Density of the sum of 2 independent random variables Just a general question about the gamma distribution.
May I know how do I find the moment generating function of $X$, given that $X\sim\Gamma(\alpha,\beta)$?
Thank you.
 A: It is entirely unclear whether by $\Gamma(\alpha,\beta)$ you mean the family of distributions in which $e^{-x/\beta}$ is one factor in the denisty function, or the one in which $e^{-\beta x}$ is one factor in the density function.  Both conventions occur.  They both give the same family of distributions, but which distribution is identified as $\Gamma(\alpha,\beta)$ is different in the two cases.  I will assuming you mean the former.
We have
$$
\int_0^\infty x^{\alpha-1} e^{-x/\gamma} \, dx = \gamma^\alpha \int_0^\infty \left(\frac x \gamma \right)^{\alpha-1} e^{-x/\gamma} \, \frac{dx}\gamma = \gamma^\alpha \int_0^\infty u^{\alpha-1} e^{-u} \, du = \gamma^\alpha\Gamma(\alpha). \tag 1
$$
Now let's look at the moment-generating function:
$$
\begin{align}
M(t) = E(e^{tX}) & = \frac{1}{\Gamma(\alpha)\beta^\alpha}\int_0^\infty e^{tx}x^{\alpha-1} e^{-x/\beta} \, dx \\[10pt]
& = \frac{1}{\Gamma(\alpha)\beta^\alpha} \int_0^\infty x^{\alpha-1}e^{-\left(\frac 1 \beta - t\right)x} \, dx. \tag 2
\end{align}
$$
Thus we have $1\left/\left(\dfrac 1 \beta - t\right)\right.$ in the role in which $\gamma$ is found in $(1)$.  Hence $(2)$ is equal to
$$
\frac{1}{\Gamma(\alpha)\beta^\alpha} \cdot \gamma^{\alpha}\Gamma(\alpha) = \left(\frac\gamma\beta\right)^\alpha = \left(\frac 1 \beta\cdot \frac{1}{\frac 1 \beta - t}\right)^\alpha = \left( \frac{1}{1-\beta t} \right)^\alpha.
$$
