Variance 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 variance of $X$ is given by:
- $\var X = \dfrac \alpha {\beta^2}$
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 Variance as Expectation of Square minus Square of Expectation:
- $\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Gamma Distribution:
- $\expect X = \dfrac \alpha \beta$
From Moment Generating Function of Gamma Distribution: Second Moment:
- $\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - t}^{\alpha + 2} }$
From Moment in terms of Moment Generating Function, we also have:
- $\expect {X^2} = \map { {M_X}} 0$
Setting $t = 0$, we obtain the second moment:
\(\ds \map {M_X} 0\) | \(=\) | \(\ds \expect {X^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - 0}^{\alpha + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\beta^{\alpha + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha \paren {\alpha + 1} } {\beta^2}\) |
So:
\(\ds \var X\) | \(=\) | \(\ds \frac {\alpha \paren {\alpha + 1} } {\beta^2} - \frac {\alpha^2} {\beta^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \alpha {\beta^2}\) |
$\blacksquare$