = + - = Finally: ( ( ) ( ) 0) ( ( ) 0) exp P N t r N r P N t tl+ - = = = = - since the number of claims in a time interval of length r does not depend on when that time interval starts (condition (2.1)). Thus inter-event times also have an exponential distribution with parameter λ .

Important information The time between claims in a Poisson process has an exponential distribution with parameter l .



Note that the inter-event time is independent of the absolute time. In other words the time until the next event has the same distribution, irrespective of the time since the last event or the number of events that have already occurred. This is referred to as the memoryless property of the exponential distribution.

Question 9.4

If reported claims follow a Poisson process with rate 5 per day (and the insurer has a 24 hour hotline), calculate:

(i) the probability that there will be fewer than 2 claims reported on a given day (ii) the probability that another claim will be reported during the next hour.

2.3 The compound Poisson process

In this section the Poisson process for the number of claims will be combined with a claim amount distribution to give a compound Poisson process for the aggregate claims. The following three important assumptions are made: the random variables X i i =

∞1 are independent and identically distributed

the random variables X i i =

∞1 are independent of ( )N t for all 0t ≥

the stochastic process N t t( )l q ≥ 0 is a Poisson process whose parameter

is denoted l . It was shown in Section 2.2 that this last assumption means that for any 0t ≥ , the random variable ( )N t has a Poisson distribution with parameter λt , so that:

P N t k t tk

kk

[ ( ) ] exp ( )!

, , ,= = − =λ λ for 0 1 2…

With these assumptions the aggregate claims process, S t t( )l q ≥ 0 , is called a

compound Poisson process with Poisson parameter l . By comparing the assumptions above with the assumptions in Section 2.3, it can be seen that the connection between the two is that if S t t( )l q ≥ 0 is a compound Poisson process

with Poisson parameter l , then, for a fixed value of t ( 0)≥ , ( )S t has a compound Poisson distribution with Poisson parameter λt .



Note the slight change in terminology here: “Poisson parameter λ ” becomes “Poisson parameter λt ” when a change is made from the process to the distribution. The common distribution function of the iX s will be denoted ( )F x and it will be assumed for the remainder of this chapter that (0) 0F = so that all claims are for positive amounts. Remember that ( )F x is defined to be ( )£P X x . In the continuous case we would find

( )F x by integrating the probability density function (pdf):

( ) ( )-•

= Úx

F x f t dt

The probability density function of the iX s, if it exists, will be denoted f x( ) and the kth moment about zero of the iX s, if it exists, will be denoted km , so that: [ ] for 1, 2, 3,k

k im E X k= = …

Important information

[ ] for 1, 2, 3,= = …kk im E X k

Whenever the common moment generating function of the X i s exists, its value at the point r will be denoted by M rX ( ) . In case you have forgotten, the definition of a moment generating function is: ( ) [ ]= rX

XM r E e Note that we are using r for the dummy variable to avoid confusion with time t . Since, for a fixed value of t , S t( ) has a compound Poisson distribution, it follows from Chapter 7 that the process ( )S t t ≥ 0 has mean λtm1, variance λtm2 , and moment generating function ( )SM r , where: ( )=exp ( ( ) 1)S XM r t M rl -



Remember from Chapter 7 that if: 1 2= + + + NS X X X where ~ ( )N Poi l , then: 1[ ] [ ]= =E S E X ml l 2

2var[ ] [ ]= =S E X ml l ( ) [ln ( )]=S N XM r M M r These formulae can be found on page 16 of the Tables. Here ( ) ~ ( )N t Poi tl , therefore 1[ ( )] =E S t tml and 2var[ ( )] =S t tml . Also

( )( ) ( ) exp ( 1)= -rN tM r t el , so:

( ) ( )( )ln ( )( ) ( ) exp ( 1) exp ( ) 1λ λ= − = −xM r

S t XM r t e t M r

For the remainder of this chapter the following (intuitively reasonable) assumption will be made concerning the rate of premium income: c m> λ 1 (2.8) so that the insurer’s premium income (per unit time) is greater than the expected claims outgo (per unit time).

Question 9.5

Why is this intuitively reasonable?

2.4 Probability of ruin in the short term

If we know the distribution of the aggregate claims S t( ) , we can often determine the probability of ruin for the discrete model over a finite time horizon directly (without reference to the models), by looking at the cashflows involved.



Example

The aggregate claims arising during each year from a particular type of annual insurance policy are assumed to follow a normal distribution with mean 0 7. P and standard deviation 2 0. P , where P is the annual premium. Claims are assumed to arise independently. Insurers are required to assess their solvency position at the end of each year. A small insurer with an initial surplus of £0.1m for this type of insurance expects to sell 100 policies at the beginning of the coming year in respect of identical risks for an annual premium of £5,000. The insurer will incur expenses of 0 2. P at the time of writing each policy. Calculate the probability that the insurer will prove to be insolvent for this portfolio at the end of the coming year. Ignore interest. Solution Using the information given, the insurer’s surplus at the end of the coming year will be:

(1) initial surplus + premiums expenses claims

0.1m 100 5,000 100 0.2 5,000 (1)

0.5m (1)

= - -

= + ¥ - ¥ ¥ -

= -

U

S

S

The distribution of S( )1 is:

S N N( ) ~ ( . , , ( . , ) ) [ . , ( . ) ]1 100 0 7 5 000 100 2 0 5 000 0 35 012 2× × × × = m m

So the probability that the surplus will be negative is:

2

[ (1) 0] [ (1) 0.5m]

( [0.35m,(0.1m) ] 0.5m)

0.5m 0.35m10.1m

1 (1.5) 1 0.93319 0.067

P U P S

P N

< = >

= >

-Ê ˆ= - FÁ ˜Ë ¯

= - F = - =

So the required probability is 6.7%.



Question 9.6

If the insurer expects to sell 200 policies during the second year for the same premium and expects to incur expenses at the same rate, calculate the probability that the insurer will prove to be insolvent at the end of the second year.

In fact, the normal distribution is probably not a very realistic distribution to use for the claim amount distribution in most portfolios, as it is symmetrical, whereas many claim amount portfolios will have skewed underlying distributions. However, it is commonly used in CT6 exam questions.

Example

The number of claims from a portfolio of policies has a Poisson distribution with parameter 30 per year. The individual claim amount distribution is lognormal with parameters 3m = and 2 1.1s = . The rate of premium income from the portfolio is 1,200 per year. If the insurer has an initial surplus of 1,000, estimate the probability that the insurer’s surplus at time 2 will be negative, by assuming that the aggregate claims distribution is approximately normal. Solution First we need the mean and variance of the aggregate claims in a two-year period. The expected number of claims will be 60. So the mean and variance are (using the formulae for the first two moments of a lognormal distribution): [ ] 3 0.55(2) 60 2,088.80E S e += = and: [ ] 6 2.2var (2) 60 218,457S e += = Ruin will occur if (2)S is greater than the initial surplus plus premiums received. So we want:

[ ] 3,400 2,088.80(2) 1,000 2 1,200 (0,1) 1 (2.8053) 0.0025218,457

P S P NÈ ˘-> + ¥ ª > = -F =Í ˙Î ˚

The probability of ruin is approximately 0.25%.



2.5 Premium security loadings

So far, we have used c to denote the rate of premium income per unit time, independent of the claims outgo. In some circumstances it is more useful to think of the rate of premium income as being related to the rate of claims outgo. For the insurer to survive, the rate at which premium income comes in needs to be greater than the rate at which claims are paid out. If this is not true, the insurer is certain to be ruined at some point. Sometimes c will be written as: c m= +( )1 1θ λ where θ ( )> 0 is the premium loading factor. The security loading is the percentage by which the rate of premium income exceeds the rate of claims outgo. So, for the Poisson process outlined above, we have: 1(1 ) ( ) (1 )c E S mq q l= + = + where q is the security loading. θ is also sometimes called the “relative security loading”. It might typically be a figure such as 0.2, ie 20%. The insurer will need to adopt a positive security loading when pricing policies, in order to cover expenses, profit, contingency margins and so on. Note that this does not mean that ruin is impossible. It is quite possible for the actual claims outgo to exceed substantially its expected value. So even in this situation the insurer’s probability of ruin is non-zero.

Question 9.7

What security loading is used in the example on page 20?



In summary:

Mean, variance and MGF of the total claim amount For a compound Poisson process S t( ) , the mean and variance of the total claim amount are given by:

[ ( )] ( )=E S t t E Xl 2var[ ( )] ( )=S t t E Xl The moment generating function of the process is given by:

( )( ) ( ) exp ( ( ) 1= -S t XM r t M rl

2.6 A technicality

In the next section a technical result will be needed concerning M rX ( ) (the moment generating function of the individual claim amount distribution), which, for convenience, will be presented here. It will be assumed throughout the remainder of this chapter that there is some number γ ( )0 < ≤ ∞γ such that M rX ( ) is finite for all r < γ and: ( )lim X

rM r

g -Æ= • (2.9)

(For example, if the X i s have a range bounded above by some finite number, then γ will be ∞ ; if the X i s have an exponential distribution with parameter α , then γ will be equal to α .) Suppose for example that claim amounts have a continuous uniform distribution on the interval (0,10) , so that they are bounded above by 10. Then the moment generating function of the claim distribution is (from the Tables):

M r erX

r( ) =

−10 110

This is defined for all positive values of r , and so in this case γ = ∞ . We can see that as r → ∞ , the limit of the MGF is infinite. If the claim distribution is Exp( )α , the MGF (as stated in the Tables) is:

M r rrX ( ) ( / )= − =

−−1 1α α

α



This tends to infinity as r tends to α from below. In the next section the following result will be needed: lim ( ( ) )X

rM r cr

gl

-Æ- = • (2.10)

If g is finite, (2.10) follows immediately from (2.9). This is because λ , c and r would all have finite values in the limit. Now it will be shown that (2.10) holds when g is infinite. This requires a little more care. First note that there is a positive number, ε say, such that: [ ] 0iP X e> > The reason for this is that all claim amounts are positive. So, if we pick a small enough number ( ε = 0 01. maybe), we’re bound to get a proportion of claims whose amount exceeds this. This probability will be denoted by π . Then: M r eX

r( ) ≥ ε π This follows by considering the claims below and above ε :

( ) ( ) ( | ) ( ) ( | ) ( )

0

= = £ £ + > >

≥ +

rX rX rXX

r

M r E e E e X P X E e X P X

e e

e e e e

p

Hence: lim ( ( ) ) lim ( )r

Xr r

M r cr e crel l pÆ• Æ•

- ≥ - = •

Here the erε term is tending to +∞ , while the −cr term is tending to −∞ . Remember that in such cases the exponential term always “wins”. So the limit is +∞ .

Important information

lim ( ( ) )Æ•

- = •Xr

M r crl



3 The adjustment coefficient and Lundberg’s inequality

This section will look at the probability of ruin and introduce the adjustment coefficient, a parameter associated with risk. The letters R and r will be used interchangeably for the adjustment coefficient.

3.1 Lundberg’s inequality

Lundberg’s inequality states that: ψ ( ) U RU≤ −exp where U is the insurer’s initial surplus and ( )Uy is the probability of ultimate ruin. R is a parameter associated with a surplus process known as the adjustment coefficient. Its value depends upon the distribution of aggregate claims and on the rate of premium income. Before defining R the importance of the result and some features of the adjustment coefficient will be illustrated. The proof of Lundberg’s inequality is not on the syllabus for Subject CT6. Don’t worry at this stage about what R actually represents. It will be defined shortly. Until then just think of it as a parameter associated with the surplus process. Note that if we can find a value for R , then Lundberg’s inequality tells us that we can find an upper bound for the probability of ruin. This is a very useful result. Figure 1 shows a graph of both exp RU- and ψ ( )U against U when claim amounts are exponentially distributed with mean 1, and when the premium loading factor is 10%. (The solution for R will be found in Section 3.2. The formula for ψ ( )U is given in Section 3.)

In fact it can be shown that the value of R in this case is 1011

.

It can be seen that, for large values of U , ψ ( )U is very close to the upper bound, so that ψ ( ) U RU≈ −exp .



Figure 1 In the actuarial literature, exp RU- is often used as an approximation to ψ ( )U . R can be interpreted as measuring risk. The larger the value of R , the smaller the upper bound for ψ ( )U will be. Hence, ψ ( )U would be expected to decrease as R increases. R is a function of the parameters that affect the probability of ruin, and R ’s behaviour as a function of these parameters can be observed. Note that R is an inverse measure of risk. Larger values of R imply smaller ruin probabilities, and vice versa. Figure 2 shows a graph of R as a function of the loading factor, θ , when: (i) the claim amount distribution is exponential with mean 10, and (ii) all claims are of amount 10.



Figure 2

Note that in both cases, R is an increasing function of θ . This is not surprising as ψ ( )U would be expected to be a decreasing function of θ , and since ψ ( ) expU RU≈ −l q , any factor causing a decrease in ψ ( )U would cause R to increase.

Question 9.8

Why would ψ ( )U be expected to be a decreasing function of θ ?

Note also that the value of R when claim amounts are exponentially distributed is less than the value of R when all claim amounts are 10. Again, this result is not surprising. Both claim amount distributions have the same mean, but the exponential distribution has greater variability. Greater variability is associated with greater risk, and hence a larger value of ψ ( )U would be expected for the exponential distribution, and a lower value of R . This example illustrates that R is affected by the premium loading factor and by the characteristics of the individual claim amount distribution. R is now defined and shown, in general, to encapsulate all the factors affecting a surplus process.



3.2 The adjustment coefficient – compound Poisson processes

The surplus process depends on the initial surplus, on the aggregate claims process and on the rate of premium income. The adjustment coefficient is a parameter associated with a surplus process which takes account of two of these factors: aggregate claims and premium income. The adjustment coefficient gives a measure of risk for a surplus process. When aggregate claims are a compound Poisson process, the adjustment coefficient is defined in terms of the Poisson parameter, the moment generating function of individual claim amounts and the premium income per unit time. The adjustment coefficient, denoted R , is defined to be the unique positive root of: ( ) 0XM r crl l- - = (3.1) So, R is given by: ( )XM R cRl l= + (3.2) Note that, although R relates to the aggregate claims, the MGF used in the definition is for the individual claim amount. It is probably not at all obvious to you at this stage why R is defined in this way. The reason is bound up with the proof of Lundberg’s inequality, which you are not required to know. Please accept the definition, so that you can find the value of R in simple cases. Note that equation (3.1) implies that the value of the adjustment coefficient depends on the Poisson parameter, the individual claim amount distribution and the rate of premium income. However, writing c m= +( )1 1θ λ gives: M r m rX ( ) ( )= + +1 1 1θ so that R is independent of the Poisson parameter and simply depends on the loading factor, θ , and the individual claim amount distribution. You can see from this equation that all the λ ’s have cancelled.

Important information The adjustment coefficient can be found by solving the equation: ( )λ λ= +XM r cr or 1( ) 1 (1 )= + +XM r m rq



Since the rate of claims outgo per unit time is λm1 (the mean of a compound Poisson distribution), then if a loading factor of θ is used, the rate of premium income will be c m= +( )1 1θ λ .

Example An insurer knows from past experience that the number of claims received per month has a Poisson distribution with mean 15, and that claim amounts have an exponential distribution with mean 500. The insurer uses a security loading of 30%. Calculate the insurer’s adjustment coefficient and give an upper bound for the insurer’s probability of ruin, if the insurer sets aside an initial surplus of 1,000. Solution The equation for the adjustment coefficient is: 1( ) 1 (1 )= + +XM r m rq

We have 1~500

Ê ˆÁ ˜Ë ¯

X Exp , so that 1( ) (1 500 )-= -XM r r (this comes from the Tables),

0.3=q , and 1 [ ] 500= =m E X . Substituting these into the equation: 1(1 500 ) 1 1.3 500 1 650-- = + ¥ = +r r r Rearranging:

2

1 (1 500 )(1 650 )

1 1 500 650 325,000

0 150 325,000

0.000462

= - +

= - + -

= -

fi =

r r

r r r

r

r

From Lundberg’s inequality, ( ) -£ rUU ey , so here: 0.000462 1,000( ) 0.630- ¥£ =U ey Note that we didn’t use the Poisson parameter in our solution.

We will see another example of finding the adjustment coefficient shortly.



Question 9.9

In the previous example we ignored the fact that r could be 0. Why?

It may not be obvious to you why R does not depend on the Poisson parameter. The basic reason is that increasing the Poisson parameter speeds up the whole process, so that claims arise more quickly. This means that ruin, if it is going to happen, will happen sooner, rather than later. However it does not affect the probability that ruin does actually occur, when we are considering ruin at any time in the future. It can be shown that there is indeed only one positive root of (3.1) as follows. Define g r M r crX( ) ( )= − −λ λ and consider the graph of g r( ) over the interval [ , ]0 γ . Note first that g( )0 0= . This is what we discovered in the previous example. Further, ( )g r is a decreasing function at 0r = since:

ddr

g r ddr

M r cX( ) ( )= −λ

so that the derivative of g r( ) at r = 0 is λm c1 − which is less than zero by assumption (3.8). Recall that ′M X ( )0 gives the mean of X , which is [ ]E X or m1 . It can also be shown that if the function g r( ) has a turning point, it must be at the minimum of the function. The second derivative is:

ddr

g r ddr

M rX2

2

2

2( ) ( )= λ

which is always strictly positive. The second derivative can be written: 2( ) ( )=¢¢ rX

XM r E X e The function in this expectation is made up of two positive factors, and hence the expectation must have a positive value.



Hence, there can only be one turning point, since any turning point is a minimum. To show that there is a turning point, note from (3.10) that

lim ( )r

g rgÆ -

= • . Since g r( ) is a decreasing function at r = 0 , it must have a

minimum turning point and so the graph of g r( ) is as shown in Figure 3.

Figure 3 Thus there is a unique positive number R satisfying equation (3.1). Equation (3.1) is an implicit equation for R . For some forms of ( )F x it is possible to solve explicitly for R ; otherwise the equation has to be solved numerically. Consider the exponential distribution where F x e x( ) = − −1 α . This is in the Tables, using a as the parameter for the exponential distribution, which avoids confusion with the Poisson parameter.



For this distribution, M rrX ( ) =

−α

α, (the Tables give this as

11

-Ê ˆ-Á ˜Ë ¯ra

) so:

2

2 0

cRR

R cR cR

R R Rc c

lala

la l a la

l la a

+ =-

fi - + - =

Ê ˆfi - - = fi = -Á ˜Ë ¯ (3.3)

since R is the positive root of (3.1).

If (1 )c q la+= , then

(1 )R aq

q=

+, since the mean of this distribution is 1

1=ma

.

Question 9.10

Write down the equation for the adjustment coefficient for personal accident claims if 90% of claims are for £10,000 and 10% of claims are for £25,000, assuming a proportional security loading of 20%. Show that this equation has a solution in the range 0 00002599 0 00002601. .< <R .

If the equation for R has to be solved numerically, it is useful to have a rough idea of R ’s value. Equation (3.2) can be used to find a simple upper bound for R as follows:

λ λ

λ

λ

λ

+ =

=

> + +

= + +

∞

∞

zz

cR M R

e f x dx

Rx R x f x dx

Rm R m

X

Rx

( )

( )

( ) ( )

( )

0

12

2 2

0

112

22

1

1

The inequality is true because all the terms in the series for eRx are positive. So eRx must always be greater than the total of the first few terms.



Alternatively, we could present this proof as:

2 2

2 2

2 2

[ ]

112

1[1] [ ]2

11 [ ] [ ]2

+ =

È ˘= + + +Í ˙Î ˚

È ˘È ˘= + + +Í ˙Í ˙Î ˚Î ˚

È ˘> + +Í ˙Î ˚

RXcR E e

E RX R X

E E RX E R X

RE X R E X

l l

l

l

l

So that ( )c m R R m− >λ λ1

12

22 , giving:

R c m m< −2 1 2( ) /λ λ (3.4) so that R m m< 2 1 2θ / when c m= ( + )1 1θ λ . Notice that if the value of R is small,

then it should be very close to this upper bound since the approximation to eRx should be good.

Question 9.11

Find an upper limit for the adjustment coefficient in the previous self assessment question, and comment on your answer.

3.3 A lower bound for R

A lower bound for R can be derived when there is an upper limit, say M , to the amount of an individual claim. For example, if individual claim amounts are uniformly distributed on (0,100) , then 100M = . The result is proved in a similar fashion to Result (3.4). The lower bound is found by applying the inequality:

exp( ) exp( ) 1x xRx RMM M

£ + - for 0 x M£ £ (3.5)



The inequality is proved through the series expansion of exp( )RM :

0

1

1

1

( )exp 1 1!

1!

( )1 for 0!

exp

j

j

j j

j

j

j

x x x RM xRMM M M j M

R M xj

Rx x Mj

Rx

•

=

• -

=

•

=

+ - = + -

= +

≥ + £ £

=

Â

Â

Â

since 1-£j jx M x if 0 £ £x M . Inequality (3.5) can be used to show that:

R

Mc m>

11log( / )λ

when individual claim amounts have a continuous distribution on (0, )M . This is the lower bound for R that we are trying to find. The starting point is the equation defining R (ie Equation 3.2):

λ λ

λ

λ λ λ

+ =

≤ + −FHG

IKJ

= + −

zz

cR Rx f x dx

xM

RM xM

f x dx

MRM m

Mm

M

M

exp ( )

exp ( )

exp

0

0

1 1

1l q

Hence, rearranging:

2

1

1 ( )(exp 1) 12 3!

c RM RMRMm RMl

£ - = + + +

( )2

11! 2!

< + + +RMRM

exp RM=



This gives 1

1 log cRM ml

Ê ˆ> Á ˜Ë ¯

as required.

If c m= +( )1 1θ λ , this is just RM

> +1 1log( )θ .

Other approximations for R can be found, particularly when R is small, by truncating the series expansion of exp( )Rx .

Question 9.12

Find a lower bound for the adjustment coefficient in the example in Question 9.10. Is the argument valid when the claim distribution is discrete?

Important information An upper bound for the adjustment coefficient:

1

2

2( )-<

c mR

ml

l or 1

2

2<

mR

mq

when 1(1 )= +c mq l

A lower bound for the adjustment coefficient:

1

1 logÊ ˆ

> Á ˜Ë ¯cR

M ml or R

M> +

1 1log( )θ

3.4 The adjustment coefficient – general aggregate claims processes

In Section 3.2 the existence of the adjustment coefficient, R, was proved for a compound Poisson aggregate claims process. In this section the existence of the adjustment coefficient for a general aggregate claims process is proved. Let Si i =

∞1 be a sequence of independent identically distributed random

variables: iS ∫ aggregate claims from a risk in time period i . c is the constant premium charged to insure this risk



The following assumptions are made: c E Si> [ ] (3.6) There is some number γ > 0 such that: [ ]lim [ ]ir S c

rE e

g-

Æ -= • (3.7)

iS has density function ( )h x , x-• < < • (3.8) In the general situation the adjustment coefficient is the positive number R that can be shown to satisfy the following:

( )[ ] 1iR S cE e - =

The proof that there is one, and only one, positive number R to satisfy this is as follows. Let f r E er S ci( ) [ ]( )= − for −∞ < <r γ . Then f ( )0 1= , ′ = − <f E S ci( ) [ ]0 0 , ′′ >f x( ) 0 , and: lim ( )

rf r

gÆ -= +•

We then use the same argument as for the compound Poisson case, based on the graph below. This graph forms part of the Core Reading. R γ

f(r)

0

1

r



Comments: Suppose iS has a compound Poisson distribution with Poisson parameter λ and claim size random variable X . Then:

( )[ ] 1

[ ]

i

i

R S c

RS Rc

E e

E e e

- =

fi =

( )fi = Rc

SM R e

[ ( ) 1]

( )

XM R Rc

X

e e

M R Rc

l

l l

-fi =

fi = +

which is the same as (3.2).



4 The effect of changing parameter values on finite and infinite time ruin probabilities

4.1 Introduction

Recall that ( )Uy was defined to be ( )( ) 0, 0< >P U t t , and ( , )U ty was defined to be

( )( , ) ( ) 0, 0= < < <U t P U ty t t . In this section the effect of changing parameter values on ψ ( , )U t and ψ ( )U will be discussed. No new theory will be introduced and the method for obtaining numerical values for ψ ( , )U t will not be discussed. Features of ψ ( , )U t , and in some cases of ψ ( )U , will be illustrated by a series of numerical examples. In these examples the same basic assumptions will be made as in previous sections. In particular, it will be assumed that the aggregate claims process is a compound Poisson process. In addition it will be assumed throughout Section 4.3, Section 4.4 and Section 4.5 that: the Poisson parameter for the number of claims is 1 (4.1)

the expected value of an individual claim is 1 (4.2)

individual claims have an exponential distribution. (4.3)

In Section 4.6, Assumptions (4.2) and (4.3) will be made, but the Poisson parameter will be allowed to vary. The implication of Assumption (4.1) is that the unit of time has been chosen to be such that the expected number of claims in a unit of time is 1. Hence ψ ( , )U 500 is the probability of ruin (given initial surplus U ) over a time period in which 500 claims are expected. The actual number of claims over this time period has a Poisson distribution (with parameter 500) and could take any non-negative integer value. The implication of Assumption (4.2) is that the monetary unit has been chosen to be equal to the expected amount of a single claim. Hence ψ ( , )20 500 is the probability of ruin (over a time period in which 500 claims are expected) given an initial surplus equal to 20 times the expected amount of a single claim. The advantage of using an exponential distribution for individual claims (Assumption (4.3)) is that both exp( )RU- and ψ ( )U can be calculated for these examples. See Section 2 and Section 4.2.



4.2 A formula for ψ ( )U when F(x) is the exponential distribution

The formula for ψ ( )U when individual claims amounts are exponentially distributed with mean 1, and when the premium loading factor is θ , is given by the following result. When ( ) 1 exp( )F x x= - - :

1( ) exp1 1

UU qyq q

Ê ˆ= -Á ˜Ë ¯+ + (4.4)

The syllabus does not require this result to be derived or memorised. This result has been stated in order to illustrate how, for this particular distribution, the ultimate ruin probability is affected by changes in parameter values.

4.3 ( , )U ty as a function of t

Question 9.13

Do you think that ( , )U ty is an increasing or decreasing function of t ?

Figure 4 shows a graph of ψ ( , )15 t for 0 500t£ £ . The premium loading factor, θ , is 0.1 so that the premium income per unit time is 1.1. Also shown in Figure 4 are ψ ( )15 (dotted line) and exp( 15 )R- (solid line) for this portfolio. These last two values are shown as lines parallel to the time axis since their values are independent of time. Here: ψ ( , )15 t has been worked out using a numerical method not described here

ψ ( )15 has been calculated using equation 4.4

exp −15Rl q is the upper bound given by Lundberg’s inequality.



Figure 4

Question 9.14

What is the value of ψ ( )15 ?

Question 9.15

What is the value of e R−15 ?

The features of note in Figure 4 are: (i) ψ ( , )15 t is an increasing function of t , (ii) for small values of t , ψ ( , )15 t increases very quickly (its value doubles as

t increases from 25 to 50 and doubles again as t increases from 50 to 100),

(iii) for larger values of t , ψ ( , )15 t increases less quickly and approaches

asymptotically the value of ψ ( )15 .



General reasoning should help you to understand (ii) and (iii). You would expect a much higher probability of ruin before time 50 than before time 25 since the overall performance of the fund could easily change in such a short time period. However, if premium rates are expected to be profitable in the long term, then at time 400, say, a significant surplus will have built up in most cases and so the probability of ruin at time 425 won’t be that much higher than at time 400. We are assuming here that accumulated surpluses stay in the fund and are not, for example, distributed to shareholders.

4.4 Ruin probability as a function of initial surplus

Question 9.16

Would you expect ( , )U ty to be an increasing or decreasing function of U ?

Figure 5 shows values of ψ ( , )U t for 0 500t£ £ and for three values of the initial surplus, 15, 20,U = and 25. The premium loading factor is 0.1 as in Figure 4. For 15U = the graph of ψ ( , )U t is as in Figure 4.

Figure 5

Question 9.17

What is the value of ψ ( )20 ?



The features of note in Figure 5 are: (i) the graphs all have the same general shape, (ii) increasing the value of U decreases the value of ψ ( , )U t for any value of

t , (iii) each of the three graphs approaches an asymptotic limit as t increases

(as has already been noted for U equal to 15 in the discussion of Figure 4). Note that ψ ( ) .20 01476= and ψ ( ) .25 0 0937= .

ψ ( )U is a non-increasing function of U . In the case of exponentially distributed individual claim amounts, the derivative with respect to U of ψ ( )U is:

ddU

U Uψ θθ

ψ( ) ( )=−+1

which is negative since θ > 0 . Hence ψ ( )U is a decreasing function of U . It is intuitively clear that ψ ( , )U t (of which ψ ( )U is a special case) should be a decreasing function of U . An increase in U represents an increase in the insurer’s surplus without any corresponding increase in claim amounts. Thus, an increase in U represents an increase in the insurer’s security and hence will reduce the probability of ruin.



4.5 Ruin probability as a function of premium loading

Question 9.18

Would you expect ( , )U ty to be an increasing or decreasing function of q ?

Figure 6 shows values of ψ ( , )15 t for 0 500t£ £ and for three values of the premium loading factor, 0.1, 0.2q = and 0.3. The graph of ψ ( , )15 t for 0.1q = is the same as the graph in Figure 4 and the same as one of the graphs in Figure 5. Figure 6 is, in many respects, similar to Figure 5. The features of note in Figure 6 are: (i) the graphs of ψ ( , )15 t all have the same general shape, (ii) increasing the value of θ decreases the value of ψ ( , )15 t for any given

value of t ; this is in fact true for any value of U , and is an obvious result since an increase in θ is equivalent to an increase in the rate of premium income with no change in the aggregate claims process,

(iii) it can be seen that when 0.2q = and 0.3, ψ ( , )15 t is more or less constant

for t greater than about 150. For 1 2t t£ , the difference ψ ψ( , ) ( , )15 152 1t t− represents the probability that ruin occurs between times t1 and t2 . Hence for these values of θ , 0.2 and 0.3, (and for this value of the initial surplus, 15, and for this aggregate claims process) ruin, if it occurs at all, is far more likely to occur before time 150, ie within the time period for 150 claims to be expected, than after time 150. This point will be discussed further in Section 4.6.

Figure 6



It is clear by general reasoning that ψ ( )U must be a non-increasing function of θ . In the case of exponential individual claim amounts, ψ ( )U is a decreasing function of θ .

dd

U U U Uθ

ψ θ ψ θ ψ( ) ( ) ( ) ( ) ( )= − + − +− −1 11 2

Question 9.19

Verify this expression for dd

Uθ

ψ ( ) .

This is clearly negative since θ , U and ψ ( )U are all positive quantities. Since the derivative is less than zero for all values of θ , ψ ( )U is a decreasing function of θ . Figure 7 shows ψ ( )10 as a function of θ .

Figure 7



4.6 Ruin probability as a function of the Poisson parameter

Figure 8 shows ψ ( , )15 10 as a function of λ for three values of the premium loading factor, 0.1, 0.2q = and 0.3. This graph is identical to Figure 6 apart from the labelling of the x-axis. This can be explained by considering the following two risks.

Figure 8 Risk 1: aggregate claims are a compound Poisson process with Poisson

parameter 1 and ( ) 1 xF x e-= - . The premium income per unit time to cover this risk is ( )1+ θ .

Risk 2: aggregate claims are a compound Poisson process with Poisson

parameter 0.5 and ( ) 1 xF x e-= - . The premium income per unit time to cover this risk is 0 5 1. ( )+ θ .

The unit of time is taken to be one year. The only difference between these risks is that twice as many claims are expected each year under Risk 1. This is reflected in the two premiums. Consider Risk 2 over a new time unit equivalent to two years. Then the distribution of aggregate claims and the premium income per unit time are now identical to the corresponding quantities for Risk 1. Hence, the probability of ruin over an infinite timespan is the same for both risks.



The solid line in Figure 9 shows an outcome of the surplus process for Risk 1 when θ = 01. . The dotted line shows the same surplus process when the unit of time is two years. This illustrates that any outcome of the surplus process that causes ultimate ruin for Risk 1 will also cause ultimate ruin for Risk 2. There is thus no difference in the probability of ultimate ruin for these two risks. It is only the time (in years) until ruin that will differ. Measuring times in years, the probability of ruin by time 1 for Risk 1 is the same as the probability of ruin by time 2 for Risk 2. This explains why Figures 6 and 8 show the same functions. For example, the value of ψ ( , )15 10 when λ = 50 (Figure 8) is the same as the value of ψ ( , )15 500 when λ = 1 (Figure 6).

Figure 9 Point (iii) in Section 4.5 will now be investigated, where it was noted that values of ψ ( , )15 t were more or less constant for values of t greater than 150 when

0.2q = and 0.3. In particular, the situation will be considered when the premium loading factor is 0.2.



Consider a second aggregate claims process, which is the same as the process considered throughout this section except that its Poisson parameter is 150 and not 1. (This second process is really identical to the original one; all that has happened is that the time unit has been changed.) Use ψ * to denote ruin probabilities for the second process and ψ to denote, as before, ruin probabilities for the original process. The change of time unit means that for any

0t ≥ : ψ ψ* ( , ) ( , )U t U t= 150 but it has no effect on the probability of ultimate ruin (put t = • in the relationship above) so that: ψ ψ* ( ) ( )U U= The point made in (iii) above was that: ψ ψ( , ) ( )15 150 15≈ From this and the previous two relations it can be seen that: ª* (15,1) * (15)y y In words this relation says that for the second process, starting with initial surplus 15, the probability of ruin within one time period is almost equal to (actually a little less than) the probability of ultimate ruin. This conclusion depends crucially on the fact that ψ * ( , )15 1 is a continuous time probability of ruin. To see this, consider ψ * ( , )15 1 , which is just the probability that for the second process the surplus at the end of one time period is negative. ψ * ( , )15 1 can be calculated approximately by assuming that the aggregate claims in one time period, which will be denoted S*(1), have a normal distribution. Recall that individual claims have an exponential distribution with mean 1 and that the number of claims in one time period has a Poisson distribution with mean 150. From this: [ (1)] 150E S* = and var[ (1)] 300S* = These are calculated as λm1 150 1 150= × = and λm2 150 2 300= × = , where

2( ) 2=E X for an Exp( )1 distribution.



Now, using tables of the normal distribution:

[ ][ ]

* (15,1) * (1) 15 1.2 150

( * (1) 150) /17.32 45 /17.32

0.005

P S

P S

y = > + ¥

= - >

ª

Recall that if 2~ ( , )X N m s , then -= XZ ms

has a standard normal distribution,

ie ~ (0,1)Z N . Probabilities for this distribution can be looked up in the Tables. From Figure 6 it can be seen that the value of ψ ( , )15 150 , and hence of ψ * ( , )15 1 , is about 0.07 which is very different from the (approximate) value of the discrete time probability of ruin ψ * ( , )15 1 calculated above.

4.7 Concluding remarks

When individual claim amounts are exponentially distributed with mean 1, first note that if θ = 0 , then ψ ( )U = 1 irrespective of the value of U . We’re thinking here of substituting θ = 0 into equation 4.4. This result is in fact true for any form of ( )F x . (It trivially follows that if θ < 0 , then ψ ( )U = 1.) In other words a positive premium loading is essential if ultimate ruin is not to be certain. Also note that throughout this section it has been assumed that individual claim amounts are exponentially distributed with mean 1. This mean could be measured in units of £100, £1,000 or perhaps £1,000,000. The parameter of the exponential distribution can be set to 1 without loss of generality provided that the monetary unit is correctly specified. In simple terms, the probability of ruin when U is £1 is the same as the probability of ruin when U is 100 pence. It can be said that: ψ ( )U when F x e x( ) = − −1 α is the same as: ψ α( )U when F x e x( ) = − −1 In other words, if the expected claims per unit time increase by a factor α , so too must the initial surplus if the probability of ultimate ruin is to be unchanged.



5 Reinsurance and ruin

5.1 Introduction

One of the options open to an insurer who wishes to reduce the variability of aggregate claims from a risk is to effect reinsurance. A reduction in variability would be expected to increase an insurer’s security, and hence reduce the probability of ruin. A reinsurance arrangement could be considered optimal if it minimises the probability of ruin. As it is difficult to find explicit solutions for the probability of ruin, the effect of reinsurance on the adjustment coefficient will be considered instead. If a reinsurance arrangement can be found that maximises the value of the adjustment coefficient, the upper bound for the probability of ultimate ruin will be minimised. As the adjustment coefficient is a measure of risk, it seems a reasonable objective to maximise its value. In the following, the effect on the adjustment coefficient of proportional and of excess of loss reinsurance arrangements will be considered. Throughout this section we will use the notation =X individual claim amount, =Y amount paid by the direct insurer and =Z amount paid by the reinsurer.

5.2 Proportional reinsurance

Let us consider the idea of a proportional reinsurance approach by way of an example:

Example Consider the insurer in the example on page 19. This insurer is investigating the possibility of using proportional reinsurance. He has approached a reinsurer, who uses a security loading of 50% to calculate his reinsurance premiums. If the insurer decides to reinsure 20% of each risk in the portfolio, estimate the effect the reinsurance will have on his probability of ruin at Time 2. Again you can assume that the aggregate claim distribution is approximately normal.



Solution We first need to calculate the reinsurance premium. Since the reinsurer takes responsibility for 20% of each risk, and uses a loading factor of 50%, the reinsurance premium (per annum) will be: 3.55

1(1 ) 1.5 30 0.2 313.32RRP m eq la= + = ¥ ¥ ¥ = So over a two year period, the insurer will pay 626.64 for the reinsurance. We now use (2)netS for the insurers aggregate payments (net of reinsurance). We need the mean and variance of (2)netS , which are, using the formulae for a compound Poisson distribution as before: 3.55[ (2)] 60 0.8 1,671.04netE S e= ¥ ¥ = and: 2 8.2var[ (2)] 60 0.8 139,812.49netS e= ¥ ¥ = So ruin will occur if: (2) 1,000 2,400 626.64 2,773.36netS > + - = Using a normal distribution approach as before, we have:

[ ] 2,773.36 1,671.04(2) 2,773.36 (0,1)139,812.49

1 (2.9481) 0.0016

netP S P NÈ ˘-> = >Í ˙Î ˚

= -F =

or about 0.16%.

Question 9.20

Comment on the usefulness of reinsurance in this context.



You should remember that if we have a retained proportion a then: =Y Xa (1 )= -Z Xa Hence: [ ] [ ] [ ]= =E Y E X E Xa a [ ] [(1 ) ] (1 ) [ ]= - = -E Z E X E Xa a

Question 9.21

Write down expressions for var[ ]Y and var[ ]Z .

5.3 Excess of loss reinsurance

We can apply the same type of logic if the insurer decides to buy excess of loss reinsurance. You might like to think about the effect on the probability of ruin if the insurer in the previous example purchases excess of loss reinsurance with an individual retention of 2,000, say, and a security loading of 50% as before. If we have a retention limit M , and no upper limit, then:

<Ï

= Ì ≥Ó

X X MY

M X M

0 <Ï= Ì - ≥Ó

X MZ

X M X M

Y can be also be written as min( , )X M , and Z can also be written as max(0, )-X M .

Example Calculate [ ]E Y if X has an exponential distribution with parameter 0.01, and the insurer has an excess of loss reinsurance arrangement with retention limit M .



Solution The formula for an expectation is ( )Ú

x

xf x dx . We have to calculate [ ]E Y by carrying

out two integrals, to allow for the two different ranges of X :

0

0.01 0.01

0

[ ] ( ) ( )

0.01 0.01

•

•- -

= +

= +

Ú Ú

Ú Ú

M

M

Mx x

M

E Y xf x dx Mf x dx

xe dx Me dx

Using integration by parts:

( )

0.01 0.01 0.01

0 0

0.01 0.01 0.010 0

0.01 0.01 0.01

0.01 0.01

0.01 0.01 0.010.01 0.01 0.01

10.01

1 10.01 0.01

1 (1 ) 100 10.01

M Mx x x

M

MMx x xM

M M M

M M

x e e dx M e

xe e Me

Me e Me

e e

∞− − −

∞− − −

− − −

− −

⎡ ⎤ ⎡ ⎤= − − − + −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡ ⎤⎡ ⎤ ⎡ ⎤= − − + −⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

= − − + +

= − = −

∫

Question 9.22

Calculate var[ ]Z (in terms of M ) if ~ (0,100)X U , where the insurer has an excess of loss reinsurance arrangement with retention limit M , 0 100< <M .



5.4 Maximising the adjustment coefficient under proportional reinsurance

First consider the effect of proportional reinsurance with retention α on the insurer’s adjustment coefficient. Throughout Section 5.4 the insurer’s premium income per unit time, before payment of the reinsurance premium, will be written as ( )1 1+ θ λm , which represents the expected aggregate claims per unit time for the compound Poisson process with a loading factor θ . It will also be assumed that the reinsurance premium is calculated as ( )( )1 1 1+ −ξ α λm . Since the reinsurer pays proportion 1− α of each claim, ( )1 1− α λm represents the reinsurer’s expected claims per unit time. Thus, ξ is the premium loading factor used by the reinsurer. Hence, the insurer’s premium income, net of reinsurance, is: [( ) ( )( )]1 1 1 1+ − + −θ ξ α λm (5.1)

Question 9.23

Explain this formula.

It will also be assumed that ξ θ≥ . If this were not true, it would be possible for the insurer to pass the entire risk on to the reinsurer and to make a certain profit. This of course ignores commission, expenses and other adjustments to the theoretical risk premium. For the insurer’s premium income, net of reinsurance, to be positive: 1 1 1+ > + −θ ξ α( )( ) ie α ξ θ ξ> − +( ) / ( )1

Question 9.24

What range of values is possible for α if θ = 0 2. and ξ = 0 4. ?

There is, however, a more important constraint on the insurer. The insurer’s net of reinsurance premium income per unit time must exceed the expected aggregate claims per unit time. Otherwise ultimate ruin is certain (as noted in Section 4.7). Net of reinsurance, the insurer’s expected aggregate claims per unit time are αλm1.



Thus: ( ) ( )( )1 1 1+ − + − >θ ξ α α or:

α θξ

> −1 (5.2)

Question 9.25

So what is the range of possible values of α now, given the figures in the previous question?

Equation (5.2) specifies the insurer’s minimum retention level since: 1 1− ≥ − +θ ξ ξ θ ξ/ ( ) / ( ) when θ ξ≤ the only case of interest. If the premium loading factors are equal, then inequality (5.2) becomes α > 0 . In this case there exists a risk sharing arrangement and any retention level is possible. If, however, ξ θ> then the insurer has to retain part of the risk. Same loadings First consider the case where both the insurer and the reinsurer use θ as the premium loading factor. The adjustment coefficient will be found as a function of the retention level α , when 0.1( ) 1 xF x e-= - . The distribution of the insurer’s individual claims net of reinsurance is exponential with parameter 01. / α . This can be seen by noting that if Y X= α , then:

P Y y P X y y[ ] [ / ] exp . / ≤ = ≤ = − −α α1 01 Note that the assumptions for the claims process and the adjustment coefficient equation apply equally well in the presence of reinsurance, provided that we use the net premium and the net claim amounts in the adjustment coefficient equation.

Question 9.26

What will be the general equation for R , the direct insurer’s adjustment coefficient, when there is reinsurance?



Hence, the equation defining R (see formula (4.2)) is:

λ θ λ α λ α α+ + = −∞z( ) ( . / ) . /1 10 01 0 10

R e e dxRx x

⇒ + + =−

1 1 10 11 10

( )θ αα

RR

(5.3)

⇒ =+

R θθ α( )1 10

for 0 1< ≤α (5.4)

It can be seen that R is a decreasing function of α . This is sensible as the smaller the retention, the smaller the risk for the insurer and so ψ ( )U would be expected to increase, and R to decrease, with α . Different loadings Now consider what happens when ξ θ> . For the rest of Section 5.4 assume that: the individual claim amount distribution is 0.1( ) 1 xF x e-= -

the insurer’s premium loading factor is θ = 01. .

Case A – 0.2=x Suppose first that the reinsurer’s premium loading factor is ξ = 0 2. , so that the insurer’s (net) premium income per unit time is ( )12 1α λ− . This comes from Equation 5.1. Equation (5.2) shows that the insurer must retain at least 50% of each claim. Hence, a value of α will be sought in the interval [0.5,1] that maximises the value of R . The equation defining R is:

λ α λ λα

+ − =−

( )12 11 10

RR

Question 9.27

Derive this formula for R .



This follows from (5.3) as only the premium is different − which leads to:

22 1

10(12 )R a

a a-=-

for 0 5 1. < ≤α (5.5)

The right hand side is based on the MGF of the net claim amounts which have an Exp( . / )01 α distribution.

Question 9.28

Derive this formula for R .

As when the loading factors were equal, the adjustment coefficient depends on the retention level. The value of α that maximises R in (5.5) is sought. Differentiate R with respect to α (using the quotient rule for differentiation) to give:

2

2 220(12 ) (2 1)10(24 1)

100(12 )dRd

a a a aa a a

- - - -=-

The quotient rule is 2

-Ê ˆ =Á ˜Ë ¯

du dvv ud u d dd v v

a aa

.

Alternatively the algebra is a little easier if you write R = −−

110

112 1α α

.

Now the denominator is always positive for values of α in [0.5,1] , so there will be a turning point of the function when: 20 12 2 1 10 24 12( ) ( ) ( )α α α α− = − − ie when: 24 24 1 02α α− + = The roots of this quadratic are 0.9564 and 0.0436, and so the turning point which is of interest is 0.9564.



Remember that α must lie in the range (0.5,1). Consider the following values:

α R 0.5 0.9564 1.0

0 0.00911 0.00909

This shows that R has a maximum in [0.5,1] at 0.9564. Alternatively you can show that the second derivative is negative. Figure 10 shows R as a function of α (as given by (5.5)) for values of α greater than 0.84. This range of α values has been chosen to highlight the important features of the graph. The dotted line shows the value of R when α = 1 (ie no reinsurance).

Figure 10 It can be seen from Figure 10 that there is a range of values for α , β α< < 1, such that if the retention level is in this range, the value of the adjustment coefficient exceeds the value when α = 1. The value of β can be calculated from (5.5) by setting the value of R at β equal to the value of R at 1, giving

0.9167b = . (You can check for yourself that β = 11 12/ .) The arrow in Figure 10 indicates the value of β .



In terms of maximising the adjustment coefficient, the optimal retention level is α = .0 9564 . It should be noted, however, that optimality in one sense does not imply optimality in another. For example, if the insurer does not effect reinsurance, then the expected profit per unit time is θλm1 (ie λ , since θ = 01. and m1 10= ). θλm1 is just the “loading” bit. If the insurer effects reinsurance with retention level 0.9564, then the expected profit per unit time is 0.9128λ (ie premium income, from (5.1), less expected claims). The expected profit per unit time is now found in terms of α and λ . It has already been calculated from (5.1), that with θ = 01. , ξ = 0 2. and m1 10= , the insurer’s net premium income is ( )12 1α λ− . The insurer’s expected claims per unit time are 10αλ . Hence, the expected profit per unit time is ( )2 1α λ− . If we put α = 0 9564. , this gives 0 9128. λ , as stated above. This shows that expected profit per unit time is an increasing function of α , and if the insurer were to choose α to maximise the expected profit per unit time, the choice would be α = 1. This example illustrates a general point – the level of reinsurance is a trade-off between security and profit. Case B – 0.3=x The value of α is now found that maximises R when the reinsurer’s premium loading factor is 0.3. The calculations are very similar to the previous case. From (5.1), the insurer’s net premium income is ( )13 2α λ− , so that the equation defining R is:

λ α λ λα

+ − =−

( )13 21 10

RR

which leads to:

R =−

−

3 210 13 22

α

α α( ) for 0 67 1. < ≤α



Or R = −−

110

113 2α α

, adopting the same approach as before.

By (5.2), the insurer must retain at least 2/3 of each claim so the value of a in the range [2/3, 1] that maximises R is sought. Differentiation gives:

dRdα

α α α α

α α=

− − − −

−

30 13 2 3 2 10 26 2100 13 2

2

2 2( ) ( ) ( )

( )

and the function has a turning point when: 30 13 2 3 2 10 26 22( = ( α α α α− − −) ) ( ) ie when: 39α α2 52 4 0− + = The roots of this quadratic are 0.0820 and 1.2514, so there are no turning points in the interval [2 3 ,1] and R as a function of α in this interval increases from 0 at α = 2 3/ to 0.00909 at α = 1. Thus, the value of α which maximises the adjustment coefficient is 1. It is not always possible to increase the value of the adjustment coefficient by effecting reinsurance. Note that when an insurer effects reinsurance, this reduces the variability of the insurer’s aggregate claims. A reduction in variability is associated with an increase in the value of the adjustment coefficient. However, when ξ θ> , the insurer’s premium loading factor, net of reinsurance, decreases, and the value of the adjustment coefficient is expected to decrease with the loading factor. When the reinsurer’s premium loading factor was 0.3, the reduction in the insurer’s security caused by the reduction in the loading factor has a greater effect on the adjustment coefficient than the increase resulting from reinsurance for all values of α . Loading factor (net of reinsurance) The insurer’s premium loading factor, net of reinsurance, implied by (5.1) is now found, and shown to be an increasing function of α . The loading factor is found by dividing the expected profit per unit time by the expected claims per unit time. The expected profit per unit time is: [( ) ( )( )]1 1 1 1 1+ − + − −θ ξ α λ αλm m This is just an algebraic expression for net premiums less expected net claims.



So the loading factor is:

[(1 ) (1 )(1 ) ] /

( ) /

q q x a a a

x x q a

= + - + - -¢

= - -

Now dd

′= −

θα

ξ θ α( ) / 2 which is positive since ξ θ> , so that ′θ is an increasing

function of α . Thus, the net loading factor increases as the retention level increases.

5.5 Maximising the adjustment coefficient under excess of loss reinsurance

In this section the effect of excess of loss reinsurance on the adjustment coefficient will be considered. The following assumptions will be made for Section 5.5: the insurer’s premium income (before reinsurance) per unit time is

( )1 1+ θ λm

the reinsurance premium per unit time is ( ) ( )1+ ξ λE Z , where ξ θ( )≥ is the reinsurer’s premium loading factor, and max(0, )Z X M= - .

The insurer’s individual net claim payments are distributed as min( , )Y X M= , and the insurer’s premium income, net of reinsurance, is: c m E Z* ( ) ( ) ( )= + − +1 11θ λ ξ λ (you may see this referred to as netc ) which gives the equation defining R as:

0

* ( ) [1 ( )]M

Rx RMc R e f x dx e F Ml lÈ ˘Í ˙+ = + -Í ˙Î ˚Ú

Question 9.29

Explain where the right hand side of this equation comes from.



This is formula (4.2) with a truncated claim amount distribution as a result of the excess of loss reinsurance. To illustrate ideas, look at the situation when

~ (0,20)X U , so that ( ) 0.05f x = for 0 20x< < . Then for 0 20M< £ :

20

2( ) ( ) 0.05 10 0.025M

E Z x M dx M M= - = - +Ú

and:

M R e dx e M

Re e M

YRx RM

M

RM RM

( ) . ( . )

. ( ) ( . )

= + −

= − + −

z 0 05 1 0 05

0 05 1 1 0 05

0

The equation for R must be solved numerically for given values of θ and ξ . Figure 11 shows R as a function of M when θ ξ= = 01. . As in Section 5.2, any retention level is possible when the premium loading factors are equal. R is a decreasing function of M .

Figure 11

In Figure 13, R goes to • as M goes to 0. When θ ξ< , there is a minimum retention level for the same reason as in the previous section. Recall that the lower limit for α given by Equation 5.2 applied when we were considering proportional reinsurance.



For example, when θ = 01. and ξ = 0 2. the insurer’s net premium income, c* , is

11 12 10 0 025 2λ λ− − +. ( . )M M and this must exceed the insurer’s expected claims, net of reinsurance. The insurer’s expected net claims equal λ λE X E Z( ) ( )− ,

which gives λ( . )M M− 0 025 2 . Thus: 2 21 1.2 0.03 0.025M M M M- + - > - fi 2 40 200 0M M- + < fi 5.8579 34.1421M< < Hence, the minimum retention level is 5.8579. Similarly, when ξ = 0 4. , the minimum retention level is 10. Figure 12 shows R as a function of M for the following combinations of θ and ξ : (a) θ = 01. and ξ = 0 2. (solid line) (b) θ = 01. and ξ = 0 4. (dotted line). Without reinsurance, ie for 20M = , the insurer’s adjustment coefficient is 0.014 (irrespective of the reinsurer’s loading factor). From Figure 12, it can be seen that, for ξ = 0 2. : ( ) (20)R M R> for 9.6 20M< < ( ) (20)R M R< for 9.6M < and for ξ = 0 4. : ( ) (20)R M R< for 20R <



Figure 12 Hence, for ξ = 0 2. it is possible for the insurer to increase the value of the adjustment coefficient by effecting reinsurance, provided that the retention level is above 9.6. However, when ξ = 0 4. , the insurer should retain the entire risk in order to maximise the value of the adjustment coefficient. As in the case of proportional reinsurance, the insurer’s expected profit per unit time is reduced if reinsurance is effected.

Question 9.30

Claims occur as a Poisson process with rate λ and individual claim sizes X follow an Exp( )β distribution. The office premium includes a security loading θ1 . An individual excess of loss arrangement operates under which the reinsurer pays the excess of individual claims above an amount M in return for a premium equal to the reinsurer’s risk premium increased by a proportionate security loading θ 2 . Derive and simplify as far as possible an equation satisfied by the adjustment coefficient for the direct insurer.

Question 9.31

Use the approximation e x xx ≈ + +1 22 / to find an approximate numerical value for the adjustment coefficient for the previous example in the case where β = 0 05. , θ1 0 3= . , θ 2 0 4= . and M = 10 .




We conclude this chapter with two exam-style questions:

Exam-style question 1 A general insurance company is planning to set up a new class of travel insurance. It plans to start the business with £2 million and expects claims to occur according to a Poisson process with parameter 50. Individual claims are thought to have a gamma distribution with parameters 150a = and ¼l = . A premium loading factor of 30% is applied. Explain how each the following changes to the company’s model will affect the probability of ultimate ruin: (i) A 28% premium loading factor is applied instead. [1] (ii) Individual claims are found to have a gamma distribution with parameters

150a = and ½l = . [1] (iii) The Poisson parameter is now believed to be 60. [1] [Total 3]

Exam-style question 2 Claims occur according to a compound Poisson process at a rate of 0.2 claims per year. Individual claim amounts, X , have probability function:

( 50) 0.7

( 100) 0.3

P X

P X

= =

= =

The insurer’s surplus at time 0 is 75 and the insurer charges a premium of 120% of the expected annual aggregate claim amount at the beginning of each year. The insurer’s surplus at time t is denoted ( )U t . Find: [ (2) 0]P U < [5]



This page has been left blank so that you can keep the chapter summaries

together as a revision tool.



Chapter 9 Summary The surplus of a general insurer at future times can be modelled using aggregate claims process models, which can be used to find the probability of ruin over a finite or infinite time horizon in discrete or continuous time. Claim numbers can be modelled using a Poisson process. Total claim amounts can be modelled using a compound Poisson process. For the continuous time model with an infinite time horizon, Lundberg’s inequality, which uses a parameter r called the adjustment coefficient, provides a good approximation for the probability of ultimate ruin. The upper bound for the probability of ultimate ruin given by Lundberg’s inequality is a decreasing function of r . As a result, an insurer may want to effect a system of reinsurance that maximises the value of r . The probability of ultimate ruin also decreases if the insurer’s security loading is increased or if the insurer’s initial surplus is increased. An increase in the value of the Poisson parameter will not affect the probability of ultimate ruin. However, it will reduce the time it takes for ruin to occur. An equation for the adjustment coefficient can also be obtained when reinsurance is used, by considering the direct insurer’s net premium income and net claims outgo. In practice, the security loading used by the reinsurer will be greater than the security loading used by the direct insurer.



Chapter 9 Formulae Surplus process U t u ct S t( ) ( )= + − , t ≥ 0 (continuous time) Ruin probabilities ψ ( ) [ ( ) ]u P U t t= < 0 for some (continuous time) ψ ( , ) [ ( ) ]u t P U t t t0 00= < ≤ for some ψ h u P U t t h h h( ) [ ( ) , , , ]= < =0 2 3 for some … (discrete time) ψ h u t P U t t h h h t t( , ) [ ( ) , , , ]0 00 2 3= < = ≤ for some and … Poisson process

P N t x p t t exx

x t[ ( ) ] ( ) ( )

!= = =

−λ λ ( x = 0 1 2, , ,… )

f t eTt( ) = −λ λ ( t > 0)

Compound Poisson process M u eS t

t M uX( )

[ ( ) ]( ) = −λ 1

[ ]E S t tE X( ) ( )λ= S t tE X 2[ ( )] ( )λ=var



Adjustment coefficient For a compound Poisson process, the adjustment coefficient r is the unique positive root of the equation:

λ λ+ =cr M rX ( ) Upper and lower bounds for r :

c E XrE X

2[ / ( )]( )

λ −< r

Mc m>

11log( / )λ

If the insurer uses a security loading of θ , the equation for the adjustment coefficient is: 1 1 1+ + =( ) ( )θ m r M rX If reinsurance is effected this equation becomes:

1 1 1+ + − + =( ) ( ) ( ) ( ) ( )θ ξE X E Z r M rY where ξ denotes the reinsurer’s security loading, Y represents the amount of an individual claim paid by the insurer, net of reinsurance, and Z denotes the amount of an individual claim paid by the reinsurer. For a general aggregate claims process, the adjustment coefficient r is the unique positive root of the equation: E er S ci[ ]( )− = 1, where Si denotes the aggregate claims from a risk in time period i and c is the constant premium charged to insure this risk. Lundberg’s inequality

ψ ( )u e ru≤ −



This page has been left blank so that you can keep the chapter summaries together as a revision tool.



Chapter 9 Solutions Solution 9.1

(i) Yes (ii) No (iii) Yes, because if we expand -xe as a power series and simplify, we get:

2 31

2! 3!x x xe x- = - + - +

This gives 1 12

- - + = +xe x x

x terms in 2x and higher powers.

Solution 9.2

As the amount of initial surplus increases, ruin will become less and less likely. So the limit is zero. Solution 9.3

( , )U ty involves checking for ruin at all possible times. Since the more often we check for ruin, the more likely we are to find it, we would expect that ( , )U ty would be greater than ( , )h U ty .



Solution 9.4

(i) The expected number of claims reported on a given day is 5. So the number of claims reported on a given day has a Poisson( )5 distribution and the probability that there will be fewer than 2 claims is:

P N P N P N e e( ) ( ) ( ) .< = = + = = + =− −2 0 1 5 0 0405 5 ie 4.0% Here we have used the formula for the Poisson probability, but alternatively,

using page 176 of the Tables, ( 2) ( 1) 0.04043< = £ =P N P N . (ii) The waiting time until the next event has an Exp( )5 distribution. We need to

find a probability using the exponential distribution. To do this, we can use the cumulative distribution function:

( ) 1 tP T t e l-< = - So the probability that there will be a claim (or several claims) during the next hour ( 1

24 of a day) is:

5241

24( ) 1 0.1881P T e-< = - = ie 18.8% Note that it is unlikely that the rate would be constant over time in reality.

Solution 9.5

Otherwise the insurer would be charging premiums that were less than the amount it expected to pay out in claims. In the real world this assumption may not always be true, especially during periods of competitive pressure when premium rates are soft.



Solution 9.6

The insurer’s surplus at the end of the second year will be:

(2) initial surplus + premiums expenses claims0.1m (100 200) 5,000 (100 200) 0.2 5,000 (2)1.3m (2)

US

S

= - -= + ¥ - + ¥ ¥ -= -

+

The distribution of (2)S is:

2 2(2) ~ (300 0.7 5,000, 300 (2.0 5,000) ) [1.05m,(0.173m) ]S N N¥ ¥ ¥ ¥ = The probability that the surplus will be negative at the end of the second year is:

2

[ (2) 0] [ (2) 1.3m]

( [1.05m,(0.173m) ] 1.3m)1.3m 1.05m1

0.173m1 (1.443) 0.074

P U P S

P N

< = >

= >-Ê ˆ= -FÁ ˜Ë ¯

= -F =

So the probability of insolvency at the end of the second year is 7.4%. Solution 9.7

Using the information given in the example, we have: 3.551, 200 (1 ) 30eq= + ¥ The solution of this equation is 0.14899q = . So the security loading is about 14.9%. Solution 9.8

θ is the security loading, and ψ ( )U is the probability of ruin for a fixed level of surplus U . If θ increases the premiums we charge will increase, and we should become more secure, ie the probability of ruin should fall.



Solution 9.9

The first reason is that Core Reading defines R to be the unique positive root, so R cannot be zero. The second reason is that 0=R should always be a solution to the equation. Why? Consider the LHS. 0(0) [ ] 1= =XM E e . Consider the RHS. 11 (1 ) 0 1+ + ¥ =mq . So

0=R is always a solution, but why do we ignore it? Consider Lundberg’s inequality, ( ) -£ RUU ey . If 0=R , we have an upper bound of 1 for the probability of ruin. That should have been obvious! In practice, ignore 0=R , just as we have done in the example. Solution 9.10

The adjustment coefficient satisfies:

1 1 1+ + =( ) ( )θ m R M RX

The distribution of the individual claim sizes X is:

X =RST10 000, with probability 0.925,000 with probability 0.1

So: ( ) ( ) 0.9 10,000 0.1 25,000 11,500= = = ¥ + ¥ =ÂE X xP X x and: 10,000 25,000( ) ( ) ( ) 0.9 0.1= = = = +ÂRX Rx R R

XM R E e e P X x e e The security loading is θ = 0 2. . So the equation for the adjustment coefficient is:

1 12 11500 0 9 0110 000 25 000+ × = +. , . ., ,R e eR R

ie 1 13 800 0 9 0110 000 25 000+ = +, . ., ,R e eR R



We can show that there is a solution in the range stated by looking at the values of LHS RHS− : At R = 0 00002599. : 1 13 800 0 9 01 0 00003510 000 25 000+ − + =, ( . . ) ., ,R e eR R At R = 0 00002601. : 1 13 800 0 9 01 0 00001810 000 25 000+ − + = −, ( . . ) ., ,R e eR R Since there is a reversal of signs (and we are dealing with a continuous function), the difference must be zero at some point between these two values, ie there is a solution of the equation in the range 0 00002599 0 00002601. .< <R . Solution 9.11

Here: ( ) 11,500=E X Also: 2 2 2 2( ) ( ) 0.9 10,000 0.1 25,000 152,500,000= = = ¥ + ¥ =ÂE X x P X x So:

R mm

< =× ×

=2 2 0 2 11500

152 500 0000 00003021

2

θ . ,, ,

.

So this is a reasonable initial estimate, compared with the correct value of approximately 0.000026. Solution 9.12

RM

c mM

> = + = =1 1 1 1

25 00012 0 000007291log( / ) log( )

,log . .λ θ

The steps used in the proof are equally valid for a discrete claims distribution. However, note that the lower bound obtained here is not very close to the accurate value for R .



Solution 9.13

( , )U ty is the probability of ruin at some point between times 0 and t . This should increase with time since the longer the time period, the more chance there is of ruin. It should be intuitively obvious that 1 2( , ) ( , )<U t U ty y for 1 2<t t , since if a scenario produces ruin before time 1t , then ruin has also occurred before time 2t . Solution 9.14

ψ ( ).

..

.15 111

0 232480 1 15

11= =− ×

e

Solution 9.15

R is worked out from R =+

=×αθ

θ11 01

11.

..

So:

0.11.11515 0.2557Re e- ¥- = =

Solution 9.16

Decreasing. The bigger the initial surplus, the less chance there should be of ruin. Solution 9.17

201.10.11(20) 0.14756

1.1ey - ¥= =

This is the limit to which the middle of the three lines is tending to as t tends to ∞ . Solution 9.18

Decreasing. If everything else remains unchanged, then increasing the premium income will reduce the probability of ruin.



Solution 9.19

This is probably easiest if we start by writing:

ψθ

θ( ) exp ( )U U U=+

+ −−11

1 1o t We need to differentiate this using the product rule:

( ) = +d dv duuv u vd d dq q q

Differentiating ( )Uy with respect to θ : d

dU U U U U U

θψ θ θ θ θ θ( ) ( ) ( ) exp ( ) ( ) exp ( )= + − + + − − + + −− − − − −1 1 1 1 11 2 1 2 1o t o t

Now substituting back in each term for ψ ( )U :

dd

U U U Uθ

ψ θ ψ θ ψ( ) ( ) ( ) ( ) ( )= − + − +− −1 12 1

This is the required expression. Solution 9.20

The reinsurance has reduced the probability of ruin to some extent, ie from about 0.25% to about 0.16%. However, this result is probably quite sensitive to the assumptions made (we are near the tail of the normal distribution), and slightly different assumptions might give us very different results. We will also want to look at the effect of reinsurance on profitability. As we are paying a reinsurance premium, it is likely that the overall effect on profitability is negative (although the effect on security is positive, as we have seen). There is likely to be a trade off between security and profitability here.



Solution 9.21

2var[ ] var[ ] var[ ]= =Y X Xa a

2var[ ] var[(1 ) ] (1 ) var[ ]= - = -Z X Xa a Solution 9.22

To find var[ ]Z , we need to find 2[ ]E Z , since 2 2var[ ] [ ] [ ]= -Z E Z E Z .

100

2 2 2

0

[ ] 0 ( ) ( ) ( )= + -Ú ÚM

M

E Z f x dx x M f x dx

The pdf of the (0,100)U distribution is 1100

(see page 13 of the Tables), so:

100100 2 3 3

2 ( ) ( ) (100 )[ ]100 300 300

È ˘- - -= = =Í ˙Í ˙Î ˚

ÚM M

x M x M ME Z dx

We now need [ ]E Z :

100100 2 2( ) ( ) (100 )[ ]

100 200 200È ˘- - -= = =Í ˙Í ˙Î ˚

ÚM M

x M x M ME Z dx

So:

23 2

3 4

(100 ) (100 )var[ ]300 200

(100 ) (100 )300 40,000

Ê ˆ- -= - Á ˜Ë ¯

- -= -

M MZ

M M



Solution 9.23

The insurer will charge a premium of: 1(1 ) [ ] (1 )+ = +E X mq l q l The reinsurer will charge a premium of: (1 ) [ ]+ E Zx l But 1[ ] (1 ) [ ] (1 )= - = -E Z E X ma a , so the reinsurer’s premium is: 1(1 ) (1 )+ - mx l a So the net premium received by the insurer is the difference: 1 1 1(1 ) (1 ) (1 ) (1 ) (1 )(1 )+ - + - = + - + -m m mq l x l a q x a l Solution 9.24

α > =0 214

01429..

.

But since α cannot exceed 1, the possible range of values is 01429 1. < ≤α . Solution 9.25

α > − =1 0 20 4

05..

. . So 05 1. < ≤α .



Solution 9.26

From the previous work, we know that the equation for R is: ( )+ = Xcr M rl l With reinsurance this will become: ( )+ =net Yc r M rl l for the direct insurer. But we know that (1 ) [ ] (1 ) [ ]= + - +netc E X E Zq l x l , so the equation for R becomes: ( )(1 ) [ ] (1 ) [ ] ( )+ + - + = YE X E Z r M rl q l x l l or: ( )1 (1 ) [ ] (1 ) [ ] ( )+ + - + = YE X E Z r M rq x In the case of proportional reinsurance this is: ( )1 (1 ) [ ] (1 )(1 ) [ ] ( )+ + - + - = YE X E X r M rq x a



Solution 9.27

Using the equation from the previous question where 0.1=q , 1[ ] 10

0.1= =E X and

0.2=x , we get:

( )( )

1 1.1 10 1.2(1 ) 10 ( )

1 11 12(1 ) ( )

1 (12 1) ( )

+ ¥ - - ¥ =

fi + - - =

fi + - =

Y

Y

Y

r M r

r M r

r M r

a

a

a

But what about ( )YM r ?

1 1( ) [ ] [ ] ( ) 1

0.1 1 10

-Ê ˆ= = = = - =Á ˜Ë ¯ -rY r X

Y XrM r E e E e M r

ra aa

a

Therefore the equation is:

11 (12 1)

1 10+ - =

-r

ra

a

Solution 9.28

Dividing the previous equation through by λ gives:

1 12 1 11 10

+ − =−

( )αα

RR

Multiplying through by 1 10− αR gives: ( )( )1 12 1 10 1+ − − =α αR R R Multiplying out the left hand side and subtracting 1 from both sides gives: ( ) ( )10 120 2 1 02 2α α α− + − =R R Dividing through by R and rearranging gives the required expression.



Solution 9.29

We need ( )YM rl , where <Ï

= Ì ≥Ó

X X MY

M X M and X has pdf ( )f x . By definition,

( ) [ ]= rYYM r E e , but we need to express this as two separate integrals to take into

account the different ranges of X :

0

0

[ ] ( ) ( )

( ) ( )

•

•

= +

= +

Ú Ú

Ú Ú

MrY rx rM

M

Mrx rM

M

E e e f x dx e f x dx

e f x dx e f x dx

But the second integral is just integrating the pdf from M to • . This is the same as

( )>P X M , which can be written as 1 ( )- F M . So the right hand side of the equation is:

0

( ) (1 ( ))È ˘Í ˙+ -Í ˙Î ˚ÚM

rx rMe f x dx e F Ml



Solution 9.30

The adjustment coefficient equation is λ λ+ =cR M RX ( ) . The net rate of premium income for the direct insurer equals the rate of premiums charged to the policyholder minus the rate of premiums paid to the reinsurer:

c x M e dxM

x= + − + −∞

−z( ) ( ) ( )1 1 11 2θ λβ

θ λ β β

The second term can be integrated using the substitution y x M= − , and identifying the integral as the mean of an Exp( )β distribution. This gives:

c e M= + − + −λβ

θ θ β1 1 11 2[( ) ( ) ]

The individual net claims are the claims paid to policyholders minus the recoveries from the reinsurer. So the MGF (which is valid for all values of R β< ) is:

M R e e dx e e dxR

R eXR x x RM x

M

MR M( ) [ ]( )= + =

−−− −

∞− −zz β β

βββ β β

0

1

So the equation for the adjustment coefficient is:

λ λβ

θ θ λβ

ββ β+ + − + =−

−− − −1 1 1 11 2[( ) ( ) ] [ ]( )e R

RR eM R M

Cancelling λ ’s and multiplying through by β β( )− R gives:

M R MR R e R R e ( )1 2( ) ( )[(1 ) (1 ) ] [ ]β ββ β β θ θ β β− − −− + − + − + = −

Cancelling the β 2 ’s from both sides gives:

− + − + − + = −− − −β β θ θ ββ βR R e R R eM R M( )[( ) ( ) ] ( )1 11 2

Cancelling R’s to exclude the trivial solution gives:

− + − + − + = −− − −β β θ θ ββ β( )[( ) ( ) ] ( )R e eM R M1 11 2

The adjustment coefficient R is the smallest positive solution of this equation.



Solution 9.31

Using the values given, the adjustment coefficient equation becomes:

− + − − = −− − +0 05 0 05 13 14 0 050 5 0 5 10. ( . )( . . ) .. .R e e R

Multiplying by −20 0 5e . to clear some of the fractions gives:

e R e e R0 5 0 5 101 20 13 14. .( )( . . )− − − = Expanding the LHS and applying the approximation to the RHS:

0 90538 14 8668 1 10 50 2. .+ = + +R R R

ie − + − =0 09462 4 8668 50 02. . R R Solving this using the quadratic formula (taking the smallest positive root) gives:

R =− + − − −

−=

4 8668 4 8668 4 50 0 094622 50

0 02682. . ( )( . )

( ).



Solution to Exam-style question 1 (i) Reduction in premium loading factor Since there is a smaller loading factor, the premiums will be reduced even though claims remain the same. Hence, the probability of ultimate ruin will increase. (ii) Change in claims distribution The mean of the distribution has decreased from 600 to 300, and the variance has decreased from 2,400 to 600. Therefore the claims are smaller on average and less uncertain. Both of these factors will decrease the probability of ultimate ruin. (iii) Change in the Poisson parameter The Poisson parameter has increased so claims occur more often (but their size is unchanged). However, the premium received will also increase proportionally (as

1(1 )c mq l= + ). Hence, the timing at which ruin may occur will be earlier, but not the probability of it occurring in the first place. Therefore, the probability of ultimate ruin will be unchanged.



Solution to Exam-style question 2 The annual premium is 120% of ( )E S , where S is the aggregate claim amount in a single year. Since S has a compound Poisson distribution with Poisson parameter 0.2, we have ( ) 0.2 ( )E S E X= where: ( ) 50 0.7 100 0.3 65E X = ¥ + ¥ = Hence the annual premium is: 1.2 ( ) 1.2 0.2 ( ) 1.2 0.2 65 15.6E S E X= ¥ = ¥ ¥ = The initial surplus is 75, so the surplus at time 2 is: (2) 75 2 15.6 (2) 106.2 (2)U S S= + ¥ - = - So the probability of ruin is: [ (2) 0] [106.2 (2) 0] [ (2) 106.2] 1 [ (2) 106.2]P U P S P S P S< = - < = > = - £ Considering [ (2) 106.2]P S £ , and remembering that (2) ~ (2 0.2)N Poi ¥ we have the following scenarios where the aggregate claim amount by time 2 is less than 106.2:

number of claims amount of claim(s) probability

0 claims 0 0.4 0.67032e- =

1 claim 50 0.40.4 0.7 0.18769e- ¥ = 100 0.40.4 0.3 0.08044e- ¥ =

2 claims 50, 50 2 0.4 20.4

2 0.7 0.02628e- ¥ =

Hence [ (2) 106.2] 0.9647 [ (2) 0] 0.0353P S P U< = fi < =

CT6-10: Generalised linear models Page 35



We conclude this chapter with two exam-style questions. The first is a short question on exponential families. The second is a longer question involving MLEs.

Exam-style question 1 (Subject 106, April 2003, Question 3) (i) A random variable Y has density of exponential family form:

( )( ) exp ( , )( )

y bf y c yaq q f

fÊ ˆ-= +Á ˜Ë ¯

State the mean and variance of Y in terms of ( )b q and its derivatives and ( )a f . [1] (ii)(a) Show that an exponentially distributed random variable with mean μ has a

density that can be written in the above form. (ii)(b) Determine the natural parameter and the variance function. [3] [Total 4]

Exam-style question 2 An insurer wishes to use a generalised linear model to analyse the claim numbers on its motor portfolio. It has collected the following data on claim numbers iy ,

1, 2, ..., 35i = from three different classes of policy: Class I 1 2 0 2 1 0 0 2 2 1 Class II 1 0 1 1 0 Class III 0 0 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 0 0 0

Page 36 CT6-10: Generalised linear models


You are given that:

10

111i

iy

==Â

15

113i

iy

==Â

35

164i

iy

==Â

It wishes to use a Poisson model to analyse these data. (i) Show that the Poisson distribution is a member of the exponential family of

distributions. [2] (ii) The insurer decides to use a model (Model A) for which:

1, 2, ..., 10

log 11, 12, ..., 1516, 17, ..., 35

i

iii

am b

g

=ÏÔ= =ÌÔ =Ó

where im is the mean of the relevant Poisson distribution. Derive the likelihood function for this model, and hence find the maximum likelihood estimates for a , b and g . [4] (iii) The insurer now analyses the simpler model log im a= , 1, 2, ..., 35i = , for all

policies. Find the maximum likelihood estimate for a under this model (Model B). [2]

(iv) Show that the scaled deviance for Model A is 24.93, and find the scaled

deviance for Model B. [5] You can assume that ( ) logf y y y= is equal to zero when 0y = . (v) Compare Model A directly with Model B, by calculating an appropriate test

statistic. [2] [Total 15]



Solution 10.18

If 2~ ( , )i iY N m s , then the Pearson residuals are -i iy ms

.

From page 27, the scaled deviance is 2

21 1

( )

= =

- =Â Ân n

i ii

i i

y dms

. The deviance residuals are

given by:

( ) ( ) - -- = - =i i i ii i i i i

y ysign y d sign y m mm ms s

Hence the Pearson residuals and the deviance residuals are the same. Solution to exam-style question 1 (Subject 106, April 2003, Question 3) (i) Mean and variance We have: [ ] ( )q= ¢E Y b var[ ] ( ) ( )Y a bf q= ¢¢ (ii)(a) Exponential form The PDF of the exponential distribution with mean m is:

1( ) expm m

Ï ¸= -Ì ˝

Ó ˛

yf y

This can be written as an exponential:

1( ) exp lnm m

Ï ¸= -Ì ˝

Ó ˛

yf y

Comparing this to the standard form given in part (i), we can define:

1 1, ( ) 1, ( ) ln ln( ), ( , ) 0q f q q fm m

= - = = - = - - =a b c y

(ii)(b) Natural parameter and variance function



The natural parameter is q , so here the natural parameter is:

1m

-

The variance function is (by definition) ( )q¢¢b , so here we find:

22

1 1( ) ( )b bθ θ μθ θ

′ ′′= − = =

Solution to exam-style question 2 (i) Exponential family For the Poisson distribution, we have: ( ) / !yf y e ymm-= We wish to write this in the form:

( )( ) exp ( , )( )

y bg y c ya

q q ff

È ˘-= +Í ˙Î ˚

So, rearranging the Poisson formula:

log( ) exp log !1

yf y ym m-È ˘= -Í ˙Î ˚

We can see that this has the correct form if we write: logq m= ( )b eqq m= = ( ) 1a f f= = ( , ) log !c y yf = -



(ii) Maximum likelihood estimates Using the rearranged form for the Poisson distribution from part (i), we see that the log of the likelihood function can be written: log ( , , ) log log !I II III i i i iL y ym m m m m= - -Â Â Â (*) This now becomes, for Model A:

10 15 35 35

1 11 16 1log 10 5 20 log !i i i i

i i i iL y y y e e e ya b ga b g

= = = == + + - - - -Â Â Â Â

35

111 3 4 10 5 20 log !i

ie e e ya b ga b g

== + + - - - -Â (**)

Differentiating this log-likelihood function in turn with respect to a , b and g , we get:

log 11 10L eaa∂ = -∂

log 3 5L ebb∂ = -∂

and:

log 4 20L egg∂ = -∂

Setting each of these expressions equal to zero in turn, we find that: ˆ log1.1 0.09531a = = ˆ log 0.6 0.51083b = = - and: ˆ log 0.2 1.60944g = = - These are our maximum likelihood estimates for a , b and g .



(iii) Simpler model In this case the log-likelihood function reduces to:

35 35 35

1 1 1log 35 log ! 18 35 log !i i i

i i iL y e y e ya aa a

= = == - - = - -Â Â Â (***)

Differentiating this with respect to a , and setting the result equal to zero, we find that:

ˆ 18ˆ18 35 0 log 0.6649835

ea a Ê ˆ- = fi = = -Á ˜Ë ¯

(iv) Scaled deviance for Model A and Model B The scaled deviance for Model A is given by: 2(log log )S ASD L L= - where log SL is the value of the log likelihood function for the saturated model, and log AL is the value of the log-likelihood function for Model A. For the saturated model, we replace the im ’s with the iy ’s in Equation (*) – as it fits the observed data perfectly – so the expected results are the observed results. So:

log log log !

4 2log 2 18 4log 2 4log 2 18 15.2274

S i i i iL y y y y= - -

= ¥ - - = - = -

Â Â Â

We use the hint in the question here. logi iy y is zero when 0y = , and also when

1y = . So the only contribution to the first term is when 2y = , giving 4 lots of 2log 2 . For the log likelihood for Model A, we replace the parameters a , b and g with their

estimates a , b and g in Equation (**):

35ˆˆ ˆ

1

ˆˆ ˆlog 11 3 4 10 5 20 log !

11log1.1 3log 0.6 4log 0.2 11 3 4 4log 2

27.6944

A ii

L e e e ya b ga b g=

= + + - - - -

= + + - - - -

= -

Â



The corresponding value for log AL without the final term is –24.9218. So the scaled deviance is twice the difference in the log likelihoods: ( )2(log log ) 2 ( 15.2274) ( 27.6944) 24.93S ASD L L= - = - - - = as required. We now repeat the process for Model B. Using Equation (***), the log likelihood for Model B is:

35ˆ

1ˆlog 18 35 log !

1818log 18 4log 2 32.742235

B ii

L e yaa=

= - -

Ê ˆ= - - = -Á ˜Ë ¯

Â

The value without the final term is –29.9696. The scaled deviance is again twice the difference in the log likelihoods: ( )2(log log ) 2 ( 15.2274) ( 32.7422) 35.03S BSD L L= - = - - - = (v) Comparing A with B We can use the chi-squared distribution to compare Model A with Model B. We calculate the difference in the scaled deviances (which is just 2(log log )A BL L- ): 35.03 24.93 10.10- = This should have a chi-squared distribution with 3 1 2- = degrees of freedom, which has a critical value at the upper 5% level of 5.991. Our value is significant here, since 10.10 5.991> , so that Model A is a significant improvement over Model B. We prefer Model A here.









CT6-14: Monte Carlo simulation Page 27


Relative error A similar argument can be applied when the relative error q q q- is used to

measure the accuracy. In this case the requirement which must be satisfied is that:

/ 2ˆ

| |z

nat e

q<

or, written more accessibly: 2 2 2 2

/ 2 ˆ /( )n za t e q> Of course the exact value of q is unknown, but the estimate q may be used instead of q in this equation.

Page 28 CT6-14: Monte Carlo simulation



Exam-style question 1 (Subject 103, April 2003, Question 4) (i) Given a pseudo-random number U uniformly distributed over [0,1] , obtain an

expression in terms of U and θ for a non-negative pseudo-random variable X which has density function:

( ) = qq - xf x e [2] (ii) A sequence of simulated observations is required from the density function:

( ) = ( ) , 01

qq

->

+

xeg x k xx

where θ is a non-negative parameter and ( )k θ is a constant of integration not

involving x . (a) Describe a procedure that applies the Acceptance-Rejection method to

obtain the required observations. (b) Derive an expression involving θ and ( )k θ for the expected number of

pseudo-random variables required to generate a single observation from the density g using this method. [6]

[Total 8]



Exam-style question 2 An insurer is using the random variable X to model its claim amounts. The probability density function of X is:

5( )(500 )

xf xx

b=+

0x ≥

where b is a constant. (i) Find the value of b . [2] (ii) Show that this distribution has mean 500, and variance 500,000. [2] (iii) Find the distribution function of the distribution. [2] (iv) Comment on the ease with which we could use the inversion method to generate random numbers from this distribution. [1] (v) By using a 2-parameter Pareto distribution with 3a = and 500l = as a

bounding distribution, generate as many random numbers as possible from this three-parameter Pareto distribution. You should use the random numbers from a uniform (0,1) distribution given below (in the order given). [6]

[Total 13] For generating values from the 2-parameter Pareto: 0.74 0.82 0.16 For making the acceptance-rejection decision: 0.26 0.59 0.71



7 Appendix –Linear Congruential Generators (LCGs)

To use an LCG we need to choose three positive integers:

• a, the multiplier

• c, the increment

• m, the modulus ,m a m c> > . To obtain the desired sequence of random numbers 1 2, , , Nu u u… , first generate a

sequence of integers 1 2, , , Nx x x… in the range 0,1,2, , 1m -… starting from an initial value 0x , known as the seed. The sequence of integers is generated using the recursive rule: ( )( )1 mod , 1, 2, ,n nx ax c m n N-= + = …

ie nx is the remainder when 1nax c- + is divided by m. After that, nu is set equal to

nx m . This method generates random numbers on the interval [ )0,1 rather than [ ]0,1 . (When writing ranges, mathematicians use square brackets to indicate that the endpoint is included in the range and round brackets to indicate that it is not. So these ranges correspond to 0 1£ < and 0 1£ £ .) In fact, it is clear that it can only give numbers of the form i m where

0,1,2, , 1i m= -… . In practice this doesn’t matter, assuming m is large enough, so we won’t distinguish between the two in this chapter.

Example 14.9 Use this algorithm with parameter values a = 11 , c = 37 , m = 100 and seed 0 85x = to generate 5 random numbers from the U ( , )0 1 distribution. How suitable do you think this method would be for generating random numbers?



Solution 14.4

We can generate 3 variates from Y by using the transformation: 4 1= +Y U . By using the first members of the pairs, this gives: 1.5, 4.86, 2.548 . We need to decide whether to accept or reject each of these. We have:

( ) ( )( )

( )( )( )( )

3 1 5 132 1 53 1 42 4

- -= = = - -

¥

x xf xg x x x

Ch x

It follows that: ( ) ( ) ( )1.5 0.4375, 4.86 0.1351, 2.548 0.948924= = =g g g We only accept a value x if ( )£u g x . In this case therefore (using the second members of the pairs of random numbers) we have: 0.342 0.4375, 0.274 0.1351, 0.812 0.948924< ≥ < It follows that we accept 1.5, we reject 4.86 and we accept 2.548. Solution 14.5

Our z values are: 1 2 log 0.802 cos(2 0.450) 0.632z p= - ¥ = - 2 2 log 0.802 sin(2 0.450) 0.205z p= - ¥ = You could have used the two initial values the other way round in which case you would get different answers. Using the polar method, we have: 1 2 0.802 1 0.604v = ¥ - = 2 2 0.450 1 0.1v = ¥ - = - 2 20.604 ( 0.1) 0.374816sfi = + - =



So our z values are:

12 log 0.374816 0.604 1.382

0.374816z -= ¥ =

and: 22 log 0.374816 0.1 0.229

0.374816z -= ¥ - = -

These are our values from the standard normal distribution. If we want values from a (3,8)N distribution, then we need to find values 1x and 2x where: 3 8x z= + using the 2 z values we have just calculated. Solution 14.6

We need 2 2

22

ˆ>

zn a t

e with 2

2 1.96=za , 100,000=e and 7ˆ 10=t . Therefore:

2 14

101.96 10 38, 416

10n ¥> =

Solution to exam-style question (Subject 103, April 2003, Question 4) (i) Pseudo-random variable We can use the inverse transform method here. The distribution function for this exponential distribution is (from the Tables): ( ) 1 -= - xF x e q Using fact that ( ) ~ (0,1)F X U , we get:

1( ) 1 log(1 )XU F x e X Uqq

-= = - fi = - -



(ii)(a) Procedure We have seen in part (i) how to generate values from the distribution with PDF ( )f x . The PDF of ( )g x is similar, so we can apply the acceptance-rejection method. First, we need to find the scaling constant C , which is:

0 0 0

( ) ( ) 1 ( ) 1 ( )sup sup sup( ) 1 1> > >

= = = =+ +x x x

g x k k kCf x x x

q q qq q q

So we can use the function ( ) 1( ) 1

=+

g xCf x x

to decide whether to accept or reject each of

the values generated from the exponential distribution in part (i). We accept a particular

value x with probability 11+ x

.

The algorithm for generating pseudorandom values with PDF ( )g x is: 1. Take a value ( = x ) generated from the exponential distribution in part (i). 2. Take a new (independent) pseudorandom value (= u ) from (0,1)U .

3. If 11

<+

ux

, accept the value x ; otherwise, reject it.

4. If more values are required, go back to step 1. Otherwise, stop. (ii)(b) Expected number

The probability of accepting each exponential value is 1C

. So the expected number of

exponential values required until we accept one is C , which equals ( )qq

k .

For each cycle of the algorithm we also need a pseudorandom number for step 2. So in fact, the total expected number of pseudorandom numbers required is twice this

number, ie 2 ( )qq

k .



Solution to exam-style question 2 (i) Value for β We will use the Tables. We see that this is an example of the three-parameter Pareto distribution (see Page 15 of the Tables). Here we have: 2k = , 3a = and 500l = So the value of b is given by the constant term in the PDF given in the Tables:

3( ) (5) 500

( ) ( ) (3) (2)k

k

aa lba

G + G= =G G G G

Using the fact that ( ) ( 1)!n nG = - from Page 5 of the Tables, we find that:

3

94! 500 1.5 102! 1!

b = = ¥

We could use integration by parts or by substitution to find b using ( ) 1f x dx =Ú , but it

is clearly much quicker just to match the distribution with one given in the Tables. (ii) Mean and variance Again we will use the formulae in the Tables. Using the parameter values given above:

2 500( ) 5001 2

kE X la

¥= = =-

and: 2 2

2 2( 1) 2 4 500var( ) 500,000

( 1) ( 2) 2 1k kX a la a

+ - ¥ ¥= = =- - ¥

These are the required values. Again, integration will work here, but using the Tables will be quicker. Also, we can see from the number of marks on offer that the examiners are not expecting quantities of heavy algebra. So go for the quick approach.



(iii) Distribution function This is not in the Tables, so we will have to integrate. We have:

50 0

( ) ( ) ( )(500 )

x x tF x P X x f t dt dtt

b= £ = =+Ú Ú

Integrating by parts, using the formula on page 5 of the Tables with u tb= and

5(500 )dv tdt

-= + :

( ) 4 45

00 0

( ) 500 (500 )4 4(500 )

x xxt tF x dt t t dtt

b b b- -È ˘= = - + + +Í ˙+ Î ˚Ú Ú

( ) ( )4 3

0

1500 5004 4 3

xx x tb b- -È ˘= - + + - +Í ˙Î ˚

3 4 312 500 4(500 ) 12(500 )x

x xb b b= - -¥ + +

4 314(500 ) 12(500 )

xx x

b b= - -+ +

where 91.5 10b = ¥ as before. [Total 2] Note that the distribution function tends to 1 as x tends to infinity, as expected. You do not have to integrate by parts here if you prefer substitution. Substituting

500u t= + will also work. Alternative solution to exam-style question 2

Substituting 500u t= + in the integral, and noting that 1dudt

= , so that du in the

integral can just be replaced by dt :

500 500

5 5 4 50 500 500

( 500) 500(500 )

x u x x

u

t udt du dut u u u

b b b b= + +

=

-= = -+Ú Ú Ú



500

3 4500

5003 4

x

u ub b +È ˘= - +Í ˙Î ˚

3 4 3 4500 500

3(500 ) 4(500 ) 3 500 4 500x xb b b bÊ ˆ Ê ˆ= - + - - +Á ˜Á ˜ Ë ¯+ + ¥ ¥Ë ¯

3 45001

3(500 ) 4(500 )x xb b= - ++ +

Note to markers – this expression looks different from (but is equivalent to) the expression obtained using integration by parts. There are a number of slightly different looking but equally acceptable expressions that students might have here:

4 3( ) 14(500 ) 12(500 )

xF xx x

b b= - -+ +

or: 3 4500( ) 1

3(500 ) 4(500 )F x

x xb b= - ++ +

or: 44 500( ) 112(500 )

xF xx

b bÈ ˘+= - Í ˙+Î ˚

All of these should be given full credit. There may be other equivalent expressions too (particularly if students have not kept b for the original constant). (iv) Inverting the distribution function In order to use the inversion method to generate random values from this distribution, we will need to invert the distribution function. This will not be easy to do. We would need some form of computer assistance to solve the resulting equations. (v) Acceptance-rejection method First we want the probability density function of the two parameter Pareto distribution:

3

43 500( )

(500 )h x

x¥=+



We can now calculate C :

4

5 3( ) (500 ) 4max max max 4( ) 500(500 ) 3 500

f x x x xCh x xx

bÏ ¸+Ô Ô Ê ˆ= = ¥ = =Ì ˝ Á ˜Ë ¯++ ¥Ô ÔÓ ˛

and: ( )( )4 ( ) 500f x xg xh x x

= =+

We can now simulate the required values. We will use 0.74 to simulate a value from our two-parameter Pareto distribution, and then use the random number 0.26 to decide whether to accept or reject it. Inverting the distribution function of our two-parameter distribution:

3

1/3500( ) 1 500 (1 ) 1500

u x x ux

-Ê ˆ È ˘= - fi = - -Á ˜ Î ˚Ë ¯+

Substituting in 0.74u = , we find that 283.39x = . We will accept this if 1 (283.39)u g< , where 1 0.26u = . Since 0.26 (283.39) 0.3617g< = , we accept this random number. Repeating the process:

1/ 3500 (1 0.82) 1 385.55x -È ˘= - - =Î ˚

But 2 0.59 (385.55) 0.4354u g= > = , so we reject this random number. Finally:

1/ 3500 (1 0.16) 1 29.920x -È ˘= - - =Î ˚

But 3 0.71 (29.92) 0.0565u g= > = , so we again reject this random number. So we have a sample containing the single value 283.39.









CT6: Q&A Bank Part 2 – Questions Page 3


Question 2.4 (Exam-style)

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 2.5 (Exam-style)

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]

Page 4 CT6: Q&A Bank Part 2 – Questions


(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] Chapter 6 Question 2.6 (Exam-style)

Show that, in the usual notation for Empirical Bayes Credibility Model 1: (i) ( ) [ ( )]=E X E m q

(ii) 2( ) ( )È ˘È ˘ =Î ˚ Î ˚E X m E mq q

(iii) [ ]21var( ) ( ) var ( )È ˘= +Î ˚X E s mn

q q [6]




ijX , 1, 2, , , 1, 2, ,= =… …i N j n is a set of data where ijX represents the aggregate claims in the j th year from the i th risk in a collective. It is assumed that this set of data satisfies all the assumptions for Model 1 of Empirical Bayes Credibility Theory. (i) Carefully describe the assumptions common to the distributions of ijX and

kmX ( πj m ) in the two cases when: (a) ijX and kmX come from the same risk, so that =i k (b) ijX and kmX come from different risks, so that πi k . [6] (ii) The table below gives values of ijX for each of five risks in the collective over

each of the past five years, together with some summary statistics. Use this set of data to calculate the empirical Bayes credibility premium for risk number 1 for the coming year.

Year j

1 2 3 4 5 iX 5

2

1

1 ( )4 =

-Â ij ij

X X

Risk i

1 85 73 91 88 70 81.4 87.3 2 62 107 104 71 68 82.4 456.3 3 122 144 99 80 150 119.0 879.0 4 72 78 90 88 98 85.2 105.2 5 170 81 153 140 175 143.8 1425.7

[8] [Total 14] Question 2.8 (Exam-style)

An insurer uses Empirical Bayes Credibility Model 1 to calculate the annual premiums for a collective of risks. The insurer has data for this collective from the last five years and estimates 2[ ( )]E s q to be 620 and var[ ( )]m q to be 145. Calculate the credibility factor Z for the coming year. [2]



Question 2.9 (Developmental)

The following table gives the aggregate amounts paid out (in millions of pounds) by four companies under a certain type of fire insurance over a period of five years, together with some summary statistics for Empirical Bayes Credibility Theory Model 1:

Year Insurer 1 2 3 4 5 iX 2

1

1 ( )4 =

-Ân

ij ij

X X

1 37 43 44 50 53 45.4 39.3 2 12 17 22 23 30 20.8 45.7 3 53 58 60 59 57 57.4 7.3 4 82 81 85 89 93

Complete the table by calculating the two missing entries, and use these data to find estimates for [ ( )]E m q , 2[ ( )]E s q and var[ ( )]m q . Hence estimate the empirical Bayes credibility premium for each of the insurers for the coming year. [9]




An actuarial student is using Empirical Bayes Credibility Theory Model 2 to calculate credibility premiums for a group of insurers. He has analysed the data for six different insurers, using 10 years of past data for each insurer. He has obtained the following figures:

6 10

1 11, 498

= ==ÂÂ ij

i jP * 18.24=P

[ ]( ) 4.00=E m q [ ]var ( ) 42.1=m q 2 ( ) 62.8È ˘ =Î ˚E s q

He has just received the following information relating to a 7th insurer (Insurer I), and he wishes to update his estimates using the past ten years of claims data for Insurer I given in the table below: Year i 1 2 3 4 5 Aggregate claims iY 100 85 90 102 109

Volume iP 22 24 26 20 25 Year i 6 7 8 9 10 Aggregate claims iY 106 128 132 150 131

Volume iP 30 29 35 40 36

(i) Calculate his updated estimates for [ ]( )E m q , 2 ( )È ˘Î ˚E s q and [ ]var ( )m q , and

hence find the credibility premium for Insurer I for the coming year, given that Insurer I is expected to have a volume figure for the coming year of 38. [20]

(ii) The student also needs a credibility estimate for Insurer K, one of the six

insurers included in the original analysis. He knows that, for Insurer K, 986=Â KjY and 327=Â KjP . Explain whether the credibility premium for

Insurer K (based on the full analysis of the seven insurers) will be greater or less than the corresponding figure for Insurer I (per unit of risk volume), and give reasons for your answer. [3]

[Total 23]



Chapter 7 Question 2.11 (Bookwork)

(i) State the two conditions that must hold for a risk to be insurable. [2] (ii) List five other risk criteria that would be considered desirable by a general

insurer. [5] [Total 7] Question 2.12 (Exam-style)

(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 2.13 (Developmental)

A compound distribution 1 2 NS X X X= + + + has claim number distribution:

P N n n n( ) ( )= = + − −9 1 4 2 , n = 0 1 2, , ,… If the individual claim size distribution X is exponential with a mean of 2, what are the values of ( )E S and var( )S ? [5]

CT6: Q&A Bank Part 2 – Solutions Page 9


Solution 2.6

(i) Since the jX ’s are identically distributed, we have: ( ) ( ) [ ( )] [ ( )]= = =j jE X E X E E X E mq q [2] (ii) Since ( )m q is a function of q only, it follows that:

( ) ( ) 2( ) ( ) ( ) ( )È ˘È ˘ È ˘ È ˘= = =Î ˚ Î ˚ Î ˚ Î ˚E Xm E E Xm E m E X E mq q q q q q [2]

(iii) Expressing var( )X in terms of conditional expectations and variances gives:

[ ]2

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

1 1var( ) var[ ( )] ( ) var ( )

= +

È ˘ È ˘= + = +Í ˙ Î ˚Î ˚j j

X E X E X

E X E X E s mn n

q q

q q q q

[2] Solution 2.7

(i) Assumptions In EBCT Model 1, we assume that: (a) when =i k : (1) ijX and kmX are identically distributed but not necessarily independent. [2] (2) If iq is the risk parameter associated with risk i , ij iX q and km iX q are

independent and identically distributed. [2] (b) when πi k : If iq and kq are the values of the risk parameters associated with the two risks,

,( )ij iX q and ,( )km kX q are independent and identically distributed. [2]

Page 10 CT6: Q&A Bank Part 2 – Solutions


(ii) Empirical Bayes credibility premium First we calculate the overall mean:

1 (81.4 82.4 143.8) 102.365

= + + + =X [1]

Secondly we calculate the mean of the individual sample variances:

5 5

2

1 1

1 1 1( ) (87.3 1425.7) 590.75 4 5= =

È ˘Í ˙- = + + =Í ˙Î ˚

Â Â ij ii j

X X [1]

Thirdly we calculate the sum of the squared deviations between the individual and overall sample means:

5 522 2

1 1

2 2 2

1 1( ) 54 4

1 (81.4 143.8 5 102.36 ) 781.594

= =

È ˘Í ˙- = - ¥Í ˙Î ˚

= + + - ¥ =

Â Âi ij j

X X X X

[2]

So our estimators are: [ ( )] 102.36=E m q 2[ ( )] 590.70=E s q [1]

1var[ ( )] 781.59 590.7 663.455

= - ¥ =m q [1]

So the credibility factor is:

25 0.84885590.70[ ( )] 5

663.45var[ ( )]

= = =++

nZE sn

mqq

[1]

So the premium for the coming year for the first risk is: 1 1 (1 ) [ ( )] 0.84885 81.4 0.15115 102.36 84.57= + - = ¥ + ¥ =P Z X Z E m q [1] [Total 8]



Solution 2.8

Using the formula:

2[ ( )] / var[ ( )]=

+nZ

n E s mq q

we get:

5 0.5395 620 /145

= =+

Z [2]

Solution 2.9

For Insurer 4 we have:

430 865

= =iX [1]

and 2 2 21 1 1( ) 5 (37,080 36,980) 254 4 4

È ˘S - = S - = - =Î ˚ij i ij iX X X X [1]

So we have the following estimates:

45.4 86[ ( )] 52.4

4+ +

= = =E m Xq [1]

2

1 1

2 1( )

11[ ( )]

1 (39.3 25) 29.3254

= =

--

=

= + + =

Â ÂN n

ij ii j

X Xn

E sN

q

[1]

2 2

1 1 1

2 2 2

1 1 1( ) ( )

1 1var[ ( )]

1 1(45.4 86 4 52.4 ) 29.325 727.9753 5

= = =

- - -- -

=

= + + - ¥ - ¥ =

Â Â ÂN N n

i ij ii i j

X X X XN Nn n

m q

[2]



So the credibility factor is given by:

2

5 0.992029.325E( ( )) 5

727.975Var( ( ))

nZsn

mqq

= = =Ê ˆ Ê ˆ+Á ˜+Á ˜ Ë ¯Ë ¯

[1]

So the EBCT premiums are given by:

Insurer 1: 0.9920 45.4 0.0080 52.4 £45.46¥ + ¥ = m

Insurer 2: 0.9920 20.8 0.0080 52.4 £21.05¥ + ¥ = m

Insurer 3: 0.9920 57.4 0.0080 52.4 £57.36¥ + ¥ = m

Insurer 4: 0.9920 86.0 0.0080 52.4 £85.73¥ + ¥ = m [2] [Total 9] Solution 2.10

(i) Updated estimates and the credibility premium Since X is our estimate of [ ]( )E m q , we have:

6 10

6 101 16 10

1 1

1 1

4.00 4 1, 498 5,992= =

= =

= =

= fi = ¥ =Â Â

Â ÂÂ Â

ij iji j

ij iji j

iji j

P X

P XP

[1]

From the data given for Insurer I:

10 10

1 11,133

= == =Â ÂIj Ij Ij

j jY P X and

10

1287

==Â Ij

jP [1]

So our new estimate for [ ]( )E m q is:

7 10

1 17 10

1 1

5,992 1,133 3.99161, 498 287

= =

= =

+= = =+

Â Â

Â Â

ij iji j

iji j

P X

XP

[1]



Now consider 2 ( )È ˘Î ˚E s q . The formula for the estimator for 2 ( )È ˘

Î ˚E s q is:

2

1 1

1 1 ( )1= =

Ï ¸Ô Ô-Ì ˝-Ô ÔÓ ˛Â ÂN n

ij ij ii j

P X XN n

[1]

Since 6=N (initially) and 10=n , we know that:

6 10

2

1 1( ) 62.8 9 6 3,391.2

= =- = ¥ ¥ =Â Â ij ij i

i jP X X [1]

We now need to add on the contribution to 2

1 1( )

= =-Â Â

N n

ij ij ii j

P X X from Insurer I. This

is 10

2

1( )

=-Â Ij Ij I

jP X X . We can calculate this by rewriting it as:

10 10 10 10

2 2 2

1 1 1 1( ) 2

= = = =- = - +Â Â Â ÂIj Ij I Ij Ij I Ij Ij I Ij

j j j jP X X P X X P X X P

But since

10

110

1

=

=

=Â

Â

Ij Ijj

I

Ijj

P X

XP

, the last two terms can be added together to give:

10 10 102 2 2

1 1 1

2 2 2

( )

100 131 1,13322 36 28722 36 287

4,539.0874 4,472.7840 66.3034

= = =- = -

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

= - =

Â Â ÂIj Ij I Ij Ij I Ijj j j

P X X P X X P

[2]

So the new total for 7 10

2

1 1( )

= =-Â Â ij ij i

i jP X X is:

3,391.2 66.3034 3,457.5034+ = . [1]



Our new estimate for 2 ( )È ˘Î ˚E s q is:

1 1 3,457.5034 54.8817 9

¥ ¥ = [1]

We now want the estimate for [ ]var ( )m q . The formula for the estimator for [ ]var ( )m q is:

2 2

1 1 1 1

1 1 1 1( ) ( )* 1 1= = = =

Ê ˆ- - -Á ˜- -Ë ¯

Â Â Â ÂN n N n

ij ij ij ij ii j i j

P X X P X XP Nn N n

[1]

Substituting in the numbers we know, we get:

6 10

2

1 1

1 142.1 ( ) 62.818.24 59 = =

Ê ˆ= - -Á ˜

Ë ¯Â Â ij iji j

P X X

This tells us that:

6 10 6 10 6 102 2 2

1 1 1 1 1 1( )

59 (42.1 18.24 62.8) 49,011.536

ij ij ij ij iji j i j i j

P X X P X X P= = = = = =

- = -

= ¥ + =

Â Â Â Â Â Â

[1] It follows from this that:

6 10

2 2

1 149,011.536 4 1, 498 72,979.536

= == + ¥ =Â Â ij ij

i jP X [1]

We can now get our new estimate for this statistic by adding on the contribution from Insurer I:

7 10

2

1 172,979.536 4,539.0874 77,518.6234

= == + =Â Â ij ij

i jP X [1]



So:

7 102 2

1 1( ) 77,518.6234 3.9916 (1,498 287)

49,078.449

ij iji j

P X X= =

- = - ¥ +

=

Â Â

[1] We now need our new value for *P . Using the formula for *P , we know that:

2

1 1 1

1 1* 11 1= = =

Ê ˆÊ ˆ= - = -Á ˜Á ˜- -Ë ¯ Ë ¯

Â Â ÂN N N

i ii i

i i i

P PP P P

Nn P Nn P

Now we know that 6

11,498

== =Â i

iP P , so:

62

1118.24 1,49859 1,498

=

Ê ˆÁ ˜Á ˜= -Á ˜Á ˜Á ˜Ë ¯

Â ii

P

This gives a value for 6

2

1=Â ii

P of 631,916.32.

So 7

2

1=Â ii

P is equal to 2631,916.32 287 714, 285.32+ = . We can now find our new *P :

72

71

1

1 1 714,285.32* (1,498 287) 20.0701569 69 (1,498 287)

=

=

Ê ˆÁ ˜ Ê ˆÁ ˜= - = + - =Á ˜Á ˜ +Ë ¯Á ˜Á ˜Ë ¯

ÂÂ

ii

ii

PP P

P [2]

So our new estimate for [ ]var ( )m q is:

( )1 1 49,078.449 54.881 32.7053320.07015 69

Ê ˆ- =Á ˜Ë ¯ [1]



So our three parameter estimates are: [ ]( ) 3.9916=E m q

2 ( ) 54.881È ˘ =Î ˚E s q

[ ]var ( ) 32.7053=m q We can now find the EBCT premium for the coming year for Insurer I. The credibility factor for Insurer I is:

287 0.99418754.88128732.705

= =+

IZ [1]

So the credibility premium per unit volume is:

1,1330.994187 0.005813 3.9916 3.94799287

= ¥ + ¥ =CP [1]

So the credibility premium for the coming year is: 3.94799 38 150.024¥ = [1] [Total 20] (ii) Insurer K

The estimate for KX is 986 3.0153327

= , which is lower than the corresponding estimate

for IX , and is also lower than X . The credibility factor for Insurer K is:

[ ]

2327 0.99489

( )327

var ( )

= =È ˘Î ˚+

KZE s

m

q

q

which is bigger than the corresponding factor for Insurer I. So we will be placing more emphasis on the individual mean for Insurer K, and this will pull the estimate down more. So the credibility premium per unit of volume will be lower for Insurer K than for Insurer I. [3]



Question 3.8 (Developmental)

Individual claim amounts in a Poisson claims process with a frequency of 50 claims per year for the whole portfolio have mean £5,000 and standard deviation £2,500. If the annual premium rate is £300,000, calculate the tightest upper bound for the adjustment coefficient. [2] Question 3.9 (Developmental)

An insurer calculates the annual premiums for fire insurance of flats by increasing the risk premium by 30% and adding a £30 loading. The claim frequency is 3% and individual claim amounts can be assumed to be:

£2,000 with probability 0.9

£15,000 with probability 0.1. Calculate the insurer’s adjustment coefficient for these policies, to 2 significant figures. [8] Question 3.10 (Exam-style)

A Poisson claims process has security loading 2 / 5=q and claim size density function: 3 73 7

2 2( ) , 0- -= + >x xf x e e x (i) Derive the moment generating function (MGF) for the claim size distribution,

and state the values of t for which it is valid. [3] (ii) Calculate the value of the adjustment coefficient. [4] [Total 7]




(i)(a) Show that the adjustment coefficient for a compound Poisson claims process satisfies the inequality:

22[ / ( )]

( )-

<c E Xr

E Xl

and define what each of the symbols represents. (i)(b) Explain how this inequality provides information about the probability of

ultimate ruin for the process. [10] (ii) An insurer considers that claims of a certain type occur in accordance with a

compound Poisson process. The claim frequency for the whole portfolio is 100 per annum and individual claims have an exponential distribution with a mean of £8,000.

(a) Calculate the adjustment coefficient if the total premium rate for the

portfolio is £1,000,000 per annum. (b) Verify that the value calculated in (ii)(a) satisfies the inequality in (i)(a). (c) The insurer decides to take out excess of loss reinsurance for this

portfolio. The reinsurer has agreed to pay the excess of any individual claim above £20,000 in return for an annual premium of £80,000. Calculate the adjustment coefficient for the direct insurer when the reinsurance is in operation.

(d) Estimate the direct insurer’s probability of ultimate ruin with and without

the reinsurance arrangement, assuming that the initial surplus is £20,000 and that future premiums remain at the same level.

(e) Comment briefly on the effect of the reinsurance on the probability of

ruin. [14] [Total 24]




Claims occur on a portfolio of insurance policies according to a Poisson process with Poisson parameter l . Claim amounts, 1 2, , ,…X X are assumed to be identically distributed with moment generating function ( )XM t . The insurer calculates premiums using a loading factor ( 0)>q . The insurer’s adjustment coefficient, R , is defined to be the smallest positive root of the equation: ( )+ = Xcr M rl l where c is the insurer’s premium income rate. (i) Using the above equation for R , or otherwise, show that, provided R is small,

an approximation to R is R , where:

2 22( / )ˆ -=

+cR l ms m

where [ ]= iE Xm and 2 var[ ]= iXs . [4] (ii) Describe how the adjustment coefficient can be used to assess reinsurance

arrangements on the basis of security. [3] (iii) The Poisson parameter, l , for this portfolio is 20 and all individual claims are

for a fixed amount of £5,000. The insurer’s premium loading factor, q , is 0.15 and proportional reinsurance can be purchased from a reinsurer who calculates premiums using a loading factor of 0.25.

Calculate the maximum proportion of each claim that could be reinsured so that

the insurer’s security, measured by R , is greater than the insurer’s security without reinsurance. [9]

[Total 16]



Chapter 10 Question 3.13 (Exam-style)

In the context of Generalised Linear Models, consider the exponential distribution with density function f x( ) , where:

f x e xx( ) ( )/= >−1 0μ

μ .

(i) Show that f x( ) can be written in the form of an exponential family of

distributions. [1]

(ii) Show that the canonical parameter, θ , is given by θμ

= −1 . [1]

(iii) Determine the variance function and the dispersion parameter. [3] [Total 5] Question 3.14 (Exam-style)

The random variable iZ has a binomial distribution with parameters n and im , where 0 1im< < . 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 j , ( )b q and ( , )c y j . [4] (ii) Determine the variance function of iY . [2] [Total 6]



Rearranging this inequality gives:

P S i i< + − + =−800 000 1 52 500 1 0 90, ( ) , ( ) .½ ½

Standardising, we now have:

½ ½

9

( ) 800,000(1 ) 52,500(1 ) 748,103.05var( ) 1.0639755 10

S E S i iPS

-È ˘- + - + -<Í ˙Í ˙¥Î ˚

[2]

But ( )var( )

S E SS

- is approximately standard normal. So we must have:

½ ½

9

½ ½

½

800,000(1 ) 52,500(1 ) 748,103.05 1.281551.0639755 10

800,000(1 ) 52,500(1 ) 748,103.05 41,802.415

800,000(1 ) 789,905.46(1 ) 52,500 0

i i

i i

i i

-

-

+ - + - =¥

+ - + - =

+ - + - = [2] Solving this quadratic in ( )½1+ i gives:

( ) , . , . , ( , ),

. ( . )

½1 789 905 46 789 905 46 4 800 000 52 5002 800 000

10499 0 0625

2+ =

± − × × −×

= −

i

or

So: i = − =10499 1 010232. . ie the required interest rate is 10.23% per annum. [3]



Solution 3.8

The adjustment coefficient satisfies the inequality:

2 2 22[ / E( )] 2[300,000 / 50 5,000] 0.000064

E( ) 5,000 2,500- -< = =

+c Xr

Xl [2]

Since we don’t know the precise distribution of the individual claim amounts, this is the best we can do.

Solution 3.9

The adjustment coefficient equation is: ( )+ = rXcr E el l The average claim size is: ( ) 0.9 2,000 0.1 15,000 £3,300= ¥ + ¥ =E X [1] So the annual risk premium is: ( ) 0.03 3,300 £99= ¥ =E Xl So the annual office premium is 1.3 99 30 £158.70= ¥ + =c . [1] So the adjustment coefficient equation is: 2,000 15,0000.03 158.70 0.03(0.9 0.1 )+ = +r rr e e [1] Simplifying by dividing by 0.03 and writing 1,000=R r gives: 2 151 5.29 0.9 0.1+ = +R RR e e [1] Expanding the RHS as a series to get a first approximation:

2 2

2

1 5.29 0.9(1 2 2 ) 0.1(1 15 112.5 )

1 3.3 13.05

+ = + + + + + + +

= + + +

R R R R R

R R



So: 5.29 3.3 0.152513.05

-ª =R [1]

Evaluating 2 15( ) 0.9 0.1 1 5.29= + - -R Rf R e e R for different values of R , we get:

(0.15) 0.37 (0.1) 0.018 (0.09) 0.013

(0.094) 0.0015 (0.095) 0.00156 (0.0945) 0.000011

= = = -

= - = =

f f f

f f f [2] So R lies between 0.094 and 0.0945. So the value of the adjustment coefficient correct to 2SF is 0.000094 (in units of 1£- ). [1] [Total 8] Solution 3.10

(i) Moment generating function The moment generating function is given by:

3 7 ( 3) ( 7)3 7 3 72 2 2 2

0 0 0

( 3) ( 7)

0 0

( )

3 7 3 72 3 2 7 2(3 ) 2(7 )

• • •- - - -

• •- -

È ˘= + = +Î ˚

È ˘ È ˘= + = +Í ˙ Í ˙- - - -Í ˙ Í ˙Î ˚ Î ˚

Ú Ú ÚtX tx x x t x t x

t x t x

E e e e e dx e dx e dx

e et t t t

[2]

The first integral converges if 3<t and the second if 7<t . So the MGF is valid for values of 3<t . [1] [Total 3] (ii) Adjustment coefficient For the adjustment coefficient, we require: 1 (1 ) ( ) ( )+ + = XE X r M rq



To find ( )E X , we can differentiate the MGF:

1 13 72 2

2 23 72 2

( ) (3 ) (7 )

( ) (3 ) (7 )

3 1 5(0)18 14 21

- -

- -

= - + -

= - + -¢

= + =¢

X

X

X

M t t t

M t t t

M [1]

Alternatively we can note that this distribution is a mixture of two exponential distributions. So the mean is given by a weighted average of two exponential means, ie

51 1 1 12 3 2 7 21¥ + ¥ = . Or we could integrate from first principles. So 5

21( ) =E X and the equation for r is:

7 5 3 7 3(7 ) 7(3 ) 21 515 21 2(3 ) 2(7 ) 2(3 )(7 ) (3 )(7 )

- + - -+ ¥ = + = =- - - - - -

r r r rr r r r r r

[1]

Multiplying through by 3:

63 153(3 )(7 )

-+ =- -

rrr r

Multiplying through by (3 )(7 )- -r r and multiplying out brackets, we obtain: 2 363 9 7 63 15- - + = -r r r r Gathering up the terms, we obtain: 2 36 7 0- + =r r r [1] Factorising: (6 )(1 ) 0- - =r r r which gives 1=r as the positive solution that satisfies the inequality 3<t . So the value of the adjustment coefficient in this case is 1. [1] [Total 4]



Solution 3.11

(i)(a) Adjustment coefficient The adjustment coefficient r is the smallest positive solution of the equation: ( )+ = rXcr E el l [1] The expected claim frequency l is the expected number of claims occurring per unit of time. [1] The premium rate c is the constant amount of premium actually received per unit of time. [1] The claim size X is a random variable representing the amount of an individual claim. [1] Expanding the RHS of the equation defining the adjustment coefficient gives:

2

2( ) [1 ( ) ( ) ]2

+ = = + + +rX rcr E e r E X E Xl l l [1]

Since the individual claim sizes X take positive values, the terms on the RHS are all positive. So, ignoring terms in powers higher than 2X gives:

2

2[1 ( ) ( )]2

+ > + + rcr r E X E Xl l [1]

Subtracting a l from both sides:

2

2[ ( ) ( )]2

> + rc r r E X E Xl [1]

Dividing by r (which must be a positive number):

2[ ( ) ( )]2

> + rc E X E Xl [1]

Rearranging to get an inequality for r gives:

22[ / ( )]

( )-

<c E Xr

E Xl [1]



(i)(b) Probability of ruin The inequality in (i)(a) gives an upper limit for the adjustment coefficient. The probability of ultimate ruin starting with a surplus of u is ( ) -ª ruu ey . Combining these results gives an approximate lower limit for the probability of ultimate ruin. [1] [Total 10] (ii)(a) Adjustment coefficient The adjustment coefficient equation is: ( )+ = rXcr E el l Since individual claims have an exponential distribution with mean 8,000: ( ) ( ) 1/(1 8000 )= = -rX

XE e M r r ( 1/ 8000<t ) So the adjustment coefficient satisfies: 100 1,000,000 100 /(1 8000 )+ = -r r [1] Dividing by 100: 1 10,000 1/(1 8000 )+ = -r r So: (1 10,000 )(1 8000 ) 1+ - =r r 21 2,000 80,000,000 1+ - =r r [1] Cancelling the 1’s and factorising: 2,000 (1 40,000 ) 0- =r r The adjustment coefficient is the smallest positive solution ie 1/ 40,000 0.000025= =r [1]



(ii)(b) Verify inequality Since X has an exponential distribution: ( ) 8,000=E X and: 2 2 2 2 2( ) var( ) [ ( )] 8,000 8,000 2 8,000= + = + = ¥E X X E X So the inequality in (i)(a) states that:

22[1,000,000 /100 8,000] 0.00003125

2 8,000-< =

¥r [1]

The exact value of r (ie 0.000025) is indeed less than this. (ii)(c) Adjustment coefficient with reinsurance The adjustment coefficient equation (with the reinsurance in effect) is: ( )+ = netr X

netc r E el l [1] Since the direct insurer will still have to pay a part of every claim, the claim frequency l is unchanged. The net rate of premium income (after paying the reinsurance costs) for the direct insurer is: 1,000,000 80,000 920,000= - =netc [1]



The MGF for the net claim amount netX paid by the direct insurer is:

20,000/8000 20,000 /8000

0 20,000

20,000 20,000(1/8000 ) /8000

0 20,000

20,000(1/8000 )20,000 20,000/8,000

20,0

( ) / 8,000 / 8,000

18,000 8,000

18,000(1/ 8,000 )

1

•- -

•- - -

- --

= +

= +

-= +-

-=

Ú Ú

Ú Ú

netrX rx x r x

rx r x

rr

E e e e dx e e dx

ee dx e dx

e e er

e 00 2.520,000 2.5

1 8,000

--+

-

rre

r [2]

So the adjustment coefficient equation becomes:

20,000 2.5

20,000 2.51100 920,000 1001 8,000

--È ˘-+ = +Í ˙-Í ˙Î ˚

rrer e

r [1]

Multiplying through by 1 8,000- r and dividing through by 100 gives: 20,000 2.5 20,000 2.5(1 9, 200 )(1 8,000 ) 1 (1 8,000 )- -+ - = - + -r rr r e e r Expanding and cancelling the 1’s: 2 20,000 2.51, 200 73,600,000 8,000 -- = - rr r r e Dividing by 100r : 20,000 2.512 736,000 80 -- = - rr e



Evaluating the function 20,000 2.5( ) 80 736,000 12-= - +rf r e r for trial values of r , we find: [1]

(0.000025) 4.43(0.00003) 1.89(0.00004) 2.83(0.000033) 0.42(0.000034) 0.06

(0.000035) 0.54

=== -== -

= -

fffff

f [1] So r is approximately 0.000034 (measured in units of 1£- ). [1] (ii)(d) Probability of ruin So the probability of ruin with and without the reinsurance are approximately: 0.000025 20,000 0.61- ¥ª =without ey and 0.000034 20,000 0.51- ¥ª =with ey [1] (ii)(e) Comment So the reinsurance reduces the probability of ultimate ruin. [1] [Total 14]



Solution 3.12

(i) Approximation to R The MGF of X can be written:

2 2

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

= = + + + = + + +tXX

tX tM t E e E tX t E X E X

So: 2

2( ) 1 ( ) ( )2!

= + + +XRM R R E X E X [1]

Assuming that R is small enough for terms in 3R and higher powers to be neglected, we can write:

2

2 2( ) 1 ( )2!

ª + + +XRM R Rm m s [1]

Substituting this into the defining equation for R :

2

2 2[1 ( )]2!

+ = + + +RcR Rl l m m s

Subtracting l from both sides and rearranging, we get:

2

2 2( ) ( )2!

- = +Rc R llm m s [1]

Dividing through by R and simplifying, we get:

2 2 2 22( ) 2( / )( )

- -= =+ +

c cR lm l ml m s s m

[1]

This is the required expression. [Total 4]



(ii) Assessment of reinsurance The adjustment coefficient can be used to assess the effectiveness of different reinsurance arrangements, using Lundberg’s inequality to find an upper bound for the probability of ruin for the insurer under different reinsurance arrangements. An arrangement that produces a lower upper bound for the probability of ruin is in some sense more secure for the insurer than an arrangement that has a higher upper bound for the probability of ruin. [3] Note however that the adjustment coefficient cannot tell us anything about the relative profitability of different reinsurance arrangements. This will need to be assessed using other means. (iii) Maximum reinsurance First we consider the insurer’s security without reinsurance. The equation for the adjustment coefficient is: 11 (1 ) ( )+ + = Xm r M rq Substituting in 0.15=q , 1 5000=m and 5000( ) = r

XM r e , we get: 50001 5750+ = rr e [1] The rate of premium income is: 1(1 ) 1.15 20 5000 115,000= + = ¥ ¥ =c mq l So, using the approximation derived for the adjustment coefficient in part (i), we have:

52

2(115000 / 20 5000) 6 105000

--ª = ¥R [1]

Now assume that a proportion k of each risk is reinsured. The net premium income is now:

[1.15 5000 1.25 5000] (5750 6250 )¥ - ¥ ¥ = -k kl l [1] The MGF of the insurer’s net claim payments is now: 5000(1 )( ) -= k R

XM t e [1]



So the equation for the insurer’s adjustment coefficient is now: 5000(1 )1 (5750 6250 ) -+ - = k tk R e [1] Using the same approximation to R as before, we have:

2 2 22[(5750 6250 ) 5000(1 )] 3 5

5000 (1 ) 50000(1 )- - - -ª =

- -k k kR

k k [1]

If we want the insurer’s security with reinsurance to be greater than without reinsurance, we want the adjustment coefficient with reinsurance to be larger, ie:

52

3 5 6 1050000(1 )

-- > ¥-k

k [1]

Rearranging this inequality, we get: 23 0- <k k [1] The solution of this is 0 1/ 3< <k . So the maximum proportion to be reinsured is

1333 %. [1]

[Total 9]

CT6: Assignment X1 Solutions Page 5


Solution X1.5

(i) Bayes criterion solution

The total profit (in £000s) corresponding to each level of client-days is:

1q 2q 3q

1d 1,360 1,520 1,760

2d 1,407 1,541 1,742

3d 1,250 1,350 1,500 [1] Notice that strategy 3d is dominated by the two other strategies so we can discard it. The expected total profit (in £000s) for each level of client-days is:

1(profit under ) (1,360 0.1) (1,520 0.6) (1,760 0.3) 1,576E d = ¥ + ¥ + ¥ =

2(profit under ) (1,407 0.1) (1,541 0.6) (1,742 0.3) 1,587.9E d = ¥ + ¥ + ¥ =

The Bayes criterion solution selects the client-days with the highest expected profit. Thus the solution is 2d , with an expected profit of £1,587,900. [2] (ii) Minimax and maximax solutions Remember that strategy 3d is dominated by the two other strategies so we can discard it. The worst possible outcomes associated with 1d and 2d are: 1,360 1,407 The minimax solution is the best worst case scenario (ie maximises the minimum gain) which is 1,407, so the minimax solution is 2d . [1] The best possible outcomes associated with 1d and 2d are: 1,760 1,742 The maximax solution is the best best case scenario (ie maximises the maximum gain) which is 1,760, so the maximax solution is 1d . [1]

Page 6 CT6: Assignment X1 Solutions


Solution X1.6

The likelihood function for m is given by:

2 2 2ln ln ln1 1 12 2 2

1

1 1( )2 2

n n ix x x

nL e e e

x x

m m ms s sm

s p s p

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

Since the prior distribution for m is uninformative, we take the prior distribution to be constant. Alternatively, we could say ( ) 1f km = where k Æ• . [½] So the PDF of the posterior distribution for m is given by:

2 2ln ln1 12 2( | ) prior likelihood constant

i ix x

f x e em m

s sm- -Ê ˆ Ê ˆ- -Á ˜ Á ˜Ë ¯ Ë ¯Â Â

μ ¥ = ¥ μ [½] We now want to write this as the PDF of a normal distribution, with m as the variable. So we want it to look like:

2* 2 2

*2 2* *

1 1 ( 2 )2 2( | )f x e e

m mm m m m

s sm*

Ê ˆ-- - - +Á ˜Ë ¯μ = (*)

So, expanding the bracket in our posterior and summing the terms (ignoring any expressions that do not involve m , since these can be absorbed into the constant term):

( ) ( )

( )

2 2 22 2

2 22

1 1ln (ln ) 2 ln2 2

1 (ln ) 2 ln2

( | )i i i

i i

x x x

x x n

f x e e

e

m m ms s

m ms

m- - - - +

- - +

Â Âμ =

Â Â= [½]

( )22

22

1 2 ln2

ln2

2

i

i

x n

xnn

e

e

m ms

m ms

- - +

Ê ˆ- -Á ˜Ë ¯

Âμ

Â= [1]

Completing the square (the missing term is absorbed into the constant):

2

2ln1

2( )( | )ix

nnf x em

smÊ ˆ

- -Á ˜Ë ¯Â

μ [1] Comparing this with equation ( )* , we see that we have a normal distribution with parameters:

2

2*

lnandix

n nsm s*= =Â [½]

CT6: Assignment X2 Questions Page 1


Question X2.1

(i) In the context of Empirical Bayes Credibility Theory Model 1, the credibility factor Z is given by:

2[ ( )]var[ ( )]

=+

nZE sn

mqq

Explain how changes in the values of n, 2[ ( )]E s q and var[ ( )]m q affect the

value of Z and comment on why Z behaves in this way. [3] (ii) The table below shows the aggregate claim statistics for each of four risks over

three years:

Risk i 3

1=Â ijj

X 3

2

1( )

=-Â ij i

jX X

1 2,184 22,344 2 2,721 20,294 3 3,450 21,800 4 3,099 23,994

Use EBCT Model 1 to calculate the credibility premium for the coming year for

Risk 3. [4] (iii) Without carrying out any calculations, determine with reference to part (i) what

would happen to the credibility factor if a fifth risk was added with:

3

51

3,999=

=Â jj

X and 3

25 5

1( ) 21,734

=- =Â j

jX X [2]

[Total 9]

Page 2 CT6: Assignment X2 Questions


Question X2.2

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] 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] 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]



Assignment X2 Solutions Solution X2.1

(i) Effect on Z The credibility factor Z is calculated as: credibility premium (1 )= ¥ + - ¥iZ X Z X where n is the number of years of past data. As n increases, Z increases. So more emphasis is put on the particular risk. This makes sense, since the more information we have from the relevant risk, the less emphasis we will wish to place on the collateral data. [1]

2[ ( )]E s q is a measure of the overall variability within each of the different risks in the

group. As 2[ ( )]E s q increases, Z decreases. So more emphasis is put on the collateral data. This makes sense, since the more variable the individual risk’s experience is, the less reliable it is. So we would expect to rely more on the collateral data. [1] var[ ( )]m q is a measure of the variability between the different risks in the group. As var[ ( )]m q increases, Z increases. So more emphasis is put on the particular risk. This makes sense, since the larger the variability between the different risks, the less relevant the other risks are in assessing the premium of our particular risk. So we want to rely more on the direct data. [1] [Total 3] (ii) Credibility premium The means for each of the risks are: 1 2 3 4728 907 1,150 1,033= = = =X X X X



The estimators for the parameters are: 1

4[ ( )] (728 907 1,150 1,033) 954.5ª = + + + =E m Xq [1]

3 42 21 1

4 21 1

14

[ ( )] ( )

(11,172 10,147 10,900 11,997)

11,054

= =ª -

= + + +

=

Â Â ij ii j

E s X Xq

[1]

3 3 42 21 1 1 1

3 3 4 21 1 1

2 2 213

2 13

1 13 3

var[ ( )] ( ) ( )

(728 954.5) (907 954.5) (1,150 954.5)

(1,033 954.5) 11,054

97,941 11,054

28,962.3

= = =

È ˘Í ˙ª - - -Í ˙Î ˚

È= - + - + -Î

˘+ - - ¥˚

= ¥ - ¥

=

Â Â Âi ij ii i j

m X X X Xq

[1] So, the credibility factor is:

3 0.8871411,054328,962.3

= =+

Z

Thus the credibility premium for risk 3 is: 0.88714 1,150 0.11286 954.5 1,128¥ + ¥ = [1] [Total 4] (iii) Addition of a fifth risk

This risk has a much higher mean, so the variance of the means, var[ ( )]m q , will increase. It has a similar variance to the other risks, so the mean variance of each risk,

2[ ( )]E s q , will remain similar. [1] Hence the proportionally larger var[ ( )]m q will lead to a larger Z, because more emphasis will be put on the direct data. [1]



Question X3.6

An analyst at a general insurance company is examining claims data on a portfolio of home insurance policies in a particular region. An exponential distribution models the claim amounts and the following rating factors are used: SA sum assured, x (as a continuous variable) PT property type, iT (as a factor with 1,2, ,10i = … ) NB number of bedrooms, jB (as a factor with 1,2, ,6j = … )

The table below shows 4 models considered by the analyst and their scaled deviances for the data set.

Model Parameterised form

of the linear predictor

Number of parameters Scaled deviance

SA xa b+ 2 238.4 SA + PT 206.7

SA + PT + SA iPT 178.3 SA*PT + NB 166.2 SA*PT*NB ij ij xa b+¢¢ ¢¢ 120 58.9

(i) Complete the table. [3] (ii) On the basis of scaled deviance which model should the analyst choose? [5] (iii) What further information should the analyst consider before making her

recommendation about an appropriate choice of model? [2] [Total 10]

Page 8 CT6: Assignment X3 Questions


Question X3.7

Claims for a particular risk arrive in a Poisson process rate l . The claim sizes are independent and identically distributed with density ( )f x and are independent of the claims arrival process. Assume there is a constant (0 )< < •g g such that lim ( )Æ

= •r

M rg

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

are received continuously at constant rate with premium loading factor 0>q . (i) (a) Define the adjustment coefficient, R . (b) Define the surplus process and the probability ( )uy of ruin with initial

surplus 0>u . (c) Write down Lundberg’s inequality. [3]

(ii) Derive the adjustment coefficient if 1( )-

=x

f x e mm

, 0>x , and 0.25=q . [4]

(iii) Consider the case where ( ) ½ (1 2 )- -= +x xf x e e , 0>x , and 0.25=q . (a) Calculate the expected claim size m . (b) Calculate the corresponding adjustment coefficient, and determine an

upper bound for (15)y . (c) Compare your answers to (iii)(b) with those obtained if the claim sizes

are mistakenly assumed to be exponentially distributed with mean m , and comment briefly. [11]

[Total 18]



Question X3.8

The underwriters at a general insurance company assume that household buildings insurance claims fall into one of two tiers: “small” and “large”. The ranges for the tiers and the proportion of claims falling in each range are as follows:

Range Proportion of claims “small” (£2, , £20, )000 000 95% “large” (£20, , £200, )000 000 5%

Within each tier claim amounts are uniformly distributed. All policies are charged the same annual premium, which assumes a claim frequency of 2% per annum and incorporates a 20% security loading. The company maintains a portfolio of 10,000 policies and currently has a surplus of £1m in respect of this class of business. (i) (a) Calculate c , the premium charged for each policy.

(b) Calculate the expected size of the surplus in one year’s time. [4] (ii) Show that, if all amounts are expressed in units of £000s, the adjustment

coefficient r satisfies an equation of the form:

r r p e p e p er r r+ = + +19 14 21

22

203

200.

where the p ’s are constants whose values you should specify. [7] (iii) By expanding the right hand side of the equation in (ii) as a series and discarding

terms containing r4 or higher powers, find an approximate value for r . [4] (iv) Given that the accurate value for r is 556 10 3. × − , estimate the probability of

ultimate ruin. [2] (v) Comment on the magnitude of your probability in (iv). [1] [Total 18]











(ii) Comparing models Comparing SA+PT with SA The difference in the scaled deviances is 238.4 206.7 31.7- = This is greater than 16.92, the upper 5% critical value of a 2 2

11 2 9c c- = distribution. So SA + PT is a significant improvement over the SA model. [1] Alternatively, using the (deviance) 2 (parameters)D > ¥ D approximation, we get 31.7 2 9> ¥ so the SA+PT model is a significant improvement over the SA model. Comparing SA*PT with SA+PT The difference in the scaled deviances is 206.7 178.3 28.4- = This is greater than 16.92, the upper 5% critical value of a 2 2

20 11 9c c- = distribution. So SA*PT is a significant improvement over the SA + PT model. [1] Alternatively, using the (deviance) 2 (parameters)D > ¥ D approximation, we get 28.4 2 9> ¥ so the SA*PT model is a significant improvement over the SA+PT model. Comparing SA*PT+NB with SA*PT The difference in the scaled deviances is 178.3 166.2 12.1- = This is greater than 11.07, the upper 5% critical value of a 2 2

25 20 5c c- = distribution. So SA*PT + NB is a significant improvement over the SA*PT model. [1] Alternatively, using the (deviance) 2 (parameters)D > ¥ D approximation, we get 12.1 2 5> ¥ so the SA*PT model is a significant improvement over the SA*PT+NB model. Comparing SA*PT*NB with SA*PT+NB The difference in the scaled deviances is 166.2 58.9 107.3- = This is less than 118.7, the upper 5% critical value of a 2 2

120 25 95c c- = distribution. We have to interpolate in the tables between 2

90c and 2100c to get the figure of 118.7.

So SA*PT*NB is not a significant improvement over the SA*PT+NB model. [1] Alternatively, using the (deviance) 2 (parameters)D > ¥ D approximation, we get 107.3 2 95> ¥ so the SA*PT*NB model is not a significant improvement over the SA*PT+NB model. So the analyst should choose the SA*PT+NB model. [1]



(iii) Further information The analyst should also check • that the SA*PT+NB model is a significant improvement when the order is

different, eg add the NB factor before the PT factor [½]

• other models involving these rating factors, eg SA*NB+PT [½]

• the residuals of the proposed model (to ensure that it is a good fit to the data) [½]

• the significance of the parameters of the proposed model (to ensure that all the estimated parameters are significantly different from zero) [½]

Solution X3.7

(i)(a) Adjustment coefficient The adjustment coefficient R is the smallest strictly positive solution of the equation: 11 (1 ) ( )+ + =m R M Rq

where 10

( )•

= Úm x f x dx is the mean of the claim amount distribution. [1]

You may just prefer to remember the equation ( )+ = XcR M Rl l , and the fact that

1(1 )= +c mq l . Putting these together gives the equation quoted above. (i)(b) Surplus process and probability of ruin The surplus process is: 1( ) (1 ) ( )= + + -U t u m t S tq l Here, ( )S t is the aggregate amount of claims received up to time t . The term

1(1 )+ m tq l represents the total premiums received to time t . [½] The probability of ruin ( )uy is the probability that the surplus process becomes negative at any point in the infinite time interval 0 < < •t . ie: ( ) ( ( ) 0 for some ,0 )= < < < •u P U t t ty [½]



(i)(c) Lundberg’s inequality Lundberg’s inequality states that:

( ) -£ Ruu ey where R is the adjustment coefficient and u is the initial surplus. [1] [Total 3] (ii) Derivation The moment generating function (MGF) of this distribution is: 1( ) (1 )-= -M R Rm So the equation for R in this case is: 11 1.25 (1 )-+ = -R Rm m [1] Multiplying through by 1- Rm , we get: (1 1.25 )(1 ) 1+ - =R Rm m [1] Multiplying out the brackets and simplifying, we find that:

15

=Rm

[2]

This is the value of the adjustment coefficient in this case. [Total 4] (iii)(a) Expected claim size The expected claim size m is now:

2

0 0 0

½ (1 2 ) ½ ½ 2• • •

- - - -= + = +Ú Ú Úx x x xx e e dx x e dx x e dxm



Looking at these two integrals, we see that the first one represents the mean of an (1)Exp distribution, and the second is the mean of an (2)Exp distribution. We know

that the mean of an ( )Exp l distribution is 1/ l , so: ½ 1 ½ ½ ¾= ¥ + ¥ =m [2] Alternatively we can integrate by parts to obtain the same answer. (iii)(b) Upper bound To find the adjustment coefficient, we first need the MGF of this distribution. Using a similar logic, we can express the MGF here as the average of the MGF’s of two exponential distributions. We then use the formula for the MGF of an exponential distribution from the Tables:

0

2

0 0

1 1

( ) ( ) ½ (1 2 )

½ ½ 2

½ (1 ) ½ (1 / 2)

•- -

• •- -

- -

= = +

= +

= - + -

Ú

Ú Ú

RX Rx x xX

Rx x Rx x

M R E e e e e dx

e e dx e e dx

R R

[2] We can now write down the equation for the adjustment coefficient: 1 11 1.25 ¾ ½ (1 ) ½ (1 / 2)- -+ ¥ = - + -R R R [½] We can write this as:

4 31 0.9375(2 2 )(2 )

-+ =- -

RRR R

Multiplying through by (2 2 )(2 )- -R R , gathering terms and simplifying, we get: 3 21.875 3.625 0.75 0- + =R R R [3] Dividing through by R and solving the resulting quadratic equation gives us

0.235610=R (or a number greater than one, which cannot be the solution since one of the MGF’s is not defined for values greater than one). [1]



So the adjustment coefficient is 0.2356=R , and the upper bound for the probability of ruin is: 15 0.0292- =Re [½] (iii)(c) Exponential distribution The corresponding results for the exponential distribution are:

1 0.266675 3/ 4

= =¥

R [½]

This gives an upper bound for the probability of ruin of: 15 0.0183- =Re [½] In the exponential case the upper bound is lower, and the adjustment coefficient is higher. These both suggest that the exponential case is more secure. This is not surprising, since claims are likely to be less variable in the situation where a single claim distribution is used than in the case where the claim distribution is a mixture of two distributions. [1] [Total 11] Solution X3.8

(i)(a) Premium The risk premium per policy is:

0 02 0 95 22 0002

0 05 220 0002

. . , . , £319× × + ×FHG

IKJ = [1½]

Incorporating the security loading gives: c = × =12 319 80. £382. [½]



(i)(b) Surplus The premium for each policy contains a loading of 0 2 319 80. £63.× = During the next year the surplus will be expected to increase by the total amount of the loadings for all policies in the portfolio, which will give: 1 000 000 10 000 6380 638 000, , , . £1, ,+ × = [2] [Total 4] (ii) Adjustment coefficient The adjustment coefficient is found from the equation: λ λ+ =cr M rX ( ) We are told that λ = 0 02. and we have calculated that c = 0 3828. (in units of £000s). The moment generating function is (again working in £000s):

M r E e e dx e dxXrX

rx rx( ) ( ) . .= = +z z0 95

180 05

1802

20

20

200 [2]

Evaluating the integrals:

M r e e

re e

r

re e e

X

r r r r

r r r

( ) . .

. . . .

=−F

HGIKJ +

−FHG

IKJ

= FHGIKJ + −F

HGIKJ + −FHG

IKJ

RSTUVW

0 9518

0 05180

1 0 05180

0 9518

0 05180

0 9518

20 2 200 20

200 20 2

[1]

So now let:

p10 9518

0 05278= − = −. .

p20 9518

0 05180

0 0525= − =. . .

and p340 05

1802 778 10= = × −. .

We can write the MGF as:

M rr

p e p e p eXr r r( ) = + +

11

22

203

200o t [2]



So we find that r is the solution of the equation:

0 02 0 3828 0 02 11

22

203

200. . .+ = × + +rr

p e p e p er r ro t [1]

Dividing by 0.02 and multiplying by r gives the required equation: 2 2 20 200

1 2 319.14 r r rr r p e p e p e+ = + + [1] [Total 7] (iii) Approximate value for r Expanding the RHS as a series gives:

2 2 31 11 2 6

2 31 12 2 6

2 3 41 13 2 6

19.14 [1 (2 ) (2 ) (2 ) ]

[1 (20 ) (20 ) (20 ) ]

[1 (200 ) (200 ) (200 ) ] ( )

r r p r r r

p r r r

p r r r o r

+ = + + +

+ + + +

+ + + + + [1] Using the numerical values for the p ’s, and summing down the columns, we find that:

2 2 3 419.14 0 15.95 440.3 ( )r r r r r o r+ = + + + + [1] As we would expect, r = 0 is a root of this equation. So, to find the nontrivial root, we cancel the r terms and divide by r2 , which gives: 219.14 15.95 440.3 ( )r o r= + + ⇒ ≈ × −r 7 245 10 3. [2] [Total 4] (iv) Probability of ultimate ruin From Lundberg’s inequality the probability of ultimate ruin with an initial surplus u is ψ ( )u e ru≈ − . So, with an initial surplus of 1,000 (in units of £000s), we have: ψ ( ) exp( . , ) ..u e≈ − × × = =− −556 10 1 000 0 00383 5 56 or 0.38% [2] (v) Comment on part (iv) The probability of ultimate ruin is very low. This is because there are relatively few large claims and, as we saw in part (i)(b), the surplus can be expected to grow quite quickly because of the 20% loading incorporated in the premiums. [1]









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