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 $t = 0$:
\(\ds \map { {M_X}'} 0\) | \(=\) | \(\ds \frac {\beta^\alpha \alpha} {\paren {\beta - 0}^{\alpha + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\beta^\alpha \alpha} {\beta^{\alpha + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \alpha \beta\) |
$\blacksquare$