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. Use the Poisson process and the distribution of inter-event times to

calculate probabilities of the number of events in a given time interval
and waiting times.

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.

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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
S = X1 + X 2 + + X N , 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 S (t ) = X1 + X 2 + + X N (t ) , 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.

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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:

f (x)
lim =0
x 0 x

You can use this notation to simplify your working. For example, the function
2 3
g ( x) = 3 x + 0.5 x + 0.004 x can be rewritten as g ( x ) = 3 x + o( x ) , since
0.5 x 2 + 0.004 x3
0 as x 0 . Note that o( x) does not represent an actual number so
x
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) x2 (ii) ex (iii) e- x - 1 + 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

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.

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X i i =1 is a sequence of random variables. N (t ) t 0 and S (t ) t 0 are both

families of random variables, one for each time t 0; in other words N (t ) t 0
and S (t ) t 0 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:

N (t )
S (t ) = Â Xi
i =1

with the understanding that S(t) is zero if N(t) is zero.

The stochastic process S (t ) t 0 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.

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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 U (t ) t 0 is a stochastic process, which is known as the cash flow
process or surplus process.

Figure 1

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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:

(U ) = P [U (t ) 0, for some t , 0 t ]

(U , t ) = P [U ( ) 0, for some , 0 t]

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(U ) is the probability of ultimate ruin (given initial surplus U) and (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:

(U 2 , t ) (U1 , t ) (1.1)

(U 2 ) (U1 ) (1.2)

(U , t 1 ) (U , t 2 ) (U ) (1.3)

lim (U , t ) = (U ) (1.4)
t

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, t ) ?

u

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 (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.

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For a given interval of time, denoted h, the following two discrete time
probabilities of ruin are defined:

h (U ) = P [U (t ) < 0, for some t , t = h, 2h, 3h, ]

h (U , t )=P [U ( )<0, for some , =h, 2h, ,t - h , t ]

Note that it is assumed for convenience in the definition of 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.

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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.

h (U 2 , t ) h (U1 , t ) (1.5)

h (U 2 ) h (U1 ) (1.6)

h (U , t 1 ) h (U , t 2 ) h (U ) (1.7)

lim h (U , t ) = h (U ) (1.8)
t

h (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.

lim h (U , t ) = (U , t ) (1.10)
h 0+

lim h (U ) = (U ) (1.11)
h 0+

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

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2 The Poisson and compound Poisson processes

2.1 Introduction

In this section some assumptions will be made about the claim number process,
lN(t )qt 0
, and the claim amounts, lX i q i 1
. The claim number process will be
assumed to be a Poisson process, leading to a compound Poisson process
l q
S(t ) t 0 for aggregate claims. The assumptions made in this section will hold
for the remainder of this chapter.

2.2 The Poisson process

If you have previously studied Subject CT3, the material in this section should be
familiar to you.

Recall from Subject CT3 that we use the term “Poisson process” to describe the number
of claims arising from a time period of length t.

If the number of claims N arising from a single time period has a Poisson distribution
with parameter l then the number of claims N (t ) which arise over a time period of
length t is a Poisson process, ie N (t ) has a Poisson distribution with parameter l t .

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 0 must satisfy the l q
following conditions:

(i) N (0) = 0 , ie there are no claims at time 0

(ii) for any t 0 , 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 lN(t )qt 0

is defined to be a Poisson process with
parameter l if the following conditions are satisfied:

(i) N (0) = 0 , and N (s ) N (t ) when s t

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(ii) P N (t + h ) = r | N (t ) = r = 1 - l h + o(h )

P N (t + h ) = r + 1| N (t ) = r = l h + o(h ) (2.1)

P N (t + h) r + 1| 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) implies that there can be a maximum of one claim in a very short time
interval h. It 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 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).

When studying a Poisson process the distribution of the time to the first claim
and the times between claims is often of particular interest.

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 T1 denote the time of the first claim. Then, for a fixed
value of t , if no claims have occurred by time t , T1 t . Hence:

P (T1 t ) = P (N (t ) = 0) = exp{ - l t }

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This last step follows from the formula for the probability function of a Poisson( t)
distribution with x = 0.

And:

P (T1 t) = 1 exp{ t}

so that T1 has an exponential distribution with parameter 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 i = 2,3, , let the random variable Ti denote the time between the (i - 1) th
and the i th claims. Then:

n n +1 n
P (Tn +1 t| Â Ti = r ) = P ( Â Ti t +r | Â Ti = r )
i =1 i =1 i =1

= P (N (t + r ) = n | N (r ) = n )

= P (N (t + r ) - N (r ) = 0 | N (r ) = n )

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

P (N (t + r ) - N (r ) = 0 N (r ) = n ) = P (N (t + r ) - N (r ) = 0)

Finally:

P (N (t + r ) - N (r ) = 0) = P (N (t ) = 0) = exp{ - l t }

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 .

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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 t 0
●
● l q
the stochastic process N (t ) t 0 is a Poisson process whose parameter
is denoted l .

This last assumption means that for any t 0 , the random variable N (t ) has a
Poisson distribution with parameter t , so that:

( t )k
P [N (t ) k] exp{ t} for k 0,1, 2,
k!

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With these assumptions the aggregate claims process, lS(t )qt 0

, is called a
compound Poisson process with Poisson parameter l . By comparing the
assumptions above with the assumptions in Section 2.2, it can be seen that the
l q
connection between the two is that if S(t ) t 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 .

Note the slight change in terminology here: “Poisson parameter ” becomes

“Poisson parameter t ” when a change is made from the process to the
distribution.

The common distribution function of the X i s will be denoted F ( x ) and it will be

assumed for the remainder of this chapter that F (0) = 0 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):

x
F ( x) = Ú f (t ) dt
-

The probability density function of the X i s, if it exists, will be denoted f ( x ) and

the kth moment about zero of the X i s, if it exists, will be denoted mk , so that:

mk = E [ X ik ] for k = 1, 2, 3,

Important information

mk = E[ X ik ] for k = 1, 2, 3,

Whenever the common moment generating function of the X i s exists, its value
at the point r will be denoted by M X (r ) .

In case you have forgotten, the definition of a moment generating function is:

M X (r ) = E[e rX ]

Note that we are using r for the dummy variable to avoid confusion with time t .

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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 tm1 , variance
tm 2 , and moment generating function MS (r ) , where:

MS (r )= exp {l t (M X (r ) - 1)}

Remember from Chapter 7 that if:

S = X1 + X 2 + + XN

where N ~ Poi(l ) , then:

E[ S ] = l E[ X ] = l m1 var[ S ] = l E[ X 2 ] = l m2

M S (r ) = M N [ln M X (r )]

These formulae can be found on page 16 of the Tables.

Here N (t ) ~ Poi (l t ) , therefore E[ S (t )] = l tm1 and var[ S (t )] = l tm2 . Also

M N (t ) (r ) = exp l t (er - 1) , so:

M S (t ) ( r ) exp t (eln M x ( r ) 1) exp t M X (r ) 1

For the remainder of this chapter the following (intuitively reasonable)

assumption will be made concerning the rate of premium income:

c > m1 (2.3)

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?

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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 assess their solvency position at the end of each year.

A small insurer with an initial surplus of £0.1m 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 incurs expenses of 0.2 P at the time of writing each policy.
Calculate the probability that the insurer will prove to be insolvent 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:

U (1) = initial surplus + premiums - expenses - claims

= 0.1m + 100 5, 000 - 100 0.2 5, 000 - S (1)

= 0.5m - S (1)

The distribution of S(1) is:

S (1) ~ N (100 0.7 5,000, 100 (2.0 5,000) 2 ) . m) 2 ]

N [0.35m,( 01

So the probability that the surplus will be negative is:

P[U (1) 0] = P[ S (1) 0.5m]

= P( N [0.35m,(0.1m) 2 ] 0.5m)

Ê 0.5m - 0.35m ˆ
= 1- ÁË ˜¯ = 1 - 0.93319 = 0.067
0.1m

So the required probability is 6.7%.

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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 m = 3 and s 2 = 1.1 . 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):

E S (2) = 60e3+0.55 = 2, 088.80

and: var S (2) = 60e6+ 2.2 = 218, 457

Ruin will occur if S (2) is greater than the initial surplus plus premiums received. So we
want:

3, 400 - 2, 088.80
P S (2) 1, 000 + 2 1, 200 P N (0,1) = 1 - (2.8053) = 0.0025
218, 457

The probability of ruin is approximately 0.25%.

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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 (1 ) m1

where ( 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:

c = (1 + ) E ( S ) = (1 + )l m1

where is the security loading. 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 17?

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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 )] = l t E ( X ) var[ S (t )] = l t E ( X 2 )

The moment generating function of the process is given by:

M S (t ) (r ) = exp l t ( M X (r ) - 1

2.6 A technicality

In the next section a technical result will be needed concerning M X (r ) (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 (0 ) such that
M X (r ) is finite for all r and:

lim M X r = (2.4)
-
r

(For example, if the X i s have a range bounded above by some finite number,
then will be ; if the X i s have an exponential distribution with parameter ,
then will be equal to .)

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):

e10r 1
M X (r )
10r

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

1
M X (r ) (1 r / )
r

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This tends to infinity as r tends to from below.

In the next section the following result will be needed:

lim (l M X (r ) - cr ) = (2.5)
-
r

If is finite, (2.5) follows immediately from (2.4).

This is because , c and r would all have finite values in the limit.

Now it will be shown that (2.5) holds when is infinite. This requires a little
more care. First note that there is a positive number, say, such that:

P[ X i ] 0

The reason for this is that all claim amounts are positive.

So, if we pick a small enough number ( 0.01 maybe), we’re bound to get a
proportion of claims whose amount exceeds this.

This probability will be denoted by . Then:

M X (r ) er

This follows by considering the claims below and above :

M X (r ) = E (e rX ) = E (e rX | X ) P( X ) + E (e rX | X ) P( X )

0 + er p

Hence:

lim (l M X (r ) - cr ) lim (l e r p - cr ) =
r r

Here the e r 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 (l M X (r ) - cr ) =
r

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 CT6-09: Ruin theory Page 21

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:

(U ) exp{ RU }

where U is the insurer’s initial surplus and (U ) 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 (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 (U ) is given in Section 4.2.)

10
In fact it can be shown that the value of R in this case is .
11

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

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Page 22 CT6-09: Ruin theory

Figure 1

In the actuarial literature, exp{ -RU } is often used as an approximation to (U ) .

R can be interpreted as measuring risk. The larger the value of R , the smaller
the upper bound for (U ) will be. Hence, (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, , when:

(i) the claim amount distribution is exponential with mean 10, and

(ii) all claims are of amount 10.

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CT6-09: Ruin theory Page 23

Figure 2

Note that in both cases, R is an increasing function of . This is not surprising

as (U ) would be expected to be a decreasing function of , and since
(U ) expl q
RU , any factor causing a decrease in (U ) would cause R to
increase.

Question 9.8

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

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 (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.

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 Page 24 CT6-09: Ruin theory

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:

l M X (r ) - l - cr = 0 (3.1)

So, R is given by:

l M X (R ) = l + cR (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 (1 ) m1 gives:

M X (r ) 1 (1 )m1r

so that R is independent of the Poisson parameter and simply depends on the

loading factor, , and the individual claim amount distribution.

You can see from this equation that all the ’s have cancelled.

Important information

The adjustment coefficient can be found by solving the equation:

M X (r ) cr or M X (r ) = 1 + (1 + )m1r

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CT6-09: Ruin theory Page 25

Since the rate of claims outgo per unit time is m1 (the mean of a compound Poisson
distribution), then if a loading factor of is used, the rate of premium income will be
c (1 ) m1 .

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:

M X (r ) = 1 + (1 + )m1r

Ê 1 ˆ
We have X ~ Exp Á ˜ , so that M X (r ) = (1 - 500r ) -1 (this comes from the Tables),
Ë 500 ¯
= 0.3 , and m1 = E[ X ] = 500 . Substituting these into the equation:

(1 - 500r ) -1 = 1 + 1.3 500r = 1 + 650r

Rearranging:

1 = (1 - 500r )(1 + 650r )

1 = 1 - 500r + 650r - 325, 000r 2

0 = 150 - 325, 000r

r = 0.000462

From Lundberg’s inequality, (U ) e - rU , so here:

(U ) e -0.000462 1,000
= 0.630

Note that we didn’t use the Poisson parameter in our solution.

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

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