17
Greatest Accuracy Credibility
17.1 Introduction
In this and Chapter 18, we consider a model-based approach to the solution of the credibility problem. This approach, referred to as greatest accuracy credibility theory, is the outgrowth of a classic 1967 paper by Bühlmann [19]. Many of the ideas are also found in Whitney [131] and Bailey [9].
We return to the basic problem. For a particular policyholder, we have observed n exposure units of past claims 
. We have a manual rate 
(we no longer use M for the manual rate) applicable to this policyholder, but the past experience indicates that it may not be appropriate (
, as well as 
, could be quite different from 
). This difference raises the question of whether next year's net premium (per exposure unit) should be based on 
, on 
, or on a combination of the two.
The insurer needs to consider the following question: Is the policyholder really different from what has been assumed in the calculation of 
, or is it just random chance that is responsible for the difference between 
and 
?
While it is difficult to definitively answer that question, it is clear that no underwriting system is perfect. The manual rate 
has presumably been obtained by (a) evaluation of the underwriting characteristics of the policyholder and (b) assignment of the rate on the basis of inclusion of the policyholder in a rating class. Such a class should include risks with similar underwriting characteristics. In other words, the rating class is viewed as homogeneous with respect to the underwriting characteristics used. Surely, not all risks in the class are truly homogeneous, however. No matter how detailed the underwriting procedure, there still remains some heterogeneity with respect to risk characteristics within the rating class (good and bad risks, relatively speaking).
Thus, it is possible that the given policyholder may be different from what has been assumed. If this is the case, how should an appropriate rate for the policyholder be determined?
To proceed, let us assume that the risk level of each policyholder in the rating class may be characterized by a risk parameter 
(possibly vector valued), but that the value of 
varies by policyholder. This assumption allows us to quantify the differences between policyholders with respect to the risk characteristics. Because all observable underwriting characteristics have already been used, 
may be viewed as representative of the residual, unobserved factors that affect the risk level. Consequently, we shall assume the existence of 
, but we shall further assume that it is not observable and that we can never know its true value.
Because 
varies by policyholder, there is a probability distribution with pf 
of these values across the rating class. Thus, if 
is a scalar parameter, the cumulative distribution function 
may be interpreted as the proportion of policyholders in the rating class with risk parameter 
less than or equal to 
. (In statistical terms, 
is a random variable with distribution function 
) Stated another way, 
represents the probability that a policyholder picked at random from the rating class has a risk parameter less than or equal to 
(to accommodate the possibility of new insureds, we slightly generalize the “rating class” interpretation to include the population of all potential risks, whether insured or not).
While the 
value associated with an individual policyholder is not (and cannot be) known, we assume (for this chapter) that 
is known. That is, the structure of the risk characteristics within the population is known. This assumption can be relaxed, and we shall decide later how to estimate the relevant characteristics of 
.
Because risk levels vary within the population, it is clear that the experience of the policyholder varies in a systematic way with 
. Imagine that the experience of a policyholder picked (at random) from the population arises from a two-stage process. First, the risk parameter 
is selected from the distribution 
. Then the claims or losses X arise from the conditional distribution of X given 
, 
. Thus the experience varies with 
via the distribution given the risk parameter 
. The distribution of claims thus differs from policyholder to policyholder to reflect the differences in the risk parameters.
17.2 Conditional Distributions and Expectation
The formulation of the problem just presented involves the use of conditional distributions, given the risk parameter 
of the insured. Subsequent analyses of mathematical models of this nature will be seen to require a good working knowledge of conditional distributions and conditional expectation. A discussion of these topics is now presented.
Much of the material is of a review nature and, hence, may be quickly glossed over if you have a good background in probability. Nevertheless, there may be some material not seen before, and so this section should not be completely ignored.
Suppose that X and Y are two random variables with joint probability function (pf) or probability density function (pdf)2 
and marginal pfs 
and 
, respectively. The conditional pf of X given that 
is
If X and Y are discrete random variables, then 
is the conditional probability of the event 
under the hypothesis that 
. If X and Y are continuous, then 
may be interpreted as a definition. When X and Y are independent random variables,
and, in this case,
We observe that the conditional and marginal distributions of X are identical.
Note that
demonstrating that joint distributions may be constructed from products of conditional and marginal distributions. Because the marginal distribution of X may be obtained by integrating (or summing) y out of the joint distribution,
we find that
Formula (17.1) has an interesting interpretation as a mixed distribution (see Section 5.2.4). Assume that the conditional distribution 
is one of the usual parametric distributions, where y is the realization of a random parameter Y with distribution 
. Section 6.3 shows that if, given 
, X has a Poisson distribution with mean 
and 
has a gamma distribution, then the marginal distribution of X will be negative binomial. Also, Example 5.5 shows that if 
has a normal distribution with mean 
and variance v and 
has a normal distribution with mean 
and variance a, then the marginal distribution of X is normal with mean 
and variance 
.
Note that the roles of X and Y can be interchanged, yielding
because both sides of this equation equal the joint distribution of X and Y. Division by 
yields Bayes' theorem, namely
We now turn our attention to conditional expectation. Consider the conditional pf of X given that 
, 
. Clearly, this is a valid probability distribution, and its mean is denoted by
with the integral replaced by a sum in the discrete case.3 Clearly, (17.2) is a function of y, and it is often of interest to view this conditional expectation as a random variable obtained by replacing y by Y in the right-hand side of (17.2). Thus we can write 
instead of the left-hand side of (17.2), and so 
is itself a random variable because it is a function of the random variable Y. The expectation of 
is given by
Equation (17.3) can be proved using (17.2) as follows:
with a similar derivation in the discrete case.
It is often convenient to replace X by an arbitrary function 
in (17.2), yielding the more general definition
Similarly, 
is the conditional expectation viewed as a random variable that is a function of Y. Then, (17.3) generalizes to
To see (17.4), note that
If we choose 
, then its expected value, based on the conditional distribution of X given Y, is the variance of this conditional distribution,
Clearly, (17.5) is a function of the random variable Y.
It is instructive now to analyze the variance of X where X and Y are two random variables. To begin, note that (17.5) may be written as
Thus,
Also, because 
, we may use 
to obtain
Thus,
Finally, we have established the important formula
Formula (17.6) states that the variance of X is composed of the sum of two parts: the mean of the conditional variance plus the variance of the conditional mean.
17.3 The Bayesian Methodology
Continue to assume that the distribution of the risk characteristics in the population may be represented by 
, and the experience of a particular policyholder with risk parameter 
arises from the conditional distribution 
of claims or losses, given 
.
We now return to the problem introduced in Section 16.2. That is, for a particular policyholder, we have observed 
, where 
and 
, and are interested in setting a rate to cover 
. We assume that the risk parameter associated with the policyholder is 
(which is unknown). Furthermore, the experience of the policyholder corresponding to different exposure periods is assumed to be independent. In statistical terms, conditional on 
, the claims or losses 
are independent (although not necessarily identically distributed).
Let 
have conditional pf
Note that, if the 
are identically distributed (conditional on 
), then 
does not depend on j. Ideally, we are interested in the conditional distribution of 
, given 
, in order to predict the claims experience 
of the same policyholder (whose value of 
has been assumed not to have changed). If we knew 
, we could use 
. Unfortunately, we do not know 
, but we do know x for the same policyholder. The obvious next step is to condition on x rather than 
. Consequently, we calculate the conditional distribution of 
given 
, termed the predictive distribution, as defined in Chapter 13.
The predictive distribution of 
given 
is the relevant distribution for risk analysis, management, and decision making. It combines the uncertainty about the claims losses with that of the parameters associated with the risk process.
Here, we repeat the development in Chapter 13, noting that if 
has a discrete distribution, the integrals are replaced by sums. Because the 
are independent conditional on 
, we have
The joint distribution of X is thus the marginal distribution obtained by integrating 
out, that is,
Similarly, the joint distribution of 
is the right-hand side of (17.7) with n replaced by 
in the product. Finally, the conditional density of 
given 
is the joint density of 
divided by that of X, namely
There is a hidden mathematical structure underlying (17.8) that may be exploited. The posterior density of 
given X is
In other words, 
, and substitution in the numerator of (17.8) yields
Equation (17.10) provides the additional insight that the conditional distribution of 
given X may be viewed as a mixture distribution, with the mixing distribution being the posterior distribution 
.
The posterior distribution combines and summarizes the information about 
contained in the prior distribution and the likelihood, and consequently (17.10) reflects this information. As noted in Theorem 13.18, the posterior distribution admits a convenient form when the likelihood is derived from the linear exponential family and 
is the natural conjugate prior. When both are in place, there is an easy method to evaluate the conditional distribution of 
given X.
Note that the posterior distribution is of the same type (gamma) as the prior distribution. The concept of a conjugate prior distribution is introduced in Section 13.3. This result also implies that 
is a mixture distribution with a simple mixing distribution, facilitating evaluation of the density of 
. Further examples of this idea are found in the exercises at the end of this section.
To return to the original problem, we have observed 
for a particular policyholder and we wish to predict 
(or its mean). An obvious choice would be the hypothetical mean (or individual premium)
if we knew 
. Note that replacement of 
by 
in (17.11) yields, on taking the expectation,
so that the pure, or collective, premium is the mean of the hypothetical means. This is the premium we would use if we knew nothing about the individual. It does not depend on the individual's risk parameter, 
; nor does it use x, the data collected from the individual. Because 
is unknown, the best we can do is try to use the data, which suggest the use of the Bayesian premium (the mean of the predictive distribution):
A computationally more convenient form is
In other words, the Bayesian premium is the expected value of the hypothetical means, with expectation taken over the posterior distribution 
. Recall that in the discrete case, the integrals are replaced by sums. To prove (17.13 ), we see from (17.10) that
As expected, the revised value based on two observations is between the prior value (0.475) based on no data and the value based only on the data (0.5).
Example 17.12 is one where the random variables do not have identical distributions.
In each of Examples 17.11 and 17.12, the Bayesian estimate was a weighted average of the sample mean 
and the pure premium 
. This result is appealing from a credibility standpoint. Furthermore, the credibility factor Z in each case is an increasing function of the number of exposure units. The greater the amount of past data observed, the closer Z is to 1, consistent with our intuition.
17.4 The Credibility Premium
In Section 17.3, a systematic approach is suggested for treatment of the past data of a particular policyholder. Ideally, rather than the pure premium 
, we would like to charge the individual premium (or hypothetical mean) 
, where 
is the (hypothetical) parameter associated with the policyholder. Because 
is unknown, the hypothetical mean is impossible to determine, but we could instead condition on x, the past data from the policyholder. This leads to the Bayesian premium 
.
The major challenge with this approach is that it may be difficult to evaluate the Bayesian premium. Of course, in simple examples such as those in Section 17.3, the Bayesian premium is not difficult to evaluate numerically. But these examples can hardly be expected to capture the essential features of a realistic insurance scenario. More realistic models may well introduce analytic difficulties with respect to evaluation of 
, whether we use (17.12) or (17.13). Often, numerical integration may be required. There are exceptions, such as Examples 17.11 and 17.12.
We now present an alternative suggested by Bühlmann [19] in 1967. Recall the basic problem. We wish to use the conditional distribution 
or the hypothetical mean 
for estimation of next year's claims. Because we have observed x, one suggestion is to approximate 
by a linear function of the past data. (After all, the formula 
is of this form.) Thus, let us restrict ourselves to estimators of the form 
, where 
need to be chosen. To this end, we choose the 
s to minimize expected squared-error loss, that is,
and the expectation is over the joint distribution of 
and 
. That is, the squared error is averaged over all possible values of 
and all possible observations. To minimize Q, we take derivatives. Thus,
We shall denote by 
the values of 
that minimize (17.14). Then, equating 
to 0 yields
But 
, and so 
implies that
Equation (17.15) may be termed the unbiasedness equation because it requires that the estimate 
be unbiased for 
. However, the credibility estimate may be biased as an estimator of 
, the quantity we are trying to estimate. This bias will average out over the members of 
. By accepting this bias, we are able to reduce the overall MSE. For 
, we have
and setting this expression equal to zero yields
The left-hand side of this equation may be reexpressed as
where the second from last step follows by independence of 
and 
conditional on 
. Thus 
implies
Next, multiply (17.15) by 
and subtract from (17.16) to obtain
Equation (17.15) and the n equations (17.17) together are called the normal equations. These equations may be solved for 
to yield the credibility premium
While it is straightforward to express the solution 
to the normal equations in matrix notation (if the covariance matrix of the 
is nonsingular), we shall be content with solutions for some special cases.
Note that exactly one of the terms on the right-hand side of (17.17) is a variance term, that is, 
. The other 
terms are true covariance terms.
As an added bonus, the values 
also minimize
and
To see this, differentiate (17.19) or (17.20) with respect to 
and observe that the solutions still satisfy the normal equations (17.15) and (17.17). Thus the credibility premium (17.18) is the best linear estimator of each of the hypothetical mean 
, the Bayesian premium 
, and 
.
We now turn to some models that specify the conditional means and variances of 
and, hence, the means 
, variances 
, and covariances 
.
17.5 The Bühlmann Model
The simplest credibility model, the Bühlmann model, specifies that, for each policyholder (conditional on 
), past losses 
have the same mean and variance and are i.i.d. conditional on 
.
Thus, define
and
As discussed previously, 
is referred to as the hypothetical mean, whereas 
is called the process variance. Define
and
The quantity 
in (17.21) is the expected value of the hypothetical means, v in (17.22) is the expected value of the process variance, and a in (17.23) is the variance of the hypothetical means. Note that 
is the estimate to use if we have no information about 
(and thus no information about 
). It will also be referred to as the collective premium.
The mean, variance, and covariance of the 
may now be obtained. First,
Second,
Finally, for 
,
This result is exactly of the form of Example 17.13 with parameters 
, and 
. Thus the credibility premium is
where
and
The credibility factor Z in (17.28) with k given by (17.29) is referred to as the Bühlmann credibility factor. Note that (17.27) is of the form (16.7), and (17.28) is exactly (16.8). Now, however, we know how to obtain k, namely, from (17.29).
Formula (17.27) has many appealing features. First, the credibility premium (17.27) is a weighted average of the sample mean 
and the collective premium 
, a formula we find desirable. Furthermore, Z approaches 1 as n increases, giving more credit to 
rather than 
as more past data accumulate, a feature that agrees with intuition. Also, if the population is fairly homogeneous with respect to the risk parameter 
, then (relatively speaking) the hypothetical means 
do not vary greatly with 
(i.e. they are close in value) and hence have small variability. Thus, a is small relative to v, that is, k is large and Z is closer to zero. This observation agrees with intuition because, for a homogeneous population, the overall mean 
is of more value in helping to predict next year's claims for a particular policyholder. Conversely, for a heterogeneous population, the hypothetical means 
are more variable, that is, a is large and k is small, and so Z is closer to 1. Again this observation makes sense because, in a heterogeneous population, the experience of other policyholders is of less value in predicting the future experience of a particular policyholder than is the past experience of that policyholder.
We now present some examples.
An alternative analysis for this problem could have started with a single observation of 
. From the assumptions of the problem, S has a mean of 
and a variance of 
. While it is true that S has a gamma distribution, that information is not needed because the Bühlmann approximation requires only moments. Following the preceding calculations,
The key is to note that in calculating Z the sample size is now 1, reflecting the single observation of S. Because 
, the Bühlmann estimate is
which is n times the previous answer. That is because we are now estimating the next value of S rather than the next value of X. However, the credibility factor itself (i.e. Z) is the same whether we are predicting 
or the next value of S.
17.6 The Bühlmann–Straub Model
The Bühlmann model is the simplest of the credibility models because it effectively requires that the past claims experience of a policyholder comprise i.i.d. components with respect to each past year. An important practical difficulty with this assumption is that it does not allow for variations in exposure or size.
For example, what if the first year's claims experience of a policyholder reflected only a portion of a year due to an unusual policyholder anniversary? What if a benefit change occurred part way through a policy year? For group insurance, what if the size of the group changed over time?
To handle these variations, we consider the following generalization of the Bühlmann model. Assume that 
are independent, conditional on 
, with common mean (as before)
but with conditional variances
where 
is a known constant measuring exposure. Note that 
need only be proportional to the size of the risk. This model would be appropriate if each 
were the average of 
independent (conditional on 
) random variables each with mean 
and variance 
. In the preceding situations, 
could be the number of months the policy was in force in past year j, or the number of individuals in the group in past year j, or the amount of premium income for the policy in past year j.
As in the Bühlmann model, let
and
Then, for the unconditional moments, from (17.24) 
, and from (17.26) 
, but
To obtain the credibility premium (17.18), we solve the normal equations (17.15) and (17.17) to obtain 
. For notational convenience, define
to be the total exposure. Then, using (17.24), the unbiasedness equation (17.15) becomes
which implies
For 
, (17.17) becomes
which may be rewritten as
Then, using (17.30) and (17.31),
and so
As a result,
The credibility premium (17.18) becomes
where, with 
from (17.29),
and
Clearly, the credibility premium (17.32) is still of the form (16.7). In this case, m is the total exposure associated with the policyholder, and the Bühlmann–Straub credibility factor Z depends on m. Furthermore, 
is a weighted average of the 
, with weights proportional to 
. Following the group interpretation, 
is the average loss of the 
group members in year j, and so 
is the total loss of the group in year j. Then, 
is the overall average loss per group member over the n years. The credibility premium to be charged to the group in year 
would thus be 
for 
members in the next year.
Had we known that (17.33) would be the correct weighting of the 
to receive the credibility weight Z, the rest would have been easy. For the single observation 
, the process variance is
and so the expected process variance is 
. The variance of the hypothetical means is still a, and therefore 
. There is only one observation of 
, and so the credibility factor is
as before. Equation (17.33) should not have been surprising because the weights are simply inversely proportional to the (conditional) variance of each 
.
The assumptions underlying the Bühlmann–Straub model may be too restrictive to represent reality. In a 1967 paper, Hewitt [54] observed that large risks do not behave the same as an independent aggregation of small risks and, in fact, are more variable than would be indicated by independence. A model that reflects this observation is created in the following example.
Another generalization is provided by letting the variance of 
depend on the exposure, which may be reasonable if we believe that the extent to which a given risk's propensity to produce claims that differ from the mean is related to its size. For example, larger risks may be underwritten more carefully. In this case, extreme variations from the mean are less likely because we ensure that the risk not only meets the underwriting requirements but also appears to be exactly what it claims to be.
17.7 Exact Credibility
In Examples 17.15–17.17, we found that the credibility premium and the Bayesian premium are equal. From (17.19), we may view the credibility premium as the best linear approximation to the Bayesian premium in the sense of squared-error loss. In these examples, the approximation is exact because the two premiums are equal. The term exact credibility is used to describe the situation in which the credibility premium equals the Bayesian premium.
At first glance, it appears to be unnecessary to discuss the existence and finiteness of the credibility premium in this context, because exact credibility as defined is clearly not possible otherwise. However, in what follows, there are some technical issues to be considered, and their treatment is clearer if it is tacitly remembered that the credibility premium must be well defined, which requires that 
, 
, and 
, as is obvious from the normal equations (17.15) and (17.17). Exact credibility typically occurs in Bühlmann (and Bühlmann–Straub) situations involving linear exponential family members and their conjugate priors. It is clear that the existence of the credibility premium requires that the structural parameters 
, 
, and 
be finite.
Consider 
in this situation. Recall from (5.8) that, for the linear exponential family, the mean is
and the conjugate prior pdf is, from Theorem 13.18, given by
where the interval of support 
is explicitly identified. Also, for now, 
and k should be viewed as known parameters associated with the prior pdf 
. To determine 
, note that from (17.36) it follows that
and differentiating with respect to 
yields
Multiplication by 
results in, using (17.35),
Next, integrate both sides of (17.37) with respect to 
over the interval 
, to obtain
Therefore, it follows that
Note that, if
then
demonstrating that the choice of the symbol 
in (17.36) is not coincidental. If (17.40) holds, as is often the case, it is normally because both sides of (17.39) are equal to zero. Regardless, it is possible to have 
but 
. Also, 
may result if either 
or 
fails to be finite.
Next, consider the posterior distribution in the Bühlmann situation with
and 
given by (17.36). From Theorem 13.18 , the posterior pdf is
with
and
Because (17.41) is of the same form as (17.36), the Bayesian premium (17.13) is
with 
given by (17.43). Because 
is a linear function of the 
, the same is true of the Bayesian premium if
that is, (17.45) implies that (17.44) becomes
Clearly, for (17.45) to hold for all vectors x, both sides should be equal to zero. Also, note that (17.46) is of the form (16.7).
To summarize, posterior linearity of the Bayesian premium results (i.e. (17.46) holds) if (17.45) is true (usually with both sides equal to zero). It is instructive to note that posterior linearity of the Bayesian premium may occur even if 
. However, as long as the credibility premium is well defined (all three of 
, 
, and 
are finite), the posterior linearity of the Bayesian premium implies equality with the credibility premium, that is, exact credibility. To see this equivalence, note that, if the Bayesian premium is a linear function of 
, that is,
then it is clear that in (17.19) the quantity 
attains its minimum value of zero with 
for 
. Thus the credibility premium is 
, and credibility is exact.
The following example clarifies these concepts.
There is one last technical point worth noting. It was mentioned previously that the choice of the symbol 
as a parameter associated with the prior pdf 
is not a coincidence because it is often the case that 
. A similar comment applies to the parameter k. Because 
from (5.9), it follows from (17.37) and the product rule for differentiation that
Integrating with respect to 
over 
yields
and solving for k yields
If, in addition, (17.39) holds, then (17.40) holds, and (17.51) simplifies to
in turn simplifying to the well-known result 
if
which typically holds with both sides equal to zero.
17.8 Notes and References
In this section, one of the two major criticisms of limited fluctuation credibility has been addressed. Through the use of the variance of the hypothetical means, we now have a means of relating the mean of the group of interest, 
, to the manual, or collective, premium, 
. The development is also mathematically sound in that the results follow directly from a specific model and objective. We have also seen that the additional restriction of a linear solution is not as bad as it might be in that we still often obtain the exact Bayesian solution. There has subsequently been a great deal of effort expended to generalize the model. With a sound basis for obtaining a credibility premium, we have but one remaining obstacle: how to numerically estimate the quantities a and v in the Bühlmann formulation, or how to specify the prior distribution in the Bayesian formulation. Those matters are addressed in Chapter 18.
A historical review of credibility theory including a description of the limited fluctuation and greatest accuracy approaches is provided by Norberg [94]. Since the classic paper of Bühlmann [19], there has developed a vast literature on credibility theory in the actuarial literature. Other elementary introductions are given by Herzog [52] and Waters [130]. Other more advanced treatments are Goovaerts and Hoogstad [46] and Sundt [118]. An important generalization of the Bühlmann–Straub model is the Hachemeister [48] regression model, which is not discussed here. See also Klugman [71]. The material on exact credibility is motivated by Jewell [62]. See also Ericson [36]. A special issue of Insurance: Abstracts and Reviews (Sundt [117]) contains an extensive list of papers on credibility.
17.9 Exercises
- 17.1 Suppose that X and Z are independent Poisson random variables with means
and 
, respectively. Let 
. Demonstrate that 
is binomial. - 17.2 Suppose X is binomially distributed with parameters
and p, that is,
Suppose also that Z is binomially distributed with parameters
and p independently of X. Then, 
is binomially distributed with parameters 
and p. Demonstrate that 
has the hypergeometric distribution. - 17.3 Consider a compound Poisson distribution with Poisson mean
, where 
with 
and 
. Determine the mean and variance of X. - 17.4 Let X and Y have joint probability distribution as given in Table 17.2.
- Compute the marginal distributions of X and Y.
- Compute the conditional distribution of X given
for 
. - Compute
, 
, and 
for 
. - Compute
and 
using (17.3), (17.6), and (c).
Table 17.2 The data for Exercise 17.4.
y x 0 1 2 0 0.20 0 0.10 1 0 0.15 0.25 2 0.05 0.15 0.10 - 17.5 Suppose that X and Y are two random variables with bivariate normal joint density function
Show the following:
- The conditional density function is
Hence,
- The marginal pdf is
- The variables X and Y are independent if and only if
.
- The conditional density function is
- 17.6 Suppose that, given
, the random variable X is normally distributed with mean 
and variance 
.
- Show that
and 
. - If
and 
are independent, show that X has the same distribution as 
, where 
and Y are independent and Y conditional on 
is normally distributed with mean zero and variance 
.
- Show that
- 17.7 Suppose that
has pdf 
, and 
has pdf 
. If, given 
, X is Poisson distributed with mean 
, show that X has the same distribution as 
, where Y and Z are independent, Y is Poisson distributed with mean 
, and 
is Poisson distributed with mean 
. - 17.8 Consider a die–spinner model. The first die has one “marked” face and five “unmarked” faces, whereas the second die has four “marked” faces and two “unmarked” faces. There are three spinners, each with five equally spaced sectors marked 3 or 8. The first spinner has one sector marked 3 and four marked 8, the second has two marked 3 and three marked 8, and the third has four marked 3 and one marked 8. One die and one spinner are selected at random. If rolling the die produces an unmarked face, no claim occurs. If a marked face occurs, there is a claim and then the spinner is spun once to determine the amount of the claim.
- Determine
for each of the six die–spinner combinations. - Determine the conditional distributions
for the claim sizes for each die–spinner combination. - Determine the hypothetical means
and the process variances 
for each 
. - Determine the marginal probability that the claim
on the first iteration equals 3. - Determine the posterior distribution
of 
using Bayes' theorem. - Use (17.10) to determine the conditional distribution
of the claims 
on the second iteration given that 
was observed on the first iteration. - Use (17.13) to determine the Bayesian premium
. - Determine the joint probability that
and 
for 
, 3, 8. - Determine the conditional distribution
directly using (17.8) and compare your answer to that of (f). - Determine the Bayesian premium directly using (17.12) and compare your answer to that of (g).
- Determine the structural parameters
, and a. - Compute the Bühlmann credibility factor and the Bühlmann credibility premium to approximate the Bayesian premium
.
- Determine
- 17.9 Three urns have balls marked 0, 1, and 2 in the proportions given in Table 17.3. An urn is selected at random, and two balls are drawn from that urn with replacement. A total of 2 on the two balls is observed. Two more balls are then drawn with replacement from the same urn, and it is of interest to predict the total on these next two balls.
- Determine
. - Determine the conditional distributions
for the totals on the two balls for each urn. - Determine the hypothetical means
and the process variances 
for each 
. - Determine the marginal probability that the total
on the first two balls equals 2. - Determine the posterior distribution
using Bayes' theorem. - Use (17.10) to determine the conditional distribution
of the total 
on the next two balls drawn, given that 
was observed on the first two draws. - Use (17.13) to determine the Bayesian premium
. - Determine the joint probability that the total
on the next two balls equals 
and the total 
on the first two balls equals 2 for 
. - Determine the conditional distribution
directly using (17.8) and compare your answer to that of (f). - Determine the Bayesian premium directly using (17.12) and compare your answer to that of (g).
- Determine the structural parameters
, and a. - Determine the Bühlmann credibility factor and the Bühlmann credibility premium.
- Show that the Bühlmann credibility factor is the same if each “exposure unit” consists of one draw from the urn rather than two draws.
Table 17.3 The data for Exercise 17.9.
Urn 0s 1s 2s 1 0.40 0.35 0.25 2 0.25 0.10 0.65 3 0.50 0.15 0.35 - Determine
- 17.10 Suppose that there are two types of policyholder: type A and type B. Two-thirds of the total number of the policyholders are of type A and one-third are of type B. For each type, the information on annual claim numbers and severity are given in Table 17.4. A policyholder has a total claim amount of 500 in the past four years. Determine the credibility factor Z and the credibility premium for next year for this policyholder.
Table 17.4 The data for Exercise 17.10.
Number of claims Severity Type Mean Variance Mean Variance A 0.2 0.2 200 4,000 B 0.7 0.3 100 1,500 - 17.11 Let
represent the risk factor for claim numbers and let 
represent the risk factor for the claim severity for a line of insurance. Suppose that 
and 
are independent. Suppose also that, given 
, the claim number N is Poisson distributed and, given 
, the severity Y is exponentially distributed. The expectations of the hypothetical means and process variances for the claim number and severity as well as the variance of the hypothetical means for frequency are, respectively,
Three observations are made on a particular policyholder and we observe total claims of 200. Determine the Bühlmann credibility factor and the Bühlmann premium for this policyholder.
- 17.12 Suppose that
are independent (conditional on 
) and that
Let
- Show that
and
- Solve the normal equations for
to show that the credibility premium satisfies
where
- Show that
- 17.13 For the situation described in Exercise 13.5, determine
and the Bayesian premium 
. Why is the Bayesian premium equal to the credibility premium? - 17.14 Suppose that, for
,
a negative binomial pf with
a known quantity.- Demonstrate that the conjugate prior from Theorem 13.18 is the beta pf
where
and 
are the acceptable parameter values for 
to be a valid pdf. - Show that
if 
and 
if 
. - Show that there is no credibility premium if
. Next, show that if 
, then 
and 
. - Prove that there is no Bayesian premium if the number of observations n satisfies
and 
, and that if 
, then the Bayesian premium is linear in the 
. What happens if 
? - Show that credibility is exact if
.
- Demonstrate that the conjugate prior from Theorem 13.18 is the beta pf
- 17.15 Consider the generalization of the linear exponential family given by
If m is a parameter, this is called the exponential dispersion family. In Exercise 5.25, it is shown that the mean of this random variable is
. For this exercise, assume that m is known.- Consider the prior distribution
Determine the Bayesian premium.
- Using the same prior, determine the Bühlmann premium.
- Show that the inverse Gaussian distribution is a member of the exponential dispersion family.
- Consider the prior distribution
- 17.16 Suppose that
are independent (conditional on 
) and
Let
, 
, 
, 
, and 
.- Discuss when these assumptions may be appropriate.
- Show that
and
- Solve the normal equations for
to show that the credibility premium satisfies
- Give a verbal interpretation of the formula in (c).
- Suppose that
Show that
and that 
, where 
and 
. - Determine the Bayesian premium if
- 17.17 Suppose that, given
, the random variables 
are independent with Poisson pf
- Let
. Show that S has pf
where
has pdf 
. - Show that the Bayesian premium is
- Evaluate the distribution of S in (a) when
is a gamma distribution. What type of distribution is this?
- Let
- 17.18 Suppose that, given
, the random variables 
are independent with Poisson pf
and
has the inverse Gaussian pdf from Appendix A (with 
replaced by 
),
Define
.- Use Exercise 5.20(g) to show that the posterior distribution of
given that 
is the generalized inverse Gaussian distribution with pdf
- Use part (a) and Exercise 5.20(g) to prove that the predictive distribution of
given 
is the Sichel distribution with pdf
for
. - Use Example 7.16 to evaluate the pf
and, hence, use Exercise 17.17(b) to describe how to calculate the Bayesian premium.
- Use Exercise 5.20(g) to show that the posterior distribution of
- 17.19 Suppose that
is normally distributed with mean 
and variance v for 
. Further suppose that 
is normally distributed with mean 
and variance a. Thus,
and
Determine the posterior distribution of
and the predictive distribution of 
. Then determine the Bayesian estimate of 
. Finally, show that the Bayesian and Bühlmann estimates are equal. - 17.20 (*) Your friend selected at random one of two urns and then she pulled a ball with number 4 on it from the urn. Then she replaced the ball in the urn. One of the urns contains four balls, numbered 1–4. The other urn contains six balls, numbered 1–6. Your friend will make another random selection from the same urn.
- Estimate the expected value of the number on the next ball using the Bayesian method.
- Estimate the expected value of the number on the next ball using Bühlmann credibility.
- 17.21 The number of claims for a randomly selected insured has a Poisson distribution with parameter
. The parameter 
is distributed across the population with pdf 
. For an individual, the parameter does not change over time. A particular insured experienced a total of 20 claims in the previous two years.
- (*) Determine the Bühlmann credibility estimate for the future expected claim frequency for this particular insured.
- Determine the Bayesian credibility estimate for the future expected claim frequency for this particular insured.
- 17.22 (*) The distribution of payments to an insured is constant over time. If the Bühlmann credibility assigned for one-half year of observation is 0.5, determine the Bühlmann credibility to be assigned for three years.
- 17.23 (*) Three urns contain balls marked either 0 or 1. In urn A, 10% are marked 0; in urn B, 60% are marked 0; and in urn C, 80% are marked 0. An urn is selected at random and three balls selected with replacement. The total of the values is 1. Three more balls are selected with replacement from the same urn.
- Determine the expected total of the three balls using Bayes' theorem.
- Determine the expected total of the three balls using Bühlmann credibility.
- 17.24 (*) The number of claims follows a Poisson distribution with parameter
. A particular insured had three claims in the past three years.
- The value of
has pdf 
, 
. Determine the value of K used in Bühlmann's credibility formula. Then use Bühlmann credibility to estimate the claim frequency for this insured. - The value of
has pdf 
, 
. Determine the value of K used in Bühlmann's credibility formula. Then use Bühlmann credibility to estimate the claim frequency for this insured.
- The value of
- 17.25 (*) The number of claims follows a Poisson distribution with parameter h. The value of h has the gamma distribution with pdf
, 
. Determine the Bühlmann credibility to be assigned to a single observation. (The Bayes solution is obtained in Exercise 13.22.) - 17.26 Consider the situation of Exercise 13.24.
- Determine the expected number of claims in the second year using Bayesian credibility.
- (*) Determine the expected number of claims in the second year using Bühlmann credibility.
- 17.27 (*) One spinner is selected at random from a group of three spinners. Each spinner is divided into six equally likely sectors. The number of sectors marked 0, 12, and 48, respectively, on each spinner is as follows: spinner A, 2,2,2; spinner B, 3,2,1; and spinner C, 4,1,1. A spinner is selected at random and a zero is obtained on the first spin.
- Determine the Bühlmann credibility estimate of the expected value of the second spin using the same spinner.
- Determine the Bayesian credibility estimate of the expected value of the second spin using the same spinner.
- 17.28 The number of claims in a year has a Poisson distribution with mean
. The parameter 
has the uniform distribution over the interval (1,3).
- (*) Determine the probability that a randomly selected individual will have no claims.
- (*) If an insured had one claim during the first year, estimate the expected number of claims for the second year using Bühlmann credibility.
- If an insured had one claim during the first year, estimate the expected number of claims for the second year using Bayesian credibility.
- 17.29 (*) Each of two classes, A and B, has the same number of risks. In class A, the number of claims per risk per year has mean
and variance 
, while the amount of a single claim has mean 4 and variance 20. In class B, the number of claims per risk per year has mean 
and variance 
, while the amount of a single claim has mean 2 and variance 5. A risk is selected at random from one of the two classes and is observed for four years.
- Determine the value of Z for Bühlmann credibility for the observed pure premium.
- Suppose that the pure premium calculated from the four observations is 0.25. Determine the Bühlmann credibility estimate for the risk's pure premium.
- 17.30 (*) Let
be the outcome of a single trial and let 
be the expected value of the outcome of a second trial. You are given the information in Table 17.5. Determine the Bayesian estimate for 
.
Table 17.5 The data for Exercise 17.30.
Outcome, T 
Bühlmann estimate of 
Bayesian estimate of 
1 1/3 2.72 2.6 8 1/3 7.71 7.8 12 1/3 10.57 – - 17.31 Consider the situation of Exercise 13.25.
- Determine the expected number of claims in the second year using Bayesian credibility.
- (*) Determine the expected number of claims in the second year using Bühlmann credibility.
- 17.32 Consider the situation of Exercise 13.26.
- Use Bayesian credibility to determine the expected number of claims in the second year.
- Use Bühlmann credibility to determine the expected number of claims in the second year.
- 17.33 Two spinners,
and 
, are used to determine the number of claims. For spinner 
, there is a 0.15 probability of one claim and 0.85 of no claim. For spinner 
, there is a 0.05 probability of one claim and 0.95 of no claim. If there is a claim, one of two spinners, 
and 
, is used to determine the amount. Spinner 
produces a claim of 20 with probability 0.8 and 40 with probability 0.2. Spinner 
produces a claim of 20 with probability 0.3 and 40 with probability 0.7. A spinner is selected at random from each of 
and from 
. Three observations from the selected pair yield claim amounts of 0, 20, and 0.
- (*) Use Bühlmann credibility to separately estimate the expected number of claims and the expected severity. Use these estimates to estimate the expected value of the next observation from the same pair of spinners.
- Use Bühlmann credibility once on the three observations to estimate the expected value of the next observation from the same pair of spinners.
- (*) Repeat parts (a) and (b) using Bayesian estimation.
- (*) For the same selected pair of spinners, determine
- 17.34 (*) A portfolio of risks is such that all risks are normally distributed. Those of type A have a mean of 0.1 and a standard deviation of 0.03. Those of type B have a mean of 0.5 and a standard deviation of 0.05. Those of type C have a mean of 0.9 and a standard deviation of 0.01. There are an equal number of each type of risk. The observed value for a single risk is 0.12. Determine the Bayesian estimate of the same risk's expected value.
- 17.35 (*) You are given the following:
- The conditional distribution
is a member of the linear exponential family. - The prior distribution
is a conjugate prior for 
. 
.
, where 
is the value of a single observation.- The expected value of the process variance
.
Determine the variance of the hypothetical means
. - The conditional distribution
- 17.36 (*) You are given the following:
- X is a random variable with mean
and variance v. 
is a random variable with mean 2 and variance 4.- v is a random variable with mean 8 and variance 32.
Determine the value of the Bühlmann credibility factor Z after three observations of X.
- X is a random variable with mean
- 17.37 The amount of an individual claim has an exponential distribution with pdf
. The parameter 
has an inverse gamma distribution with pdf 
.
- (*) Determine the unconditional expected value,
. - Suppose that two claims were observed with values 15 and 25. Determine the Bühlmann credibility estimate of the expected value of the next claim from the same insured.
- Repeat part (b), but determine the Bayesian credibility estimate.
- (*) Determine the unconditional expected value,
- 17.38 The distribution of the number of claims is binomial with
and 
unknown. The parameter 
is distributed with mean 0.25 and variance 0.07. Determine the value of Z for a single observation using Bühlmann's credibility formula. - 17.39 (*) Consider four marksmen. Each is firing at a target that is 100 feet away. The four targets are 2 feet apart (i.e. they lie on a straight line at positions 0, 2, 4, and 6 in feet). The marksmen miss to the left or right, never high or low. Each marksman's shot follows a normal distribution with mean at his target and a standard deviation that is a constant times the distance to the target. At 100 feet, the standard deviation is 3 feet. By observing where an unknown marksman's shot hits the straight line, you are to estimate the location of the next shot by the same marksman.
- Determine the Bühlmann credibility assigned to a single shot of a randomly selected marksman.
- Which of the following will increase Bühlmann credibility the most?
- Revise the targets to 0, 4, 8, and 12.
- Move the marksmen to 60 feet from the targets.
- Revise the targets to 2, 2, 10, and 10.
- Increase the number of observations from the same marksman to three.
- Move two of the marksmen to 50 feet from the targets and increase the number of observations from the same marksman to two.
- 17.40 (*) Risk 1 produces claims of amounts 100, 1,000, and 20,000 with probabilities 0.5, 0.3, and 0.2, respectively. For risk 2, the probabilities are 0.7, 0.2, and 0.1. Risk 1 is twice as likely as risk 2 of being observed. A claim of 100 is observed, but the observed risk is unknown.
- Determine the Bayesian credibility estimate of the expected value of the second claim amount from the same risk.
- Determine the Bühlmann credibility estimate of the expected value of the second claim amount from the same risk.
- 17.41 (*) You are given the following:
- The number of claims for a single insured follows a Poisson distribution with mean M.
- The amount of a single claim has an exponential distribution with pdf
- M and
are independent. 
and 
.
and 
.- The number of claims and the claim amounts are independent.
- Determine the expected value of the pure premium's process variance for a single risk.
- Determine the variance of the hypothetical means for the pure premium.
- 17.42 (*) The number of claims has a Poisson distribution. For 75% of risks,
, and for 25% of risks, 
. A randomly selected risk had r claims in year 1. The Bayesian estimate of the expected number of claims in year 2 is 2.98. Determine the Bühlmann estimate of the expected number of claims in year 2. - 17.43 (*) Claim sizes have an exponential distribution with mean
. For 80% of risks, 
, and for 20% of risks, 
. A randomly selected policy had a claim of size 5 in year 1. Determine both the Bayesian and Bühlmann estimates of the expected claim size in year 2. - 17.44 (*) A portfolio has 100 risks with identical and independent numbers of claims. The number of claims for one risk has a Poisson distribution with mean
. The prior distribution is 
, 
. During year 1, 90 risks had 0 claims, 7 had 1 claim, 2 had 2 claims, and 1 had 3 claims. Determine both the Bayesian and Bühlmann estimates of the expected number of claims for the portfolio in year 2. - 17.45 (*) For a portfolio of risks, all members' aggregate losses per year per exposure have a normal distribution with a standard deviation of 1,000. For these risks, 60% have a mean of 2,000, 30% have a mean of 3,000, and 10% have a mean of 4,000. A randomly selected risk had the following experience over three years. In year 1, there were 24 exposures with total losses of 24,000. In year 2, there were 30 exposures with total losses of 36,000. In year 3, there were 26 exposures with total losses of 28,000. Determine the Bühlmann–Straub estimate of the mean aggregate loss per year per exposure for year 4.
- 17.46 (*) The number of claims for each policyholder has a binomial distribution with
and q unknown. The prior distribution of q is beta with parameters a unknown, 
, and 
. A randomly selected policyholder had two claims in year 1 and k claims in year 2. Based on the year 1 experience, the Bayesian estimate of the expected number of claims in year 2 is 2.54545. Based on years 1 and 2, the Bayesian estimate of the expected number of claims in year 3 is 3.73333. Determine k. - 17.47 In Example 17.13, if
, then 
, and the estimator is 
. That is, the data should be ignored. However, as 
increases toward 1, Z increases to 1, and the sample mean becomes the preferred predictor of 
. Explain why this is a reasonable result. - 17.48 In the following, let the random vector X represent all the past data and let
represent the next observation. Let 
be any function of the past data.
- Prove that the following is true:
where the expectation is taken over
. - Show that setting
equal to the Bayesian premium (the mean of the predictive distribution) minimizes the expected squared error,
- Show that, if
is restricted to be a linear function of the past data, then the expected squared error is minimized by the credibility premium.
- Prove that the following is true: