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 ?
Best Answer
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.