Variance of Gamma Distribution/Proof 2

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


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

for $t < \beta$.

From Variance as Expectation of Square minus Square of Expectation:


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

From Moment in terms of Moment Generating Function, we also have:


 * $\expect {X^2} = \map {M''_X} 0$

We have:

Setting $t = 0$, we obtain the second moment:

So: