Moment Generating Function of Gamma Distribution

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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:

\(\ds \map {M_X} t\) \(=\) \(\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^{\alpha - 1} e^{-\paren {\beta - t} x} \rd x\)
\(\ds \) \(=\) \(\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \paren {\frac u {\beta - t} }^{\alpha - 1} e^{-u} \frac {\rd u} {\beta - t}\) substituting $\paren {\beta - t} x = u$
\(\ds \) \(=\) \(\ds \frac {\beta^\alpha} {\map \Gamma \alpha \paren {\beta - t}^\alpha} \int_0^\infty u^{\alpha - 1} e^{-u} \rd u\)
\(\ds \) \(=\) \(\ds \frac {\beta^\alpha \map \Gamma \alpha} {\map \Gamma \alpha \paren {\beta - t}^\alpha}\) Definition of Gamma Function
\(\ds \) \(=\) \(\ds \paren {\frac \beta {\beta - t} }^\alpha\)
\(\ds \) \(=\) \(\ds \paren {\frac {\beta - t} \beta}^{-\alpha}\)
\(\ds \) \(=\) \(\ds \paren {1 - \frac t \beta}^{-\alpha}\)

$\Box$


Now take $t = \beta$.

Our integral becomes:

\(\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^{\alpha - 1} \rd x\) \(=\) \(\ds \frac {\beta^\alpha} {\alpha \map \Gamma \alpha} \bigintlimits {x^\alpha} 0 \infty\) Primitive of Power, Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds 0 + \frac {\beta^\alpha} {\alpha \map \Gamma \alpha} \lim_{x \mathop \to \infty} x^\alpha\)
\(\ds \) \(\to\) \(\ds \infty\) for $\alpha > 0$, $x^\alpha$ increases without bound as $x \mathop \to \infty$

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

$\Box$


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 $t > \beta$.

$\blacksquare$


Examples

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


First Moment

The first moment generating function of $X$ is given by:

$\map { {M_X}'} t = \dfrac {\beta^\alpha \alpha} {\paren {\beta - t}^{\alpha + 1} }$


Second Moment

The second moment generating function of $X$ is given by:

$\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - t}^{\alpha + 2} }$


Third Moment

The third moment generating function of $X$ is given by:

$\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\paren {\beta - t}^{\alpha + 3} }$


Fourth Moment

The fourth moment generating function of $X$ is given by:

$\map { {M_X}^{\paren 4} } t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} } {\paren {\beta - t}^{\alpha + 4} }$


Sources

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