Negative Binomial Distribution
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 6 – Negative Binomial Distribution
This post has exercises on negative binomial distributions, reinforcing concepts discussed in
this previous post. There are several versions of the negative binomial distribution. The exercises are to reinforce the thought process on how to use the versions of negative binomial distribution as well as other distributional quantities.
Practice Problem 6A 
The annual claim frequency for an insured from a large population of insured individuals is modeled by the following probability function.
Determine the following:

Practice Problem 6B 
The number of claims in a year for an insured from a large group of insureds is modeled by the following model. The parameter varies from insured to insured. However, it is known that is modeled by the following density function. Given that a randomly selected insured has at least one claim, determine the probability that the insured has more than one claim. 
Practice Problem 6C 
Suppose that the number of accidents per year per driver in a large group of insured drivers follows a Poisson distribution with mean . The parameter follows a gamma distribution with mean 0.6 and variance 0.24. Determine the probability that a randomly selected driver from this group will have no more than 2 accidents next year. 
Practice Problem 6D 
Suppose that the random variable follows a negative binomial distribution such that
Determine the mean and variance of . 
Practice Problem 6E 
Suppose that the random variable follows a negative binomial distribution with mean 0.36 and variance 1.44.
Determine . 
Practice Problem 6F 
A large group of insured drivers is divided into two classes – “good” drivers and “bad”drivers. Seventy five percent of the drivers are considered “good” drivers and the remaining 25% are considered “bad”drivers. The number of claims in a year for a “good” driver is modeled by a negative binomial distribution with mean 0.5 and variance 0.625. On the other hand, the number of claims in a year for a “bad” driver is modeled by a negative binomial distribution with mean 2 and variance 4. For a randomly selected driver from this large group, determine the probability that the driver will have 3 claims in the next year. 
Practice Problem 6G 
The number of losses in a year for one insurance policy is the random variable where . The random variable is modeled by a geometric distribution with mean 0.4 and variance 0.56.
What is the probability that the total number of losses in a year for three randomly selected insurance policies is 2 or 3? 
Practice Problem 6H 
The random variable follows a negative binomial distribution. The following gives further information.
Determine and . 
Practice Problem 6I 
Coin 1 is an unbiased coin, i.e. when tossing the coin, the probability of getting a head is 0.5. Coin 2 is a biased coin such that when tossing the coin, the probability of getting a head is 0.6. One of the coins is chosen at random. Then the chosen coin is tossed repeatedly until a head is obtained.
Suppose that the first head is observed in the fifth toss. Determine the probability that the chosen coin is Coin 2. 
Practice Problem 6J 
In a production process, the probability of manufacturing a defective rear view mirror for a car is 0.075. Assume that the quality status of any rear view mirror produced in this process is independent of the status of any other rear view mirror. A quality control inspector is to examine rear view mirrors one at a time to obtain three defective mirrors.
Determine the probability that the third defective mirror is the 10th mirror examined. 
Problem  Answer 

6A 

6B  
6C  0.9548 
6D  mean = 0.65, variance = 0.975 
6E  0.016963696 
6F  0.04661 
6G  
6H 

6I  0.329543 
6J  0.008799914 
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Several versions of negative binomial distribution
This post shows how to work with negative binomial distribution from an actuarial modeling perspective. The negative binomial distribution is introduced as a Poissongamma mixture. Then other versions of the negative binomial distribution follow. Specific attention is paid to the thought processes that facilitate calculation involving negative binomial distribution.
Negative Binomial Distribution as a PoissonGamma Mixture
Here’s the setting for the Poissongamma mixture. Suppose that has a Poisson distribution with mean and that is a random variable that varies according to a gamma distribution with parameters (shape parameter) and (rate parameter). Then the following is the unconditional probability function of .
The distribution described in (1) is one parametrization of the negative binomial distribution (derived here). It has two parameters and (coming from the gamma mixing weights). The following is another parametrization.
The distribution described in (2) is obtained when the gamma mixing weight has a shape parameter and a scale parameter . Since the gamma scale parameter and rate parameter is related by , (2) can be derived from (1) by setting .
Both (1) and (2) contain the ratio that is expressed using the gamma function. The next task is to simplify the ratio using a general notion of binomial coefficient.
The Poissongamma mixture is discussed in this blog post in a companion blog called Topics in Actuarial Modeling.
General Binomial Coefficient
The familiar binomial coefficient is the following:
where the top number is a positive integer and the bottom number is a nonnegative integer such that . Other notations for binomial coefficient are , and . The right hand side of the above expression can be simplified by canceling out .
The expression in (4) is obtained by canceling out in (3). Note that does not have to be an integer for the calculation in (4) to work. The bottom number has to be a nonnegative number since is involved. However, can be any positive real number as long as .
Thus the expression in (4) gives a new meaning to the binomial coefficient where is a positive real number and is a nonnegative integer such that .
For example, and . The thought process is that the numerator is obtained by subtracting 1 times from . If , this thought process would not work. For convenience, when is a positive real number.
We now use the binomial coefficient defined in (5) to simplify the ratio where is a positive real number and is a nonnegative integer. We use a key fact about gamma function: . Then for any integer , we have the following derivation.
The right hand side of the above expression is precisely the binomial coefficient when . Thus we have the following relation.
where is an integer with .
Negative Binomial Distribution
With relation (6), the two versions of Poissongamma mixture stated in (1) and (2) are restated as follows:
The above two parametrizations of negative binomial distribution are used if information about the Poissongamma mixture is known. In (7), the gamma distribution in the Poissongamma mixture has shape parameter and rate parameter . In (8), the gamma distribution has shape parameter and scale parameter . The following is a standalone version of the binomial distribution.
In (9), the negative binomial distribution has two parameters and where . In this parmetrization, the parameter is simply a real number between 0 and 1. It can be viewed as a probability. In fact, this is the case when the parameter is an integer. Version (9) can be restated as follows when is an integer.
In version (10), the parameters are (a positive integer) and a real number with . Since is an integer, the usual binomial coefficient appears in the probability function.
Version (10) has a natural interpretation. A Bernoulli trial is an random experiment that results in two distinct outcome – success or failure. Suppose that the probability of success is in each trial. Perform a series of independent Bernoulli trials until exactly successes occur where is a fixed positive integer. Let the random variable be the number of failures before the occurrence of the th success. Then (10) is the probability function for the random variable .
A special case of (10). When the parameter is 1, the negative binomial distribution has a special name.
The distribution in (11) is said to be a geometric distribution with parameter . The random variable defined by (11) can be interpreted as the number of failures before the occurrence of the first success when performing a series of independent Bernoulli trials. Another important property of the geometric distribution is that it is the only discrete distribution with the memoryless property. As a result, the survival function of the geometric distribution is where .
More about Negative Binomial Distribution
The probability functions of various versions of the negative binomial distribution have been developed in (1), (2), (7), (8), (9), (10) and (11). Other distributional quantities can be derived from the Poissongamma mixture. We derive the mean and variance of the negative binomial distribution.
Suppose that the negative binomial distribution is that of version (8). The conditional random variable has a Poisson distribution with mean and the random variable has a gamma distribution with a shape parameter and a scale parameter . Note that and . Furthermore, we have the conditional mean and conditional variance
The following derives the mean and variance of .
The above mean and variance are for parametrization in (8). To obtain the mean and variance for the other parametrizations, make the necessary translation. For example, to get (7), plug into the above mean and variance. For (9), let . Then solve for and plug that into the above mean and variance. Version (10) should have the same formulas as for (9). To get (11), set . The following table lists the negative binomial mean and variance.
Version  Mean  Variance 

(7)  
(8)  
(9) and (10)  
(11) 
The table shows that the variance of the negative binomial distribution is greater than its mean (regardless of the version). This stands in contrast with the Poisson distribution whose mean and the variance are equal. Thus the negative binomial distribution would be a suitable model in situations where the variability of the empirical data is greater than the sample mean.
Modeling Claim Count
The negative binomial distribution is a discrete probability distribution that takes on the nonnegative integers . Thus it can be used as a counting distribution, i.e. a model for the number of events of interest that occur at random. For example, the described above can be a good model for the frequency of loss, i,e, the random variable of the number of losses, either arising from a portfolio of insureds or from a particular insured in a given period of time.
The Poissongamma model has a great deal of flexibility. Consider a large population of individual insureds. The number of losses (or claims) in a year for each insured has a Poisson distribution with mean . From insured to insured, there is uncertainty in the mean annual claim frequency . However, the random variable varies according to a gamma distribution. As a result, the annual number of claims for an “average” insured or a randomly selected insured from the population will follow a negative binomial distribution.
Thus in a Poissongamma model, the claim frequency for an individual in the population follows a Poisson distribution with unknown gamma mean. The weighted average of these conditional Poisson claim frequencies is a negative binomial distribution. Thus the average claim frequency over all individuals has a negative binomial distribution.
The table in the preceding section shows that the variance of the negative binomial distribution is greater than the mean. This is in contrast to the fact that the variance and the mean of a Poisson distribution are equal. Thus the unconditional claim frequency is more dispersed than its conditional distributions. The increased variance of the negative binomial distribution reflects the uncertainty in the parameter of the Poisson mean across the population of insureds. The uncertainty in the parameter variable has the effect of increasing the unconditional variance of the mixture distribution of . Recall that the variance of a mixture distribution has two components, the weighted average of the conditional variances and the variance of the conditional means. The second component represents the additional variance introduced by the uncertainty in the parameter .
We present two examples. More examples to come at the end of the post.
Example 1
For a given insured driver in a large portfolio of insured drivers, the number of collision claims in a year has a Poisson distribution with mean . The Poisson mean follows a gamma distribution with mean 4 and variance 80. For a randomly selected insured driver from this portfolio,
 what is the probability of having exactly 2 collision claims in the next year?
 what is the probability of having at most one collision claim in the next year?
The number of collision claims in a year is a Poissongamma mixture and thus is a negative binomial distribution. From the given gamma mean and variance, we can determine the parameters of the gamma distribution. In this example, we use the parametrization of (8). Expressing the gamma mean and variance in terms of the shape and scale parameters, we have and . These two equations give and . The probabilities are calculated based on (8).
The answer for the first question is . The answer for the second question is . Thus there is a closed to 65% chance that an insured driver has at most one claim in a year.
Example 2
For an automobile insurance company, the distribution of the annual number of claims for a policyholder chosen at random is modeled by a negative binomial distribution that is a Poissongamma mixture. The gamma distribution in the mixture has a shape parameters of and scale parameter . What is the probability that a randomly selected policyholder has more than two claims in a year?
Since the gamma shape parameter is 1, the unconditional number of claims in a year is a geometric distribution with parameter . The following is the desired probability.
A Recursive Formula
The probability functions described in (1), (2), (7), (8), (9), (10) and (11) describe clearly how the negative binomial probabilities are calculated based on the two given parameters. The probabilities can also be calculated recursively. Let where . We introduce a recursive formula that allows us to compute the value if is known. The following is the form of the recursive formula.
In (12), the numbers and are constants. Note that the formula (12) calculates probabilities for all . It turns out that the initial probability is determined by the constants and . Thus the constants and completely determines the probability distribution represented by . Any discrete probability distribution that satisfies this recursive relation is said to be a member of the (a,b,0) class of distributions.
We show that the negative binomial distribution is a member of the (a,b,0) class of distributions. First, assume that the negative binomial distribution conforms to the parametrization in (8) with parameters and . Then let and be defined as follows:
.
Let the initial probability be . We claim that the probabilities generated by the formula (12) are identical to the ones calculated from (8). To see this, let’s calculate a few probabilities using the formula.
The above derivation demonstrates that formula (12) generates the same probabilities as (8). By adjusting the constants and , the recursive formula can also generate the probabilities in the other versions of the negative binomial distribution. For the negative binomial version (9) with parameters and , the and should be defined as follows:
With the initial probability , the recursive formula (12) will generate the same probabilities as those from version (9).
More Examples
Example 3
Suppose that an insured will produce claims during the next exposure period is
where . Furthermore, the parameter varies according to a distribution with the following density function:
What is the probability that a randomly selected insured will produce more than 2 claims during the next exposure period?
Note that the claim frequency for an individual insured has a Poisson distribution with mean . The given density function for the parameter is that of a gamma distribution with and rate parameter . Thus the number of claims in an exposure period for a randomly selected (or “average” insured) will have a negative binomial distribution. In this case the parametrization (7) is the most useful one to use.
The following calculation gives the relevant probabilities to answer the question.
Summing the three probabilities gives . Then . There is a 19.42% chance that a randomly selected insured will have more than 2 claims in an exposure period.
Example 4
The number of claims in a year for each insured in a large portfolio has a Poisson distribution with mean . The parameter follows a gamma distribution with mean 0.75 and variance 0.5625.
Determine the proportion of insureds that are expected to have less than 1 claim in a year.
Setting and gives and . Thus the parameter follows a gamma distribution with shape parameter and scale parameter . This is an exponential distribution with mean 0.75. The problems asks for the proportion of insured with . Thus the answer is . Thus about 74% of the insured population are expected to have less than 1 claim in a year.
Example 5
Suppose that the number of claims in a year for an insured has a Poisson distribution with mean . The random variable follows a gamma distribution with shape parameter and scale parameter .
One thousand insureds are randomly selected and are to be observed for a year. Determine the number of selected insureds expected to have exactly 3 claims by the end of the oneyear observed period.
With this being a Poissongamma mixture, the number of claims in a year for a randomly selected insured has a negative binomial distribution. Using (8) and based on the gamma parameters given, the following is the probability function of negative binomial distribution.
The following gives the calculation for .
With , about 149 of the randomly selected insureds will have 3 claims in the observed period.
Example 6
Suppose that the annual claims frequency for an insured in a large portfolio of insureds has a distribution that is in the (a,b,0) class. Let be the probability that an insured has claims in a year.
Given that , and , determine the probability that an insured has no claims in a oneyear period.
Given , and , find . Based on the recursive relation (12), we have the following two equations of and .
Solving these two equations gives and . Plugging and into the recursive relation gives the answer.
.
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Revised Nov 2, 2018.