Author: Dan Ma
Insurance payment per payment vs payment per loss
This post picks up where this previous post leaves off. The previous post is the second part of a 3part discussion on mathematical model of insurance payment. The previous post ends with pointing out the difference between payment per loss and payment per payment. This post focuses on the insurance payment per payment and points out how the two payments are related. Another previous post is on how to calculate the variance of the insurance payment per loss.
Comparing the Two Payments
The insurance coverage being considered here is that losses are paid subject to an ordinary deductible . If a loss is less than the deductible, the coverage pays nothing and if a loss exceeds the deductible, the coverage pays the loss less the deductible. This is called the insurance payment per loss variable and is described as follows:
(1)……
The L in the subscript of indicates that the variable is the payment per loss. The variable includes the probability as well as the probability . Thus the average is the average over all losses, hence the name payment per loss.
The other payment variable of interest is the payment per payment called . This variable captures the losses that require a claim payment, i.e. all losses exceeding the deductible. For , losses below the deductible are not considered and are truncated.
(2)……
The variable is truncated and shifted. It is truncated below at and is shifted to the left by the amount . The expected value is the average of all losses that require a payment, or the average of all payments that are made, hence the name payment per payment. The average is also called the mean excess loss function.
(3)……
When the is understood, we may use the notation instead of . When the random variable is a distribution for insurance losses, the function is interpreted as mean excess loss – the average portion of a loss that is in excess of the deductible. If represents a distribution of lifetime, then the expectation is called the mean residual life or complete expectation of life. It is then notated by or .
We now compare the probability density functions (pdfs) and the cumulative distribution functions (cdfs) of the two payment variables. Let and be the pdf and cdf of the loss random variable , respectively.
(4)……
(5)……
Both pdfs are a function of the pdf of the loss . Assuming that the distribution of is a continuous distribution, the pdf is a mixed distribution (part discrete and part continuous). It has a point mass at with probability , representing the scenario that no payment is made for a small loss. The pdf is a continuous one if is continuous. It is the result of divided by since it is a conditional distribution (only losses exceeding the deductible are considered). The cdfs of the payment per loss and the payment per payment are given by the following:
(6)……
(7)……
The cdf has a jump at , reflecting the scenario that there is no payment for any loss less than the deductible . Note that the size of the jump is . The cdf has no jump at since losses below the deductible are not accounted for in the variable .
Once the pdfs are derived, the calculation of the mean and variance of each of the two variables is possible. Once again, assuming the loss distribution is a continuous one, the following integrals give the calculation. When the loss distribution is discrete, simply replace integrals with summations.
(8)……
(9)……
(10)……
(11)……
(12)……
(13)……
Comparing (8) and (11) and comparing (9) and (12) give the following relations.
(14)…………or……
(15)…………or……
The relations (14) and (15) show that only one set of calculation needs to be made. Once the first and second moments of one variable are known, the moments of the other variable are obtained by adjusting with , the probability of having a claim.
Demonstrating the Calculation
We work one example to illustrate the above calculation.
Example 1
The loss has the density function where . The coverage has a deductible of 3. Compute the mean and variance of the insurance payment per loss. Use the per loss results to obtain the mean and variance of the payment per payment. Write out the cdf of and the cdf of .
First, obtain the density function of the payment per loss.
The following calculates the mean and variance.
The probability of a claim is . Dividing the results by this probability gives the following results.
To contrast, , the average payment for loss if there is no deductible. By imposing a deductible of 3, on average the amount of 5 – 2.2295 = 2.7705 per loss is shifted to the insured. The average payment per payment 2.84375 is higher than the average payment per loss 2.2295 because the per payment calculation centers on the loss exceeding the deductible.
The cdf of the original loss is where . The cdfs of and are given by:
With the cdf of set up, we can determine the probabilities of claim payments. For example, if a claim has to be paid, what is the probability that the claim payment is between 3 and 5?
Using Limited Expectation
The above discussion focuses on developing the probability distribution of the payment per loss and the probability distribution of the payment per payment, as well as how the two distributions relate. Once the distribution of a payment variable is established, we can then evaluate various distributional quantities regarding the payment in question – e.g. mean and variance.
Another way to calculate is through the limited loss random variable .
(16)……
It is clear that . In words, buying a policy with a deductible of and another policy with a policy limit of equals full coverage. The expectation is called the limited expectation. Immediately we have the following relationship.
(17)…………or……
One advantage of the relationship (17) is that the limited expectation is provided in a table of distributions (table). For several distributions such as exponential, lognormal, Pareto and a few others, the formulas of are quite easy to implement. When the loss distribution is one of these distributions, (17) provides another way to compute the expected payment per loss. Combining (17) and (14) gives the following formula for the expected payment per payment (or mean excess loss).
(18)……
Franchise Deductible
The deductible in the above discussion is an ordinary deductible. The key characteristic of ordinary deductible is that the first dollars are not paid by the insurer. A franchise deductible works like an ordinary deductible except that when the loss exceeds the deductible, the policy pays the loss in full.
(19)……
Instead of building a new set of formulas, we can simply relate the expected payment under the franchise deductible with the expected payment under the ordinary deductible.
(20)……expected payment per loss under franchise deductible =
(21)……expected payment per payment under franchise deductible =
The payment under a franchise deductible is larger than the payment under an ordinary deductible. By how much? The deductible . The addition of in (20) reflects the additional benefit when . The expected payment per payment is already conditioned on . Thus is unconditionally added to in (21). The variance can be calculated using basic principle. Example 2 shows how.
To distinguish between the two deductibles, we adopt the convention that deductible means ordinary deductible. If franchise deductible is used, “franchise” would have to be explicitly stated.
More Examples
Example 2
Work Example 1 assuming a franchise deductible.
To keep things straight, we let to denote the payment per loss and to denote the payment per payment (the notation is applicable just for this example). Then and are greater than their counterparts for ordinary deductible according to (20) and (21).
To find the variance, it is helpful to work with the pdfs.
For franchise deductible, there is no need to shift the pdf by the deductible. There is a point mass at like before. The pdf continues with the interval (3, 10). With this pdf, the following gives the calculation of the variance.
We can leverage the per loss variance to obtain the per payment variance.
The cdfs are given by the following:
If the insurer has to pay a claim, what is the probability that it is between 6 and 8?
Note that this answer is the same one as in Example 1 since there is no shifting with a franchise deductible.
Example 3
The discussion is not complete without mentioning exponential distribution. Suppose that the pdf of the loss is where , i.e. the pdf for an exponential distribution with mean . The deductible has an ordinary deductible of . The expected payment per loss and the variance of payment per loss are:
The above results are obtained by using the pdf of the payment per loss , which can be obtained according to (4). The results for payment per payment are obtained by dividing by .
The mean and variance of are identical to the original exponential loss distribution. This is to be expected. Because the exponential distribution is memoryless, the payment variable is identical to the original loss , an exponential distribution with mean . For models of insurance payments with a deductible, knowing that the loss distribution is exponential simplifies the calculation greatly.
Example 4
This example illustrates how to obtain the mean payment per loss using the table approach – using limited expectation given in this table. Suppose that the loss distribution is a lognormal distribution with parameters and . The coverage has a deductible of 200. Compute the expected payment per loss .
Using the formula for and the formula for the limited expectation found in the table, the expected payment per loss is given by:
Example 5
This example continues Example 4. We now tackle the variance of the payment per loss. For the lognormal loss distribution, it is difficult to find the pdf of the payment per loss . In the table, there is no direct formula for . However, there is a way to use the limited expectations and to derive . First, let’s break down the components that go into .
The in above is the pdf of the lognormal distribution. Of course we are not to evaluate the above two integrals directly. However, they can be expressed using limited expectations as follows:
With respect to the lognormal parameters and and the deductible 200, the limited expectation is given by the following. The first limited expectation is already calculated in Example 4.
The calculation continues:
Example 6
This example discusses the Pareto distribution as the loss distribution. The Pareto distribution is a twoparameter family of distributions. The pdf and cdf are given below, along with the mean and the variance.
For the Pareto mean to exist, the parameter must be greater than 1. For the variance to exist, the parameter must be greater than 2. When the coverage has an ordinary deductible , consider the insurance payment per loss and the insurance payment per loss . In terms of these two payment variables, the Pareto distribution is very tractable. To find mean payment per loss, either start with the pdf of or use the limited expectation from the table. Because the payment per payment is also a Pareto distribution, there are shortcuts for finding the variance of payment per loss (see Example 3 here).
When the loss is a Pareto distribution with parameters and , the payment per payment has a Pareto distribution with parameters and . Let’s examine the expected value of , or the mean excess loss.
Note that the mean excess loss is an increasing function of . This means that the higher the deductible, the larger the expected claim if such a large loss occurs! If a random loss is modeled by such a distribution, it is a catastrophic risk situation. In general, an increasing mean excess loss function is an indication of a heavy tailed distribution. Thus the Pareto distribution is considered a heavy tailed distribution and is a suitable candidate for modeling situations that have potential extreme large losses.
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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.
In contrast, the next post is a discussion on the insurance payment per payment.
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.
In contrast, the next post is a discussion on the insurance payment per payment.
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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 multiplied by integers 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 an expression for some constants and .
(1)……….
Multiplying (1) by gives the following.
(1a)……….
Note that the righthand side of (1a) is a linear expression of . This provides a way to fit observations to (a,b,0) distributions.
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 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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Practice Problem Set 11 – (a,b,0) class
The practice problems in this post focus on counting distributions that belong to the (a,b,0) class, reinforcing the concepts discussed in this blog post in a companion blog.
The (a,b,1) class is a generalization of (a,b,0) class. It is discussed here. A practice problem set on the (a,b,1) class is found here.
Notation: for where is the counting distribution being focused on.
Practice Problem 11A 
Suppose that claim frequency follows a negative binomial distribution with parameters and . The following is the probability function.
Evaluate the negative binomial distribution in two ways.

Practice Problem 11B 
Suppose that follows a distribution in the (a,b,0) class. You are given that Evaluate the probability that is at least 1. 
Practice Problem 11C 
The following information is given about a distribution from the (a,b,0) class. What is the form of the distribution? Evaluate . 
Practice Problem 11D 
For a distribution from the (a,b,0) class, the following information is given. Determine the variance of this distribution. 
Practice Problem 11E 
For a distribution from the (a,b,0) class, you are given that and . Find the value of . 
Practice Problem 11F 
You are given that the distribution for the claim count satisfies the following recursive relation: Determine . 
Practice Problem 11G 
Suppose that the random variable is from the (a,b,0) class. You are given that and . Calculate the probability that is at least 3. 
Practice Problem 11H 
For a distribution from the (a,b,0) class, you are given that
Determine . 
Practice Problem 11I 
For a distribution from the (a,b,0) class, you are given that and . Evaluate its mean. 
Practice Problem 11J 
The random variable follows a distribution from the (a,b,0) class. You are given that and . Evaluate . 
Practice Problem 11K 
The random variable follows a distribution from the (a,b,0) class. Suppose that and . Determine . 
Practice Problem 11L 
Given that a discrete distribution is a member of the (a,b,0) class. Which of the following statement(s) are true?
B. ……….. 2 only C. ……….. 3 only D. ……….. 1 and 2 only E. ……….. 1 and 3 only 
Practice Problem 11M 
Given that a discrete distribution is a member of the (a,b,0) class, determine the variance of the distribution if and . 
Practice Problem 11N 
For a distribution in the (a,b,0) class, and . Furthermore, the mean of the distribution is 1. Determine . 
Problem  Answer 

11A 

11B 

11C 

11D 

11E 

11F 

11G 

11H 

11I 

11J 

11K 

11L 

11M 

11N 

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Practice Problem Set 10 – value at risk and tail value at risk
In actuarial applications, an important focus is on developing loss distributions for insurance products. It is also critical to employ risk measures to evaluate the exposure to risk. This post provides practice problems on two risk measures that are useful from an actuarial perspective. They are: valueatrisk (VaR) and tailvalueatrisk (TVaR).
Practice problems in this post are to reinforce the concepts of VaR and TVaR discussed in this blog post in a companion blog.
Most of the practice problems refer to parametric distributions highlighted in a catalog for continuous parametric models.
Practice Problem 10A 
Losses follow a paralogistic distribution with shape parameter and scale parameter .
Determine the VaR at the security level 99%. 
Practice Problem 10B 
Annual aggregate losses for an insurer follow an exponential distribution with mean 5,000. Evaluate VaR and TVaR for the aggregate losses at the 99% security level. 
Practice Problem 10C 
For a certain line of business for an insurer, the annual losses follow a lognormal distribution with parameters and . Evaluate the valueatrisk and the tailvalueatrisk at the 95% security level. 
Practice Problem 10D 
Annual losses follow a normal distribution with mean 1000 and variance 250,000. Compute the tailvaluerisk at the 95% security level. 
Practice Problem 10E 
An insurance company models its liability insurance business using a Pareto distribution with shape parameter and scale parameter . Evaluate the valueatrisk and the tailvalueatrisk at the 99.5% security level. 
Practice Problem 10F 
Losses follow an inverse exponential distribution with parameter . Calculate the valueatrisk at the 99% security level. 
Practice Problem 10G 
Losses follow a mixture of two exponential distributions with equal weights where one exponential distribution has mean 10 and the other has mean 20. Evaluate the valueatrisk and the tailvalueatrisk at the 95% security level. 
Practice Problem 10H 
Losses follow a mixture of two Pareto distributions with equal weights where one Pareto distribution has shape parameter and scale parameter and the other has shape parameter and scale parameter . Evaluate the valueatrisk and the tailvalueatrisk at the 99% security level. 
Practice Problem 10I 
Losses follow a Weibull distribution with parameters and . Determine the valueatrisk at the security level 99.5%. 
Practice Problem 10J 
Losses follow an inverse Pareto distribution with parameters and . Determine the valueatrisk at the security level 99%. 
Problem  Answer 

10A 

10B 

10C 

10D 

10E 

10F 

10G 

10H 

10I 

10J 

Daniel Ma Math
Daniel Ma Mathematics
Actuarial exam
2018 – Dan Ma
Practice Problem Set 9 – Expected Insurance Payment – Additional Problems
This practice problem set is to reinforce the 3part discussion on insurance payment models (Part 1, Part 2 and Part 3). The practice problems in this post are additional practice problems in addition to Practice Problem Set 7 and Practice Problem Set 8.
Practice Problem 9A 
Losses follow a distribution that is a mixture of two equally weighted Pareto distributions, one with parameters and and the other with with parameters and . An insurance coverage for these losses has an ordinary deductible of 1000. Calculate the expected payment per loss. 
Practice Problem 9B 
Losses, prior to any deductible being applied, follow an exponential distribution with mean 17.5. An insurance coverage has a deductible of 8. Inflation of 15% impacts all claims uniformly from the current year to next year.
Determine the percentage change in the expected claim cost per loss from the current year to next year. 
Practice Problem 9C 
Losses follow a distribution that has the following density function.
An insurance policy is purchased to cover these losses. The policy has a deductible of 3. Calculate the expected insurance payment per payment. 
Practice Problem 9D 
Losses follow a distribution with the following density function.
An insurance coverage pays losses up to a maximum of 100,000. Determine the average payment per loss. 
Practice Problem 9E 
You are given the following information.
Determine the average insurance payment per loss. 
Practice Problem 9F 
You are given the following information.
Determine the proportion of the losses that exceed 1,000. 
Practice Problem 9G 
Losses follow a uniform distribution on the interval . The insurance coverage has a deductible of 250. Determine the variance of the insurance payment per loss. 
Practice Problem 9H 
Losses follow an exponential distribution with mean 500. An insurance coverage that is designed to cover these losses has a deductible of 1,000. Determine the coefficient of variation of the insurance payment per loss. 
Practice Problem 9I 
Losses are modeled by an exponential distribution with mean 3,000. An insurance policy covers these losses according to the following provisions.
Determine the expected insurance payment per loss. 
Practice Problem 9J 
You are given the following information.
Determine the insurance company’s expected claim cost per claim after the effective date of the reinsurance policy. 
Problem  Answer 

9A 

9B 

9C 

9D 

9E 

9F 

9G 

9H 

9I 

9J 

Daniel Ma actuarial
Dan Ma actuarial
2017 – Dan Ma