21
Compound distributions
21.1 Introduction
In earlier parts of this book, we concentrated on the present value of the benefits paid on a single insurance or annuity contract. The insurer, of course, is interested in the total benefits paid on an entire portfolio of policies. An obvious way to handle this is simply to obtain the present value of the total amount paid on all policies in the portfolio, as the sum of the individual random variables. This is known as the individual risk model. There is another method for estimating the total amount paid on a group of policies, known as the collective risk model, which has advantages in certain cases. In this chapter, we deal with a static version of this model, covering a 1-year time period. Chapter 23, concerned with ruin theory, will involve a dynamic multi-period version of the collective risk model. The combined subject matter of these chapters has traditionally been referred to as risk theory in the actuarial literature. The collective risk model is particularly useful for casualty insurance such as automobile, home, or health policies. The following are three main ways in which such contracts differ from life insurance:
- In a given period, there can be several claims under a single policy. Clearly, you can have several accidents or several visits to the doctor, even in a relatively short period. However, no matter how long the period, you can only die once.
- The amount of each claim can vary substantially. A collision claim under an automobile policy can range from a small amount for a dented fender, to the complete cost of the vehicle. A health claim may involve a single visit to a doctor, or it may involve prolonged treatment, drugs, and hospital care costing a large amount of money. By contrast, although the amount paid on a life insurance policy can vary by time of death, we do not have variation of the amount for different ‘kinds’ of dying.
- Such contracts are usually written for a relatively short period such as a year, and are then renewed if the insured wishes to continue. They do not have the long-term nature of the life contracts we have discussed. Consequently, the effect of interest is not so important. To simplify the mathematics, the effects of interest will be ignored in the models of the subsequent chapters.
The collective risk model views total claims as a compound distribution, which we will now examine. To motivate the idea, consider the following game. Toss two coins, and for each head that comes up, throw a die. What is the distribution of the total? First, we identify the range. The possible totals can range from 0, which occurs if you toss two tails, to 12, which occurs if you toss two heads, and get a 6 on each of the two throws of the die. After an elementary but somewhat tedious calculation, we can arrive at the following distribution, which the reader should verify before proceeding any further. The probabilities of 0 to 12 respectively, in multiples of 1/144, are 36, 12, 13, 14, 15, 16, 17, 6, 5, 4, 3, 2, 1.
We could complicate this problem tremendously. Instead of two coins, toss 1000. Instead of a simple die throw, choose a much more complicated random variable, possibly one with a continuous distribution. It may become impossible to actually calculate the exact distribution as we did above, but we still may want to say something. At the very least we want to compute the mean and variance of the resulting distribution, or possibly higher moments. We may want to calculate the moment generating function. We may be able to find a known distribution that closely approximates the one we are interested in.
What is the relation of this game to insurance? The collective risk model identifies two main factors that influence the total claims. One is the claim frequency, that is, the number of claims that will occur over a certain period. This will be a discrete random variable taking nonnegative integers as values. The second factor is the amount that will be paid, given that a claim has occurred. This is known as the severity of the claim. We have observed that in any fixed period under a life insurance policy, the claim severity is normally just a constant, but under other types of insurance, it will vary substantially. Even though the insurer is ultimately interested in the total payout, it has been found advantageous to first model frequency and severity separately and to then combine the results to determine total claims. One reason for this is that changed conditions can affect these factors in different ways. For example, requiring automobile passengers to wear seat belts has little effect on the frequency of car accidents, but it certainly tends to reduce the claim payments for personal injuries. On the other hand, the introduction of daytime headlights is likely to have little effect on the severity of claims, but it might well reduce the number of accidents. Another example involves the effect of seasonal differences. It may be that people drive faster during summer months, when the weather is better, so a typical summer accident is more serious than one in the winter. By contrast, one might well expect more accidents in the winter.
We will now describe the formal model. We have a fixed period, a collection of policies, and we want to predict S, the total claims from all policies over that period. We let N denote the frequency of claims and X the severity, both of which we model as random variables. Throughout the discussion, we make some standard assumptions, which are reasonable in most insurance situations, although they may not always hold exactly. We assume that the severity of claims is independent of the frequency and that the severity of any one claim is independent of that of others. We also postulate that the severity follows the same distribution over the period. This will normally hold for a sufficiently small period, although it may not for longer periods due to seasonal differences such as that alluded to above.
Let Xi denote the amount of the ith claim. Our last assumption says that there is a single severity distribution given by a random variable X, and we assume that each Xi is distributed as X. The total amount of claims is then simply given by the independent sum
We have encountered sums of random variables before, but the above is quite different, since the number of summands is random, rather than a fixed integer.
We can now observe that the simple coin–dice problem we mentioned at the beginning is really an example of this type, where N takes the values 0, 1, 2 with respective probabilities 1/4, 1/2, 1/4, and X takes the value 1, 2, 3, …, 6 with equal probabilities.
The distribution of S is known as a compound distribution, which is usually prefaced by referring to the distribution of N. For example, if N is Poisson, we call S a compound Poisson distribution. This comes from the fact that when X is a random variable that takes the value 1 with certainty, then the resulting compound distribution is just N itself.
To summarize the goals in this chapter, we will be given N and X, and our object is to investigate the compound distribution S as given by (21.21). We will denote this distribution by the symbol ⟨N, X⟩.
21.2 The mean and variance of S
As mentioned, although it may be difficult to calculate the distribution of S exactly, it is quite simple to find its mean and variance, given the corresponding quantities for N and X. Throughout this chapter, we will let p denote the probability function of N. From the law of total expectation (A.29)
When N = n, S is just the sum of n independent copies of a random variable with the distribution of X. By using the fact that N and X are independent, we can write
leading to
an intuitively obvious result. Similarly,
The second moment of any random variable is the sum of the variance plus the square of the mean, so that
We use here the fact that the variance of a sum of independent random variables is the sum of the variances. The last two formulas yield
and subtracting the term E(S)2 = E(N)2E(X)2, we obtain
There is an informative explanation of the above formula. Variance represents uncertainty, and this decomposes the uncertainty in the value of S into two parts. The first term gives the uncertainty resulting from the severity, and second term gives the uncertainty arising from the frequency.
21.3 Generating functions
The same conditioning technique as used above can be employed to deduce the moment generating function (m.g.f) MS(t) and the probability generating function (p.g.f.) PS(t) (see Sections A.9 and A.10). When N = n, we have that S = X1 + X2 + ⋅⋅⋅ + Xn, and so
where we invoke independence in order to write the expectation of a product as a product of expectations. Since each Xi is distributed as the random variable X,
and
from which we conclude that
When X takes nonnegative integer values, we can use A.34 and A.35 to conclude that
giving us the nice result that
21.4 Exact distribution of S
In the previous two sections, we considered the problem of getting partial information about S through moments and generating functions, but it is natural to ask if we can find the exact distribution of this random variable. The answer is that it is easy enough to write down a formula for this, but in all but some very simple cases, it is not at all easy to actually use the formula to calculate numbers.
What is the probability that S takes a value less than or equal to s? Once again we use the conditioning technique. If N = n, then the answer is just the probability that X1 + X2 + ⋅⋅⋅ + Xn ⩽ s, which is just the n-fold convolution of FX at the point s, which we have denoted by F*n(s) (see Section A.12). It then follows that
or similarly, by using density/probability functions,
where f*0(k) takes a value of 1 for k = 0 and zero elsewhere, and f*1 is just f.
Example 21.1 Suppose that N takes the values 0, 1, 2 with probabilities 0.5, 0.3, 0.2, respectively, and X takes the values 1, 2, 3 with probabilities 0.4, 0.2, 0.4, respectively. Find the distribution of S.
Solution. We form the following table, in which for each row, the entries are multiplied by the weights in the bottom row to get the totals in the far right hand column:
| k | f*(0)(k) | f*(1)(k) | f*(2)(k) | fS(k) |
| 0 | 1 | 0 | 0 | 0.500 |
| 1 | 0 | 0.4 | 0 | 0.120 |
| 2 | 0 | 0.2 | 0.16 | 0.092 |
| 3 | 0 | 0.4 | 0.16 | 0.152 |
| 4 | 0 | 0 | 0.36 | 0.072 |
| 5 | 0 | 0 | 0.16 | 0.032 |
| 6 | 0 | 0 | 0.16 | 0.032 |
| Weights | 0.5 | 0.3 | 0.2 |
21.5 Choosing a frequency distribution
Given a certain portfolio of insurance policies, how does the insurer select appropriate distributions in order to model the aggregate claims S? In this section, we focus on the claim frequency N. We could conceivably do this by strictly empirical means. We might use past data from similar policies to try to estimate a distribution. This is a statistical problem that we do not concentrate on in this book. There are, however, many advantages to choosing a distribution from one of several well-known families of discrete distributions. We then have nice mathematical expressions for the distribution. These families are based on one or more parameters. The estimation procedure is confined to choosing just these parameters from the observed data rather than the entire distribution. Three families that play major roles in modelling claim frequency are the binomial, Poisson, and negative binomial distributions. Details and notation are given in Sections A.11.1, A.11.2 and A.11.3 respectively.
Is the binomial a suitable distribution for claim frequency? To illustrate, suppose that our period of time is 30 days, and our observed data show that on average we can expect 10 claims over each 30-day period. Suppose also that we now assume a time homogeneity for frequency, which is analogous to the assumption for severity that we made as part of our general postulates. That is, we assume that the rate of claims remains constant over the period. (This assumption may not be completely realistic in certain cases. For automobile insurance, for example, there are more chances of an accident occurring during the rush hour than in the middle of the night.) As a final condition, suppose we assume that we will get at most one claim per day. We then can look upon this as 30 repeated trials. Each day we either get a claim, which constitutes ‘success’, or no claim which constitutes ‘failure’. In order that the expected number of claims equals our estimated value of 10, we must take the probability of a claim each day to be 1/3. So indeed, under our assumptions, we can model N by Bin(30, 1/3).
However, what if we decide that our limit of one claim a day is not really an accurate assumption, and that we may well experience more on some days? We could get a more accurate model by assuming that there would be no more than one claim each half-day period. We would still get a binomial distribution, but now with m = 60, and we have to change p to 1/6 to preserve an expectation of 10. Perhaps, however, even this half-day limitation is not quite accurate, and we should replace it by an hour, or perhaps a minute, or even a second. In fact, why not allow complete freedom and take the limiting distribution? This leads immediately to the the Poisson distribution which is one of the most common distributions used for modelling claim frequency. It arises in a natural way from our independence and time-homogeneous assumptions, by taking the limit of binomials.
It is worthwhile to note that the variance of a compound Poisson distribution has a particularly simple form. If S ∼ ⟨N, X⟩, where N ∼ Poisson(λ), then from A.42, E(N) = Var(N) = λ, so that
The negative binomial is also a popular choice for modelling insurance claims, but the reason is not immediate. It stems from the following idea. Suppose we assume that each insured individual produces claims according to a Poisson distribution, but that the parameter of this distribution can differ according to this individual. For a simple example, suppose that automobile drivers are classified as either good or bad. Assume that the claims of the good drivers are distributed as Poisson(1), while the claims of the bad drivers are distributed as Poisson(2). Assume, furthermore, that 60% of drivers are good and the rest are bad, but that the insurer has no way of distinguishing between the classes. It would then be reasonable to model the claims frequency N as a mixture of the two distributions Poisson(1) and Poisson(2), with respective weights 0.6 and 0.4 (see Section A.13). Of course, this is oversimplified and we could strive for a more sophisticated model by considering a mixing distribution involving several different values. We may even consider letting the parameter vary continuously over an entire interval of positive numbers and take a continuous mixing distribution. Remarkably, it turns out that if we take the gamma distribution (see Section A.11.6) for this purpose, our mixed Poisson is a negative binomial. Precisely, if we want a mixture of random variables 
, in which Λ has a Gamma(α, β) distribution, then the resulting mixed distribution N satisfies
Applying (A.53) repeatedly gives
By comparing with (A.45) we see that N ∼ Negbin(α, (β + 1)− 1).
21.6 Choosing a severity distribution
What distributions are suitable for measuring claim amounts? For many types of insurance, claims can assume a large number of values, and it is usually convenient to model claims by a continuous distribution. We will discuss some possibilities. See Sections A11.5–A11.8 for details and notation.
Is a normal distribution a suitable one for modelling severity? A possible drawback is that the claim distribution will almost always be positive-valued (it is not usual for the policyholder to pay the insurer), and the normal of course takes values over the entire real line. This by itself is not a major concern. If the mean of a normal is sufficiently high relative to its standard deviation, there will be so little chance of a negative value that for all practical purposes, we may as well consider it as positive-valued.
A more major difficulty is that the normal density is not the right shape for most applications, as it does not give sufficiently high weight to lower valued claims. We usually want a distribution that has a greater concentration of mass on the left. (This could be described as a distribution with the mean greater than the median.) The family of gamma distributions does have this general shape that we want, and provide a popular choice for severity modelling.
Another important criterion for selecting a severity distribution is tail behaviour. For any distribution X, the function sX(t) = P(X > t) approaches 0 as t approaches ∞. However, the rate at which convergence to 0 occurs will differ. We say that the distribution X has heavier right tails than the distribution of Y if sY(t)/sX(t) approaches 0 (or equivalently, by L’Hôpital’s rule, if fY(t)/fX(t) approaches 0). A heavier-tailed distribution therefore gives more weight to large values. If one desires a heavier-tailed distribution than the gamma, the Pareto distribution (see Section A.11.8.) is a possible choice. Using L’Hôpital’s rule to find the limit of the ratio of density functions, we can verify that this is heavier-tailed than any gamma distribution. The heavy-tailed feature of this distribution is further revealed by the fact that for large enough k, the kth moment becomes infinite.
21.7 Handling the point mass at 0
When X is discrete, then S is clearly discrete, but what happens when X is continuous? Provided that N takes the value 0 with positive probability, we will have a distribution of mixed type. Since S = 0 whenever N = 0, the distribution for S will have what is known as a point mass at the point 0. It is often convenient to split off the continuous part. Let
(see Section A.8) which will be continuous when X is.( Note that this is different from S+ = S|S ⩾ 0). Then S can be considered as a mixture of S+ and 0 (the random variable that always takes the value 0), with respective weights 1 − p(0) and p(0). It follows from (A.67) that
Since the zero random variable is always less than or equal to s,
Similarly, since the m.g.f. of the zero random variable is identically equal to 1, the m.g.f. of S+ is given by
Example 21.2 Suppose that N has a Geom(p) distribution and X has an Exp(λ) distribution. What is the distribution of S?
Solution. For a general severity distribution X,
and since p(0) = 1 − p,
By substituting MX(t) = λ/(λ − t) from (A.56), we have
which from (A.56) again is the m.g.f. of an Exp(λ(1 − p)) distribution. Invoking Theorem A.2, we know that the distribution of S is a mixture of 0 and an Exp(λ(1 − p)) distribution with weights 1 − p and p, respectively.
21.8 Counting claims of a particular type
21.8.1 One special class
Suppose we divide our claims into two groups, ‘special’ and ‘non-special,’ and we are interested in knowing the number of special claims, as well as the total. Denote the special claim frequency by N1. Our problem is to deduce the distribution of N1 given the distribution of N. The special claims can be anything at all – those of high amount, those of low amount, those divisible by 79. It does not really matter. All we need to know is the probability of a special claim and we can write down a formula for the distribution of N1. What is the probability that N1 = k, given that the probability of a special claim is π? In order to get k special claims, there must be at least k claims in total. That is, N must take the value of k + r for some nonnegative integer r. From the law of total probability (A.30),
Given that we have k + r claims in total, the number of special claims out of these is distributed as Bin(k + r, π), so we can substitute in the above to get
Example 21.3 N takes the values 0, 1, 2, 3, 4 with probabilities 0.3, 0.1, 0.3, 0.2, 0.1 respectively, and X takes the values 1 to 100 with equal probabilities. What is the probability that we have exactly 2 claims for an amount less than or equal to 60?
Solution. The special claims are those for an amount less than or equal to 60, so π = 0.6.
While formula (21.9) may be good for calculating individual probabilities, it can be tedious to calculate the entire distribution. It is often better to proceed by calculating the p.g.f. of N1. We do this by making the ingenious observation that N1 is itself a compound distribution. In fact, N1 ∼ ⟨N, δπ⟩, where δπ takes the value 1 with probability π and 0 with probability 1 − π. This is clear, since we can count special claims by simply assigning a value of 0 whenever we get a non-special claim. We note that δπ is a Bernoulli random variable, that is, it has a binomial distribution with m = 1 and, therefore, its p.g.f. is 1 − π + πt. From (21.21),
This formula allows us to show that for our three major counting distributions, Poisson, binomial, and negative binomial, N1 is of the same type as N, but with a changed parameter.
If N ∼ Poisson(λ), then
showing that N1 ∼ Poisson(λπ).
If N ∼ Bin(m, p), then
showing that N1 ∼ Bin(m, pπ).
The last case is a bit tricker. Suppose N ∼ Negbin(r, p). Can we expect N1 to be a negative binomial with changed parameters? Motivated by the binomial case, we might try leaving r the same and modifying p. We cannot, however, take πp for the new value of p since that would not give us the correct value of πp/(1 − p) for E(N1) that we must have by formula (A.46). We will, however, at least get the right mean if we multiply α by π, where α = p/(1 − p) is the alternate parameter to p mentioned in Section 21.5. This indeed is the right answer. To verify this, it is convenient to express the probability function of N in terms of this new parameter. We can write
So if M is the negative binomial with the first parameter r and the second (modified) parameter απ, then
Therefore, reverting to our original parameter,
21.8.2 Special classes in the Poisson case
Suppose now that we have two special classes, with the number of special claims in the two classes denoted by N1 and N2, respectively. We can write down a formula similar to (21.9) for the joint distribution of N1 and N2. If the probability of a claim in the first class is π1 and that of a claim in the second class is π2, then
What do you suppose the covariance of N1 and N2 will be? We would naturally expect this to be negative. A high value for one type of claim would seem to indicate that there are fewer of the other type. Indeed, this will be the case for most distributions, but remarkably, for the particular case where N is Poisson, the two distributions are independent. If N ∼ Poisson(λ), we obtain from the above
The third term above equals 1, since it is the sum over all nonnegative integers of the probability function for a Poisson(λ(1 − π1 − π2)) distribution. We know from above that Ni ∼ Poisson(λπi) for i = 1, 2, so
and the independence result is proved. We can similarly show that if we have r special classes, with Nr denoting the number of claims in class r, then the collection (N1, N2, …, Nr) will be independent.
This allows us to write certain compound Poisson distributions in an alternate form, which is sometimes useful. Suppose that X takes finitely many values, say x1, x2, …, xn. Let πi denote the probability that X = xi. Let Ni be the number of claims for amount xi. We can then write
This does not use the fact that N is Poisson and is true for any frequency distribution. The problem is that, in general, this formulation is of little use. We may not be able to easily identify the distribution of the various Ni and, even if we can, such as in the binomial or negative binomial cases, they will not be independent. In general, it can be very difficult to deal with dependent sums. In the Poisson case, however, we know that Ni ∼ Poisson(λπi) and that the Ni are independent. The above expression is often easier to deal with than formula (21.7) if we want to compute the exact distribution for the compound Poisson distribution where X is finite-valued. We need only compute a single n-fold convolution.
21.9 The sum of two compound Poisson distributions
The sum of two independent Poisson distributions is itself Poisson distributed as shown by Example A.1 in Section A.11.2. We now show that the same statement is true for compound Poisson distributions. Given two independent compound Poisson distributions, S1 = ⟨N1, X1⟩ and S2 = ⟨N2, X2⟩, where N1 ∼ Poisson(λ1) and N2 ∼ Poisson(λ2), let S = S1 + S2. Then
It follows that S ∼ ⟨N, X⟩, where N ∼ Poisson(λ1 + λ2) and X is a mixture of X1 and X2 with respective weights λ1/(λ1 + λ2) and λ2/(λ1 + λ2).
21.10 Deductibles and other modifications
Up to now we have assumed that the insurer pays the totality of any loss as given by the random variable X. In practice, the insurer often only covers part of the loss, leaving the insured to pay the remainder. The prime motivation is to make the insured party partially responsible, so that they have an interest in taking steps to avoid loss.
This has major implications when we look at the statistical problem of inferring details about loss distributions from the data furnished by insurers on their claims experience. In practice, this data will give the amounts actually paid on claims, rather than the actual losses. In order to estimate loss distributions, one needs to understand clearly the relationship between the amount of the loss and the amount actually paid under the common types of modifications.
21.10.1 The nature of a deductible
One of the most common modification devices is a deductible. Under this arrangement, the insurer only pays the losses that are above some amount d fixed in advance. The purchaser of the insurance pays for the first d units of loss, and of course if the loss is less than d, the insurer pays nothing, and the insured is fully responsible. This has an added advantage to the insurer of preventing an undue expense involved in processing small claims.
If the original severity distribution is given by the random variable X, and there is a deductible of d, the amount actually paid by the insurer is
It is not always easy to describe the exact distribution of this random variable given the distribution of X, but it is relatively simple to compute its expectation. This is given for continuous X by
where the second expression follows from the first by integrating by parts (or directly from (A.15).
Another random variable of interest, which is associated with the above, is
By looking separately at the case where X is less than or greater than d, it easily follows that, in general,
so that
For continuous X, it follows from (21.10) by calculating directly that
Example 21.4 X takes the values 100, 200, 300, 400 with probabilities 0.4, 0.3, 0.2, 0.1, respectively. Describe the distributions of (X − d)+ and (X∧d), for d = 230. Find E(X − 230)+ and E(X∧230).
Solution. (X − 230)+ takes the value 0 with probability 0.7, 70 with probability 0.2, and 170 with probability 0.1, while X∧230 takes the value 100 with probability 0.4, 200 with probability 0.3, and 230 with probability 0.3. Calculating directly,
It is often convenient to compute E(X − d)+ from formula (21.21). This is particularly true when X is infinite and discrete. We can compute E(X∧d) by a finite sum, as opposed to the infinite series involved in computing E(X − d)+ directly. Here is a typical example.
Example 21.5 Suppose X ∼ Geom(p). Find E(X − 1/2)+.
Solution. Since X is always greater than 1/2 unless it is equal to 0, we know that X∧1/2 takes the value 0 with probability 1 − p and the value 1/2 with probability p. So E(X∧1/2) = p/2. Since E(X) = p/1 − p, we conclude that
21.10.2 Some calculations in the discrete case
Suppose that X takes integer values. The integrals involving fX in (21.10) and (21.12) must be replaced by summations with the same limits, and with fX now equal to the probability function. They apply only to integer values of d. The expressions involving sX, however, remain valid as is and apply to any value of d. Of course, in this case, sX is a step function and the integral may be rewritten as a sum. For example, if k ⩽ d < k + 1 for some integer k, then
To illustrate, look again at Example 21.4, except we take a unit to be 100, so now X takes the values 1, 2, 3, 4. We have s(2) = 0.3, s(3) = 0.1, s(4) = 0, and
as above.
It is quite simple to compute all values of E(X − d)+ in the case of an integer-valued X. When d is an integer, (21.13) just says that
and we get the recursion formula
where we start the recursion with E(X − 0)+ = E(X). Noninteger values of d are computed exactly by linear interpolation, since (21.13) immediately implies that, for d = k + r where k is an integer and 0 < r < 1,
Example 21.6 Suppose that the probability function of X takes the values f(0) = 0.2, f(1) = 0.2, f(2) = 0.3, f(3) = 0.1, f(4) = 0.2. Find E(X − d)+ for d = 0, 1, 2, 3, 4, and 2.6.
Solution. We first calculate s(0) = 0.8, s(1) = 0.6, s(2) = 0.3, s(3) = 0.2, s(4) = 0, and
so that E(X − 0)+ = 1.9, E(X − 1)+ = 1.9 − 0.8 = 1.1, E(X − 2)+ = 1.1 − 0.6 = 0.5, E(X − 3)+ = 0.5 − 0.3 = 0.2, E(X − 4)+ = 0.2 − 0.2 = 0. This final value of 0 serves as a check that we have done the recursion correctly. Finally, we have
21.10.3 Some calculations in the continuous case
In general, (X − d)+ will have a point mass at 0, as it will take a value of 0 with probability FX(d). Therefore, if X is continuous, (X − d)+ will be a mixed distribution. In such cases, we may want to proceed as in Section 21.7 and consider the random variable (X − d)+ = (X − d)|X > d. We then have that (X − d)+ is a mixture of 0 and (X − d)+) with weights FX(d) and sX(d), respectively. It follows that
Example 21.7 If X ∼ Exp(λ), what is the distribution of (X − d)+?
Solution. If Y denotes X − d+, then
so that Y has exactly the same distribution as X. This only happens with an exponential distribution. The result at first glance seems surprising. It says that no matter how high the deductible is, the excess of the loss over the deductible is distributed as the original loss. The point to keep in mind is that Y is conditioned on the loss being above the deductible, which of course will have a very small chance of occurring for high values of d.
Example 21.8 If X ∼ Pareto(θ, α), what is the distribution of (X − d)+?
Solution. If Y denotes (X − d)+,
and we see that
In general, we will not be able to easily identify the distributions associated with the deductible d as we did in the above two examples, although in some cases, we may be able to compute expectations. This is true for the gamma distribution with first parameter 2.
Example 21.9 If X ∼ Gamma (2, β), find E[(X − d)+] and E[(X − d)+)].
Solution. Integrating by parts,
so that
From (21.21),
and from (21.21),
As a check, both these expectations reduce to E(X) when d = 0.
21.10.4 The effect on aggregate claims
Up to now we have focused on the severity distribution X. We now turn our attention to the effect of a deductible on the aggregate claim distribution S. It is important to distinguish two situations.
In one case, the deductible is applied directly to aggregate claims. This could arise from a reinsurance arrangement. It is often the case that a party known as a reinsurer agrees to cover part of the losses of the original insurer in return for a premium. In one common type of arrangement, the reinsurer would impose a deductible on the aggregate claims for a certain portfolio. This is known as stop-loss reinsurance. In such a case, we are interested in the random variable (S − d)+, which will be the amount paid by the reinsurer.
In the second situation, the deductible d is applied to each individual claim. There are two ways to proceed here. The obvious way is to simply note that in place of the distribution S ∼ ⟨N, X⟩, which applies without the deductible, we are now interested in the distribution
There is an alternate representation for S′ that is useful with certain distributions. Let us motivate this by asking the following deep philosophical question. Is a claim for an amount of 0 really a claim? The simple answer is that it either is or is not, depending on which way you want it. In the first representation of S, there will be claims for zero amount, namely those that are less than the deductible and for which nothing is reimbursed. Suppose we decide not to count these as claims. Our severity will then be distributed as (X − d)+ rather than (X − d)+, since we now only consider a claim to have occurred if it is over the deductible. We must then, however, also change the distribution N to count only those claims for an amount above the deductible. We know how to do this from Section 21.8. The special claims in this case are those for an amount greater than d. Under this approach, we have
To use (21.17) effectively, we have to know that claim frequency and severity remain independent when we make these transformations. A general proof of this fact becomes somewhat involved in notation, and we will not present it, but the idea is straightforward as the following example illustrates.
Example 21.10 Suppose that N takes values of 0,1,2, and and let p(i) be the probability that N = i. Suppose that X takes the values x1, x2, x3, where x1 ⩽ d and x2 and x3 are greater than d. Let N1 denotes the number of claims for an amount greater than d. Show that
Solution. Let a, b, c denote respectively the probability that X takes the values x1,x2, x3.
For the event on the left to occur we need either one claim for an amount of x2 or two claims, where one is of amount x2 and the other is of amount x1. The required probability then is p(1)b + p(2)2ab. Now, for N1 = 1 we need either one claim for an amount of either x2 or x3, or two claims where one is for x1 and the other is for either x2 or x3. So P[N1 = 1] = p(1)(b + c) + p(2)2a(b + c)]. Moreover P[(X − 2)+ = x2 − d] = b/b + c. Multiplying the last two quantities gives the first.
Use of (21.17) in place of (21.16) works particularly well when N is one of three basic cases of Poisson, binomial, or negative binomial, where we know what ⟨N, δπ⟩ is, and when we know what (X − d)+ is, as in the exponential or Pareto distributions for severity.
Example 21.11 Suppose that N ∼ Poisson(2) and X ∼ Exp(3). A deductible of 2 is applied to each claim. Find the variance of the resulting distribution of aggregate claims.
Solution. We know that (X − 2)+ has the same distribution as X and therefore a second moment of 
. We replace the original N by a Poisson(2e− 6) distribution, and by using (21.21), the resulting variance is 
.
21.10.5 Other modifications
Another method of modifying the original claim amount is to set a maximum value m. The insurer will pay at most m regardless of the actual value of the loss. If X is the original severity distribution, the amount paid on a claim would then be X∧m. There could be both a deductible d and and a maximum m imposed. The amount paid on a claim in this situation would be X∧(m + d) − X∧d. Yet another modification is for the insurer to pay only a certain percentage of the loss. The amount paid on a claim will now be αX, for some 0 ⩽ α ⩽ 1. In doing calculations where all these modifications are present, it is useful to keep in mind the fact that
Example 21.12 A policy will cover 80% of all losses in excess of 100, with the further provision that a maximum payment of 900 will be made regardless of the amount of the loss. Express the amount paid on a claim with terms of the form X∧d.
Solution. The amount paid is 0.8(X − 100)+ provided that 0.8(X − 100) ⩽ 900, which will occur for X ⩽ 1225. We can express this as
21.11 A recursion formula for S
21.11.1 The positive-valued case
We suppose that X takes positive integers as values, and we seek a recursion formula to compute the probability function of S. It turns out that this is possible provided that the probability function of N satisfies a certain recursion. The required property is that for some constants a and b,
To develop our recursion formula, we first need some identities for convolutions of f = fX. By definition, we know that for all n and all positive integers x,
There is, however, another curious identity relating convolutions.
Proposition 21.1 For all n and all positive integer values of x,
Proof. Let X1, X2, …, Xn be independent and each distributed as X, and let A = ∑ni = 1Xi. Then, for any positive integer i ⩽ x,
since in order for the event in the numerator to occur, X1 = i and the other n − 1 random variables add up to x − i. It follows that
There is, however, nothing special about X1, and, by symmetry, we get exactly the same equality, with X1 replaced by Xj for j = 2, 3, …, n. Add up this last equality for all n values of j. The left hand side is just E(A|A = x), which is simply x. The right hand side of each equation is a constant that gets multiplied by n in the sum. Equating and rearranging gives the stated identity.
In the remainder of this section, we will let g denote the probability function of S.
Theorem 21.1 (The recursion formula) Suppose that p satisfies (21.21). Then, for all positive integers k,
Proof. For any positive integer k, we have
Consider the term multiplying a in the above, which is
By applying (21.19) to each summand (after the first), we can write this as
The sum of the first row, excluding the leading term p(0)f(k), is
Similarly, the sum of the ith row is just f(i)g(k − i). The leading term f(k)p(0) equals f(k)g(0), since, given the restriction of strictly positive values for X, the only way for S to be equal to 0 is if N = 0. The sum of the entire array is then simply
Next, consider the term multiplying b in (21.21). We do exactly what we did above except we use the identity in Proposition 21.1 in place of (21.21). In this case, the term from (21.20) introduces a coefficient of 1/n in the column of the array involving p(n − 1). Had we used (21.21), we could not have conveniently summed along rows. The beauty of the other identity is that it introduces a term n/k in the column involving p(n − 1) that conveniently cancels with the 1/n. The sum in this case is the same as (21.21), except that it is multiplied by 1/k and if(i) replaces f(i), giving
Substituting in (21.20) gives the recursion formula.
The starting value for the recursion is given by g(0) = p(0), as indicated above.
Of course, in order that this formula be useful, we need to know that there are distributions of N that satisfy the given condition on p. Fortunately, this occurs for our three main families.
If N ∼ Poisson(λ), then
so (21.18) holds with a = 0, b = λ.
If N ∼ Bin(m, p), then
so that (21.18) holds with a = −p/(1 − p), b = (m + 1)p/(1 − p).
If N ∼ Negbin(r, p), then
so that (21.18) holds with a = p, b = (r − 1)p.
What are the other possibilities? It turns out that there are none, and that, remarkably, only these three families satisfy the required recurrence relation.
Theorem 21.2 Suppose that N is a nonnegative-valued random variable satisfying (21.21). Then:
- if a = 0, N ∼ Poisson(b);
- if a > 0, N ∼ Negbin(b/a + 1, a);
- If a < 0, N ∼ Bin(m, −a/(1 − a)) for some positive integer m.
Proof.
- If a = 0, then clearly p(k) = (bk/k!)p(0). This gives
so that p(0) = e− b. Substituting this into the expression for p(k), we see that N ∼ Poisson(b).
- Suppose a > 0. Note first that p(0) cannot be 0. For, if so, then all p(k) would be 0, and we would not have a probability distribution. Let r = (b/a) + 1. Then r ⩾ 0, for, otherwise we have p(1) = (a + b)p(0) < 0. Writing b = (r − 1)a, we calculate inductively p(1) = rap(0), p(2) = [r(r + 1)/2]a2p(0), …, p(k) = [r(r + 1)… (r + k − 1)/k!]akp(0), …. This gives
so that p(0) = (1 − a)r, showing that a < 1. By substituting this into the expression for p(k), we see that N ∼ Negbin(r, a).
- Suppose that a < 0. Since a + (b/k) will become negative for sufficiently large k, it must necessarily become zero for some value of k. If not, we would eventually get negative probabilities. Therefore, we must have that b > 0 and a = −b/(m + 1) for some positive integer m. This means that p(k) = 0 for k > m. We note that a + b/k = −a(m − k + 1)/k, so that p(1) = −amp(0), p(2) = ( − a)2[m (m − 1)/2!]p(0), …, p(k) = ( − a)k[(m(m − 1)…m − k + 1)/k!]p(0), …. Then
Let p = −a/(1 − a). Then, − a = p/(1 − p), so that p(0) = (1 − p)m, and by substituting this into the expression for p(k), we see that N ∼ Bin(m, p).
Example 21.13 Go back to the coin–dice problem that started this chapter and illustrate that you can find these probabilities by recursion.
Solution. In this case N ∼ Bin(2, 0.5), so we have a = −1, b = 3. Moreover, f(i) = 1/6 for i = 1, 2, …, 6. We will do the first few calculations here to illustrate the procedure.
21.11.2 The case with claims of zero amount
There is a more general recursion formula that allows for the possibility that X can take a value of 0. This is
which reduces to that given above when f(0) = 0. This can be derived by a suitable modification of the proof of Theorem 21.1. We note that there will be two extra rows in the first array, which we used to compute the sum of the term multiplying a. One row at the beginning will have terms of the form f(0)f*r(k), and one at the end will have terms of the form f(k)f*r(0). The beginning row will just sum to f(0)g(k). The ending row will combine with the leading term to sum to f(k)g(0). In this case, g(0) is not equal to p(0). In the second calculation, when we compute the summation multiplying b, we only get this row at the end, in view of the extra coefficient of i, which makes entries in the new first row equal to 0. The final conclusion is that
which leads to (21.21).
The disadvantage of using (21.22) is that we have a more complicated calculation for the initial value than before:
An easier procedure is to follow the alternate method we mentioned for handling per-claim deductibles in the previous section. That is, we simply get rid of the zero claims by counting only the positive ones. From Section 21.8, we know now to modify N in all the relevant cases. We must also replace X by X+, but that is done simply by multiplying each probability by 1/(1 − f(0)).
Example 21.14 Suppose N ∼ Negbin(0.5, 0.4). X takes the values 0,1,2 with probabilities 0.25, 0.35, 0.40, respectively. Write down a recursion formula for computing g.
Solution. We follow the second procedure. The probability of a nonzero claim is 0.75. Recall that in the negative binomial, the ratio p/(1 − p) gets multiplied by this probability and changes from 2/3 to 1/2. So the new value of p is 1/3, and we have a = 1/3, b = −1/6. The random variable X+ takes the values 1, 2 with probabilities 7/15, 8/15, respectively. The recursion becomes
Notes and references
The reader is cautioned that some authors use an alternative definition of the negative binomial random variable. In the formulation in terms of repeated trials, they would count the total number, rather than just the successes. Their random variable would then be equal to N + r, where N is the definition that we have adopted. Moreover, parameters chosen for the different distributions are not standardized and different choices are made by various authors. We indicated an alternate choice for the negative binomial. Another example occurs with the exponential distribution. While we chose to parametrize this by the hazard rate, some may use the mean and take the parameter to be the reciprocal of ours. This is carried forward to the gamma distributions. So for example, what we call a Gamma(α, β) distribution could be termed a Gamma(α, β− 1) distribution by others.
Exercises
- 21.1 If N ∼ Binomial(9, 1/3) and X ∼ Gamma(2, 0.5), find E(S) and Var(S).
- 21.2 If N takes the values 0, 1, 2, 3 with probabilities 0.3, 0.4, 0.2, and 0.1 respectively, and X takes values 1, 2, 3, 4, 5 each with probability 0.2, find the probability that S = 4.
- 21.3 If N takes values 0, 1, 2, 3 with probabilities 0.5, 0.3, 0.1, 0.1 respectively, and X takes values 1, 2, 3, 4, 5, 6 with probabilities 0.2, 0.2, 0.2, 0.2, 0.1, 0.1 respectively, find the probability that S = 6.
- 21.4 Suppose that N is negative binomial with mean 4 and variance 12, and X is exponentially distributed. Let N1 denote the number of claims that are less than the average claim amount. Find the variance of N1.
- 21.5 The frequency of accidents for automobile drivers over a certain period follows a Poisson distribution. Good drivers can expect to have on average one accident over that period, while bad drivers can expect to have two accidents. It is estimated that 80% of drivers are good and 20% are bad. If an accident occurs, the claim amount is exponentially distributed with a mean of 100. Calculate the expected value and variance of the aggregate claims over this period.
- 21.6 Suppose that N ∼ Negbin(2, 0.8) and X ∼ Gamma(4, 3). Find E(S) and Var(S).
- 21.7 Suppose that N takes the values 0, 1, 2 with probabilities 0.5, 0.3, 0.2 respectively, and X takes values 1, 2, 3, with probabilities 0.3, 0.6, 0.1 respectively.
- Find the probability that S = 3.
- Let Ni = the number of claims of size i, for i = 1, 2, 3. Are N1 and N2 independent?
- 21.8 Suppose that N is a continuous mixture of Poisson(λ) distributions where λ ∼ Gamma (3, 2). Find E(N) and Var(N).
- 21.9 Suppose that N takes the values 0, 1, 2, 3 with probabilities 0.4, 0.3, 0.2, 0.1 respectively, and that X takes the values 10 with a probability of 0.5, 20 with a probability of 0.3, 30 with a probability of 0.1, and various other values, all higher than 30, with a total probability of 0.1. (You are not given these values.)
- Find the probability that S = 30.
- Find E[(S − 15)+], given that E(X) = 20.
- 21.10 Suppose N ∼ Poisson(2) and X ∼ Exp(3).
- Find E(S) and Var(S).
- Find MS(1), where MS is the m.g.f. of S.
- 21.11 The number of customers arriving at a restaurant is Poisson distributed with a mean of 15 per hour. The amount that each customer spends is exponentially distributed with an average of 20. The restaurant is open 16 hours each day and the daily expenses are 4500. Using a normal approximation, estimate the probability that on a given day, revenue will cover expenses.
- 21.12 Suppose that N is a continuous mixture of Poisson distributions where the mean is itself a random variable. Find P(N = 0) in each of the following cases.
- The mean is uniformly distributed on the interval [1, 3].
- The mean has a Gamma (2, 3) distribution.
- 21.13 Suppose we have two mutually exclusive special classes of claims N1 and N2. Show, by an example, that if N has a binomial distribution, then N1 and N2 need not be independent.
- 21.14 Suppose that the probability of n claims is 1/2n + 1 and the probability is e− 2x that a given claim will be greater than x. What is the probability that the aggregate claims will be less than or equal to log (10)?
- 21.15 A policy will cover 75% of all losses in excess of 240 with the further provision that a maximum payment of 2700 will be made regardless of the loss. Express the amount paid on a claim with terms of the form X∧d, where X is the actual loss.
- 21.16 Each hour, vehicles pass a certain point on a highway in accordance with a Poisson distribution. The expected number of vehicles that pass during the hour is four. Assume that one half of all passing vehicles are trucks, and one quarter are sports cars. Find the probability that, in a given hour, the passing vehicles include exactly two trucks and exactly one sports car. There is no restriction on the number of vehicles other than trucks and sports cars (so, for example, the event of two trucks, one sport car and seven other vehicles would satisfy the given condition).
- 21.17 Suppose that N is geometric with mean = 1, and X takes the values 10 or 20 with equal probability. Find E(S − 30)+.
- 21.18 For a certain insurer, N has a Poisson distribution, and X is exponentially distributed. Each claim is subject to a deductible of d. If d = 2, the expected amount paid by the insurer is equal to 100. If d = 3, this expected payout reduces to 50. What is the expected payout if d = 1?
- 21.19 The density function of X is given by
In order to reduce the expected amount paid, the insurer is considering two possibilities. One is to introduce a deductible of two per claim. The other is to pay only a maximum of 7 per claim. Which scheme should they adopt if they want to minimize the expected payout?
- 21.20 The manufacturer of a television set costing 1000, offers a guarantee to repair or replace the set for free for the first year. The number of defective sets follows a Poisson distribution with mean 4. Half the defective sets require replacement and half require a repair costing 500. The manufacturer purchases an insurance policy that will cover the total cost of this guarantee above 1500. (So, for example, if there were four defective sets and each required replacement the insurer would pay 2500 to the manufacturer.) Find the expected amount that the insurer will pay.
- 21.21 For a certain collection of contracts, N ∼ Poisson(4), while X ∼ Exp(3). Suppose that each individual claim is subject to a deductible of 2. If S′ is the total amount actually paid on all claims, find the variance of S′.
- 21.22 For a certain firm, the number of losses of a certain type has a Poisson(2) distribution. The amount of a loss takes a value of 100, 200 or 300, with probabilities 0.5, 0.3, 0.2, respectively. The firm purchases an insurance policy that will cover all losses above an aggregate deductible of 200. (So, for example, if there was one loss of 100 and three losses of 300, the insurer would pay 800.) What is the expected reimbursement by the insurer?
- 21.23 A manager of a certain office is offered a bonus each month if the total expenses are under 1000. The bonus is half the difference between 1000 and the expenses. So, for example, if expenses were 800, the bonus would be 100. Find the expected value of the bonus in each of the following cases.
- Expenses are exponentially distributed with a mean of 2000.
- Expenses are uniformly distributed on the interval [0, 4000].
- 21.24 The distribution of N is a mixture of Poisson(λ) distributions, where λ follows a Gamma distribution with α = 10 and β = 2. Moreover, X has a Pareto distribution with θ = 4, α = 3. A per-claim deductible of 2 is applied. Find the expectation and variance of the aggregate payments made on all claims.
- 21.25 You are the manufacturer of a product that gives guarantees against failure. Each month there is a 50% chance that there will be exactly one failure, a 30% chance that there will be exactly two failures, and a 20% chance that there will be exactly three failures. Moreover, 20% of failures will be complete, requiring a full reimbursement of 800, while 80% will require only partial reimbursement of 400. Each month you purchase insurance that will provide all reimbursements for that period above a total of 1000. (So, for example, if there were three complete failures, the insurer would pay 1400.) What is the expected amount of reimbursement that the insurer will pay each month?
- 21.26 For a certain insurer, the frequency of claims has a negative binomial distribution with an expected value of 16 and the claim severity distribution is Pareto (600, 2). The insurer is planning to introduce a per-claim deductible of 200. If this is done, what would be the reduction in the expected value of aggregate claims?
- 21.27 E(X∧d)/E(X) is known as the loss elimination ratio (LER), since it gives the proportion of the risk to the insurer that is eliminated by a deductible of d. For each of the following distributions, find the LER in terms of d and the parameters. (a) X ∼ Exp(λ), (b) X ∼ Pareto(Θ, α), (c) X ∼ Gamma(2, β). In each case verify that your answers have the correct limits as d approaches 0 or ∞.
- 21.28 Suppose that the severity distribution changes from X to (1 + r)X due to inflation.
- Show that if the deductible d is increased to (1 + r)d, the LER is unchanged. What happens to the LER if d is unchanged?
- Suppose that X is exponentially distributed and for a certain value of d the LER is 0.3. If r = 0.10 and d is unchanged, what is the new LER?
- 21.29 Suppose that N is a continuous mixture of Poisson(λ) distributions, where λ ∼ Gamma (2, 1) and X takes the value 1 with probability 0.6 and 2 with probability 0.4. If fS(10) = c and fS(11) = d, find a formula for fS(12) in terms of c and d.
- 21.30 For a certain collection of contracts, X takes values 1, 2, 3, each with equal probabilities. You know that the insurer uses a recursion formula to calculate g(s) = P[S = s] and you are trying to determine what distribution is being used for N. All you have to go on is a scrap of paper on which the following appears:
where x, y, and z are numbers that have been smudged and are unreadable. You can, however, read enough of the numbers to definitely conclude that y is strictly less than 2x. What is the distribution of N and why? (Just identify the basic type. You do not have enough information to determine the parameters.)
- 21.31 Compute gS(x) for x = 0, 1, …, 5 for the following three compound distributions, each with claim amount distribution given by fX(1) = 0.7 and fX(2) = 0.3: (a) Poisson with λ = 4.5; (b) negative binomial with r = 4.5 and p = 0.5; (c) binomial with m = 9 and p = 0.5.
- 21.32 Suppose you have 11 boxes, numbered 1 to 11, and each box contains three balls, of which two are numbered 1 and one is numbered 2. A coin has a probability of 0.6 of coming up heads. You are going to toss the coin 11 times, and if a head occurs on the ith toss, you are going to take box number i and randomly select a ball. Let g denote the probability function of the random variable S, the total of all the selected balls. Given that g(10) = 0.1386 and g(11) = 0.1055, find g(12).
- 21.33 Aggregate claims S follow a compound Poisson distribution with λ = log (4) and with the probability function of X given by fX(k) = 2− k/(klog (2)), k = 1, 2, …. What is the distribution of S?
- 21.34 Suppose that (Ni), i = 1, 2, 3, is an independent family of random variables where Ni ∼ Poisson(i). If S = 3N1 + 2N2 + 5N3, find distributions N and X so that S ∼ ⟨N, X⟩.