Gaussian Variance

The variance of a distribution $f(t)$ is defined as its second central moment:

$$\sigma^2 \isdef \int_{-\infty}^\infty (t-\mu)^2 f(t)dt$$ (D.43)

where $\mu$ is the mean of $f(t)$ .

To show that the variance of the Gaussian distribution is $\sigma^2$ , we write, letting $g\isdef 1/\sqrt{2\pi\sigma^2}$ ,

$$\begin{eqnarray*} \int_{-\infty}^\infty (t-\mu)^2 f(t) dt &\isdef & g \int_{-\infty}^\infty (t-\mu)^2 e^{-\frac{(t-\mu)^2}{2\sigma^2}} dt\\ &=&g \int_{-\infty}^\infty \nu^2 e^{-\frac{\nu^2}{2\sigma^2}} d\nu\\ &=&g \int_{-\infty}^\infty \underbrace{\nu}_{u} \cdot \underbrace{\nu e^{-\frac{\nu^2}{2\sigma^2}} d\nu}_{dv}\\ &=& \left. g \nu (-\sigma^2)e^{-\frac{\nu^2}{2\sigma^2}} \right\vert _{-\infty}^{\infty} \\ & & - g \int_{-\infty}^\infty (-\sigma^2) e^{-\frac{\nu^2}{2\sigma^2}} d\nu \\ &=&\sigma^2 \end{eqnarray*}$$

where we used integration by parts and the fact that $\nu f(\nu)\to 0$ as $\left\vert\nu\right\vert\to\infty$ .