Moment-generating Function of a Binomial Random Variable

Date: 10/27/1999 at 16:52:28
From: J. Lockett
Subject: Probability (Moment-generating Functions)

Find the moment-generating function (MGF) of a binomial random 
variable if:

     X = Bin(p)

that is, X is a binomial with parameter P. Find Mx(t).

I started with the theorems on MGF about discrete random variables, 
M(t) = E(e^tx)

     M sub 1|t = 0 = E(x)
     M sub 2|t = 0 = E(x^2)
     M sub 3 etc.

I just am not sure how to apply this to binomial random variables.

Date: 10/27/1999 at 17:06:13
From: Doctor Anthony
Subject: Re: Probability (Moment-generating Functions)

In fact the binomial distribution requires two parameters, n and p.

The mgf is given by  E[e^(tx)]

So for the binomial distribution

     Mx(t) = SUM(k=0 to n)[e^(tk).C(n,k).p^k.q^(n-k)]

           = [q + p.e^t]^n

           = [q + p(1 + t + t^2/2! + ...)]^n     and q+p=1

           = [1 + p(t + t^2/2! + ...)]^n

           = 1 + np(t + t^2/2! + ...)
              + n(n-1)/2! p^2(t + t^2/2! + ...)^2 + ...

and from this the coefficient of t/1! is np and the coefficient of 
t^2/2! is np+n(n-1)p^2

Thus E(x) = np and

     E(x^2) = np + n(n-1)p^2

            = np + n^2.p^2 - np^2 

            = np(1-p) + n^2.p^2

            = npq + n^2.p^2

            = npq + [E(x)}^2     and so

     E(x^2) - [E(x)}^2 = npq 

Thus the mean of the distribution is np and the variance is npq.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/