Variance of Gamma Distribution

Theorem
Let $X \sim \Gamma \left({\alpha, \beta}\right)$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.

Then:


 * $\displaystyle \operatorname{var} \left({X}\right) = \frac \alpha {\beta^2}$

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


 * $\displaystyle f_X\left({x}\right) = \frac { \beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\Gamma \left({\alpha}\right)}$

From Variance as Expectation of Square minus Square of Expectation:


 * $\displaystyle \operatorname{var} \left({X}\right) = \int_0^\infty x^2 f_X \left({x}\right) \rd x - \left({\mathbb E \left[{X}\right]}\right)^2$

So: