The Poisson distribution is related to the exponential distribution. Suppose an event can occur several times within a given unit of time. When the total number of occurrences of the event is unknown, we can think of it as a random variable. This random variable has a Poisson distribution if and only if the time elapsed between two successive occurrences of the event has an exponential distribution and it is independent of previous occurrences.

A classical example of a random variable having a Poisson distribution is the number of phone calls received by a call center. If the time elapsed between two successive phone calls has an exponential distribution and it is independent of the time of arrival of the previous calls, then the total number of calls received in one hour has a Poisson distribution.

The concept is illustrated by the plot above, where the number of phone calls received is plotted as a function of time. The graph of the function makes an upward jump each time a phone call arrives. The time elapsed between two successive phone calls is equal to the length of each horizontal segment and it has an exponential distribution. The number of calls received in 60 minutes is equal to the length of the segment highlighted by the vertical curly brace and it has a Poisson distribution.

The following sections provide a more formal treatment of the main characteristics of the Poisson distribution.

A Poisson random variable is characterized as follows:

Definition
Let
be a discrete random
variable. Let its
support be the set of
positive integer
numbers:Let
.
We say that
has a **Poisson distribution** with parameter
if its probability mass
function
iswhere
is the factorial of
.

The relation between the Poisson distribution and the exponential distribution is summarized by the following proposition:

Proposition The number of occurrences of an event within a unit of time has a Poisson distribution with parameter if and only if the time elapsed between two successive occurrences of the event has an exponential distribution with parameter and it is independent of previous occurrences.

Proof

Denote by:and by the number of occurrences of the event. Since if and only if (convince yourself of this fact), the proposition is true if and only if:for any . To verify that the equality holds, we need to separately compute the two probabilities. We start with:Denote by the sum of waiting times:Since the sum of independent exponential random variables with common parameter is a Gamma random variable with parameters and , then is a Gamma random variable with parameters and , i.e. its probability density function is:where and the last equality stems from the fact that we are considering only integer values of . We need to integrate the density function to compute the probability that is less than :The last integral can be computed integrating by parts times:Multiplying by , we obtain:Thus, we have obtained:Now, we need to compute the probability that is greater than or equal to :which is exactly what we needed to prove.

The expected value of a Poisson random variable is:

Proof

It can be derived as follows:

The variance of a Poisson random variable is:

Proof

It can be derived thanks to the usual variance formula ():

The moment generating function of a Poisson random variable is defined for any :

Proof

Using the definition of moment generating function:where:is the usual Taylor series expansion of the exponential function. Furthermore, since the series converges for any value of , the moment generating function of a Poisson random variable exists for any .

The characteristic function of a Poisson random variable is:

Proof

Using the definition of characteristic function:where:is the usual Taylor series expansion of the exponential function (note that the series converges for any value of ).

The distribution function of a Poisson random variable is:where is the floor of , i.e. the largest integer not greater than .

Proof

Using the definition of distribution function:

Values of are usually computed by computer algorithms. For example, the MATLAB command:

`poisscdf(x,lambda)`

returns the value of the distribution function at the point
`x`

when the parameter of the distribution is equal to
`lambda`

.

Below you can find some exercises with explained solutions:

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