Let $X_1, X_2,\dots, X_n$ be iid sample from $X\sim Gamma(\alpha, \beta)$. Use the moment generating function to show that the sample mean $\bar{X}$ also has a Gamma distribution.
Is my proof correct as follows?
Note the MGF of Gamma distribution is given by
$$
M_{X}(t)=(1-\beta t)^{-\alpha},
$$
for $\alpha, \beta>0$ and $t<1/\beta$.
The MGF of $\bar{X}$ is
$$
M_{\bar{X}}(t)=E[\exp(t\bar{X})]=(E[e^{tX_1/n}])^n=(1-\beta t/n)^{-\alpha n}
$$
Take $\beta'=\beta/n$ and $\alpha'=\alpha n$. It has the forms of the MGF of Gamma, so we prove it.
Best Answer
Yes, the answer is correct (do note that there are two parametrizations of the Gamma distribution so it may be good the denote which one is used).
The one thing I would add would be the following: For two r.v. $X_1$ and $X_2$ with mgfs $M_{X_1}(s)$ and $M_{X_2}(s)$ and cdfs $F_{X_1}(x)$ and $F_{X_2}(x)$, it holds that if \begin{equation} M_{X_1}(s) = M_{X_2}(s) \Rightarrow F_{X_1}(x) = F_{X_2}(x) \end{equation}
which I know you are going for but it's nice to see it stated explicitly.