Use of the Moment Generating Function for the Binomial Distribution

A histogram of a binomial distribution. C.K.Taylor

The mean and the variance of a random variable X with a binomial probability distribution can be difficult to calculate directly. Although it can be clear what needs to be done in using the definition of the expected value of X and X2, the actual execution of these steps is a tricky juggling of algebra and summations. An alternate way to determine the mean and variance of a binomial distribution is to use the moment generating function for X.

Binomial Random Variable

We start with our random variable X and describe the probability distribution more specifically. We perform n independent Bernoulli trials, each of which has probability of success p and probability of failure 1 - p. Thus the probability mass function is

f (x) = C(n , x)px(1 – p)n - x

Here the term C(n , x) denotes the number of combinations of n elements taken x at a time, and x can take the values 0, 1, 2, 3, . . ., n.

Moment Generating Function

We use this probability mass function to obtain the moment generating function of X:

M(t) = Σx = 0n etxC(n,x)>)px(1 – p)n - x.

We see that we can combine the terms with exponent of x:

M(t) = Σx = 0n (pet)xC(n,x)>)(1 – p)n - x.

Furthermore, by use of the binomial formula, the above expression is simply:

M(t) = [(1 – p) + pet]n.

Calculation of the Mean

In order to find the mean and variance, we need to know both M’(0) and M’’(0). We begin by calculating our derivatives, and then we evaluate each of them at t = 0.

We see that the first derivative of the moment generating function is:

M’(t) = n(pet)[(1 – p) + pet]n - 1.

From this we can calculate the mean of our probability distribution. M(0) = n(pe0)[(1 – p) + pe0]n - 1 = np. This matches the expression that we obtained directly from the definition of the mean.

Calculation of the Variance

The calculation of the variance is performed in a similar manner. We first differentiate the moment generating function again, and then we evaluate this derivative at t = 0. Here we see that

M’’(t) = n(n - 1)(pet)2[(1 – p) + pet]n - 2 + n(pet)[(1 – p) + pet]n - 1.

To calculate the variance of this random variable we need to find M’’(t). Here we have M’’(0) = n(n - 1)p2 +np. The variance σ2 of our distribution is

σ2 = M’’(0) – [M’(0)]2 = n(n - 1)p2 +np - (np)2 = np(1 - p).

Although this method is somewhat involved, it is not as complicated as calculating the mean and variance directly from the probability mass function.