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
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
This is discrete, so it's the sum instead
$\displaystyle E(e^{tX})=.2\sum_{x=1}^{\infty} e^{tx}(.8)^{x-1}$
$\displaystyle = .25\sum_{x=1}^{\infty} e^{tx}(.8)^x$
$\displaystyle =.25\sum_{x=1}^{\infty} (.8e^t)^x$
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.