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 img. We have a manual rate img (we no longer use M for the manual rate) applicable to this policyholder, but the past experience indicates that it may not be appropriate (img, as well as img, could be quite different from img). This difference raises the question of whether next year's net premium (per exposure unit) should be based on img, on img, 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 img, or is it just random chance that is responsible for the difference between img and img?

While it is difficult to definitively answer that question, it is clear that no underwriting system is perfect. The manual rate img 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 img (possibly vector valued), but that the value of img 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, img may be viewed as representative of the residual, unobserved factors that affect the risk level. Consequently, we shall assume the existence of img, but we shall further assume that it is not observable and that we can never know its true value.

Because img varies by policyholder, there is a probability distribution with pf img of these values across the rating class. Thus, if img is a scalar parameter, the cumulative distribution function img may be interpreted as the proportion of policyholders in the rating class with risk parameter img less than or equal to img. (In statistical terms, img is a random variable with distribution function img) Stated another way, img represents the probability that a policyholder picked at random from the rating class has a risk parameter less than or equal to img (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 img value associated with an individual policyholder is not (and cannot be) known, we assume (for this chapter) that img 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 img.

Because risk levels vary within the population, it is clear that the experience of the policyholder varies in a systematic way with img. Imagine that the experience of a policyholder picked (at random) from the population arises from a two-stage process. First, the risk parameter img is selected from the distribution img. Then the claims or losses X arise from the conditional distribution of X given img, img. Thus the experience varies with img via the distribution given the risk parameter img. The distribution of claims thus differs from policyholder to policyholder to reflect the differences in the risk parameters.

img

img

img

17.2 Conditional Distributions and Expectation

The formulation of the problem just presented involves the use of conditional distributions, given the risk parameter img 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 img and marginal pfs img and img, respectively. The conditional pf of X given that img is

equation

If X and Y are discrete random variables, then img is the conditional probability of the event img under the hypothesis that img. If X and Y are continuous, then img may be interpreted as a definition. When X and Y are independent random variables,

equation

and, in this case,

equation

We observe that the conditional and marginal distributions of X are identical.

Note that

equation

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,

equation

we find that

Formula (17.1) has an interesting interpretation as a mixed distribution (see Section 5.2.4). Assume that the conditional distribution img is one of the usual parametric distributions, where y is the realization of a random parameter Y with distribution img. Section 6.3 shows that if, given img, X has a Poisson distribution with mean img and img has a gamma distribution, then the marginal distribution of X will be negative binomial. Also, Example 5.5 shows that if img has a normal distribution with mean img and variance v and img has a normal distribution with mean img and variance a, then the marginal distribution of X is normal with mean img and variance img.

Note that the roles of X and Y can be interchanged, yielding

equation

because both sides of this equation equal the joint distribution of X and Y. Division by img yields Bayes' theorem, namely

equation

We now turn our attention to conditional expectation. Consider the conditional pf of X given that img, img. 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 img instead of the left-hand side of (17.2), and so img is itself a random variable because it is a function of the random variable Y. The expectation of img is given by

Equation (17.3) can be proved using (17.2) as follows:

equation

with a similar derivation in the discrete case.

img

It is often convenient to replace X by an arbitrary function img in (17.2), yielding the more general definition

equation

Similarly, img 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

equation

If we choose img, 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

equation

Thus,

equation

Also, because img, we may use img to obtain

equation

Thus,

equation

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.

img

img

17.3 The Bayesian Methodology

Continue to assume that the distribution of the risk characteristics in the population may be represented by img, and the experience of a particular policyholder with risk parameter img arises from the conditional distribution img of claims or losses, given img.

We now return to the problem introduced in Section 16.2. That is, for a particular policyholder, we have observed img, where img and img, and are interested in setting a rate to cover img. We assume that the risk parameter associated with the policyholder is img (which is unknown). Furthermore, the experience of the policyholder corresponding to different exposure periods is assumed to be independent. In statistical terms, conditional on img, the claims or losses img are independent (although not necessarily identically distributed).

Let img have conditional pf

equation

Note that, if the img are identically distributed (conditional on img), then img does not depend on j. Ideally, we are interested in the conditional distribution of img, given img, in order to predict the claims experience img of the same policyholder (whose value of img has been assumed not to have changed). If we knew img, we could use img. Unfortunately, we do not know img, but we do know x for the same policyholder. The obvious next step is to condition on x rather than img. Consequently, we calculate the conditional distribution of img given img, termed the predictive distribution, as defined in Chapter 13.

The predictive distribution of img given img 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 img has a discrete distribution, the integrals are replaced by sums. Because the img are independent conditional on img, we have

equation

The joint distribution of X is thus the marginal distribution obtained by integrating img out, that is,

Similarly, the joint distribution of img is the right-hand side of (17.7) with n replaced by img in the product. Finally, the conditional density of img given img is the joint density of img divided by that of X, namely

There is a hidden mathematical structure underlying (17.8) that may be exploited. The posterior density of img given X is

In other words, img, and substitution in the numerator of (17.8) yields

Equation (17.10) provides the additional insight that the conditional distribution of img given X may be viewed as a mixture distribution, with the mixing distribution being the posterior distribution img.

The posterior distribution combines and summarizes the information about img 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 img is the natural conjugate prior. When both are in place, there is an easy method to evaluate the conditional distribution of img given X.

img

img

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 img is a mixture distribution with a simple mixing distribution, facilitating evaluation of the density of img. 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 img for a particular policyholder and we wish to predict img (or its mean). An obvious choice would be the hypothetical mean (or individual premium)

if we knew img. Note that replacement of img by img in (17.11) yields, on taking the expectation,

equation

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, img; nor does it use x, the data collected from the individual. Because img 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 img. Recall that in the discrete case, the integrals are replaced by sums. To prove (17.13 ), we see from (17.10) that

equation

img

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).

img

img

Example 17.12 is one where the random variables do not have identical distributions.

img

In each of Examples 17.11 and 17.12, the Bayesian estimate was a weighted average of the sample mean img and the pure premium img. 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 img, we would like to charge the individual premium (or hypothetical mean) img, where img is the (hypothetical) parameter associated with the policyholder. Because img 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 img.

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 img, 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 img or the hypothetical mean img for estimation of next year's claims. Because we have observed x, one suggestion is to approximate img by a linear function of the past data. (After all, the formula img is of this form.) Thus, let us restrict ourselves to estimators of the form img, where img need to be chosen. To this end, we choose the imgs to minimize expected squared-error loss, that is,

and the expectation is over the joint distribution of img and img. That is, the squared error is averaged over all possible values of img and all possible observations. To minimize Q, we take derivatives. Thus,

equation

We shall denote by img the values of img that minimize (17.14). Then, equating img to 0 yields

equation

But img, and so img implies that

Equation (17.15) may be termed the unbiasedness equation because it requires that the estimate img be unbiased for img. However, the credibility estimate may be biased as an estimator of img, the quantity we are trying to estimate. This bias will average out over the members of img. By accepting this bias, we are able to reduce the overall MSE. For img, we have

equation

and setting this expression equal to zero yields

equation

The left-hand side of this equation may be reexpressed as

equation

where the second from last step follows by independence of img and img conditional on img. Thus img implies

Next, multiply (17.15) by img 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 img to yield the credibility premium

While it is straightforward to express the solution img to the normal equations in matrix notation (if the covariance matrix of the img 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, img. The other img terms are true covariance terms.

As an added bonus, the values img also minimize

and

To see this, differentiate (17.19) or (17.20) with respect to img 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 img, the Bayesian premium img, and img.

img

We now turn to some models that specify the conditional means and variances of img and, hence, the means img, variances img, and covariances img.

17.5 The Bühlmann Model

The simplest credibility model, the Bühlmann model, specifies that, for each policyholder (conditional on img), past losses img have the same mean and variance and are i.i.d. conditional on img.

Thus, define

equation

and

equation

As discussed previously, img is referred to as the hypothetical mean, whereas img is called the process variance. Define

and

The quantity img 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 img is the estimate to use if we have no information about img (and thus no information about img). It will also be referred to as the collective premium.

The mean, variance, and covariance of the img may now be obtained. First,

Second,

(17.25) equation

Finally, for img,

This result is exactly of the form of Example 17.13 with parameters img, and img. 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 img and the collective premium img, a formula we find desirable. Furthermore, Z approaches 1 as n increases, giving more credit to img rather than img as more past data accumulate, a feature that agrees with intuition. Also, if the population is fairly homogeneous with respect to the risk parameter img, then (relatively speaking) the hypothetical means img do not vary greatly with img (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 img is of more value in helping to predict next year's claims for a particular policyholder. Conversely, for a heterogeneous population, the hypothetical means img 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.

img

img

img

An alternative analysis for this problem could have started with a single observation of img. From the assumptions of the problem, S has a mean of img and a variance of img. 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,

equation

The key is to note that in calculating Z the sample size is now 1, reflecting the single observation of S. Because img, the Bühlmann estimate is

equation

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 img 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 img are independent, conditional on img, with common mean (as before)

equation

but with conditional variances

equation

where img is a known constant measuring exposure. Note that img need only be proportional to the size of the risk. This model would be appropriate if each img were the average of img independent (conditional on img) random variables each with mean img and variance img. In the preceding situations, img 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

equation

and

equation

Then, for the unconditional moments, from (17.24) img, and from (17.26) img, but

equation

To obtain the credibility premium (17.18), we solve the normal equations (17.15) and (17.17) to obtain img. For notational convenience, define

equation

to be the total exposure. Then, using (17.24), the unbiasedness equation (17.15) becomes

equation

which implies

For img, (17.17) becomes

equation

which may be rewritten as

Then, using (17.30) and (17.31),

equation

and so

equation

As a result,

equation

The credibility premium (17.18) becomes

where, with img from (17.29),

equation

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, img is a weighted average of the img, with weights proportional to img. Following the group interpretation, img is the average loss of the img group members in year j, and so img is the total loss of the group in year j. Then, img is the overall average loss per group member over the n years. The credibility premium to be charged to the group in year img would thus be img for img members in the next year.

Had we known that (17.33) would be the correct weighting of the img to receive the credibility weight Z, the rest would have been easy. For the single observation img, the process variance is

equation

and so the expected process variance is img. The variance of the hypothetical means is still a, and therefore img. There is only one observation of img, and so the credibility factor is

(17.34) equation

as before. Equation (17.33) should not have been surprising because the weights are simply inversely proportional to the (conditional) variance of each img.

img

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.

img

Another generalization is provided by letting the variance of img 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.

img

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 img, img, and img, 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 img, img, and img be finite.

Consider img 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 img is explicitly identified. Also, for now, img and k should be viewed as known parameters associated with the prior pdf img. To determine img, note that from (17.36) it follows that

equation

and differentiating with respect to img yields

equation

Multiplication by img results in, using (17.35),

Next, integrate both sides of (17.37) with respect to img over the interval img, to obtain

equation

Therefore, it follows that

Note that, if

then

demonstrating that the choice of the symbol img 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 img but img. Also, img may result if either img or img fails to be finite.

Next, consider the posterior distribution in the Bühlmann situation with

equation

and img 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 img given by (17.43). Because img is a linear function of the img, 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 img. However, as long as the credibility premium is well defined (all three of img, img, and img 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 img, that is,

equation

then it is clear that in (17.19) the quantity img attains its minimum value of zero with img for img. Thus the credibility premium is img, and credibility is exact.

The following example clarifies these concepts.

img

There is one last technical point worth noting. It was mentioned previously that the choice of the symbol img as a parameter associated with the prior pdf img is not a coincidence because it is often the case that img. A similar comment applies to the parameter k. Because img from (5.9), it follows from (17.37) and the product rule for differentiation that

equation

Integrating with respect to img over img yields

equation

and solving for k yields

If, in addition, (17.39) holds, then (17.40) holds, and (17.51) simplifies to

(17.52) equation

in turn simplifying to the well-known result img if

equation

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, img, to the manual, or collective, premium, img. 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

  1. 17.1 Suppose that X and Z are independent Poisson random variables with means img and img, respectively. Let img. Demonstrate that img is binomial.
  2. 17.2 Suppose X is binomially distributed with parameters img and p, that is,
    equation

    Suppose also that Z is binomially distributed with parameters img and p independently of X. Then, img is binomially distributed with parameters img and p. Demonstrate that img has the hypergeometric distribution.

  3. 17.3 Consider a compound Poisson distribution with Poisson mean img, where img with img and img. Determine the mean and variance of X.
  4. 17.4 Let X and Y have joint probability distribution as given in Table 17.2.
    1. Compute the marginal distributions of X and Y.
    2. Compute the conditional distribution of X given img for img.
    3. Compute img, img, and img for img.
    4. Compute img and img 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
  5. 17.5 Suppose that X and Y are two random variables with bivariate normal joint density function
    equation

    Show the following:

    1. The conditional density function is
      equation

      Hence,

      equation
    2. The marginal pdf is
      equation
    3. The variables X and Y are independent if and only if img.
  6. 17.6 Suppose that, given img, the random variable X is normally distributed with mean img and variance img.
    1. Show that img and img.
    2. If img and img are independent, show that X has the same distribution as img, where img and Y are independent and Y conditional on img is normally distributed with mean zero and variance img.
  7. 17.7 Suppose that img has pdf img, and img has pdf img. If, given img, X is Poisson distributed with mean img, show that X has the same distribution as img, where Y and Z are independent, Y is Poisson distributed with mean img, and img is Poisson distributed with mean img.
  8. 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.
    1. Determine img for each of the six die–spinner combinations.
    2. Determine the conditional distributions img for the claim sizes for each die–spinner combination.
    3. Determine the hypothetical means img and the process variances img for each img.
    4. Determine the marginal probability that the claim img on the first iteration equals 3.
    5. Determine the posterior distribution img of img using Bayes' theorem.
    6. Use (17.10) to determine the conditional distribution img of the claims img on the second iteration given that img was observed on the first iteration.
    7. Use (17.13) to determine the Bayesian premium img.
    8. Determine the joint probability that img and img for img, 3, 8.
    9. Determine the conditional distribution img directly using (17.8) and compare your answer to that of (f).
    10. Determine the Bayesian premium directly using (17.12) and compare your answer to that of (g).
    11. Determine the structural parameters img, and a.
    12. Compute the Bühlmann credibility factor and the Bühlmann credibility premium to approximate the Bayesian premium img.
  9. 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.
    1. Determine img.
    2. Determine the conditional distributions img for the totals on the two balls for each urn.
    3. Determine the hypothetical means img and the process variances img for each img.
    4. Determine the marginal probability that the total img on the first two balls equals 2.
    5. Determine the posterior distribution img using Bayes' theorem.
    6. Use (17.10) to determine the conditional distribution img of the total img on the next two balls drawn, given that img was observed on the first two draws.
    7. Use (17.13) to determine the Bayesian premium img.
    8. Determine the joint probability that the total img on the next two balls equals img and the total img on the first two balls equals 2 for img.
    9. Determine the conditional distribution img directly using (17.8) and compare your answer to that of (f).
    10. Determine the Bayesian premium directly using (17.12) and compare your answer to that of (g).
    11. Determine the structural parameters img, and a.
    12. Determine the Bühlmann credibility factor and the Bühlmann credibility premium.
    13. 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
  10. 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
  11. 17.11 Let img represent the risk factor for claim numbers and let img represent the risk factor for the claim severity for a line of insurance. Suppose that img and img are independent. Suppose also that, given img, the claim number N is Poisson distributed and, given img, 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,
    img img img
    img img

    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.

  12. 17.12 Suppose that img are independent (conditional on img) and that
    equation

    Let

    equation
    1. Show that
      equation

      and

      equation
    2. Solve the normal equations for img to show that the credibility premium satisfies
      equation

      where

      equation
  13. 17.13 For the situation described in Exercise 13.5, determine img and the Bayesian premium img. Why is the Bayesian premium equal to the credibility premium?
  14. 17.14 Suppose that, for img,
    equation

    a negative binomial pf with img a known quantity.

    1. Demonstrate that the conjugate prior from Theorem 13.18 is the beta pf
      equation

      where img and img are the acceptable parameter values for img to be a valid pdf.

    2. Show that img if img and img if img.
    3. Show that there is no credibility premium if img. Next, show that if img, then img and img.
    4. Prove that there is no Bayesian premium if the number of observations n satisfies img and img, and that if img, then the Bayesian premium is linear in the img. What happens if img?
    5. Show that credibility is exact if img.
  15. 17.15 Consider the generalization of the linear exponential family given by
    equation

    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 img. For this exercise, assume that m is known.

    1. Consider the prior distribution
      equation

      Determine the Bayesian premium.

    2. Using the same prior, determine the Bühlmann premium.
    3. Show that the inverse Gaussian distribution is a member of the exponential dispersion family.
  16. 17.16 Suppose that img are independent (conditional on img) and
    equation

    Let img, img, img, img, and img.

    1. Discuss when these assumptions may be appropriate.
    2. Show that
      equation

      and

      equation
    3. Solve the normal equations for img to show that the credibility premium satisfies
      equation
    4. Give a verbal interpretation of the formula in (c).
    5. Suppose that
      equation

      Show that img and that img, where img and img.

    6. Determine the Bayesian premium if
      equation
  17. 17.17 Suppose that, given img, the random variables img are independent with Poisson pf
    equation
    1. Let img. Show that S has pf
      equation

      where img has pdf img.

    2. Show that the Bayesian premium is
      equation
    3. Evaluate the distribution of S in (a) when img is a gamma distribution. What type of distribution is this?
  18. 17.18 Suppose that, given img, the random variables img are independent with Poisson pf
    equation

    and img has the inverse Gaussian pdf from Appendix A (with img replaced by img),

    equation

    Define img.

    1. Use Exercise 5.20(g) to show that the posterior distribution of img given that img is the generalized inverse Gaussian distribution with pdf
      equation
    2. Use part (a) and Exercise 5.20(g) to prove that the predictive distribution of img given img is the Sichel distribution with pdf
      equation

      for img.

    3. Use Example 7.16 to evaluate the pf
      equation

      and, hence, use Exercise 17.17(b) to describe how to calculate the Bayesian premium.

  19. 17.19 Suppose that img is normally distributed with mean img and variance v for img. Further suppose that img is normally distributed with mean img and variance a. Thus,
    equation

    and

    equation

    Determine the posterior distribution of img and the predictive distribution of img. Then determine the Bayesian estimate of img. Finally, show that the Bayesian and Bühlmann estimates are equal.

  20. 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.
    1. Estimate the expected value of the number on the next ball using the Bayesian method.
    2. Estimate the expected value of the number on the next ball using Bühlmann credibility.
  21. 17.21 The number of claims for a randomly selected insured has a Poisson distribution with parameter img. The parameter img is distributed across the population with pdf img. For an individual, the parameter does not change over time. A particular insured experienced a total of 20 claims in the previous two years.
    1. (*) Determine the Bühlmann credibility estimate for the future expected claim frequency for this particular insured.
    2. Determine the Bayesian credibility estimate for the future expected claim frequency for this particular insured.
  22. 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.
  23. 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.
    1. Determine the expected total of the three balls using Bayes' theorem.
    2. Determine the expected total of the three balls using Bühlmann credibility.
  24. 17.24 (*) The number of claims follows a Poisson distribution with parameter img. A particular insured had three claims in the past three years.
    1. The value of img has pdf img, img. Determine the value of K used in Bühlmann's credibility formula. Then use Bühlmann credibility to estimate the claim frequency for this insured.
    2. The value of img has pdf img, img. Determine the value of K used in Bühlmann's credibility formula. Then use Bühlmann credibility to estimate the claim frequency for this insured.
  25. 17.25 (*) The number of claims follows a Poisson distribution with parameter h. The value of h has the gamma distribution with pdf img, img. Determine the Bühlmann credibility to be assigned to a single observation. (The Bayes solution is obtained in Exercise 13.22.)
  26. 17.26 Consider the situation of Exercise 13.24.
    1. Determine the expected number of claims in the second year using Bayesian credibility.
    2. (*) Determine the expected number of claims in the second year using Bühlmann credibility.
  27. 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.
    1. Determine the Bühlmann credibility estimate of the expected value of the second spin using the same spinner.
    2. Determine the Bayesian credibility estimate of the expected value of the second spin using the same spinner.
  28. 17.28 The number of claims in a year has a Poisson distribution with mean img. The parameter img has the uniform distribution over the interval (1,3).
    1. (*) Determine the probability that a randomly selected individual will have no claims.
    2. (*) If an insured had one claim during the first year, estimate the expected number of claims for the second year using Bühlmann credibility.
    3. If an insured had one claim during the first year, estimate the expected number of claims for the second year using Bayesian credibility.
  29. 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 img and variance img, 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 img and variance img, 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.
    1. Determine the value of Z for Bühlmann credibility for the observed pure premium.
    2. Suppose that the pure premium calculated from the four observations is 0.25. Determine the Bühlmann credibility estimate for the risk's pure premium.
  30. 17.30 (*) Let img be the outcome of a single trial and let img 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 img.

    Table 17.5 The data for Exercise 17.30.

    Outcome, T img Bühlmann estimate of img Bayesian estimate of img
     1 1/3  2.72 2.6
     8 1/3  7.71 7.8
    12 1/3 10.57
  31. 17.31 Consider the situation of Exercise 13.25.
    1. Determine the expected number of claims in the second year using Bayesian credibility.
    2. (*) Determine the expected number of claims in the second year using Bühlmann credibility.
  32. 17.32 Consider the situation of Exercise 13.26.
    1. Use Bayesian credibility to determine the expected number of claims in the second year.
    2. Use Bühlmann credibility to determine the expected number of claims in the second year.
  33. 17.33 Two spinners, img and img, are used to determine the number of claims. For spinner img, there is a 0.15 probability of one claim and 0.85 of no claim. For spinner img, there is a 0.05 probability of one claim and 0.95 of no claim. If there is a claim, one of two spinners, img and img, is used to determine the amount. Spinner img produces a claim of 20 with probability 0.8 and 40 with probability 0.2. Spinner img produces a claim of 20 with probability 0.3 and 40 with probability 0.7. A spinner is selected at random from each of img and from img. Three observations from the selected pair yield claim amounts of 0, 20, and 0.
    1. (*) 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.
    2. Use Bühlmann credibility once on the three observations to estimate the expected value of the next observation from the same pair of spinners.
    3. (*) Repeat parts (a) and (b) using Bayesian estimation.
    4. (*) For the same selected pair of spinners, determine
      equation
  34. 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.
  35. 17.35 (*) You are given the following:
    1. The conditional distribution img is a member of the linear exponential family.
    2. The prior distribution img is a conjugate prior for img.
    3. img.
    4. img, where img is the value of a single observation.
    5. The expected value of the process variance img.

    Determine the variance of the hypothetical means img.

  36. 17.36 (*) You are given the following:
    1. X is a random variable with mean img and variance v.
    2. img is a random variable with mean 2 and variance 4.
    3. 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.

  37. 17.37 The amount of an individual claim has an exponential distribution with pdf img. The parameter img has an inverse gamma distribution with pdf img.
    1. (*) Determine the unconditional expected value, img.
    2. 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.
    3. Repeat part (b), but determine the Bayesian credibility estimate.
  38. 17.38 The distribution of the number of claims is binomial with img and img unknown. The parameter img is distributed with mean 0.25 and variance 0.07. Determine the value of Z for a single observation using Bühlmann's credibility formula.
  39. 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.
    1. Determine the Bühlmann credibility assigned to a single shot of a randomly selected marksman.
    2. Which of the following will increase Bühlmann credibility the most?
      1. Revise the targets to 0, 4, 8, and 12.
      2. Move the marksmen to 60 feet from the targets.
      3. Revise the targets to 2, 2, 10, and 10.
      4. Increase the number of observations from the same marksman to three.
      5. Move two of the marksmen to 50 feet from the targets and increase the number of observations from the same marksman to two.
  40. 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.
    1. Determine the Bayesian credibility estimate of the expected value of the second claim amount from the same risk.
    2. Determine the Bühlmann credibility estimate of the expected value of the second claim amount from the same risk.
  41. 17.41 (*) You are given the following:
    1. The number of claims for a single insured follows a Poisson distribution with mean M.
    2. The amount of a single claim has an exponential distribution with pdf
      equation
    3. M and img are independent.
    4. img and img.
    5. img and img.
    6. The number of claims and the claim amounts are independent.
      1. Determine the expected value of the pure premium's process variance for a single risk.
      2. Determine the variance of the hypothetical means for the pure premium.
  42. 17.42 (*) The number of claims has a Poisson distribution. For 75% of risks, img, and for 25% of risks, img. 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.
  43. 17.43 (*) Claim sizes have an exponential distribution with mean img. For 80% of risks, img, and for 20% of risks, img. 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.
  44. 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 img. The prior distribution is img, img. 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.
  45. 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.
  46. 17.46 (*) The number of claims for each policyholder has a binomial distribution with img and q unknown. The prior distribution of q is beta with parameters a unknown, img, and img. 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.
  47. 17.47 In Example 17.13, if img, then img, and the estimator is img. That is, the data should be ignored. However, as img increases toward 1, Z increases to 1, and the sample mean becomes the preferred predictor of img. Explain why this is a reasonable result.
  48. 17.48 In the following, let the random vector X represent all the past data and let img represent the next observation. Let img be any function of the past data.
    1. Prove that the following is true:
      equation

      where the expectation is taken over img.

    2. Show that setting img equal to the Bayesian premium (the mean of the predictive distribution) minimizes the expected squared error,
      equation
    3. Show that, if img is restricted to be a linear function of the past data, then the expected squared error is minimized by the credibility premium.

Notes

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