use the formulae for the moments given in the Tables: 0.2 ( ) 0.25 0.8
kq E N p = = and: 2 2 0.2 var( ) 0.3125 0.8 kq N p = = Page 40 CT6-08:
Risk models (2) IFE: 2010 Examinations The Actuarial Education
Company Alternatively, we could derive the moment

an insurance company becoming ruined. In Section 3 we will introduce
the adjustment coefficient, a parameter associated with risk, and
Lundbergs inequality. Section 4 considers the effect of changing
parameter values on the probability of ruin for an insu

; b = 100 Show that the company must sell at least 884 policies in a
year to be at least 99% sure that the premium income will exceed the
claims and expenses outgo. (ii) Now suppose that the values of ,
and b are not known with certainty but could be any

M t tT tC C C CC C From the information given in the question, since
Ci has a compound Poisson distribution it has moment generating
function: ( ) exp ( ( ) 1) exp 0.02( ( ) 1) = -= - [ ] [ ] C iX X i M t qM t M t
The random variable X takes the value m w

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, Xi i l q = 1. The claim number process
will be assumed to be a Poisson process, leading to a

that will be in any given range of values. As in the previous example,
it is assumed for simplicity that there is no uncertainty about the
moments or distribution of the individual claim amounts, ie about
F(x). The uncertainty about the value of can be mo

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 t . CT6-09: Ruin theory Page

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 Xi 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. Page 4

parameter . Important information The time between claims in a
Poisson process has an exponential distribution with parameter l .
Page 16 CT6-09: Ruin theory IFE: 2010 Examinations The Actuarial
Education Company Note that the inter-event time is independ

than on the average, ie the standard deviation is reduced by a greater
percentage than the mean. This is very often the case for excess-ofloss reinsurance. This was a difficult question, with lots of things
going on. If you got the main steps more or less

process. Of course, if you have previously studied Subject CT4, you
should be familiar with these ideas already. It can be seen that: = =
( ) 1 ( ) N t i i St X with the understanding that S(t) is zero if N(t) is zero.
The stochastic process cfw_ 0 ( )

you can keep the chapter summaries together as a revision tool. CT608: Risk models (2) Page 31 The Actuarial Education Company IFE:
2010 Examinations Chapter 8 Solutions Solution 8.1 Under this
arrangement, if the gross amount of an individual claim is X

to compare the answers to the above example with those to the first
example in the previous section. The values of the mean are in all
cases the same, as are the variances when a single policy is considered
(part (i). The difference occurs when variances

2010 Examinations (i) Now put: = 1; = 0 01 . ; b = 100 into these
formulae to show that: E[S] = 70n and var(S) = 127.802n Hence, S has
approximately a normal distribution with mean 70n and standard
deviation 127.80 n . The premium income is 80n and the sm

claims and expenses outgo. CT6-08: Risk models (2) Page 25 The
Actuarial Education Company IFE: 2010 Examinations 4 Exam-style
question Exam-style question 1 (Subject 106, April 2003, Q9 (part) A
portfolio consists of 500 independent risks. For the ith ri

Examinations 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

] var[ ] TYY Y = + + 1 2 500 " Since [ ] EY q i i = m and 2 2 var[ ] (1 ) Yq q
q ii i i = +- s m , we have: 500 1 [ ] m = = i i ET q 500 500 2 2 1 1 var[ ]
(1 ) s m = = =+ - i ii i i T q qq since the risks are independent. CT608: Risk models (2) Page 37 T

briefly review the key areas of Part 2, or maybe re-read the summaries
at the end of Chapters 5 to 8. Question and Answer Bank You should
now be able to answer the questions in Part 2 of the Question and
Answer Bank. We recommend that you work through sev

amount of the claim. Between claims the surplus increases at constant
rate c per unit time. The model being used for the insurers surplus
incorporates many simplifications, as will any model of a complex
real-life operation. Some important simplifications

compound Poisson distribution and so: [] ( ) E R E S np = = l l i l l l s = +
= + + 2 2 2 22 2 2 2 1 var [ ] ( ) ( var[ ] [ ] ) ( )( expcfw_ ) R ES S ES i ii p sn s
n Example Each year an insurance company issues a number of
household contents insurance p

Page 19 The Actuarial Education Company IFE: 2010 Examinations
3.4 Variability in claim numbers and claim amounts and parameter
uncertainty This section contains two more examples. The first is a
rather complicated example involving uncertainty over claim

compound Poisson distribution. CT6-08: Risk models (2) Page 35 The
Actuarial Education Company IFE: 2010 Examinations Now the
variance of I S is given by 2 var( )I S m = m , where 2 2 m EY = ( ) , and:
1,000 3 3 22 2 4 4 0 1,000 1,000 2 3 5 4 4 0 1,000 3

policyholder in this part of the portfolio is equally likely to be a good
driver or a bad driver but that it cannot be known whether a
particular policyholder is a good driver or a bad driver. Solution
Let i , i = 1, 2, . , n be the Poisson parameter of t

nn ( ) ( ) [ ( ) ( )] ( ) + = 1 + (2.4) and this identity holds for n =
1,2,3,. Now divide (2.4) by h , and let h go to zero from above to get
the differentialdifference equation: 1 ( ) [ ( ) ( )] n nn d pt p t pt dt = - l (2.5) Remember that the definiti

Tables, where the parameter is lt . This will be proved by deriving and
solving a differential-difference equation. For a fixed value of 0 t >
and a small positive value of h , condition on the number of claims at
time t and write: 1 1 ( ) ( )[ ( )] ( )[1

( ) = d fx f 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 ( ) ( ) Gs t ts cs ( , ) = 1 + where c(s) is some function of s.
As we are integrating with

portfolio. It may be helpful to think of this as a model of part of a
motor insurance portfolio. The policies in the whole portfolio have
been subdivided according to their values for rating factors such as
age of driver, type of car and even past claims

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 represents the average rate
of occurrence of claims (eg 50 per day), whic

individual claims can be subject to a reinsurance agreement, either
proportional or excess of loss. The individual risk model considers the
payments made under each policy (risk) separately. The risks are
assumed to be independent and the number of risks

calculate the skewness by subtraction either. Page 32 CT6-08: Risk
models (2) IFE: 2010 Examinations The Actuarial Education
Company Solution 8.3 W is the random variable: W X MX M = > | ie
the reinsurers conditional claim amount variable that we used in

practical situations, finding an exact value for the probability of ruin is
impossible. In some cases there are useful approximations to ( ) u ,
even if calculation of an exact value is not possible. 1.4 The probability
of ruin in discrete time The two pr

portfolio. Then N has a Poisson distribution with parameter 0.4n and S
can be written: S = i N = 1 (Xi + Yi) where cfw_ X Y i ii + = 1 is a
sequence of independent and identically distributed random
variables, independent of N. From this it can be seen t

insurer pays net of reinsurance. But: ( ) ( ) ( ) 375 ES S ES ES -= - = R R m
So the expected profit is 62 5. , and the percentage reduction in the
expected profit (which was 100 without reinsurance) is 37.5%. (iii)
Now consider the variance of the profit

the whole portfolio in one year in this approximate model. (ii)
Determine the mean and variance of T , and compare your answers to
those in part (i). [4] Assume that 0.02 qi = for all i , and if a claim
occurs, it is of size m with probability one. (iii)

0 . 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 x 0 ? (i) 2 x (ii) x e (iii)
1 - - + x e x For the purpose