Variance of Gamma Distribution/Proof 3

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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

From Expectation of Power of Gamma Distribution‎, we have:

$\expect {X^n} = \dfrac {\alpha^{\overline n} } {\beta^n}$

where $\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.


Hence:

\(\ds \var X\) \(=\) \(\ds \expect {X^2} - \paren {\expect X}^2\) Variance as Expectation of Square minus Square of Expectation
\(\ds \) \(=\) \(\ds \dfrac {\alpha^{\overline 2} } {\beta^2} - \paren {\dfrac {\alpha^{\overline 1} } \beta}^2\)
\(\ds \) \(=\) \(\ds \dfrac {\alpha \paren {\alpha + 1} } {\beta^2} - \paren {\dfrac \alpha \beta}^2\) Definition of Rising Factorial
\(\ds \) \(=\) \(\ds \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2}\)
\(\ds \) \(=\) \(\ds \frac \alpha {\beta^2}\)

$\blacksquare$