Variance of Gaussian Distribution

Theorem
Let $X \sim N \left({\mu, \sigma^2}\right)$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the Gaussian distribution.

Then:


 * $\operatorname{var} \left({X}\right) = \sigma^2$

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


 * $f_X \left({x}\right) = \dfrac 1 {\sigma \sqrt{2 \pi} } \, \exp \left({-\dfrac { \left({x - \mu}\right)^2} {2 \sigma^2} }\right)$

From Variance as Expectation of Square minus Square of Expectation:


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

So: