1.5 Poisson distribution

Poisson distribution

Some experiments results in counting the number of times particular events occur in given times of on given physical objects. For example, we would count the number of phone calls arriving at a switch board between 9 and 10 am, the number of flaws in 100 feet of wire, the number of customers that arrive at a ticket window between 12 noon and 2 pm, or the number of defects in a 100-foot roll of aluminum screen that is 2 feet wide. Each count can be looked upon as a random variable associated with an approximate Poisson process provided the conditions in the definition below are satisfied.

POISSON PROCCESS

Let the number of changes that occur in a given continuous interval be counted. We have an approximate Poisson process with parameter $\lambda >0$ if the following are satisfied:

• The number of changes occurring in nonoverlapping intervals are independent.
• The probability of exactly one change in a sufficiently short interval of length h is approximately $\lambda h$ .
• The probability of two or more changes in a sufficiently short interval is essentially zero.

Suppose that an experiment satisfies the three points of an approximate Poisson process. Let X denote the number of changes in an interval of "length 1" (where "length 1" represents one unit of the quantity under consideration). We would like to find an approximation for $P\left(X=x\right)$ , where x is a nonnegative integer. To achieve this, we partition the unit interval into n subintervals of equal length 1/ n . If N is sufficiently large (i.e., much larger than x ), one shall approximate the probability that x changes occur in this unit interval by finding the probability that one change occurs exactly in each of exactly x of these n subintervals. The probability of one change occurring in any one subinterval of length 1/ n is approximately $\lambda \left(1/n\right)$ by condition (2). The probability of two or more changes in any one subinterval is essentially zero by condition (3). So for each subinterval, exactly one change occurs with a probability of approximately $\lambda \left(1/n\right)$ . Consider the occurrence or nonoccurrence of a change in each subinterval as a Bernoulli trial. By condition (1) we have a sequence of n Bernoulli trials with probability p approximately equal to $\lambda \left(1/n\right)$ . Thus an approximation for $P\left(X=x\right)$ is given by the binomial probability $$\frac{n!}{x!\left(n-x\right)!}{\left(\frac{\lambda }{n}\right)}^x{\left(1-\frac{\lambda }{n}\right)}^{n-x}.$$

In order to obtain a better approximation, choose a large value for n . If n increases without bound, we have that

$$\underset{n\to \infty }{\lim }\frac{n!}{x!\left(n-x\right)!}{\left(\frac{\lambda }{n}\right)}^x{\left(1-\frac{\lambda }{n}\right)}^{n-x}=\underset{n\to \infty }{\lim }\frac{n\left(n-1\right)\mathrm{...}\left(n-x+1\right)}{n^x}\frac{{\lambda }^x}{x!}{\left(1-\frac{\lambda }{n}\right)}^n{\left(1-\frac{\lambda }{n}\right)}^{-x}.$$

Now, for fixed x , we have

$$\begin{array}{l}\underset{n\to \infty }{\lim }\frac{n\left(n-1\right)\mathrm{...}\left(n-x+1\right)}{n^x}=\underset{n\to \infty }{\lim }\left[1\left(1-\frac{1}{n}\right)\mathrm{...}\left(1-\frac{x-1}{n}\right)\right]=\mathrm{1,}\\ \underset{n\to \infty }{\lim }{\left(1-\frac{\lambda }{n}\right)}^n=e^{-\lambda },\end{array}$$

and $$\underset{n\to \infty }{\lim }{\left(1-\frac{\lambda }{n}\right)}^{-x}=1.$$

Thus, $$\underset{n\to \infty }{\lim }\frac{n!}{x!\left(n-x\right)!}{\left(\frac{\lambda }{n}\right)}^x{\left(1-\frac{\lambda }{n}\right)}^{n-x}=\frac{{\lambda }^x{e}^{-\lambda }}{x!}=P\left(X=x\right),$$ approximately. The distribution of probability associated with this process has a special name.

POISSON DISTRIBUTION

We say that the random variable X has a Poisson distribution if its p.d.f. is of the form $$f\left(x\right)=\frac{{\lambda }^x{e}^{-\lambda }}{x!},x=\mathrm{0,1,2,...,}$$ where $\lambda >0$ .

It is easy to see that $f\left(x\right)$ enjoys the properties pf a p.d.f. because clearly $f\left(x\right)\ge 0$ and, from the Maclaurin’s series expansion of $e^{\lambda }$ , we have

$${\displaystyle \sum _{x=0}^{\infty }\frac{{\lambda }^x{e}^{-\lambda }}{x!}}=e^{-\lambda }{\displaystyle \sum _{x=0}^{\infty }\frac{{\lambda }^x}{x!}=e^{-\lambda }{e}^{\lambda }=1.}$$

To discover the exact role of the parameter $\lambda >0$ , let us find some of the characteristics of the Poisson distribution . The mean for the Poisson distribution is given by

$$E\left(X\right)={\displaystyle \sum _{x=0}^{\infty }x\frac{{\lambda }^x{e}^{-\lambda }}{x!}}=e^{-\lambda }{\displaystyle \sum _{x=1}^{\infty }x\frac{{\lambda }^x}{\left(x-1\right)!}},$$ because $\left(0\right)f\left(0\right)=0$ and $x/x!=1/\left(x-1\right)!$ , when $x>0$ .

If we let $k=x-1$ , then $$E\left(X\right)=e^{-\lambda }{\displaystyle \sum _{k=0}^{\infty }\frac{{\lambda }^{k+1}}{k!}=}\lambda e^{-\lambda }{\displaystyle \sum _{k=0}^{\infty }\frac{{\lambda }^k}{k!}=\lambda e^{-\lambda }{e}^{\lambda }}=\lambda .$$

That is, the parameter $\lambda$ is the mean of the Poisson distribution . On the Figure 1 is shown the p.d.f. and c.d.f. of the Poisson Distribution for $\lambda =1,\lambda =4,\lambda =10.$

To find the variance, we first determine the second factorial moment $E\left[X\left(X-1\right)\right]$ . We have,

$$E\left[X\left(X-1\right)\right]={\displaystyle \sum _{x=0}^{\infty }x\left(x-1\right)\frac{{\lambda }^x{e}^{-\lambda }}{x!}=e^{-\lambda }{\displaystyle \sum _{x=2}^{\infty }\frac{{\lambda }^x}{\left(x-2\right)!}}},$$ because $\left(0\right)\left(0-1\right)f\left(0\right)=\mathrm{0,}\left(1\right)\left(1-1\right)f\left(1\right)=0$ , and $x\left(x-1\right)/x!=1/\left(x-2\right)!$ , when $x>1$ .

If we let $k=x-2$ , then $$E\left[X\left(X-1\right)\right]=e^{-\lambda }{\displaystyle \sum _{k=0}^{\infty }\frac{{\lambda }^{k+2}}{k!}={\lambda }^2{e}^{-\lambda }{\displaystyle \sum _{k=0}^{\infty }\frac{{\lambda }^x}{k!}={\lambda }^2}}{e}^{-\lambda }{e}^{\lambda }={\lambda }^2.$$

Thus, $$Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2=E\left[X\left(X-1\right)\right]+E\left(X\right)-{\left[E\left(X\right)\right]}^2={\lambda }^2+\lambda -{\lambda }^2=\lambda .$$

That is, for the Poisson distribution, $\mu ={\sigma }^2=\lambda$ .

If X has a Poisson distribution with parameter $\lambda$ , we saw that with n large,

$$P\left(X=x\right)\approx \left(\begin{array}{l}n\\ x\end{array}\right){\left(\frac{\lambda }{n}\right)}^x{\left(1-\frac{\lambda }{n}\right)}^{n-x},$$ where, $p=\lambda /n$ so that $\lambda =np$ in the above binomial probability. That is, if X has the binomial distribution $b\left(n,p\right)$ with large n , then

$$\frac{{\left(np\right)}^x{e}^{-np}}{x!}=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}.$$

This approximation is reasonably good if n is large. But since $\lambda$ was fixed constant in that earlier argument, p should be small since $np=\lambda$ . In particular, the approximation is quite accurate if $n\ge 20$ and $p\le 0.05$ , and it is very good if $n\ge 100$ and $np\le 10$ .