Moment-generating Function of a Binomial Random VariableDate: 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/ |
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