How to Calculate the Variance of a Poisson Distribution

mathematical function
The probability mass function for the Poisson distribution includes the number e and a parameter denoted by the Greek letter lambda. Its domain is the set of nonnegative integers. C.K.Taylor

The variance of a distribution of a random variable is an important feature. This number indicates the spread of a distribution, and it is found by squaring the standard deviation. One commonly used discrete distribution is that of the Poisson distribution. We will see how to calculate the variance of the Poisson distribution with parameter λ.

The Poisson Distribution

Poisson distributions are used when we have a continuum of some sort and are counting discrete changes within this continuum. This occurs when we consider the number of people who arrive at a movie ticket counter in the course of an hour, keep track of the number of cars traveling through an intersection with four way stop or count the number of flaws occurring in a length of wire.

If we make a few clarifying assumptions in these scenarios, then these situations match the conditions for a Poisson process. We then say that the random variable, which counts the number of changes, has a Poisson distribution.

The Poisson distribution actually refers to an infinite family of distributions. These distributions come equipped with a single parameter λ. The parameter is a positive real number that is closely related to the expected number of changes observed in the continuum. Furthermore, we will see that this parameter is equal to not only the mean of the distribution but also the variance of the distribution.

The probability mass function for a Poisson distribution is given by:

f(x) = (λx e)/x!

In this expression, the letter e is a number and is the mathematical constant with a value approximately equal to 2.718281828. The variable x can be any nonnegative integer.

Calculating the Variance

To calculate the mean of a Poisson distribution, we use this distribution's moment generating function. We see that:

M( t ) = E[etX] = Σ etXf( x) = ΣetX λx e)/x!

We now recall the Maclaurin series for eu. Since any derivative of the function eu is eu, all of these derivatives evaluated at zero give us 1. The result is the series eu = Σ un/n!.

By use of the Maclaurin series for eu, we can express the moment generating function not as a series, but in a closed form. We combine all terms with the exponent of x. Thus M(t) = eλ(et - 1).

We now find the variance by taking the second derivative of M and evaluating this at zero. Since M’(t) =λetM(t), we use the product rule to calculate the second derivative:

M’’(t)=λ2e2tM’(t) + λetM(t)

We evaluate this at zero and find that M’’(0) = λ2 + λ. We then use the fact that M’(0) = λ to calculate the variance.

Var(X) = λ2 + λ – (λ)2 = λ.

This shows that the parameter λ is not only the mean of the Poisson distribution but is also its variance.