Variance of the Standard Normal Distribution

Question

I'm having a hard time computing the variance of the standard normal distribution($u=0$ and $\sigma=1$). $$Var(X)=\int_{-\infty}^{\infty}x^2 e^{-x^2/2}dx$$ I want to substitute $y -> -x^2/2$

Which means I should get to $$Var(X)=\int_{-\infty}^{\infty}-ye^ydy$$

But this doesn't compute to 1 and I'm stuck here. Am I doing something wrong ? Could someone show me the right way to compute the proof, step by step ?

Answer 1

If you substitute $y\gets x^2/2$ you must also substitute $\mathsf d y \gets x~\mathsf d x$

$$\int x\,\mathrm e^{-x^2/2}\cdot x~\mathsf d x= \int \sqrt{y\,}~\mathrm e^{-y}~\mathsf d y$$

PS: you must also look at the interval of integration and whether any folding may happen.

PPS: Also the density function for $X\sim\mathcal N(0,1^2)$ is $f_X(x)=\frac 1{\sqrt{2\pi}}\mathrm e^{-x^2/2}$

Answer 2

When you are dealing with the transformation which are not one-to-one, you need to be careful. Note that the integrand is an even function, we have

$$ \begin{align} Var[X] &= \int_{-\infty}^{\infty} x^2 \frac {1} {\sqrt{2\pi}} e^{-x^2/2}dx \\ &= \frac {2} {\sqrt{2\pi}} \int_0^{\infty} x^2 e^{-x^2/2}dx \\ &= \frac {2} {\sqrt{2\pi}} \int_0^{\infty} 2y e^{-y} \frac {\sqrt{2}} {2\sqrt{y}}dy \\ & = \frac {2} {\sqrt{\pi}} \int_0^{\infty} y^{1/2} e^{-y} dy \\ & = \frac {2} {\sqrt{\pi}} \Gamma\left(\frac {1} {2} + 1\right) \\ & = \frac {2} {\sqrt{\pi}} \frac {1} {2}\Gamma\left(\frac {1} {2}\right) \\ & = 1 \end{align}$$

So just the last two steps require you some knowledge about the Gamma function.