The numbers and , both positive, are fixed constants and are the parameters of the distribution. The symbol is the gamma function. This density function describes how the potential gamma observations distribute across the positive x-axis. Baked into this gamma probability density function are two pieces of information about the Poisson distribution. We can simply read off the information from the parameters and in the density function.
Poisson-Gamma Mixture
For a reason that will be given shortly, the parameters and in (1) gives a negative binomial distribution. The following is the probability function of this negative binomial distribution.
If the parameter is a positive integer, then (2) has a nice interpretation in the context of a series of independent Bernoulli trials. Consider performing a series of independent trials where each trial has one of two distinct outcomes (called success or failure). Assume that the probability of a success in each trial is . In this case, the quantity in (2) is the probability of having failures before the occurrence of the th success.
Note that when is a positive integer, the binomial coefficient has the usual calculation . When is not an integer but is only a positive number, the binomial coefficient is calculated as follows:
For this new calculation to work, does not have to be a positive integer. It only needs to be a positive real number. Then the distribution in (2) does not have a natural interpretation in terms of performing a series of independent Bernoulli trials. It is simply a counting distribution, i.e. a random discrete variable modeling the number of occurrences of a type of random events.
What does this have to do with gamma distribution and Poisson distribution? When a conditional random variable has a Poisson distribution such that its mean is an unknown random quantity but follows a gamma distribution with parameters and as described in (1), the unconditional distribution for has a negative binomial distribution as described in (2). In other words, the mixture of Poisson distributions with gamma mixing weights is a negative binomial distribution.
There is an insurance interpretation of the Poisson-gamma mixture. Suppose that in a large pool of insureds, the annual claim frequency of an insured is a Poisson distribution with mean . The quantity varies from insured to insured but is supposed to follow a gamma distribution. Here we have a family of conditional Poisson distributions where each one is conditional on the characteristics of the particular insured in question (a low risk insured has a low value and a high risk insured has a high value). The “average” of the conditional Poisson distributions will be a negative binomial distribution (using the gamma distribution to weight the parameter ). Thus the claim frequency for an “average” insured in the pool should be modeled by a negative binomial distribution. For a randomly selected insured from the pool, if we do not know anything about this insured, we can use the unconditional negative binomial distribution to model the claim frequency.
A detailed discussion of the negative binomial distribution is found here and here. The notion of mixture distributions and Poisson-gamma mixture in particular are discussed here. Many distributions applicable in actuarial applications are mixture distributions (see here for examples).
Waiting Times in a Poisson Process
The parameter in (1) is called the shape parameter while the parameter is called the rate parameter. The name of rate parameter will be made clear shortly. For this interpretation of gamma distribution to work, the shape parameter must be a positive integer.
The gamma distribution as described in (1) is intimately connected to a Poisson process through the rate parameter . In a Poisson process, event of a particular interest occurs at random at the rate of per unit time. Two questions are of interest. How many times will this event occur in a given time interval? How long do we have to wait to observe the first occurrence, the second occurrence and so on? The gamma distribution as described in (1) gives an answer to the second question when is a positive integer.
A counting process is a random experiment that observes the random occurrences of a certain type of events of interest in a time period. A Poisson process is a counting process that satisfies three simplifying assumptions that deal with independence and uniformity in time.
A good example of a Poisson process is the well known experiment in radioactivity conducted by Rutherford and Geiger in 1910. In this experiment, -particles were emitted from a polonium source and the number of -particles were counted during an interval of 7.5 seconds (2,608 many such time intervals were observed). In these 2,608 intervals, a total of 10,097 particles were observed. Thus the mean count per period of 7.5 seconds is 10097 / 2608 = 3.87. In this experiment, -particles are observed at a rate of 3.87 per unit time (7.5 seconds).
In general, consider a Poisson process with a rate of per unit time, i.e. the event that is of interest occurs at the rate of per unit time. There are two random variables of interest. One is which is the number of occurrences of the event of interest from time 0 to time . The other is which is the waiting time until the th occurrence of the event of interest where is a positive integer.
What can we say about the random variables ? First, the random variable , the number of occurrences of the event of interest in a unit time interval, has a Poisson distribution with mean . The following is the probability function.
For any positive real number , the random variable , which is a discrete random variable, follows a Poisson distribution with mean . The following is the probability function.
For a more detailed discussion of why and are Poisson, see here.
We now discuss the distributions for and . The random variable would be the mean time to the first occurrence. It has an exponential distribution with mean . The following is the probability density function.
The random variable has a gamma distribution with shape parameter and rate parameter . The following is the probability density function.
The density functions in (5) and (6) are derived from the Poisson distributions in (3) and (4). For example, for (the first occurrence taking place after time ) to happen, there is no occurrence in the time interval . Thus . Note that is a Poisson random variable with mean . The following derives the density function in (5).
Note that the cumulative distribution function is that for an exponential distribution. Taking the derivative gives the density function in (5). Extending the same reasoning, for (the th occurrence taking place after time ) to happen, there must be at most occurrences in the time interval . Once again, we are relating to the Poisson random variable . More specifically .
Taking the derivative of (8) gives the density function in (6). See here for a more detailed discussion of the relation between gamma distribution and Poisson distribution in a Poisson process.
Evaluating Gamma Survival Function
The relation (7) shows that the gamma survival function is the cumulative distribution function (CDF) of the corresponding Poisson distribution. Consider the following integral.
The integrand in the above integral is the density function of a gamma distribution (with the shape parameter being a positive integer). The limits of the integral are from to . Conceptually the integral is identical to (7) since the integral gives , which is the survival function of the gamma distribution in question. According to (7), the integral in (9) has a closed form as follows.
The above is a combination of (7) and (9) and is a way to evaluate the right tail of the gamma distribution when the shape parameter is a positive integer . Instead of memorizing it, we can focus on the thought process.
Given the integral in (9), note the shape parameter and the rate parameter . Consider a Poisson process with rater parameter . In this Poisson process, the random variable , the number of occurrences of the event of interest in a unit time interval, has a Poisson distribution with mean . The random variable , the number of occurrences of the event of interest in a time interval of length , has a Poisson distribution with mean . Then the integral in (9) is equivalent to , the probability that there are at most occurrences in the time interval .
The above thought process works even if the gamma distribution has no relation to any Poisson process, e.g. it is just the model for size of insurance losses. We can still pretend there is a Poisson relationship and obtain a closed form evaluation of the survival function. It must be emphasized that this evaluation of the gamma survival function in closed form is possible only when the shape parameter is a positive integer. If it is not, we then need to evaluate the gamma survival function using software.
Remarks
The discussion here presents two relationships between the gamma distribution and the Poisson distribution. In the first one, mixing conditional Poisson distributions with gamma mixing weights produces a negative binomial distribution. Taking the shape parameter and rate parameter in (1) produces the negative binomial distribution as in (2). This is one interpretation of the two gamma parameters.
In the second interpretation, the rate parameter of the gamma model is intimately tied to the rate parameter of a Poisson process. The interplay between the number of occurrences of the event of interest and the waiting time until the th occurrence connects the gamma distribution with the Poisson distribution. For a fuller discussion of this interplay, see here.
The gamma distribution is mathematically derived from the gamma function. See here for an introduction to the gamma distribution.
The three assumptions for a Poisson process that are alluded to above allow us to obtain a binomial distribution when dividing a time interval into subintervals (for a sufficiently large ). As the subintervals get more and more granular, the binomial distributions have Poisson distribution as the limiting distribution. The derivation from binomial to Poisson is discussed here, here and here.
Practice problems can be found here in a companion blog for practice problems.
Dan Ma gamma distribution
Dan Ma Poisson distribution
Dan Ma math
Daniel Ma gamma distribution
Daniel Ma mathematics
Daniel Ma Poisson distribution
2018 – Dan Ma
]]>The catalog lists the models according to their mathematical properties (e.g. how they are derived and how they are related to the gamma distribution). The models have been discussed quite extensively in the blog for Topics in Actuarial Modeling. The catalog puts all the models in one place so to speak. It has the links to the blog posts that describe the models. Just click on a link if anyone is interested in knowing more about a particular parametric model. The parametric models in the catalog include the following:
This list is by no means comprehensive or exhaustive. It should be a good resource to begin the modeling process (or the study of the actuarial modeling process). In fact, the modeling process is part of the exam syllabus of the Society of Actuaries. The catalog of models is found here.
actuarial modeling
parametric severity models
Dan Ma math
Daniel Ma mathematics
2017 – Dan Ma
]]>In the above diagram, the red curve is the density function for the Pareto Type I distribution while the green curve is the density function for the Pareto Type II distribution. For both distributions, the shape parameter is and the scale parameter is .
Note that the green density curve is the result of shifting the red curve horizontally 5 units to the left. The following are the density functions.
Green Curve
From the diagram and the density functions, we see that the green density curve is the result of shifting the red curve to the left by 5. Otherwise, both curve have the same distributional shape.
Since Pareto Type II is a shifting of Pareto Type I, they share similar mathematical properties. For example, they are both heavy tailed distributions. Thus they are appropriate for modeling extreme losses (in insurance applications) or financial fiasco or financial ruin (in financial applications). Which one to use depends on where the starting point of the random variable is.
However, the mathematical calculations are different to a large degree since the Pareto Type II density function is the result of a horizontal shift.
We would like to introduce blog posts in two companion blogs that discuss the Pareto distribution in greater details. This blog post is from a companion blog called Topics in Actuarial Modeling. It describes Pareto Type I and Type II in greater details. This blog post from another companion blog called Actuarial Modeling Practice has a side-by-side comparison of Pareto Type I and Pareto Type II. It also discusses the two Pareto types from a calculations stand point. This blog post has practice problems on the two Pareto types.
Another important mathematical property of Pareto Type II is that it is the mixture of exponential distributions with gamma mixing weights (exponential-gamma mixture). This is also discussed in the highlighted blog posts.
2017 – Dan Ma
]]>________________________________________________________________________
Three versions
The negative binomial distribution has two parameters and , where is a positive real number and . The first two versions arise from the case that is a positive integer, which can be interpreted as the random experiment of a sequence of independent Bernoulli trials until the th success (the trials have the same probability of success ). In this interpretation, there are two ways of recording the random experiment:
The other parameter is the probability of success in each Bernoulli trial. The notation is the binomial coefficient where and are non-negative integers and is defined as:
With this in mind, the following are the probability functions of the random variables and .
The thought process for (1) is that for the event to happen, there can only be successes in the first trials and one additional success occurring in the last trial (the th trial). The thought process for (2) is that for the event to happen, there are trials ( failures and successes). In the first trials, there can be only failures (or equivalently successes). Note that . Thus knowing the mean of will derive the mean of , a fact we will use below.
Instead of memorizing the probability functions (1) and (2), it is better to understand and remember the thought processes involved. Because of the natural interpretation of performing Bernoulli trials until the th success, it is a good idea to introduce the negative binomial distribution via the distributions described by (1) and (2), i.e., the case where the parameter is a positive integer. When , the random experiment is a sequence of independent Bernoulli trials until the first success (this is called the geometric distribution).
Of course, (1) and (2) can also simply be used as counting distributions without any connection with a series of Bernoulli trials (e.g. used in an insurance context as the number of losses or claims arising from a group of insurance policies).
The binomial coefficient in (0) is defined when both numbers are non-negative integers and that the top one is greater than or equal to the bottom one. However, the rightmost term in (0) can be calculated even when the top number is not a non-negative integer. Thus when is any real number, the rightmost term (0) can be calculated provided that the bottom number is a positive integer. For convenience we define . With this in mind, the binomial coefficient is defined for any real number and any non-negative integer .
The third version of the negative binomial distribution arises from the relaxation of the binomial coefficient just discussed. With this in mind, the probability function in (2) can be defined for any positive real number :
where .
Of course when is a positive integer, versions (2) and (3) are identical. When is a positive real number but is not an integer, the distribution cannot be interpreted as the number of failures until the occurrence of th success. Instead, it is used as a counting distribution.
________________________________________________________________________
The probabilities sum to one
Do the probabilities in (1), (2) or (3) sum to one? For the interpretations of (1) and (2), is it possible to repeatedly perform Bernoulli trials and never get the th success? For , is it possible to never even get a success? In tossing a fair coin repeatedly, soon enough you will get a head and even if is a large number, you will eventually get number of heads. Here we wish to prove this fact mathematically.
To show that (1), (2) and (3) are indeed probability functions, we use a fact concerning Maclaurin’s series expansion of the function , a fact that is covered in a calculus course. In the following two results, is a fixed positive real number and is any non-negative integer:
The result (4) is to rearrange the binomial coefficient in probability function (3) to another binomial coefficient with a negative number. This is why there is the word “negative” in negative binomial distribution. The result (5) is the Maclaurin’s series expansion for the function . We first derive these two facts and then use them to show that the negative binomial probabilities in (3) sum to one. The following derives (4).
To derive (5), let . Based on a theorem that can be found in most calculus text, the function has the following Maclaurin’s series expansion (Maclaurin’s series is simply Taylor’s series with center = 0).
where . Now, filling in the derivatives , we have the following derivation.
We can now show that the negative binomial probabilities in (3) sum to one. Let .
________________________________________________________________________
The moment generating function
We now derive the moment generating function of the negative binomial distribution according to (3). The moment generation function is over all real numbers for which is defined. The following derivation does the job.
The above moment generating function works for the negative binomial distribution with respect to (3) and thus to (2). For the distribution in (1), note that . Thus . The moment generating function of (1) is simply the above moment generating function multiplied by the factor . To summarize, the moment generating functions for the three versions are:
The domain of the moment generating function is the set of all that for which or is defined and is positive. Based on the form that it takes, we focus on making sure that . This leads to the domain .
________________________________________________________________________
The mean and the variance
With the moment generating function derived in the above section, we can now focus on finding the moments of the negative binomial distribution. To find the moments, simply take the derivatives of the moment generating function and evaluate at . For the distribution represented by the probability function in (3), we calculate the following:
After taking the first and second derivatives and evaluate at , the first and the second moments are:
The following derives the variance.
The above formula is the variance for the three versions (1), (2) and (3). Note that . In contrast, the variance of the Poisson distribution is identical to its mean. Thus in the situation where the variance of observed data is greater than the sample mean, the negative binomial distribution should be a better fit than the Poisson distribution.
________________________________________________________________________
The independent sum
There is an easy consequence that follows from the moment generating function derived above. The sum of several independent negative binomial distributions is also a negative binomial distribution. For example, suppose are independent negative binomial random variables (version (3)). Suppose each has parameters and (the second parameter is identical). The moment generating function of the independent sum is the product of the individual moment generating functions. Thus the following is the moment generating function of .
where . The moment generating function uniquely identifies the distribution. The above is that of a negative binomial distribution with parameters and according to (3).
A special case is that the sum of independent geometric distributions is a negative binomial distribution with the parameter being . The following is the moment generating function of the sum of independent geometric distributions.
________________________________________________________________________
Consider a counting process that describes the occurrences of a certain type of events of interest in a unit time interval subject to three simplifying assumptions (discussed below). We are interested in counting the number of occurrences of the event of interest in a unit time interval. As a concrete example, consider the number of cars arriving at an observation point in a certain highway in a period of time, say one hour. We wish to model the probability distribution of how many cars that will arrive at the observation point in this particular highway in one hour. Let be the random variable described by this probability distribution. We wish to konw the probability that there are cars arriving in one hour. We start with using a binomial distribution as an approximation to the probability . We will see that upon letting , the is a Poisson probability.
Suppose that we know , perhaps an average obtained after observing cars at the observation points for many hours. The simplifying assumptions alluded to earlier are the following:
Assumption 1 means that a large number of cars arriving in one period does not imply fewer cars will arrival in the next period and vice versa. In other words, the number of cars that arrive in any one given moment does affect the number of cars that will arrive subsequently. Knowing how many cars arriving in one minute will not help predict the number of cars arriving at the next minute. Assumption 2 means that the rate of cars arriving is dependent only on the length of the time interval and not on when the time interval occurs (e.g. not on whether it is at the beginning of the hour or toward the end of the hour). The assumptions 2 and 3 allow us to think of a very short period of time as a Bernoulli trial. Thinking of the arrival of a car as a success, each short time interval will result in only one success or one failure.
To start, we can break up the hour into 60 minutes (into 60 Bernoulli trials). We then consider the binomial distribution with and . So the following is an approximation to our desired probability distribution.
Conceivably, there can be more than 1 car arriving in a minute and observing cars in a one-minute interval may not be a Bernoulli trial. For a one-minute interval to qualify as a Bernoulli trial, there is either no car arriving or 1 car arriving in that one minute. So we can break up an hour into 3,600 seconds (into 3,600 Bernoulli trials). We now consider the binomial distribution with and .
It is also conceivable that more than 1 car can arrive in one second and observing cars in one-second interval may still not qualify as a Bernoulli trial. So we need to get more granular. We can divide up the hour into equal subintervals, each of length . The assumptions 2 and 3 ensure that each subinterval is a Bernoulli trial (either it is a success or a failure; one car arriving or no car arriving). Assumption 1 tells us that all the subintervals are independent. So breaking up the hour into moments and counting the number of moments that are successes will result in a binomial distribution with parameters and . So we are ready to proceed with the following approximation to our probability distribution .
As we get more granular, . We show that the limit of the binomial probability in is the Poisson distribution with parameter . We show the following.
In the derivation of , we need the following two mathematical tools. The statement is one of the definitions of the mathematical constant e. In the statement , the integer in the numerator is greater than the integer in the denominator. It says that whenever we work with such a ratio of two factorials, the result is the product of with the smaller integers down to . There are exactly terms in the product.
The following is the derivation of .
In , we have . The reason being that the numerator is a polynomial where the leading term is . Upon dividing by and taking the limit, we get 1. Based on , we have . For the last limit in the derivation we have .
We conclude with some comments. As the above derivation shows, the Poisson distribution is at heart a binomial distribution. When we divide the unit time interval into more and more subintervals (as the subintervals get more and more granular), the resulting binomial distribution behaves more and more like the Poisson distribution.
The three assumtions used in the derivation are called the Poisson postulates, which are the underlying assumptions that govern a Poisson process. Such a random process describes the occurrences of some type of events that are of interest (e.g. the arrivals of cars in our example) in a fixed period of time. The positive constant indicated in Assumption 2 is the parameter of the Poisson process, which can be interpreted as the rate of occurrences of the event of interest (or rate of changes, or rate of arrivals) in a unit time interval, meaning that the positive constant is the mean number of occurrences in the unit time interval. The derivation in shows that whenever a certain type of events occurs according to a Poisson process with parameter , the counting variable of the number of occurrences in the unit time interval is distributed according to the Poisson distribution as indicated in .
If we observe the occurrences of events over intervals of length other than unit length, say, in an interval of length , the counting process is governed by the same three postulates, with the modification to Assumption 2 that the rate of changes of the process is now . The mean number of occurrences in the time interval of length is now . The Assumption 2 now states that for any very short time interval of length (and that is also a subinterval of the interval of length under observation), the probability of having one occurrence of event in this short interval is . Applyng the same derivation, it can be shown that the number of occurrences () in a time interval of length has the Poisson distribution with the following probability mass function.
]]>A random experiment resulting in two distinct outcomes (success or failure) is called a Bernoulli trial (e.g. head or tail in a coin toss, whether or not the birthday of a customer is the first of January, whether an insurance claim is above or below a given threshold etc). Suppose a series of independent Bernoulli trials are performed until reaching the rth success where the probability of success in each trial is . Let be the number of failures before the occurrence of the rth success. The following is the probablity mass function of .
Be definition, the survival function and cdf of are:
For each positive integer , let be the number of successes in performing a sequence of independent Bernoulli trials where is the probability of success. In other words, has a binomial distribution with parameters and .
If the random experiment requires more than failures to reach the rth success, there are at most successes in the first trails. Thus the survival function of is the same as the cdf of a binomial distribution. Equivalently, the cdf of is the same as the survival function of a binomial distribution. We have the following:
Remark
The relation is analogous to the relationship between the Gamma distribution and the Poisson distribution. Recall that a Gamma distribution with shape parameter and scale parameter , where is a positive integer, can be interpreted as the waiting time until the nth change in a Poisson process. Thus, if the nth change takes place after time , there can be at most arrivals in the time interval . Thus the survival function of this Gamma distribution is the same as the cdf of a Poisson distribution. The relation is analogous to the following relation.
A previous post on the negative binomial distribution is found here.
]]>The negative binomial distribution arises naturally from a probability experiment of performing a series of independent Bernoulli trials until the occurrence of the rth success where r is a positive integer. From this starting point, we discuss three ways to define the distribution. We then discuss several basic properties of the negative binomial distribution. Emphasis is placed on the close connection between the Poisson distribution and the negative binomial distribution.
________________________________________________________________________
Definitions
We define three versions of the negative binomial distribution. The first two versions arise from the view point of performing a series of independent Bernoulli trials until the rth success where r is a positive integer. A Bernoulli trial is a probability experiment whose outcome is random such that there are two possible outcomes (success or failure).
Let be the number of Bernoulli trials required for the rth success to occur where r is a positive integer. Let is the probability of success in each trial. The following is the probability function of :
The idea for is that for to happen, there must be successes in the first trials and one additional success occurring in the last trial (the th trial).
A more common version of the negative binomial distribution is the number of Bernoulli trials in excess of r in order to produce the rth success. In other words, we consider the number of failures before the occurrence of the rth success. Let be this random variable. The following is the probability function of :
The idea for is that there are trials and in the first trials, there are failures (or equivalently successes).
In both and , the binomial coefficient is defined by
where is a positive integer and is a nonnegative integer. However, the right-hand-side of can be calculated even if is not a positive integer. Thus the binomial coefficient can be expanded to work for all real number . However must still be nonnegative integer.
For convenience, we let . When the real number , the binomial coefficient in can be expressed as:
where is the gamma function.
With the more relaxed notion of binomial coefficient, the probability function in above can be defined for all real number r. Thus the general version of the negative binomial distribution has two parameters r and , both real numbers, such that . The following is its probability function.
Whenever r in is a real number that is not a positive integer, the interpretation of counting the number of failures until the occurrence of the rth success is no longer important. Instead we can think of it simply as a count distribution.
The following alternative parametrization of the negative binomial distribution is also useful.
The parameters in this alternative parametrization are r and . Clearly, the ratio takes the place of in . Unless stated otherwise, we use the parametrization of .
________________________________________________________________________
What is negative about the negative binomial distribution?
What is negative about this distribution? What is binomial about this distribution? The name is suggested by the fact that the binomial coefficient in can be rearranged as follows:
The calculation in can be used to verify that is indeed a probability function, that is, all the probabilities sum to 1.
In , we take . The step above uses the following formula known as the Newton’s binomial formula.
________________________________________________________________________
The Generating Function
By definition, the following is the generating function of the negative binomial distribution, using :
where . Using a similar calculation as in , the generating function can be simplified as:
As a result, the moment generating function of the negative binomial distribution is:
________________________________________________________________________
Independent Sum
One useful property of the negative binomial distribution is that the independent sum of negative binomial random variables, all with the same parameter , also has a negative binomial distribution. Let be an independent sum such that each has a negative binomial distribution with parameters and . Then the sum has a negative binomial distribution with parameters and .
Note that the generating function of an independent sum is the product of the individual generating functions. The following shows that the product of the individual generating functions is of the same form as , thus proving the above assertion.
________________________________________________________________________
Mean and Variance
The mean and variance can be obtained from the generating function. From and , we have:
Note that . Thus when the sample data suggest that the variance is greater than the mean, the negative binomial distribution is an excellent alternative to the Poisson distribution. For example, suppose that the sample mean and the sample variance are 3.6 and 7.1. In exploring the possibility of fitting the data using the negative binomial distribution, we would be interested in the negative binomial distribution with this mean and variance. Then plugging these into produces the negative binomial distribution with and .
________________________________________________________________________
The Poisson-Gamma Mixture
One important application of the negative binomial distribution is that it is a mixture of a family of Poisson distributions with Gamma mixing weights. Thus the negative binomial distribution can be viewed as a generalization of the Poisson distribution. The negative binomial distribution can be viewed as a Poisson distribution where the Poisson parameter is itself a random variable, distributed according to a Gamma distribution. Thus the negative binomial distribution is known as a Poisson-Gamma mixture.
In an insurance application, the negative binomial distribution can be used as a model for claim frequency when the risks are not homogeneous. Let has a Poisson distribution with parameter , which can be interpreted as the number of claims in a fixed period of time from an insured in a large pool of insureds. There is uncertainty in the parameter , reflecting the risk characteristic of the insured. Some insureds are poor risks (with large ) and some are good risks (with small ). Thus the parameter should be regarded as a random variable . The following is the conditional distribution of (conditional on ):
Suppose that has a Gamma distribution with scale parameter and shape parameter . The following is the probability density function of .
Then the joint density of and is:
The unconditional distribution of is obtained by summing out in .
Note that the integral in the fourth step in is 1.0 since the integrand is the pdf of a Gamma distribution. The above probability function is that of a negative binomial distribution. It is of the same form as . Equivalently, it is also of the form with parameter and .
The variance of the negative binomial distribution is greater than the mean. In a Poisson distribution, the mean equals the variance. Thus the unconditional distribution of is more dispersed than its conditional distributions. This is a characteristic of mixture distributions. The uncertainty in the parameter variable has the effect of increasing the unconditional variance of the mixture distribution of . 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 (see The variance of a mixture).
________________________________________________________________________
The Poisson Distribution as Limit of Negative Binomial
There is another connection to the Poisson distribution, that is, the Poisson distribution is a limiting case of the negative binomial distribution. We show that the generating function of the Poisson distribution can be obtained by taking the limit of the negative binomial generating function as . Interestingly, the Poisson distribution is also the limit of the binomial distribution.
In this section, we use the negative binomial parametrization of . By replacing for , the following are the mean, variance, and the generating function for the probability function in :
Let r goes to infinity and goes to zero and at the same time keeping their product constant. Thus is constant (this is the mean of the negative binomial distribution). We show the following:
The right-hand side of is the generating function of the Poisson distribution with mean . The generating function in the left-hand side is that of a negative binomial distribution with mean . The following is the derivation of .
We now focus on the limit in the exponent.
The middle step in uses the L’Hopital’s Rule. The result in is obtained by combining and .
________________________________________________________________________
Reference
Example 1
Suppose that the arrivals of customers in a gift shop at an airport follow a Poisson distribution with a mean of per 10 minutes. Furthermore, suppose that each arrival can be classified into one of three distinct types – type 1 (no purchase), type 2 (purchase under $20), and type 3 (purchase over $20). Records show that about 25% of the customers are of type 1. The percentages of type 2 and type 3 are 60% and 15%, respectively. What is the probability distribution of the number of customers per hour of each type?
Example 2
Roll a fair die times where is random and follows a Poisson distribution with parameter . For each , let be the number of times the upside of the die is . What is the probability distribution of each ? What is the joint distribution of ?
In Example 1, the stream of customers arrive according to a Poisson distribution. It can be shown that the stream of each type of customers also has a Poisson distribution. One way to view this example is that we can split the Poisson distribution into three Poisson distributions.
Example 2 also describes a splitting process, i.e. splitting a Poisson variable into 6 different Poisson variables. We can also view Example 2 as a multinomial distribution where the number of trials is not fixed but is random and follows a Poisson distribution. If the number of rolls of the die is fixed in Example 2 (say 10), then each would be a binomial distribution. Yet, with the number of trials being Poisson, each has a Poisson distribution with mean . In this post, we describe this Poisson splitting process in terms of a “random” multinomial distribution (the view point of Example 2).
________________________________________________________________________
Suppose we have a multinomial experiment with parameters , , , where
Suppose that follows a Poisson distribution with parameter . For each , let be the number of occurrences of the type of outcomes in the trials. Then are mutually independent Poisson random variables with parameters , respectively.
The variables have a multinomial distribution and their joint probability function is:
where are nonnegative integers such that .
Since the total number of multinomial trials is not fixed and is random, is not the end of the story. The following is the joint probability function of :
To obtain the marginal probability function of , , we sum out the other variables () in and obtain the following:
Thus we can conclude that , , has a Poisson distribution with parameter . Furrthermore, the joint probability function of is the product of the marginal probability functions. Thus we can conclude that are mutually independent.
________________________________________________________________________
Example 1
Let be the number of customers per hour of type 1, type 2, and type 3, respectively. Here, we attempt to split a Poisson distribution with mean 30 per hour (based on 5 per 10 minutes). Thus are mutually independent Poisson variables with means , , , respectively.
Example 2
As indicated earlier, each , , has a Poisson distribution with mean . According to , the joint probability function of is simply the product of the six marginal Poisson probability functions.
The above distribution is said to be a Poisson distribution with parameter . The Poisson distribution is usually used to model the random number of events occurring in a fixed time interval. As will be shown below, . Thus the parameter is the rate of occurrence of the random events; it indicates on average how many events occur per unit of time. Examples of random events that may be modeled by the Poisson distribution include the number of alpha particles emitted by a radioactive substance counted in a prescribed area during a fixed period of time, the number of auto accidents in a fixed period of time or the number of losses arising from a group of insureds during a policy period.
Each of the above examples can be thought of as a process that generates a number of arrivals or changes in a fixed period of time. If such a counting process leads to a Poisson distribution, then the process is said to be a Poisson process.
We now discuss some basic properties of the Poisson distribution. Using the Taylor series expansion of , the following shows that is indeed a probability distribution.
The generating function of the Poisson distribution is (see The generating function). The mean and variance can be calculated using the generating function.
The Poisson distribution can also be interpreted as an approximation to the binomial distribution. It is well known that the Poisson distribution is the limiting case of binomial distributions (see [1] or this post).
One application of is that we can use Poisson probabilities to approximate Binomial probabilities. The approximation is reasonably good when the number of trials in a binomial distribution is large and the probability of success is small. The binomial mean is and the variance is . When is small, is close to 1 and the binomial variance is approximately . Whenever the mean of a discrete distribution is approximately equaled to the mean, the Poisson approximation is quite good. As a rule of thumb, we can use Poisson to approximate binomial if and .
As an example, we use the Poisson distribution to estimate the probability that at most 1 person out of 1000 will have a birthday on the New Year Day. Let and . So we use the Poisson distribution with . The following is an estimate using the Poisson distribution.
Another useful property is that the independent sum of Poisson distributions also has a Poisson distribution. Specifically, if each has a Poisson distribution with parameter , then the independent sum has a Poisson distribution with parameter . One way to see this is that the product of Poisson generating functions has the same general form as (see The generating function). One interpretation of this property is that when merging several arrival processes, each of which follow a Poisson distribution, the result is still a Poisson distribution.
For example, suppose that in an airline ticket counter, the arrival of first class customers follows a Poisson process with a mean arrival rate of 8 per 15 minutes and the arrival of customers flying coach follows a Poisson distribution with a mean rate of 12 per 15 minutes. Then the arrival of customers of either types has a Poisson distribution with a mean rate of 20 per 15 minutes or 80 per hour.
A Poisson distribution with a large mean can be thought of as an independent sum of Poisson distributions. For example, a Poisson distribution with a mean of 50 is the independent sum of 50 Poisson distributions each with mean 1. Because of the central limit theorem, when the mean is large, we can approximate the Poisson using the normal distribution.
In addition to merging several Poisson distributions into one combined Poisson distribution, we can also split a Poisson into several Poisson distributions. For example, suppose that a stream of customers arrives according to a Poisson distribution with parameter and each customer can be classified into one of two types (e.g. no purchase vs. purchase) with probabilities and , respectively. Then the number of “no purchase” customers and the number of “purchase” customers are independent Poisson random variables with parameters and , respectively. For more details on the splitting of Poisson, see Splitting a Poisson Distribution.
Reference
Note that the derivative of at each order is a multiple of a Poisson probability. Thus the Poisson distribution is coded by the function . Because of this reason, such a function is called a generating function (or probability generating function). This post discusses some basic facts about the generating function (gf) and its cousin, the moment generating function (mgf). One important characteristic is that these functions generate probabilities and moments. Another important characteristic is that there is a one-to-one correspondence between a probability distribution and its generating function and moment generating function, i.e. two random variables with different cumulative distribution functions cannot have the same gf or mgf. In some situations, this fact is useful in working with independent sum of random variables.
————————————————————————————————————
The Generating Function
Suppose that is a random variable that takes only nonegative integer values with the probability function given by
The idea of the generating function is that we use a power series to capture the entire probability distribution. The following defines the generating function that is associated with the above sequence , .
Since the elements of the sequence are probabilities, we can also call the generating function of the probability distribution defined by the sequence in . The generating function is defined wherever the power series converges. It is clear that at the minimum, the power series in converges for .
We discuss the following three properties of generating functions:
The Poisson generating function at the beginning of the post is an example demonstrating property 1 (see Example 0 below for the derivation of the generating function). In some cases, the probability distribution of an independent sum can be deduced from the product of the individual generating functions. Some examples are given below.
————————————————————————————————————
Generating Probabilities
We now discuss the property 1 indicated above. To see that generates the probabilities, let’s look at the derivatives of :
By letting above, all the terms vanishes except for the constant term. We have:
Thus the generating function is a compact way of encoding the probability distribution. The probability distribution determines the generating function as seen in . On the other hand, and demonstrate that the generating function also determines the probability distribution.
————————————————————————————————————
Generating Moments
The generating function also determines the moments (property 2 indicated above). For example, we have:
Note that . Thus the higher moment can be expressed in terms of and where .
————————————————————————————————————
More General Definitions
Note that the definition in can also be interpreted as the mathematical expectation of , i.e., . This provides a way to define the generating function for random variables that may take on values outside of the nonnegative integers. The following is a more general definition of the generating function of the random variable , which is defined for all where the expectation exists.
————————————————————————————————————
The Generating Function of Independent Sum
Let be independent random variables with generating functions , respectively. Then the generating function of is given by the product .
Let be the generating function of the independent sum . The following derives . Note that the general form of generating function is used.
The probability distribution of a random variable is uniquely determined by its generating function. In particular, the generating function of the independent sum that is derived in is unique. So if the generating function is of a particular distribution, we can deduce that the distribution of the sum must be of the same distribution. See the examples below.
————————————————————————————————————
Example 0
In this example, we derive the generating function of the Poisson distribution. Based on the definition, we have:
Example 1
Suppose that are independent random variables where each has a Bernoulli distribution with probability of success . Let . The following is the generating function for each .
Then the generating function of the sum is . The following is the binomial expansion:
By definition , the generating function of is:
Comparing and , we have
The probability distribution indicated by and is that of a binomial distribution. Since the probability distribution of a random variable is uniquely determined by its generating function, the independent sum of Bernoulli distributions must ave a Binomial distribution.
Example 2
Suppose that are independent and have Poisson distributions with parameters , respectively. Then the independent sum has a Poisson distribution with parameter .
Let be the generating function of . For each , the generating function of is . The key to the proof is that the product of the has the same general form as the individual .
The generating function in is that of a Poisson distribution with mean . Since the generating function uniquely determines the distribution, we can deduce that the sum has a Poisson distribution with parameter .
Example 3
In rolling a fair die, let be the number shown on the up face. The associated generating function is:
The generating function can be further reduced as:
Suppose that we roll the fair dice 4 times. Let be the sum of the 4 rolls. Then the generating function of is
The random variable ranges from 4 to 24. Thus the probability function ranges from to . To find these probabilities, we simply need to decode the generating function . For example, to find , we need to find the coefficient of the term in the polynomial . To help this decoding, we can expand two of the polynomials in .
Based on the above polynomials, there are three ways of forming . They are: , , . Thus we have:
To find the other probabilities, we can follow the same decoding process.
————————————————————————————————————
Remark
The probability distribution of a random variable is uniquely determined by its generating function. This fundamental property is useful in determining the distribution of an independent sum. The generating function of the independent sum is simply the product of the individual generating functions. If the product is of a certain distributional form (as in Example 1 and Example 2), then we can deduce that the sum must be of the same distribution.
We can also decode the product of generating functions to obtain the probability function of the independent sum (as in Example 3). The method in Example 3 is quite tedious. But one advantage is that it is a “machine process”, a pretty fool proof process that can be performed mechanically.
The machine process is this: Code the individual probability distribution in a generating function . Then raise it to . After performing some manipulation to , decode the probabilities from .
As long as we can perform the algebraic manipulation carefully and correctly, this process will be sure to provide the probability distribution of an independent sum.
————————————————————————————————————
The Moment Generating Function
The moment generating function of a random variable is on all real numbers for which the expected value exists. The moments can be computed more directly using an mgf. From the theory of mathematical analysis, it can be shown that if exists on some interval , then the derivatives of of all orders exist at . Furthermore, it can be show that .
Suppose that is the generating function of a random variable. The following relates the generating function and the moment generating function.
————————————————————————————————————
Reference