Four posts have already been devoted to describing three models for “per loss” insurance payout. These are mixed distributions modeling the amount the insurer pays out for each random loss. They can also be viewed as mixtures. We now turn our attention to the mixed distributions modeling the “per period” payout for an insurance policy. That is, the mixed distributions we describe here are to model the total amount of losses paid out for each insurance policy in a given policy period. This involves the uncertain random losses as well as uncertain claim frequency. In other words, there is a possiblity of having no losses. When there are losses in a policy period, the number of losses can be uncertain (there can be only one loss or multiple losses). The links to the previous posts on mixed distributions are found at the end of this post.

The following is the general setting of the insurance problem we discuss in this post.

- The random variable is the size of the random loss that is covered in an insurance contract. We assumme that is a continuous random variable. Naturally, the support of is the set of nonnegative numbers (or some appropriate subset).
- Let be the “per loss” payout paid to the insured by the insurer. The variable could refect the coverage modification such as deductible and/or policy cap or other policy provisions that are applicable in the insurance contract.
- Let be the number of claims in a given policy period. In this post, we assume that has only two possibilities: or . In other words, each policy has at most one claim in a period. Let .
- Let be the total amount paid to the insured by the insurer during a fixed policy period.

The total claims variable is the mixture of and . The conditional variable is a point mass representing “no loss”. On the other hand, we assume that . Thus is a mixture of a point mass at the origin and the “per loss” payout variable .

We first have a general discussion of the stated insurance setting. Then we discuss several different cases based on four coverage modifications that can be applied in the insurance contract. In each case, we illustrate with the exponential distribution. The four cases are:

*Case 1*. . There is no coverage modification. The insurer pays the entire loss amount.
*Case 2*. The insurance contract has a cap and the cap amount is .
*Case 3*. The insurance contract is an excess-of-loss policy. The deductible amount is .
*Case 4*. The insurance contract has a deductible and a policy cap where .

**General Discussion**

The total payout is the mixture of a point mass at and the “per loss” payout . The following is the distribution :

where

Since the distribution of is a mixture, we have a wealth of information available for us. For example, the following lists the mean, higher moments, variance, the moment generating function and the skewness.

The Derivations:

for al integers

The following is another way to derive using the total variance formula:

The above derivations are based on the idea of mixtures. The two conditional variables are and . The mixing weights are and . For more basic information on distributions that are mixtures, see this post (Basic properties of mixtures).

We now discuss the four specific cases based on the variations on the coverage modifications that can be placed on the “per loss” variable .

**Case 1**

This is the case that the insurance policy has no coverage modification. The insurer pays the entire random loss. Thus . The following is the payout rule of :

This is a mixed distribution consisting of a point mass at the origin (no loss) and the random loss . In this case, the “per loss” variable . Thus is a mixture of of the following two distributions.

**Case 1 – Distribution Function**

The following shows as a mixture, the explicit rule of and the density of .

.

**Case 1 – Basic Properties**

Using basic properties of mixtures stated in the general case, we obtain the following:

for all integers

**Case 1 – Exponential Example**

If the unmodified random loss has an exponential distribution, we have the following results:

for all integers

**Case 2**

This is the case that the insurance policy has a policy cap. The “per loss” payout amount is capped at the amount . The following is the payout rule of :

**Case 2 – Per Loss Variable **

The following lists out the information we need for . For more information about the “per loss” payout for an insurance contract with a policy cap, see the post An insurance example of a mixed distribution – I.

for all integers

**Case 2 – Distribution Function**

Since is a mixture, the distribution of is a mixture of a point mass at the origin (no loss) and the mixture . As in the general case discussed above, the distribution function is a weighted average of and where is the distribution function of the point mass at . The following shows the distribution function and the density function of .

.

**Case 2 – Basic Properties**

To obtain the basic properties such as , , and , just take the weighted average of the point mass and the “per loss” of this case. In other words, they are obtained by weighting the point mass (of no loss) with the “per loss variable .

**Case 2 – Exponential Example**

If the unmodified loss has an exponential distribution, we have the following results:

where

**Case 3**

This is the case that the insurance policy is an excess-of-loss policy. The insurer agrees to pay the insured the amount of the random loss in excess of a fixed amount . The following is the payout rule of :

**Case 3 – Per Loss Variable **

The following lists out the information we need for . For more information about the “per loss” payout for an insurance contract with a deductible, see the post An insurance example of a mixed distribution – II.

for all integer

**Case 3 – Distribution Function**

Since is a mixture, the distribution of is a mixture of a point mass at the origin (no loss) and the mixture . As in the general case discussed above, the distribution function is a weighted average of and where is the distribution function of the point mass at . The following shows the distribution function and the density function of .

.

Note that the point mass of is made up of two point masses, one from having no loss and one from having losses less than the deductible.

**Case 3 – Basic Properties**

The basic properties of as a mixture are obtained by applying the general formulas with the specific information about the “per loss” in this case. In other words, they are obtained by weighting the point mass (of no loss) with the “per loss variable .

**Case 3 – Exponential Example**

If the unmodified loss has an exponential distribution, then we have the following results:

**Case 4**

This is the case that the insurance policy has both a policy cap and a deductible. The “per loss” payout amount is capped at the amount and is positive only when the loss is in excess of the deductible . The following is the payout rule of :

**Case 4 – Per Loss Variable **

The following lists out the information we need for . For more information about the “per loss” payout for an insurance contract with a deductible and a policy cap, see the post An insurance example of a mixed distribution – III.

for all integer

**Case 4 – Distribution Function**

Since is a mixture, the distribution of is a mixture of a point mass at the origin (no loss) and the mixture . As in the general case discussed above, the distribution function is a weighted average of and where is the distribution function of the point mass at . The following shows the distribution function and the density function of .

.

Note that the point mass of is made up of two point masses, one from having no loss and one from having losses less than the deductible.

**Case 4 – Basic Properties**

The basic properties of as a mixture are obtained by applying the general formulas with the specific information about the “per loss” in this case. In other words, they are obtained by weighting the point mass (of no loss) with the “per loss variable .

**Case 4 – Exponential Example**

If the unmodified loss has an exponential distribution, then we have the following results:

Another view of :

where is the in Case 2.

Also, it can be shown that:

where is the in Case 2.

Here's the links to the previous discussions of mixed distributions:

An insurance example of a mixed distribution – I

An insurance example of a mixed distribution – II

An insurance example of a mixed distribution – III

Mixed distributions as mixtures