Variance of Gaussian Distribution/Proof 2

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


 * $M_X \left({t}\right) = \exp \left({\mu t + \dfrac 1 2 \sigma^2 t^2}\right)$

From Variance as Expectation of Square minus Square of Expectation:


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

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


 * $\mathbb E \left[{X^2}\right] = M''_X \left({0}\right)$

We have:

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

So: