pdf of the sample mean from gamma distribution pdf of gamma distribution is
$f(x)=\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$ and for $\bar{X}$ has following pdf
$f_\bar{X}(x)=\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}\bar{x}^{n\alpha-1}e^{-n\beta\bar{x}}$, is it true? How to find $f_\bar{X}(x)$?
 A: As suggested by user365239, moment generating functions are an ideal trick to find distributions for sums of independent random variables. The moment generating function is defined as
$$m_X(t)=\mathbb{E}\left[e^{tX}\right]$$
In the case of a Gamma distribution, this evaluates to
$$m_X(t)=\left(1-\frac{t}{\beta}\right)^{-\alpha}$$
What happens if you consider $\bar{X}=\frac{X_1+X_2+\ldots+X_n}{n}$? In that case the moment generating function is
$$m_{\bar{X}}(t)=\mathbb{E}\left[e^{t\frac{X_1+X_2+\ldots+X_n}{n}}\right]=\mathbb{E}\left[e^{\frac{t}{n}(X_1+X_2+\ldots+X_n)}\right]=\mathbb{E}\left[e^{\frac{t}{n}X_1}\right]\mathbb{E}\left[e^{\frac{t}{n}X_2}\right]\ldots\mathbb{E}\left[e^{\frac{t}{n}X_n}\right]$$
The last step is because of the independence of the variables. Because they are all identically distributed we have
$$m_{\bar{X}}(t)=\left(1-\frac{t}{n\beta}\right)^{-\alpha}\left(1-\frac{t}{n\beta}\right)^{-\alpha}\ldots\left(1-\frac{t}{n\beta}\right)^{-\alpha}=\left(1-\frac{t}{n\beta}\right)^{-n\alpha}$$
But this is indeed the moment generating function of a Gamma distributed random variable with parameters $n\alpha$ and $n\beta$.
