# Thread: finding the mgf function

1. ## finding the mgf function

Can someone help me with this problem? I don't understand how to do it, I know that the mgf is definied as the integral from negative to positive infinity e^(tx)f(x) but I don't understand how to do it for this question

2. This is discrete, so it's the sum instead

$E(e^{tX})=.2\sum_{x=1}^{\infty} e^{tx}(.8)^{x-1}$

$= .25\sum_{x=1}^{\infty} e^{tx}(.8)^x$

$=.25\sum_{x=1}^{\infty} (.8e^t)^x$

Now use the geometric sum from calc2, that's why this is a geometric random variable

You can prove directly that the mean is 1/p.
So with p=.2 so the mean is indeed 5.