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:

\(\ds \var X\) \(=\) \(\ds \frac {\alpha \paren {\alpha + 1} } {\beta^2} - \frac {\alpha^2} {\beta^2}\)
\(\ds \) \(=\) \(\ds \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2}\)
\(\ds \) \(=\) \(\ds \frac \alpha {\beta^2}\)

$\blacksquare$