UEA Actuarial Methods CMP-5001B 2015/2016
Actuarial methods – Calculations
Lecture 1
Introduction
These six lectures run alongside the sessions at Aviva. They develop in more detail some of the
calculations that may be used in General Insurance. They also develop applications for some of the
statistical techniques you studied last semester. The material in these six lectures will be examined in a
two hour exam in May/June. There should be no conflict between the material taught here and that
taught in other courses. If you feel there is a difference in emphasis please talk to me. Aviva are
approaching the subject from the viewpoint of practical experience whereas I am sticking more rigorously
to the approach in the institute/faculty of actuaries syllabus and course (CT6).
This first lecture will look at some statistical distributions and their applications in general insurance.
In the second lecture we look at risk models.
The third lecture will take a simple look at ruin theory which will be developed more fully in the final year.
In the fourth lecture we will develop some techniques based on Bayesian statistics.
The fifth lecture will continue looking at Bayesian methods and specifically credibility theory.
The sixth lecture will talk about run off triangles. This session is where there will be a degree of overlap
with the Aviva course and I will be relying on the fact that you have been taught the chain ladder method
in their session eight, and spend more time on developing other methods.
Background to General Insurance
Features of Insurance
The policyholder must have an interest in the risk being insured. This is a primary difference between
insurance and a bet. So you can take out a policy to cover your own home but not to cover the home of
your accident prone friend.
The risk must be relatively easy to calculate in financial terms. The relevant calculation basis will be set
out in the contract eg if your TV explodes you may get a new replacement of a similar type or you may get
only the money to replace with a similar model of a similar age, depending on the wording of your
contract.
Ideally all risks insured will be independent of one another. This will not always be possible, an insurer
will not refuse to insure a property because they already insure the neighbour.
Events that are being insured should be relatively rare. You cannot insure against the grass in your
garden growing.
Insurance groups large numbers of similar (but still ideally independent) risks this should reduce the
variance and enable you to estimate the potential cost more easily. If you are insuring static caravans
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UEA Actuarial Methods CMP-5001B 2015/2016
against fire your ideal portfolio would include a large number of caravans spread over a large number of
caravan parks rather than the same number of caravans spread over 10 caravan parks in one county.
The insurer needs to set a limit on the liability they can incur under each contract. This may be by writing
a limit in the contract or by the use of reinsurance which we will talk about later on today. The limit could
be the replacement cost of an appliance.
The insurer will try to prevent moral hazards. So the terms of a motor insurance policy will discourage
you from driving into a wall to write the car off, for example they will not replace your 10 year old car
with a brand new one.
Clearly it will not be possible to meet all the above criteria in every insurance contract.
Contracts will normally be for a fixed time period, very often this will be a year but for example travel
insurance may only cover one trip, although even that will usually cover you for some risks from the time
you book the trip.
Usually the claims are not for a fixed amount as losses can vary, for example if your home is burgled,
payment will be based on what has actually been stolen not on a set amount for any burglary.
A claim occurring does not usually bring the policy to an end, in fact you may claim more than once in a
year, you may have a stone hitting the car windscreen and claim for new glass and then be involved in an
accident a few weeks later.
Although the policy itself is usually for a fixed term claims may occur at any time during the term or even
after this term. This is particularly the case for say industrial injuries, it may be many years before
damage due to inhalation of asbestos comes to light for example. Estimating these future claims is a large
part of the work we will be looking at in this course.
Reinsurance
Reinsurance is a type of insurance that the insurer takes out to protect itself against excessive payouts to
its policyholders. There are special insurance companies known as reinsurance companies who specialise
in this business.
We will in this course be especially interested in two types of reinsurance known as excess of loss
reinsurance and proportional reinsurance. It is important you understand how they work because it is
possible they may crop up in questions across the range of calculation techniques we develop.
Excess of loss reinsurance
The insurer and reinsurer agree what is known as a retention limit. The insurer pays the full claim value if
the value is less than the retention limit. If the claim amount is more than the retention limit then the
insurer pays the retention limit and the reinsurer the excess over this limit. So for example the insurer
has a retention limit of £40,000 on motor insurance claims, the majority of these will be less than £40,000
so the insurer pays the full amount but if somebody crashes a brand new Ferrari and writes it off then the
insurer will pay out £40,000 and the reinsurer the value say £150,000 less the retention so in this case
they will pay £110,000.
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Proportional reinsurance
The other form of reinsurance we will consider is proportional where it is agreed that the insurer and
reinsurer will each pay the same proportion of every claim. So if the proportion is 80% insurer 20%
reinsurer then for a repair costing £1500 the insurer will pay £1200 and the reinsurer £300. For the
£150,000 Ferrari the insurer will pay £120,000 and the reinsurer £30,000.
Policy excess
An excess on a policy is a way insurers discourage small claims. The insured has to pay the first £X of any
claim. The excess will vary according to the type of policy but typically it will be at least £50. This means
it is not worth claiming for very minor repairs. If you as the insured want to reduce the cost of your
insurance one way of doing this is to agree to a larger excess but you will only save if you do not make a
claim.
Loss Distributions
In general insurance there are a number of situations where we need to models to enable us to make
estimates and this is often done by applying statistical distributions. We will look at some of the
distributions which may be useful. Some of these you will have already met in your statistics modules,
others are more specific to Actuarial work you will find key information for 19 distributions (7 discrete,
12 continuous) in the tables (yellow pages 6-15). We will not look at all of them in this course and some
you will already be familiar with anyway. Whilst it is useful to know they are there in the constraints of an
exam you will not want to spend your time leafing through these.
Probably the most fundamental need of any insurer is estimating the total amount that will be paid out in
claims during say a year or a month. It might be possible to model this directly. Alternatively we could
model separately the number of claims and the amount of individual claims. Immediately we see why we
need both discrete and continuous distributions. The number of claims is clearly discrete whereas the
amount may be regarded as continuous.
We could assume we have claims from a known distribution but often we will need to estimate the
parameters of the distribution from the data we have. It is usually assumed that claims distributions are
positively skewed and long tailed.
Defining our distributions
Discrete distributions have a probability function 𝑝(𝑥) which gives the probability that 𝑥 takes each
possible discrete value where the values may be finite or infinite. If we look at tossing a coin 10 times and
observing the number of heads then x can only take integer values between 0 and 10. Whereas if we look
at the number of claims on a very large insurance portfolio we may regard this as infinite especially for
general insurance when usually a policyholder can make any number of claims on a policy. ∑∞
−∞ 𝑝(𝑥) = 1
Continuous distributions have a probability density function (PDF) 𝑓(𝑥) this can be regarded as analogous
to the probability function. We tend to regard distributions of claim amounts as continuous even if in
practice an amount can only be defined down to pence. Again it is possible to have distributions which are
finite or infinite if there is a maximum total claim per annum then this is finite if there is no maximum
∞
then it may be regarded as infinite. ∫−∞ 𝑓(𝑥) = 1
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Both discrete and continuous distributions have what the tables describe as the distribution function
𝐹(𝑥). This is perhaps more usefully (and commonly) known as the cumulative distribution function. It
gives the probability that a function has a value of 𝑎 or less. Because we know the total probability will be
1 we can also find the probability that it is greater than 𝑎 as 1 − 𝐹(𝑎). The formula for the cumulative
𝑎
distribution function may be complex it is ∑𝑥−∞ 𝑝(𝑥) or ∫−∞ 𝑓(𝑥) in these cases values may be tabulated.
You will be familiar with the use of normal distribution tables. If we want the probability a value lies
𝑏
between a and b we want to find 𝐹(𝑏) − 𝐹(𝑎) or ∑𝑏𝑎 𝑝(𝑥) or ∫𝑎 𝑓(𝑥).
To complete the definition of the distribution we need to specify a parameter or parameters. One of the
major areas we will be looking at in this subject is estimating these parameters.
The tables also include probability generating functions we will not use these in this subject:
∞
Once we have a distribution of claims then we can use it to calculate the average claim ∫0 𝑥 𝑓(𝑥) or to find
𝑏
the probability that a claim is between a & b ∫𝑎 𝑓(𝑥). This latter calculation could also be done using the
cumulative distribution function 𝐹(𝑏) − 𝐹(𝑎).
Moments
We are very familiar with the first two moments of a distribution, the mean and the variance.
The tables give formulae for these in terms of the parameters.
We may need to calculate these from first principles in which case we use:
𝑏
𝐸(𝑥) = ∫𝑎 𝑥𝑓(𝑥)𝑑𝑥 or 𝐸(𝑥) = ∑𝑏𝑎 𝑥𝑝(𝑥) as appropriate
And 𝑉𝑎𝑟(𝑥) = 𝐸(𝑥 2 ) − (𝐸(𝑥))2
For a number of distributions they also give us a moment generating function.
A moment generating function is 𝑀𝑋 (𝑡) = 𝐸(𝑒 𝑡𝑋 ) = 1 + 𝑡𝑚1 +
moment about the origin.
𝑡 2 𝑚2
2!
+
𝑡 3 𝑚3
3!
+ ⋯ where 𝑚𝑗 is the 𝑗 𝑡ℎ
Moment generating functions do not always exist but if they do then to get the 𝑗 𝑡ℎ moment about the
origin we differentiate 𝑗 times with respect to 𝑡 and set 𝑡 = 0.
Useful distributions
We start by looking at some useful continuous distributions we will look at some more distributions
including discrete ones next time.
In general we will use these to look at claim size.
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Exponential distribution
𝑓(𝑥) = 𝜆𝑒 −𝜆𝑥
𝐹(𝑥) = 1 − 𝑒 −𝜆𝑥
1
1
The mean of the exponential is 𝜆 and variance 𝜆2
exponential lambda = 0.5
0.5
0.4
0.3
exponential lambda
= 0.5
0.2
0.1
0
0.10.91.72.53.34.14.95.76.57.38.18.99.7
The shape of the exponential is appropriate for larger claims but in reality people do not tend to claim
very small amounts. In fact many policies contain a policy excess clause as mentioned earlier meaning
that they do not accept claims less than £50 or £100 say. In this case they will deduct the excess from any
claim they do pay (for this reason it is sometimes known as a deductible).
There may also be reinsurance which means the insurance has an upper limit on the amount they pay.
The exponential distribution may not be the distribution of choice for insurers but it is the distribution of
choice when asking questions as it is relatively easy to calculate with. Many questions in this course will
require you to find the mean (and sometimes variance) of what is known as a curtate exponential
distribution, in other words just a portion of the distribution.
To do this you will need to use integration by parts which you should be familiar with:
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − 𝑣𝑑𝑢
Let us consider the situation where an insurer wants to find the mean claim size. They assume that claims
are exponentially distributed and if they settled every claim in full the mean claim size would be £200. In
fact there excess of loss insurance meaning that the reinsurer will pay the value of the claim over and
above £500. Find the expected value of the insurers portion of the claim.
500
In this case we can find the expected value as ∫0
∞
𝑥𝑓(𝑥) + ∫500 500𝑓(𝑥)
Which for the exponential distribution becomes:
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500
∫
𝑥𝜆𝑒
∞
−𝜆𝑥
+ ∫ 500𝜆𝑒 −𝜆𝑥
0
500
1
1
Now we know the mean claim size is £200 and 𝐸(𝑋) = 𝜆 so 200 = 𝜆
Giving 𝜆 = 0.005
500
∫
∞
0.005𝑥𝑒 −0.005𝑥 + ∫ 500 ∗ 0.005𝑒 −0.005𝑥
0
500
The second integral is straightforward and we will leave it for now except for bringing constant outside.
The first integral can be evaluated using integration by parts:
Let 𝑢 = 𝑥 and 𝑑𝑣 = 0.005𝑒 −0.005𝑥 so
𝑣 = ∫ 𝑑𝑣 = ∫ 0.005𝑒 −0.005𝑥 = − 𝑒 −0.005𝑥
And 𝑑𝑢 = 1
So our formula becomes
500
∞
500
[−𝑥𝑒 −0.005𝑥 ]
+ ∫ 𝑒 −0.005𝑥 + 2.5 ∫ 𝑒 −0.005𝑥
0
0
500
We can then proceed with our new integral and the second integral from our original equation
500
1
500
−1 −0.005𝑥 ∞
+ [−
𝑒 −0.005𝑥 ]
+ 2.5 [
𝑒
]
0
0.005
0
0.005
500
1
1
−1 −∞
1
[−500𝑒 −0.005∗500 + 0] + [−
𝑒 −0.005∗500 +
𝑒 0 ] + 2.5 [
𝑒
+
𝑒 −0.005∗500 ]
0.005
0.005
0.005
0.005
1
1
1
[−500𝑒 −2.5 ] + [−
𝑒 −2.5 +
𝑒 −2.5 ]
] + 2.5 [
0.005
0.005
0.005
1
1
−500𝑒 −2.5 −
𝑒 −2.5 +
+ 500𝑒 −2.5
0.005
0.005
First and last term cancel out
1
1
−
𝑒 −2.5 = 200 − 16.417 = 183.58
0.005 0.005
[−𝑥𝑒 −0.005𝑥 ]
You will normally end up with terms cancelling out as above.
If you are given a policy excess E then the lower limit on your first integral will be E and the amount
payable will be 𝑥 − 𝐸.
If you want a variance you will need to calculate 𝐸(𝑥 2 ) which will mean you need to integrate by part
twice.
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UEA Actuarial Methods CMP-5001B 2015/2016
Normal Distribution
The normal distribution is not in reality a particularly useful distribution for claim amounts. However as
it is relatively easy to use it is often used in questions to show understanding.
𝑓(𝑥) =
1
𝜎√2𝜋
𝑒
1 𝑥−𝜇 2
− (
)
2 𝜎
Where 𝑚𝑒𝑎𝑛 = 𝜇 and 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝜎 2
𝐹(𝑥) is tabulated for a standard normal distribution where 𝜇 = 0 and 𝜎 2 = 1
You will be pleased to know you will not be expected to integrate this to find the mean of a curtate
distribution
Normal mu=0 sigma =1
0.5
0.4
0.3
Normal
0.2
0.1
0
-3 -2.6-2.2-1.8-1.4 -1 -0.6-0.2 0.2 0.6 1 1.4 1.8 2.2 2.6 3
The type of question you may get asked is what retention limit should be set that if we want 95% of claims
to be less than the retention limit.
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Lognormal distribution
The lognormal distribution is defined as X has a lognormal distribution if Log(X) is normally distributed.
This is potentially a realistic claim distribution and can be used in a similar way to the normal
distribution.
𝑓(𝑥) =
1 1 −1(𝑙𝑛(𝑥)−𝜇)
𝑒 2 𝜎
𝜎 √2𝜋 𝑥
2
you would not be expected to remember this.
In this case the parameters do not represent the mean and variance which can be calculated as
1 2
2
2
Mean = 𝑒 𝜇+2𝜎 variance = 𝑒 2𝜇+𝜎 (𝑒 𝜎 − 1)
lognormal mu=0.5 sigma=0.5
1
0.8
0.6
lognormal
0.4
0.2
0.1
0.7
1.3
1.9
2.5
3.1
3.7
4.3
4.9
5.5
6.1
6.7
7.3
7.9
8.5
9.1
9.7
0
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Gamma distribution
The Gamma distribution is defined as
𝜆𝛼
𝑓(𝑥) =
𝑥 𝛼−1 𝑒 −𝜆𝑥
Γ(𝛼)
Reminder from your statistics: Γ(1) = 1, Γ(𝛼) = 𝛼 − 1! if 𝛼 is integer
F(x) needs to be tabulated.
𝛼
𝛼
Mean =𝜆 Variance = 𝜆2
Gamma alpha=1 gamma =0.5
0.5
0.4
0.3
Gamma alpha=1
gamma =0.5
0.2
0.1
0
0.10.71.31.92.53.13.74.34.95.56.16.77.37.98.59.19.7
There are few questions in this particular part of the course using the Gamma distribution but we will use
it later on.
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Uniform Distribution
Whilst the uniform distribution is not really a realistic claims distribution it can be useful to demonstrate
calculations because of its simplicity. It has two parameters 𝑎 and 𝑏 where these are the lower and upper
limits respectively.
1
𝑓(𝑥) =
𝑏−𝑎
𝑥−𝑎
𝐹(𝑥) =
𝑏−𝑎
1
1
Mean = 2 (𝑎 + 𝑏) Variance = 12 (𝑏 − 𝑎)2
Uniform a=0 b=100
0.012
0.01
0.008
Uniform a=0
b=100
0.006
0.004
0.002
0
8
16
24
32
40
48
56
64
72
80
88
96
0
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Pareto distribution
The Pareto distribution can be described with varying numbers of parameters. The tables show both two
and three parameters we are only concerned with the two parameter version in this course. The course
actually specifies the two parameter version differently from the book. I set out below the course version
i.e. the one you need to know.
𝛼𝜆𝛼
𝑓(𝑥) = (𝜆+𝑥)𝛼+1
𝜆 𝛼
)
𝜆+𝑥
𝐹(𝑥) = 1 − (
The tail of the Pareto distribution is heavier than the tail of the exponential this can be useful to an insurer
who wishes to be cautious as it errs on the side of there being a higher claim payments.
The mean of the Pareto is
𝜆
and the
𝛼−1
𝛼𝜆2
(𝛼−2)
variance is (𝛼−1)2
Pareto alpha= 0.3 & lambda = 0.6
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0.1
0.7
1.3
1.9
2.5
3.1
3.7
4.3
4.9
5.5
6.1
6.7
7.3
7.9
8.5
9.1
9.7
Pareto alpha= 0.3 &
lambda = 0.6
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Weibull distribution
The Weibull distribution gives a tail somewhere between the exponential and the Pareto if the parameter
𝛾 < 1. It is a very flexible distribution. If 𝛾 > 1 then the tail is lighter than the exponential
The distribution is defined as
𝛾
𝑓(𝑥) = 𝑐𝛾𝑥 𝛾−1 𝑒 −𝑐𝑥
𝛾
𝐹(𝑥) = 1 − 𝑒 −𝑐𝑥
The moments of the Weibull distribution are
𝑟 1
𝐸(𝑋 𝑟 ) = Γ (1 + 𝛾) 𝑟 for r= 1,2,3.....
𝑐𝛾
Weibull c=0.3 gamma=0.6
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
Weibull c=0.3
gamma=0.6
0.1 0.7 1.3 1.9 2.5 3.1 3.7 4.3 4.9 5.5 6.1 6.7 7.3 7.9 8.5 9.1 9.7
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Fitting a distribution
Whilst questions may say claims are distributed according to a known distribution with known
parameters in reality we may assume a distribution and are unlikely to know the actual parameters. We
will therefore need to use our data to estimate the parameters. We will look at three main methods used
for fitting a distribution, the method of moments, maximum likelihood and method of percentiles.
Different methods will be more suitable for fitting different distributions.
Method of moments
In this intuitively straightforward method of estimation we can calculate the parameters of any
distribution by setting sample moments about 0, in other words we do not use the variance as our second
moment but 𝐸(𝑋 2 ) to distribution moments. If the distribution has several parameters then you will need
to do this for several moments for which you could use the moment generating function and then solve
simultaneous equations.
This method can be used very simply for the exponential distribution although it may not be a particularly
good estimator for a one parameter distribution. It can also be used for many of the other distributions as
an illustration we will look at the lognormal distribution as this is more difficult than some.
Example
We believe claims are log normally distributed but do not know the parameters of the distribution. In the
last year the mean claim size has been 200 and the variance of claims has been 2500. Use the method of
moments to fit a lognormal distribution.
So we have
1 2
200 = 𝑒 𝜇+2𝜎 and
2
2
2500 = 𝑒 2𝜇+𝜎 (𝑒 𝜎 − 1)
2
1 2
2
Now we note that 𝑒 2𝜇+𝜎 = (𝑒 𝜇+2𝜎 )
2
So 2500 = 2002(𝑒 𝜎 − 1)
2
2500 = 40000𝑒 𝜎 − 40000
42500
2
= 𝑒𝜎
40000
Taking natural logs
0.06062 = 𝜎 2
𝜎 = 0.24622
Substituting into first equation
1
200 = 𝑒 𝜇+20.06062
Taking natural logs
5.29832 = 𝜇 + 0.03031
𝜇 = 5.26801
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Method of Percentiles
This method is similar to the method of moments, but rather than equating moments you equate
percentiles for example you could equate the quartiles. For certain distributions where method of
moments and maximum likelihood are difficult this can prove useful. If there is only one parameter then
you would equate the median. If there are two parameters then you would normally use the upper and
lower quartile, but the method is not uniquely defined so other percentiles may be used.
We will look at an example for the normal distribution.
Example
We have observed a number of claims on a class of business where we believe claim values are normally
distributed. 25% of the claims are below £300 and 25% are above £500. Estimate the parameters of the
distribution using the method of percentiles.
Using the standard normal distribution tables on pages 160-161 of the tables we find the value of 𝑥 for
which 75% of observations are to the left of 𝑥. This falls somewhere between 0.67 and 0.68. We could
then interpolate to find a closer value. However very helpfully if you turn over to page 162 they have
actually done this for us for common percentages and we can look up 25% in the tail as 0.6745. (It is up to
you which of these methods you use. The latter is quicker but as you may not have this table available in
your statistics exam you may prefer to use the former method for consistency)
300−𝜇
500−𝜇
So we have 𝜎 = −0.6745 and 𝜎 = 0.6745
300 − 𝜇 = −0.6745𝜎 and 500 − 𝜇 = 0.6745𝜎
Adding these two equations gives
800 − 2𝜇 = 0
So 𝜇 = 400 as we might expect
Therefore
500 − 400 = 0.6745𝜎
And hence 𝜎 = 148
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Method of maximum likelihood
Put simply if we want to estimate 𝜃 for each observed value we calculate the probability of observing 𝑥𝑖
given that the distribution has parameter 𝜃. Technically this is equivalent to the likelihood of 𝜃 given 𝑥𝑖 .
If we have a sample size of n we multiply these n probabilities together and find 𝜃 to give this a maximum
probability.
In formulae we calculate a likelihood function
𝑛
𝐿(𝜃) = ∏ 𝑓(𝑥𝑖 | 𝜃)
1
It can be more practical to deal with addition rather than multiplication so we can take logs
𝑛
𝑙𝑜𝑔𝐿(𝜃) = ∑ 𝑙𝑜𝑔𝑓(𝑥𝑖 | 𝜃)
1
Maximising either of these will give the maximum likelihood estimator for 𝜃.
To maximise we need to differentiate the expression and set the differential equal to zero. It is important
to check that we have found a maximum and not a minimum.
As an example we will look at estimating the parameter of an exponential distribution.
Example
We believe claim values are exponentially distributed we want to estimate the parameter 𝜆 of the
distribution. This year we have had 200 claims for a total of £100,000.
The Likelihood function will be
𝐿 = 𝜆𝑒 −𝜆𝑥1 ∗ 𝜆𝑒 −𝜆𝑥2 ∗ 𝜆𝑒 −𝜆𝑥3 ∗ … 𝜆𝑒 −𝜆𝑥200
200
𝐿 = ∏ 𝜆𝑒 −𝜆𝑥𝑖
1
200
200
𝑙𝑜𝑔𝐿 = ∑ 𝑙𝑜𝑔𝜆 − ∑ 𝜆𝑥𝑖
1
1
200
𝑙𝑜𝑔𝐿 = 200𝑙𝑜𝑔𝜆 − 𝜆 ∗ ∑ 𝑥𝑖
1
𝑙𝑜𝑔𝐿 = 200𝑙𝑜𝑔𝜆 − 𝜆 ∗ 100000
Differentiating we get
200
− 100000
𝜆
Equating this to zero gives
200
= 100000
𝜆
Or 𝜆 = 0.002
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UEA Actuarial Methods CMP-5001B 2015/2016
1
This gives a mean of 0.002 = 500 which is what we would expect.
Consider now the case where there is an excess of loss reinsurance treaty with the reinsurer paying claims
above £1000. In this case we have 180 claims below £1000 totalling £80,000 and 20 claims where the
insurer pays £1000 (and reinsurer pays balance)
The probability that a claim is greater than £1000 is 𝑒 −𝜆1000 (from 1 − 𝐹(𝑥))
So likelihood function becomes
𝐿 = 𝜆𝑒 −𝜆𝑥1 ∗ 𝜆𝑒 −𝜆𝑥2 ∗ 𝜆𝑒 −𝜆𝑥3 ∗ … 𝜆𝑒 −𝜆𝑥180 * 𝑒 −𝜆1000 ∗ … 𝑒 −𝜆1000
180
𝐿 = ∏ 𝜆𝑒 −𝜆𝑥𝑖 ∗ 𝑒 −20𝜆1000
1
180
180
𝑙𝑜𝑔𝐿 = ∑ 𝑙𝑜𝑔𝜆 − ∑ 𝜆𝑥𝑖 − 20000𝜆
1
1
180
𝑙𝑜𝑔𝐿 = 180𝑙𝑜𝑔𝜆 − 𝜆 ∗ ∑ 𝑥𝑖 − 20000𝜆
1
𝑙𝑜𝑔𝐿 = 180𝑙𝑜𝑔𝜆 − 𝜆 ∗ 80000 − 20000𝜆
Differentiating we get
180
− 100000
𝜆
Equating this to zero gives
180
= 100000
𝜆
Or 𝜆 = 0.0018
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UEA Actuarial Methods CMP-5001B 2015/2016
Effects of reinsurance
Without reinsurance we can say expected level of claim is
∞
∫0 𝑥 𝑓(𝑥)
If we have excess of loss reinsurance with retention R then expected level of claim will be
𝑅
∞
∫0 𝑥 𝑓(𝑥) +R ∫𝑅 𝑓(𝑥)
If we also allow for k% inflation in claim sizes but leave the retention unchanged we have expected claims
of
𝑅/(1+𝑘)
∞
(1 + 𝑘)𝑥 𝑓(𝑥) +R ∫𝑅/(1+𝑘) 𝑓(𝑥)
∫0
In the case of excess of loss reinsurance method of moments cannot be employed so we need to use
method of percentiles or maximum likelihood.
If we have proportional reinsurance where the insurer is liable for L% the expected level of claim will be
∞
𝐿
∫ 𝑥 𝑓(𝑥)
100 0
Effects of policy excess
This works very similarly to excess of loss reinsurance expected claims if excess is E will be
𝐸
∞
0 ∫0 𝑓(𝑥) + ∫𝐸 (𝑥 − 𝐸)𝑓(𝑥)
Exercises
1. April 2005 question 5
2. April 2007 question 3 (This question is not difficult in fact it is similar to 1. but the language is a
bit different)
3. September 2009 question 5
4. September 2010 question 7
5. An insurer believes claims on a particular class of business are normally distributed with mean
£600 and variance £80,000. They have found that it costs them £100 to administer each claim and
on this basis they decide to impose a policy excess of £100 what percentage of claims will now not
be paid.
If you want something a bit more challenging try these too
6. September 2006 question 8
7. September 2008 question 11
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