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millwallcrazy
May11-10, 01:12 PM
I know that hte MGF is = the E[e^tx]

How do i show that if i take a sample (X1;X2; : : :Xn) from the exponential density f(x) = A*e^(-Ax), then the sum Z = sum(Xi) has the gamma density?

I found that the MGF for the exponential was A/(t-A) if that helps

Thanks

statdad
May12-10, 08:06 AM
Note that


e^{t\sum X_i}} = \prod{e^{tX_i}}


This will relate the form of the mgf of the sum to the individual mgfs of the sample.