Moment Generating Function of Gamma Distribution

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

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


 * $\map {M_X} t = \begin {cases} \paren {1 - \dfrac t \beta}^{-\alpha} & t < \beta \\ \text {does not exist} & t \ge \beta \end {cases}$

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


 * $\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$

From the definition of a moment generating function:


 * $\ds \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$

First take $t < \beta$.

Then:

Now take $t = \beta$.

Our integral becomes:

So $\expect {e^{\beta X} }$ does not exist.

Finally take $t > \beta$.

We have that $-\paren {\beta - t}$ is positive.

As a consequence of Exponential Dominates Polynomial, we have:
 * $x^{\alpha - 1} < e^{-\paren {\beta - t} x}$

for sufficiently large $x$.

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

We conclude that the integral is divergent.

Hence $\expect {e^{t X} }$ does not exist for $t > \beta$.

Examples
In the below, $t < \beta$ throughout.