Let $X_1,\ldots,X_n$ be IID from EXP($\lambda=1/ \theta$) so $E[X_1\mid\theta]=\theta$. Use MGF technique to find distribution of $\sum_{i=1}^{n} X_i$ I ended up getting $\left(\dfrac{\lambda}{\lambda-t}\right)^n \sim \operatorname{Gamma}(n, \lambda)$
Is this right, I haven't done the process in a while so if this is wrong I would like to see the process. Thanks. 
 A: Let
$$
Y = \sum_{i=1}^n X_i
$$
Then the moment-generating function for $Y$ is
$$
M_Y(t)=\text{E}\left[e^{tY}\right] = \text{E}\left[e^{t\sum_{i=1}^n X_i}\right] = 
\text{E}\left[\prod_{i=1}^n e^{t X_i}\right] = \prod_{i=1}^n \text{E}\left[e^{t X_i}\right]
= \prod_{i=1}^n M_{X_i}(t) = (M_{X_1}(t))^n
$$
Above, we've made liberal use of the fact that they are IID, namely when we switched products and expectations, and when we let $M_{X_1}$ stand in for all the moment-generating functions.  This is a standard result from probability: "The MGF of a sum is the product of the MGFs", IF the variables are independent.  So what is the MGF for the exponential distribution?  I'll just denote the exponential RVs by $X$ now, dropping the subscripts.  Wikipedia tells me that this is
$$
M_X(t) = (1-\theta t)^{-1},
$$
so that
$$
M_Y(t) = (1-\theta t)^{-n},
$$
which you recognize as the MGF for $Y\sim\Gamma(n,\theta)$.  If you didn't know that fact, you could explicitly work backwards from the MGF to a density function by the following integral
$$
\frac{1}{2\pi} \int_{-\infty}^\infty M_Y(it) e^{-itx} dt =
\frac{1}{2\pi} \int_{-\infty}^\infty \frac{e^{-itx}}{(1-i\theta t)^{n}} dt 
$$ 
But again, you kind of have to know that this integral represents the PDF of the Gamma distribution.  I'm sure you could manipulate that with the help of some integral tables to get it to give you the Gamma PDF.
