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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \alpha {\beta^2}\) |
$\blacksquare$