The Negative Binomial Distribution

A counting distribution is a discrete distribution with probabilities only on the nonnegative integers. Such distributions are important in insurance applications since they can be used to model the number of events such as losses to the insured or claims to the insurer. Though playing a prominent role in statistical theory, the Poisson distribution is not appropriate in all situations, since it requires that the mean and the variance are equaled. Thus the negative binomial distribution is an excellent alternative to the Poisson distribution, especially in the cases where the observed variance is greater than the observed mean.

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.

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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 X_1 be the number of Bernoulli trials required for the rth success to occur where r is a positive integer. Let p is the probability of success in each trial. The following is the probability function of X_1:

\displaystyle (1) \ \ \ \ \ P(X_1=x)= \binom{x-1}{r-1} p^r (1-p)^{x-r} \ \ \ \ \ \ \ x=r,r+1,r+2,\cdots

The idea for (1) is that for X_1=x to happen, there must be r-1 successes in the first x-1 trials and one additional success occurring in the last trial (the xth 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 X_2 be this random variable. The following is the probability function of X_2:

\displaystyle (2) \ \ \ \ \ P(X_2=x)=\binom{x+r-1}{x} p^r (1-p)^x \ \ \ \ \ \ \ x=0,1,2,\cdots

The idea for (2) is that there are x+r trials and in the first x+r-1 trials, there are x failures (or equivalently r-1 successes).

In both (1) and (2), the binomial coefficient is defined by

\displaystyle (3) \ \ \ \ \ \binom{y}{k}=\frac{y!}{k! \ (y-k)!}=\frac{y(y-1) \cdots (y-(k-1))}{k!}

where y is a positive integer and k is a nonnegative integer. However, the right-hand-side of (3) can be calculated even if y is not a positive integer. Thus the binomial coefficient \displaystyle \binom{y}{k} can be expanded to work for all real number y. However k must still be nonnegative integer.

\displaystyle (4) \ \ \ \ \ \binom{y}{k}=\frac{y(y-1) \cdots (y-(k-1))}{k!}

For convenience, we let \displaystyle \binom{y}{0}=1. When the real number y>k-1, the binomial coefficient in (4) can be expressed as:

\displaystyle (5) \ \ \ \ \ \binom{y}{k}=\frac{\Gamma(y+1)}{\Gamma(k+1) \Gamma(y-k+1)}

where \Gamma(\cdot) is the gamma function.

With the more relaxed notion of binomial coefficient, the probability function in (2) above can be defined for all real number r. Thus the general version of the negative binomial distribution has two parameters r and p, both real numbers, such that 0<p<1. The following is its probability function.

\displaystyle (6) \ \ \ \ \ P(X=x)=\binom{x+r-1}{x} p^r (1-p)^x \ \ \ \ \ \ \ x=0,1,2,\cdots

Whenever r in (6) 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.

\displaystyle (6a) \ \ \ \ \ P(X=x)=\binom{x+r-1}{x} \biggl(\frac{\alpha}{\alpha+1}\biggr)^r \biggl(\frac{1}{\alpha+1}\biggr)^x \ \ \ \ \ \ \ x=0,1,2,\cdots

The parameters in this alternative parametrization are r and \alpha>0. Clearly, the ratio \frac{\alpha}{\alpha+1} takes the place of p in (6). Unless stated otherwise, we use the parametrization of (6).
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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 (6) can be rearranged as follows:

\displaystyle \begin{aligned}(7) \ \ \ \ \ \binom{x+r-1}{x}&=\frac{(x+r-1)(x+r-2) \cdots r}{x!} \\&=(-1)^x \frac{(-r-(x-1))(-r-(x-2)) \cdots (-r)}{x!} \\&=(-1)^x \frac{(-r)(-r-1) \cdots (-r-(x-1))}{x!} \\&=(-1)^x \binom{-r}{x} \end{aligned}

The calculation in (7) can be used to verify that (6) is indeed a probability function, that is, all the probabilities sum to 1.

\displaystyle \begin{aligned}(8) \ \ \ \ \ 1&=p^r p^{-r}\\&=p^r (1-q)^{-r} \\&=p^r \sum \limits_{x=0}^\infty \binom{-r}{x} (-q)^x \ \ \ \ \ \ \ \ (8.1) \\&=p^r \sum \limits_{x=0}^\infty (-1)^x \binom{-r}{x} q^x \\&=\sum \limits_{x=0}^\infty \binom{x+r-1}{x} p^r q^x \end{aligned}

In (8), we take q=1-p. The step (8.1) above uses the following formula known as the Newton’s binomial formula.

\displaystyle (9) \ \ \ \ \ (1+t)^w=\sum \limits_{k=0}^\infty \binom{w}{k} t^k

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The Generating Function
By definition, the following is the generating function of the negative binomial distribution, using :

\displaystyle (10) \ \ \ \ \ g(z)=\sum \limits_{x=0}^\infty \binom{r+x-1}{x} p^r q^x z^x

where q=1-p. Using a similar calculation as in (8), the generating function can be simplified as:

\displaystyle (11) \ \ \ \ \ g(z)=p^r (1-q z)^{-r}=\frac{p^r}{(1-q z)^r}=\frac{p^r}{(1-(1-p) z)^r}; \ \ \ \ \ z<\frac{1}{1-p}

As a result, the moment generating function of the negative binomial distribution is:

\displaystyle (12) \ \ \ \ \ M(t)=\frac{p^r}{(1-(1-p) e^t)^r}; \ \ \ \ \ \ \ t<-ln(1-p)

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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 p, also has a negative binomial distribution. Let Y=Y_1+Y_2+\cdots+Y_n be an independent sum such that each X_i has a negative binomial distribution with parameters r_i and p. Then the sum Y=Y_1+Y_2+\cdots+Y_n has a negative binomial distribution with parameters r=r_1+\cdots+r_n and p.

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 (11), thus proving the above assertion.

\displaystyle (13) \ \ \ \ \ h(z)=\frac{p^{\sum \limits_{i=1}^n r_i}}{(1-(1-p) z)^{\sum \limits_{i=1}^n r_i}}
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Mean and Variance
The mean and variance can be obtained from the generating function. From E(X)=g'(1) and E(X^2)=g'(1)+g^{(2)}(1), we have:

\displaystyle (14) \ \ \ \ \ E(X)=\frac{r(1-p)}{p} \ \ \ \ \ \ \ \ \ \ \ \ \ Var(X)=\frac{r(1-p)}{p^2}

Note that Var(X)=\frac{1}{p} E(X)>E(X). 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 (14) produces the negative binomial distribution with r=3.7 and p=0.507.
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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 N has a Poisson distribution with parameter \theta, 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 \theta, reflecting the risk characteristic of the insured. Some insureds are poor risks (with large \theta) and some are good risks (with small \theta). Thus the parameter \theta should be regarded as a random variable \Theta. The following is the conditional distribution of N (conditional on \Theta=\theta):

\displaystyle (15) \ \ \ \ \ P(N=n \lvert \Theta=\theta)=\frac{e^{-\theta} \ \theta^n}{n!} \ \ \ \ \ \ \ \ \ \ n=0,1,2,\cdots

Suppose that \Theta has a Gamma distribution with scale parameter \alpha and shape parameter \beta. The following is the probability density function of \Theta.

\displaystyle (16) \ \ \ \ \ g(\theta)=\frac{\alpha^\beta}{\Gamma(\beta)} \theta^{\beta-1} e^{-\alpha \theta} \ \ \ \ \ \ \ \ \ \ \theta>0

Then the joint density of N and \Theta is:

\displaystyle (17) \ \ \ \ \ P(N=n \lvert \Theta=\theta) \ g(\theta)=\frac{e^{-\theta} \ \theta^n}{n!} \ \frac{\alpha^\beta}{\Gamma(\beta)} \theta^{\beta-1} e^{-\alpha \theta}

The unconditional distribution of N is obtained by summing out \theta in (17).

\displaystyle \begin{aligned}(18) \ \ \ \ \ P(N=n)&=\int_0^\infty P(N=n \lvert \Theta=\theta) \ g(\theta) \ d \theta \\&=\int_0^\infty \frac{e^{-\theta} \ \theta^n}{n!} \ \frac{\alpha^\beta}{\Gamma(\beta)} \ \theta^{\beta-1} \ e^{-\alpha \theta} \ d \theta \\&=\int_0^\infty \frac{\alpha^\beta}{n! \ \Gamma(\beta)} \ \theta^{n+\beta-1} \ e^{-(\alpha+1) \theta} d \theta \\&=\frac{\alpha^\beta}{n! \ \Gamma(\beta)} \ \frac{\Gamma(n+\beta)}{(\alpha+1)^{n+\beta}} \int_0^\infty \frac{(\alpha+1)^{n+\beta}}{\Gamma(n+\beta)} \theta^{n+\beta-1} \ e^{-(\alpha+1) \theta} d \theta \\&=\frac{\alpha^\beta}{n! \ \Gamma(\beta)} \ \frac{\Gamma(n+\beta)}{(\alpha+1)^{n+\beta}} \\&=\frac{\Gamma(n+\beta)}{\Gamma(n+1) \ \Gamma(\beta)} \ \biggl( \frac{\alpha}{\alpha+1}\biggr)^\beta \ \biggl(\frac{1}{\alpha+1}\biggr)^n \\&=\binom{n+\beta-1}{n} \ \biggl( \frac{\alpha}{\alpha+1}\biggr)^\beta \ \biggl(\frac{1}{\alpha+1}\biggr)^n \ \ \ \ \ \ \ \ \ n=0,1,2,\cdots \end{aligned}

Note that the integral in the fourth step in (18) 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 (6a). Equivalently, it is also of the form (6) with parameter r=\beta and p=\frac{\alpha}{\alpha+1}.

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 N is more dispersed than its conditional distributions. This is a characteristic of mixture distributions. The uncertainty in the parameter variable \Theta has the effect of increasing the unconditional variance of the mixture distribution of N. 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 \Theta (see The variance of a mixture).

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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 r \rightarrow \infty. Interestingly, the Poisson distribution is also the limit of the binomial distribution.

In this section, we use the negative binomial parametrization of (6a). By replacing \frac{\alpha}{\alpha+1} for p, the following are the mean, variance, and the generating function for the probability function in (6a):

\displaystyle \begin{aligned}(19) \ \ \ \ \ \ &E(X)=\frac{r}{\alpha} \\&\text{ }\\&Var(X)=\frac{\alpha+1}{\alpha} \ \frac{r}{\alpha}=\frac{r(\alpha+1)}{\alpha^2} \\&\text{ } \\&g(z)=\frac{1}{[1-\frac{1}{\alpha}(z-1)]^r} \ \ \ \ \ \ \ z<\alpha+1 \end{aligned}

Let r goes to infinity and \displaystyle \frac{1}{\alpha} goes to zero and at the same time keeping their product constant. Thus \displaystyle \mu=\frac{r}{\alpha} is constant (this is the mean of the negative binomial distribution). We show the following:

\displaystyle (20) \ \ \ \ \ \lim \limits_{r \rightarrow \infty} [1-\frac{\mu}{r}(z-1)]^{-r}=e^{\mu (z-1)}

The right-hand side of (20) is the generating function of the Poisson distribution with mean \mu. The generating function in the left-hand side is that of a negative binomial distribution with mean \displaystyle \mu=\frac{r}{\alpha}. The following is the derivation of (20).

\displaystyle \begin{aligned}(21) \ \ \ \ \ \lim \limits_{r \rightarrow \infty} [1-\frac{\mu}{r}(z-1)]^{-r}&=\lim \limits_{r \rightarrow \infty} e^{\displaystyle \biggl(ln[1-\frac{\mu}{r}(z-1)]^{-r}\biggr)} \\&=\lim \limits_{r \rightarrow \infty} e^{\displaystyle \biggl(-r \ ln[1-\frac{\mu}{r}(z-1)]\biggr)} \\&=e^{\displaystyle \biggl(\lim \limits_{r \rightarrow \infty} -r \ ln[1-\frac{\mu}{r}(z-1)]\biggr)} \end{aligned}

We now focus on the limit in the exponent.

\displaystyle \begin{aligned}(22) \ \ \ \ \ \lim \limits_{r \rightarrow \infty} -r \ ln[1-\frac{\mu}{r}(z-1)]&=\lim \limits_{r \rightarrow \infty} \frac{ln(1-\frac{\mu}{r} (z-1))^{-1}}{r^{-1}} \\&=\lim \limits_{r \rightarrow \infty} \frac{(1-\frac{\mu}{r} (z-1)) \ \mu (z-1) r^{-2}}{r^{-2}} \\&=\mu (z-1) \end{aligned}

The middle step in (22) uses the L’Hopital’s Rule. The result in (20) is obtained by combining (21) and (22).

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Reference

  1. Klugman S.A., Panjer H. H., Wilmot G. E. Loss Models, From Data to Decisions, Second Edition., Wiley-Interscience, a John Wiley & Sons, Inc., New York, 2004

The generating function

Consider the function g(z)=\displaystyle e^{\alpha (z-1)} where \alpha is a positive constant. The following shows the derivatives of this function.

\displaystyle \begin{aligned}. \ \ \ \ \ \ &g(z)=e^{\alpha (z-1)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g(0)=e^{-\alpha} \\&\text{ } \\&g'(z)=e^{\alpha (z-1)} \ \alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g'(0)=e^{-\alpha} \ \alpha \\&\text{ } \\&g^{(2)}(z)=e^{\alpha (z-1)} \ \alpha^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(2)}(0)=2! \ \frac{e^{-\alpha} \ \alpha^2}{2!} \\&\text{ } \\&g^{(3)}(z)=e^{\alpha (z-1)} \ \alpha^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(3)}(0)=3! \ \frac{e^{-\alpha} \ \alpha^3}{3!} \\&\text{ } \\&\ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \\&\text{ } \\&g^{(n)}(z)=e^{\alpha (z-1)} \ \alpha^n \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(n)}(0)=n! \ \frac{e^{-\alpha} \ \alpha^n}{n!} \end{aligned}

Note that the derivative of g(z) at each order is a multiple of a Poisson probability. Thus the Poisson distribution is coded by the function g(z)=\displaystyle e^{\alpha (z-1)}. 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.

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The Generating Function
Suppose that X is a random variable that takes only nonegative integer values with the probability function given by

(1) \ \ \ \ \ \ P(X=j)=a_j, \ \ \ \ j=0,1,2,\cdots

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 a_j, .

(2) \ \ \ \ \ \ g(z)=a_0+a_1 \ z+a_2 \ z^2+ \cdots=\sum \limits_{j=0}^\infty a_j \ z^j

Since the elements of the sequence a_j are probabilities, we can also call g(z) the generating function of the probability distribution defined by the sequence in (1). The generating function g(z) is defined wherever the power series converges. It is clear that at the minimum, the power series in (2) converges for \lvert z \lvert \le 1.

We discuss the following three properties of generating functions:

  1. The generating function completely determines the distribution.
  2. The moments of the distribution can be derived from the derivatives of the generating function.
  3. The generating function of a sum of independent random variables is the product of the individual 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.

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Generating Probabilities
We now discuss the property 1 indicated above. To see that g(z) generates the probabilities, let’s look at the derivatives of g(z):

\displaystyle \begin{aligned}(3) \ \ \ \ \ \ &g'(z)=a_1+2 a_2 \ z+3 a_3 \ z^2+\cdots=\sum \limits_{j=1}^\infty j a_j \ z^{j-1} \\&\text{ } \\&g^{(2)}(z)=2 a_2+6 a_3 \ z+ 12 a_4 \ z^2=\sum \limits_{j=2}^\infty j (j-1) a_j \ z^{j-2} \\&\text{ } \\&g^{(3)}(z)=6 a_3+ 24 a_4 \ z+60 a_5 \ z^2=\sum \limits_{j=3}^\infty j (j-1)(j-2) a_j \ z^{j-3} \\&\text{ } \\&\ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \\&\text{ } \\&g^{(n)}(z)=\sum \limits_{j=n}^\infty j(j-1) \cdots (j-n+1) a_j \ z^{j-n}=\sum \limits_{j=n}^\infty \binom{j}{n} n! \ a_j \ z^{j-n} \end{aligned}

By letting z=0 above, all the terms vanishes except for the constant term. We have:

(4) \ \ \ \ \ \ g^{(n)}(0)=n! \ a_n=n! \ P(X=n)

Thus the generating function is a compact way of encoding the probability distribution. The probability distribution determines the generating function as seen in (2). On the other hand, (3) and (4) demonstrate that the generating function also determines the probability distribution.

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Generating Moments
The generating function also determines the moments (property 2 indicated above). For example, we have:

\displaystyle \begin{aligned}(5) \ \ \ \ \ \ &g'(1)=0 \ a_0+a_1+2 a_2+3 a_3+\cdots=\sum \limits_{j=0}^\infty j a_j=E(X) \\&\text{ } \\&g^{(2)}(1)=0 a_0 + 0 a_1+2 a_2+6 a_3+ 12 a_4+\cdots=\sum \limits_{j=0}^\infty j (j-1) a_j=E[X(X-1)] \\&\text{ } \\&E(X)=g'(1) \\&\text{ } \\&E(X^2)=g'(1)+g^{(2)}(1) \end{aligned}

Note that g^{(n)}(1)=E[X(X-1) \cdots (X-(n-1))]. Thus the higher moment E(X^n) can be expressed in terms of g^{(n)}(1) and g^{(k)}(1) where k<n.
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More General Definitions
Note that the definition in (2) can also be interpreted as the mathematical expectation of z^X, i.e., g(z)=E(z^X). 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 X, which is defined for all z where the expectation exists.

(6) \ \ \ \ \ \ g(z)=E(z^X)

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The Generating Function of Independent Sum
Let X_1,X_2,\cdots,X_n be independent random variables with generating functions g_1,g_2,\cdots,g_n, respectively. Then the generating function of X_1+X_2+\cdots+X_n is given by the product g_1 \cdot g_2 \cdots g_n.

Let g(z) be the generating function of the independent sum X_1+X_2+\cdots+X_n. The following derives g(z). Note that the general form of generating function (6) is used.

\displaystyle \begin{aligned}(7) \ \ \ \ \ \ g(z)&=E(z^{X_1+\cdots+X_n}) \\&\text{ } \\&=E(z^{X_1} \cdots z^{X_n}) \\&\text{ } \\&=E(z^{X_1}) \cdots E(z^{X_n}) \\&\text{ } \\&=g_1(z) \cdots g_n(z) \end{aligned}

The probability distribution of a random variable is uniquely determined by its generating function. In particular, the generating function g(z) of the independent sum X_1+X_2+\cdots+X_n that is derived in (7) 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.

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Example 0
In this example, we derive the generating function of the Poisson distribution. Based on the definition, we have:

\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)&=\sum \limits_{j=0}^\infty \frac{e^{-\alpha} \alpha^j}{j!} \ z^j \\&\text{ } \\&=\sum \limits_{j=0}^\infty \frac{e^{-\alpha} (\alpha z)^j}{j!}  \\&\text{ } \\&=\frac{e^{-\alpha}}{e^{- \alpha z}} \sum \limits_{j=0}^\infty \frac{e^{-\alpha z} (\alpha z)^j}{j!} \\&\text{ } \\&=e^{\alpha (z-1)} \end{aligned}

Example 1
Suppose that X_1,X_2,\cdots,X_n are independent random variables where each X_i has a Bernoulli distribution with probability of success p. Let q=1-p. The following is the generating function for each X_i.

. \ \ \ \ \ \ g(z)=q+p z

Then the generating function of the sum X=X_1+\cdots+X_n is g(z)^n=(q+p z)^n. The following is the binomial expansion:

\displaystyle \begin{aligned}(8) \ \ \ \ \ \ g(z)^n&=(q+p z)^n \\&\text{ } \\&=\sum \limits_{j=0}^n \binom{n}{j} q^{n-j} \ p^j \ z^j  \end{aligned}

By definition (2), the generating function of X=X_1+\cdots+X_n is:

(9) \ \ \ \ \ \ g(z)^n=\sum \limits_{j=0}^\infty P(X=j) \ z^j

Comparing (8) and (9), we have

\displaystyle (10) \ \ \ \ \ \ P(X=j)=\left\{\begin{matrix}\displaystyle \binom{n}{j} p^j \ q^{n-j}&\ 0 \le j \le n\\{0}&\ j>n \end{matrix}\right.

The probability distribution indicated by (8) and (10) 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 X_1,X_2,\cdots,X_n are independent and have Poisson distributions with parameters \alpha_1,\alpha_2,\cdots,\alpha_n, respectively. Then the independent sum X=X_1+\cdots+X_n has a Poisson distribution with parameter \alpha=\alpha_1+\cdots+\alpha_n.

Let g(z) be the generating function of X=X_1+\cdots+X_n. For each i, the generating function of X_i is g_i(z)=e^{\alpha_i (z-1)}. The key to the proof is that the product of the g_i has the same general form as the individual g_i.

\displaystyle \begin{aligned}(11) \ \ \ \ \ \ g(z)&=g_1(z) \cdots g_n(z) \\&\text{ } \\&=e^{\alpha_1 (z-1)} \cdots e^{\alpha_n (z-1)} \\&\text{ } \\&=e^{(\alpha_1+\cdots+\alpha_n)(z-1)} \end{aligned}

The generating function in (11) is that of a Poisson distribution with mean \alpha=\alpha_1+\cdots+\alpha_n. Since the generating function uniquely determines the distribution, we can deduce that the sum X=X_1+\cdots+X_n has a Poisson distribution with parameter \alpha=\alpha_1+\cdots+\alpha_n.

Example 3
In rolling a fair die, let X be the number shown on the up face. The associated generating function is:

\displaystyle. \ \ \ \ \ \ g(z)=\frac{1}{6}(z+z^2+z^3+z^4+z^5+z^6)=\frac{z(1-z^6)}{6(1-z)}

The generating function can be further reduced as:

\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)&=\frac{z(1-z^6)}{6(1-z)} \\&\text{ } \\&=\frac{z(1-z^3)(1+z^3)}{6(1-z)} \\&\text{ } \\&=\frac{z(1-z)(1+z+z^2)(1+z^3)}{6(1-z)} \\&\text{ } \\&=\frac{z(1+z+z^2)(1+z^3)}{6}  \end{aligned}

Suppose that we roll the fair dice 4 times. Let W be the sum of the 4 rolls. Then the generating function of Z is

\displaystyle. \ \ \ \ \ \  g(z)^4=\frac{z^4 (1+z^3)^4 (1+z+z^2)^4}{6^4}

The random variable W ranges from 4 to 24. Thus the probability function ranges from P(W=4) to P(W=24). To find these probabilities, we simply need to decode the generating function g(z)^4. For example, to find P(W=12), we need to find the coefficient of the term z^{12} in the polynomial g(z)^4. To help this decoding, we can expand two of the polynomials in g(z)^4.

\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)^4&=\frac{z^4 (1+z^3)^4 (1+z+z^2)^4}{6^4} \\&\text{ } \\&=\frac{z^4 \times A \times B}{6^4} \\&\text{ } \\&A=(1+z^3)^4=1+4z^3+6z^6+4z^9+z^{12} \\&\text{ } \\&B=(1+z+z^2)^4=1+4z+10z^2+16z^3+19z^4+16z^5+10z^6+4z^7+z^8  \end{aligned}

Based on the above polynomials, there are three ways of forming z^{12}. They are: (z^4 \times 1 \times z^8), (z^4 \times 4z^3 \times 16z^5), (z^4 \times 6z^6 \times 10z^2). Thus we have:

\displaystyle. \ \ \ \ \ \  P(W=12)=\frac{1}{6^4}(1+4 \times 16+6 \times 10)=\frac{125}{6^4}

To find the other probabilities, we can follow the same decoding process.

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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 g(z). Then raise it to n. After performing some manipulation to g(z)^n, decode the probabilities from g(z)^n.

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.

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The Moment Generating Function
The moment generating function of a random variable X is M_X(t)=E(e^{tX}) on all real numbers t 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 M_X(t) exists on some interval -a<t<a, then the derivatives of M_X(t) of all orders exist at t=0. Furthermore, it can be show that E(X^n)=M_X^{(n)}(0).

Suppose that g(z) is the generating function of a random variable. The following relates the generating function and the moment generating function.

\displaystyle \begin{aligned}. \ \ \ \ \ \ &M_X(t)=g(e^t) \\&\text{ } \\&g(z)=M_X(ln z)  \end{aligned}

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Reference

  1. Feller W. An Introduction to Probability Theory and Its Applications, Third Edition, John Wiley & Sons, New York, 1968