Variance of Gamma Distribution/Proof 2

Proof
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 $t < \beta$.

From Variance as Expectation of Square minus Square of Expectation:


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

From Expectation of Gamma Distribution, we have:


 * $\expect X = \dfrac \alpha \beta$

From Moment Generating Function of Gamma Distribution: Second Moment:


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

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:

So: