Proof that $\sum_{i=1}^nX_i \sim \operatorname{Gamma}(n)$ How to prove that $\sum_{i=1}^n X_i$ has a $\operatorname{Gamma}(n)$ distribution, where $X_1,\ldots,X_n$ are independent standard exponentials?
 A: $$M_{X_i}(t) =\mathbb{E}[\exp(X_it)]= \frac{1}{1-t}$$
$$M_{\sum_{i=1}^nX_i}(t) =\mathbb{E}[\exp(\sum_{i=1}^nX_it)]=\mathbb{E}[\prod_{i=1}^n\exp(X_it)] =\prod_{i=1}^n\mathbb{E}[\exp(X_it)]=\left(\frac{1}{1-t}\right)^n$$
which is the mgf of Gamma($n$)
A: Lets find the characteristic function for $\sum_{k=1}^nX_k$

$\phi_{\sum_{k=1}^n X_k}(t) = \prod_{k=1}^n\phi_{X_k}(t)=\prod_{k=1}^n\frac{1}{1-it}=\frac{1}{(1-it)^n} $ - the characteristic function for Gamma(n)
A: Define $S_k := \displaystyle\sum_{i=1}^{k} X_i$.
$S_1 = X_1 \sim \text{Expo}(1) = \text{Gamma}(1,1)$
Suppose $S_k \sim \text{Gamma}(k,1)$ for all natural numbers $k \leq n-1$. We will show that the random variable $S_n\sim \text{Gamma} (n,1)$. We can write $S_n$ in this way: $S_n = S_{n-1} + X_n$ where $S_{n-1}$ and $X_n$ are independent. So the density of $S_n$ is:
$\displaystyle f_{S_n} (x) = \int_0^x f_{S_{n-1}} (y)f_{X_n} (x-y) dy = \int_0^x \frac{1}{\Gamma (n-1)} y^{n-2}e^{-y}e^{-(x-y)} dy = \frac{1}{\Gamma (n)} x^{n-1}e^{-x} $. Therefore, $S_n\sim \text{Gamma} (n,1)$.
