# Variance of Gamma Distribution

## 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 1

From the definition of the Gamma distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$
$\ds \var X = \int_0^\infty x^2 \map {f_X} x \rd x - \paren {\expect X}^2$

So:

 $\ds \var X$ $=$ $\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^{\alpha + 1} e^{-\beta x} \rd x - \paren {\frac \alpha \beta}^2$ Expectation of Gamma Distribution $\ds$ $=$ $\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \paren {\frac t \beta}^{\alpha + 1} e^{-t} \frac {\d t} \beta - \frac {\alpha^2} {\beta^2}$ substituting $t = \beta x$ $\ds$ $=$ $\ds \frac {\beta^\alpha} {\beta^{\alpha + 2} \map \Gamma \alpha} \int_0^\infty t^{\alpha + 1} e^{-t} \rd t - \frac {\alpha^2} {\beta^2}$ $\ds$ $=$ $\ds \frac {\map \Gamma {\alpha + 2} } {\beta^2 \map \Gamma \alpha} - \frac {\alpha^2} {\beta^2}$ Definition of Gamma Function $\ds$ $=$ $\ds \frac {\map \Gamma {\alpha + 2} - \alpha^2 \map \Gamma \alpha} {\beta^2 \map \Gamma \alpha}$ $\ds$ $=$ $\ds \frac {\alpha \paren {\alpha + 1} \map \Gamma \alpha - \alpha^2 \map \Gamma \alpha} {\beta^2 \map \Gamma \alpha}$ Gamma Difference Equation $\ds$ $=$ $\ds \frac {\alpha \map \Gamma \alpha \paren {\alpha + 1 - \alpha} } {\beta^2 \map \Gamma \alpha}$ $\ds$ $=$ $\ds \frac \alpha {\beta^2}$

$\blacksquare$

## Proof 2

By Moment Generating Function of Gamma Distribution, the moment generating function of $X$ is given by:

$\ds \map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha}$

for $t < \beta$.

$\ds \var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Gamma Distribution, we have:

$\expect X = \dfrac \alpha \beta$
$\map { {M_X}''} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - t}^{\alpha + 2} }$

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:

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

$\blacksquare$

## Proof 3

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 $\ds$ $=$ $\ds \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2}$ $\ds$ $=$ $\ds \frac \alpha {\beta^2}$

$\blacksquare$