Show that for a random variable $X \sim f(x):=(2\pi)^{-\frac{1}{2}}exp(\frac{-x^2}{2})$:
$1)\ \ E[X] = 0$
$2)\ \ Var(X) = 1$
I already proved $1)$ without any difficulties, because $$E[X]=(2\pi)^{-\frac{1}{2}}\int_\mathbb R x e^{\frac{-x^2}{2}}$$ is easy to calculate. However, I got stuck with $2)$, because I don't know how to calculate $$E[X^2]=(2\pi)^{-\frac{1}{2}}\int_\mathbb R x^2 e^{\frac{-x^2}{2}}$$
Any help is greatly appreciated!
Do the following,
$-\frac{\partial}{\partial a} exp(-ax^2) = x^2exp(-ax^2)$
Integrate both sides,
$-\frac{\partial}{\partial a}\int exp(-ax^2) dx = \int x^2exp(-ax^2) dx$
$-\frac{\partial}{\partial a}\sqrt{\frac{\pi}{a}} = \int x^2exp(-ax^2) dx $
Take the derivative, set a = 1/2.