Month: January 2019
Calculating the variance of insurance payment
The post supplements a threepart discussion on the mathematical models of insurance payments: part 1, part 2 and part 3. This post focuses on the calculation of the variance of insurance payments.
There are three practice problem sets for the 3part discussion on the mathematical models of insurance payments – problem set 7, problem set 8 and problem set 9. Problems in these problem sets are on calculation of expected payments. We present several examples in this post on variance of insurance payment. A practice problem set will soon follow.
Coverage with an Ordinary Deductible
To simplify the calculation, the only limit on benefits is the imposition of a deductible. Suppose that the loss amount is the random variable . The deductible is . Given that a loss has occurred, the insurance policy pays nothing if the loss is below and pays if the loss exceeds . The payment random variable is denoted by or and is explicitly described as follows:
(1)……
The subscript L in is to denote that this variable is the payment per loss. This means that its mean, , is the average payment over all losses. A related payment variable is which is defined as follows:
(2)……
The variable is a truncated variable (any loss that is less than the deductible is not considered) and is also shifted (the payment is the loss less the deductible). As a result, is a conditional distribution. It is conditional on the loss exceeding the deductible. The subscript P in indicates that the payment variable is the payment per payment. This means that its mean, , is the average payment over all payments that are made, i.e. average payment over all losses that are eligible for a claim payment.
The focus of this post is on the calculation of (the average payment over all losses) and (the variance of payment per loss). These two quantities are important in the actuarial pricing of insurance. If the policy were to pay each loss in full, the average amount paid would be , the average of the loss distribution. Imposing a deductible, the average amount paid is , which is less than . On the other hand, , the variance of the payment per loss, is smaller than , the variance of the loss distribution. Thus imposing a deductible not only reduces the amount paid by the insurer, it also reduces the variability of the amount paid.
The calculation of and can be done by using the pdf of the original loss random variable .
(3)……
(4)……
(5)……
The above calculation assumes that the loss is a continuous random variable. If the loss is discrete, simply replace integrals by summation. The calculation in (3) and (4) can also be done by integrating the pdf of the payment variable .
(6)……
(7)……
(8)……
It will be helpful to also consider the pdf of the payment per payment variable .
(9)……
Three Approaches
We show that there are three different ways to calculate and .
 Using basic principle.
 Considering as a mixture.
 Considering as a compound distribution.
Using basic principle refers to using (3) and (4) or (7) and (8). The second approach is to treat as a mixture of a point mass of 0 with weight and the payment per payment with weight . The third approach is to treat as a compound distribution where the number of claims is a Bernoulli distribution with and the severity is the payment . We demonstrate these approaches with a series of examples.
Examples
Example 1
The random loss has an exponential distribution with mean 50. A coverage with a deductible of 25 is purchased to cover this loss. Calculate the mean and variance of the insurance payment per loss.
We demonstrate the calculation using the three approaches discussed above. The following gives the calculation based on basic principles.
In the above calculation, we perform a change of variable via . We now do the second approach. Note that the variable also has an exponential distribution with mean 50 (this is due to the memoryless property of the exponential distribution). The point mass of 0 has weight and the variable has weight .
In the third approach, the frequency variable is Bernoulli with and . The severity variable is . The following calculates the compound variance.
Note that the average payment per loss is , a substantial reduction from the mean if the policy pays each loss in full. The standard deviation of is , which is a reduction from 50, the standard deviation of original loss distribution. Clearly, imposing a deductible (or other limits on benefits) has the effect of reducing risk for the insurer.
When the loss distribution is exponential, approach 2 and approach 3 are quite easy to implement. This is because the payment per payment variable has the same distribution as the original loss distribution. This happens only in this case. If the loss distribution is any other distribution, we must determine the distribution of before carrying out the second or the third approach.
We now work two more examples that are not exponential distributions.
Example 2
The loss distribution is a uniform distribution on the interval . The insurance coverage has a deductible of 20. Calculate the mean and variance of the payment per loss.
The following gives the basic calculation.
The mean and variance of the loss distribution are 50 and (if the coverage pays for each loss in full). By imposing a deductible of 20, the mean payment per loss is 32 and the variance of payment per loss is 682.67. The effect is a reduction of risk since part of the risk is shifted to the policyholder.
We now perform the calculation using the the other two approaches. Note that the payment per payment has a uniform distribution on the interval . The following calculates according to the second approach.
For the third approach, the frequency is a Bernoulli variable with and the severity variable is , which is uniform on .
Example 3
In this example, the loss distribution is a Pareto distribution with parameters and . The deductible of the coverage is 500. Calculate the mean and variance of the payment per loss.
Note that the payment per payment also has a Pareto distribution with parameters and . This information is useful for implementing the second and the third approach. First the calculation based on basic principles.
Now, the mixture approach (the second approach). Note that .
Now the third approach, which is to calculate the compound variance.
Remarks
For some loss distributions, the calculation of the variance of , the payment per loss, can be difficult mathematically. The required integrals for the first approach may not have closed form. For the second and third approach to work, we need to have a handle on the payment per payment . In many cases, the pdf of is not easy to obtain or its mean and variance are hard to come by (or even do not exist). For these examples, we may have to find the variance numerically. The examples presented are some of the distributions that are tractable mathematically for all three approaches. These three examples are such that the second and third approaches represent shortcuts for find variance of because have a known form and requires minimal extra calculation. For other cases, it is possible that the second or third approach is doable but is not shortcut. In that case, any one of the approaches can be used.
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Practice Problem Set 12 – (a,b,1) class
The practice problems in this post are to reinforce the concepts of (a,b,0) class and (a,b,1) class discussed this blog post and this blog post in a companion blog. These two posts have a great deal of technical details, especially the one on (a,b,1) class. The exposition in this blog post should be more accessible.
Notation: for whenever is the counting distribution that is one of the (a,b,0) distributions – Poisson, binomial and negative binomial distribution. The notation is the probability that a zerotruncated distribution taking on the value . Likewise is the probability that a zeromodified distribution taking on the value .
Practice Problem 12A 

Practice Problem 12B 
This problem is a continuation of Problem 12A. The following is the probability generating function (pgf) of the Poisson distribution in Problem 12A.

Practice Problem 12C 
Consider a negative binomial distribution with and .

Practice Problem 12D 
The following is the probability generating function (pgf) of the negative binomial distribution in Problem 12C.

Practice Problem 12E 
This is a continuation of Problem 12C and Problem 12D.

Practice Problem 12F 
This problem is similar to Problem 12E.

Practice Problem 12G 
Suppose that the following three probabilities are from a zerotruncated (a,b,0) distribution.

Practice Problem 12H 
Consider a zeromodified distribution. The following three probabilities are from this zeromodified distribution.

Practice Problem 12I 
For a distribution from the (a,b,0) class, you are given that
Determine . 
Practice Problem 12J 
Generate an extended truncated negative binomial (ETNB) distribution with and . Note that this is to start with a negative binomial distribution with parameters and and then derive its zerotruncated distribution. The parameters and will not give a distribution but over look this point and go through the process of creating a zerotruncated distribution. In particular, determine the following.

Problem  Answer 

12A 

12B 

12C 

12D 

21E 

12F 

12G 

12H 

12I 

12J 

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The (a,b,0) and (a,b,1) classes
This post is on two classes of discrete distributions called the (a,b,0) class and (a,b,1) class. This post is a followup on two previous posts – summarizing the two posts and giving more examples. The (a,b,0) class is discussed in details in this post in a companion blog. The (a,b,1) class is discussed in details in this post in a companion blog.
Practice problems for the (a,b,0) class is found here. The next post is a practice problem set on the (a,b,1) class.
The (a,b,0) Class
A counting distribution is a discrete probability distribution that takes on the nonnegative integers (0, 1, 2, …). Counting distributions are useful when we want to model occurrences of a certain random events. The three commonly used counting distributions would be the Poisson distribution, the binomial distribution and the negative binomial distribution. All three counting distributions can be generated recursively. For these three distributions, the ratio of any two consecutive probabilities can be expressed as a linear quantity.
To make the last point in the preceding paragraph clear, let’s set some notations. For any integer , let be the probability that the counting distribution in question takes on the value . For example, if we are considering the counting random variable , then . Let’s look at the situation where the ratio of any two consecutive values of can be expressed as a linear expression for some constants and .
(1)……….
Any counting distribution that satisfies the recursive relation (1) is said to be a member of the (a,b,0) class of distributions. Note that the recursion starts at . Does that mean can be any probability value we assign? The value of is fixed because all the must sum to 1.
The three counting distribution mentioned above – Poisson, binomial and negative binomial – are all members of the (a,b,0) class. In fact the (a,b,0) class essentially has three distributions. In other words, any member of (a,b,0) class must be one of the three distributions – Poisson, binomial and negative binomial.
An (a,b,0) distribution has its usual parameters, e.g. Poisson has a parameter , which is its mean. So we need to way to translate the usual parameters to and from the parameters and . This is shown in the table below.
Table 1
Distribution  Usual Parameters  Probability at Zero  Parameter a  Parameter b 

Poisson  0  
Binomial  and  
Negative binomial  and  
Negative binomial  and  
Geometric  0  
Geometric  0 
Table 1 provides the mapping to translate between the usual parameters and the recursive parameters and .
Example 1
Let and . Let the initial probability be . Generate the first 4 probabilities according to the recursion formula (1)
Note that the sum of to is 1. So this has to be a binomial distribution and not Poisson or negative binomial. The binomial parameters are and . According to Table 1, this translate to and . The initial probability is .
Example 2
This example generates several probabilities recursively for the negative binomial distribution with and . According to Table 1, this translates to and . The following shows the probabilities up to .
The above probabilities can also be computed using the probability function given below.
For the (a,b,0) class, it is not just about calculating probabilities recursively. The parameters and also give information about other distributional quantities such as moments and variance. For a more detailed discussion of the (a,b,0) class, refer to this post in a companion blog.
The (a,b,1) Class
If the (a,b,0) class is just another name for the three distributions of Poisson, binomial and negative binomial, what is the point of (a,b,0) class? Why not just work with these three distributions individually? Sure, generating the probabilities recursively is a useful concept. The probability functions of the three distributions already give us a clear and precise way to calculate probabilities. The notion of (a,b,0) class leads to the notion of (a,b,1) class, which gives a great deal more flexibility in the modeling counting distributions. It is possible that the (a,b,0) distributions do not adequately describe a random counting phenomenon being observed. For example, the sample data may indicate that the probability at zero may be larger than is indicated by the distributions in the (a,b,0) class. One alternative is to assign a larger value for and recursively generate the subsequent probabilities for . This recursive relation is the defining characteristics of the (a,b,1) class.
A counting distribution is a member of the (a,b,1) class of distributions if the following recursive relation holds for some constants and .
(2)……….
Note that the recursion begins at . Can the values for and be arbitrary? The initial probability is an assumed value. The probability is the value such that the sum is .
The (a,b,1) class gives more flexibility in modeling. For example, the initial probability is in the negative binomial distribution in Example 2. If this is felt to be too small, then a larger value for can be assigned and then let the remaining probabilities be generated by recursion. We demonstrate how this is done using the same (a,b,0) distribution in Example 2.
Before we continue with Example 2, we comment that there are two subclasses in the (a,b,1) class. The subclasses are distinguished by whether or . The (a,b,1) distributions are called zerotruncated distributions in the first case and are called zeromodified distributions in the second case.
Because there are three related distributions, we need to establish notations to keep track of the different distributions. We use the notations established in this post. The notation refers to the probabilities for an (a,b,0) distribution. From this (a,b,0) distribution, we can derive a zerotruncated distribution whose probabilities are notated by . From this zerotruncated distribution, we can derive a zeromodified distribution whose probabilities are denoted by . For example, for the negative binomial distribution in Example 2, we derive a zerotruncated negative binomial distribution (Example 3) and from it we derive a zeromodified negative binomial distribution (Example 4).
Example 3
In Example 3, we calculated the (a,b,0) probabilities up to . We now calculate the probabilities for the corresponding zerotruncated negative binomial distribution. For a zerotruncated distribution, the value of zero is not recorded. So is simply divided by .
(3)……….
The sum of , , must be 1 since is a probability distribution. The (a,b,0) is . Then , which means . The following shows the zerotruncated probabilities.
The above are the first 6 probabilities of the zerotruncated negative binomial distribution with and or with the usual parameters and . The above can also be calculated recursively by using (2). Just calculate P_1^T$ and the rest of the probabilities can be generated using the recursion relation (2).
Example 4
From the zerotruncated negative binomial distribution in Example 3, we generate a zeromodified negative binomial distribution. If the original is considered too small,e.g. not big enough to account for the probability of zero claims, then we can assign a larger value to the zero probability. Let’s say 0.10 is more appropriate. So we set . Then the rest of the must sum to , or 0.9 in this example. The following shows how the zeromodified probabilities are related to the zerotruncated probabilities.
(4)……….
The following gives the probabilities for the zeromodified negative binomial distribution.
…(assumed value)
The same probabilities can also be obtained by using the original (a,b,0) probabilities directly as follows:
(5)……….
ETNB Distribution
Examples 2, 3 and 4 show, starting with with an (a,b,0) distribution, how to derive a zerotruncated distribution and from it a zeromodified distribution. In these examples, we start with a negative binomial distribution and the derived distributions are zerotruncated negative binomial distribution and zeromodified negative binomial distribution. If the starting distribution is a Poisson distribution, then the same process would produce a zerotruncated Poisson distribution and a zeromodified Poisson distribution (with a particular assumed value of ).
There are members of the (a,b,1) class that do not originate from a member of the (a,b,0) class. Three such distributions are discussed in this post on the (a,b,1) class. We give an example discussing one of them.
Example 5
This example demonstrates how to work with the extended truncated negative binomial distribution (ETNB). The usual negative binomial distribution has two parameters and in one version ( and in another version). Both parameters are positive real numbers. To define an ETNB distribution, we relax the parameter to include the possibility of in addition to . Of course if , then we just have the usual negative binomial distribution. So we focus on the new situation of .
Let’s say and . We take these two parameters and generate the “negative binomial” probabilities, from which we generate the zerotruncated probabilities as shown in Example 3. Now the parameters and do not belong to a legitimate negative binomial distribution. In fact the resulting are negative values. So this “negative binomial” distribution is just a device to get things going.
According to Table 1, and translate to and . We generate the “negative binomial” probabilities using the recursive relation (1). Don’t be alarmed that the probabilities are negative.
The initial is greater than 1 and the other so called probabilities are negative. But they are just a device to get the ETNB probabilities. Using the formula stated in (3) gives the following zerotruncated ETNB probabilities.
The above gives the first 5 probabilities of the zerotruncated ETNB distribution with parameters and . It is an (a,b,1) distribution that does not originate from any (legitimate) (a,b,0) distribution.
Practice Problems
The next post is a practice problem set on the (a,b,1) class.
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