Moment Generating Function of Gamma Distribution

Theorem
Let $X \sim \Gamma \left({\alpha, \beta}\right)$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.

Then the moment generating function of $X$, $M_X$, is given by:


 * $\displaystyle M_X \left({t}\right) = \begin{cases} \left({1 - \frac t \beta}\right)^{-\alpha} & t < \beta \\ \text{does not exist} & t \ge \beta \end{cases}$

Proof
From the definition of the Gamma distribution, $X$ has probability density function:


 * $\displaystyle f_X\left({x}\right) = \frac { \beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\Gamma \left({\alpha}\right)}$

From the definition of a moment generating function:


 * $\displaystyle M_X \left({t}\right) = \mathbb E \left[{ e^{t X} }\right] = \int_0^\infty e^{tx} f_X \left({x}\right) \rd x$

First take $t < \beta$.

Then:

Now take $t = \beta$.

Our integral becomes:

So $\mathbb E \left[{ e^{\beta X} }\right]$ does not exist.

Finally take $t > \beta$.

Note that $-\left({\beta - t}\right)$ is therefore positive.

As a consequence of Exponential Dominates Polynomial, we have $x^{\alpha - 1} < e^{-\left({\beta - t}\right) x}$ for sufficiently large $x$.

Therefore, in this case, the integrand increases without bound.

We conclude that the integral is divergent, hence $\mathbb E \left[{ e^{tX} }\right]$ does not exist for $t > \beta$.