Expectation of Gamma Distribution/Proof 2

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Theorem

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

The expectation of $X$ is given by:

$\expect X = \dfrac \alpha \beta$


Proof

By Moment Generating Function of Gamma Distribution, the moment generating function of $X$ is given by:

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

for $t < \beta$.

From Moment in terms of Moment Generating Function:

$\expect X = \map { {M_X}'} 0$

From Moment Generating Function of Gamma Distribution: First Moment:

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

Hence setting $t = 0$:

\(\ds \map { {M_X}'} 0\) \(=\) \(\ds \frac {\beta^\alpha \alpha} {\paren {\beta - 0}^{\alpha + 1} }\)
\(\ds \) \(=\) \(\ds \frac {\beta^\alpha \alpha} {\beta^{\alpha + 1} }\)
\(\ds \) \(=\) \(\ds \frac \alpha \beta\)

$\blacksquare$