0
$\begingroup$

I'm not exactly sure how to go about this, could I use the definition of the MGF or something along those lines?

$\endgroup$
1
$\begingroup$

Note that if $X\sim \text{Gamma}(N,\lambda)$ then $$ X\stackrel{d}{=}X_1+\dotsb+X_N $$ where $X_i\sim \text{Exp}(\lambda)$ are i.i.d exponential random variables with rate parameter equal to $\lambda$. It follows by the law of total expectation that $$ M_{X}(t)=Ee^{tX}=g_{N}(M_{X_1}(t)) $$ where $g_N$ is the probability generating function of $N$ and $M_{X_1}$ is the moment generating function of $X_1$.

You can do something similar using characteristic functions and then invert using the inversion formula.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.