If $X \sim N(0,1)$, why is $E(X^2)=1$?

Question

If $X$ is a normally distributed with mean $0$ and variance $1$, expectation of $X$ equals $0$ but why is $E(X^2)=1$?

Answer 1

For any random variable $X$ with a finite second moment, $$ \operatorname{Var}X=\operatorname EX^2-(\operatorname EX)^2 $$ so that $\operatorname{Var}X=\operatorname EX^2$ if $\operatorname EX=0$.

In our case, $\operatorname EX^2=\operatorname{Var} X=1$.

Answer 2

Another way of seeing this: If $X$ is standard normal, then $X^2$ is chi-squared with one degree of freedom. The expected value of a chi-squared being equal to the degrees of freedom, the result follows.

Answer 3

Hint:

Compute $$ \int_{-\infty}^{\infty} x^2 \left(\frac{1}{\sqrt{2\pi}} {\rm e}^{-\frac{x^2}{2}}\right)dx$$

Hint to hint:

Note that $$ \int x^2 {\rm e}^{-\frac{x^2}{2}}dx = \frac{\sqrt{\pi}}{\sqrt{2}} {\rm erf}\left( \frac{x}{\sqrt{2}}\right) - x {\rm e}^{-\frac{x^2}{2}} + \mathcal{C},$$ with erf being the error function defined by: $$ {\rm erf}(z) =\frac{2}{\sqrt{\pi}} \int_0^{z} {\mathrm e}^{-t^2} dt, $$ which can be viewed as special function (that is, it has clear, proven properties and numerical computer implementation).

Answer 4

The definition of variance for a random variable $X$, $var(X) = E[(X - \mu)^2]$, since $\mu = 0$, it's obvious that $var(X) = E[X^2]$.