TITLE: expected value calculation for squared normal distribution
QUESTION [23 upvotes]: I need help with the following problem. Suppose $Z=N(0,s)$ i.e. normally distributed random variable with standard deviation $\sqrt{s}$. I need to calculate $E[Z^2]$. My attempt is to do something like
\begin{align}
E[Z^2]=&\int_0^{+\infty} y \cdot Pr(Z^2=y)dy\\
=& \int_0^{+\infty}y\frac{1}{\sqrt{2\pi s}}e^{-\frac y{2s}}dy\\
=&\frac{1}{\sqrt{2\pi s}}\int_0^{\infty}ye^{-\frac y{2s}}dy.
\end{align}
By using integration by parts we get
$$\int_0^{\infty}ye^{-\frac y{2s}}dy=\int_0^{+\infty}2se^{-\frac y{2s}}dy=4s^2.$$
Hence $E[Z^2]=\frac{2s\sqrt{2s}}{\sqrt{\pi}},$ which does not coincide with the answer in the text. Can someone point the mistake?

REPLY [91 votes]: This is old, but I feel like an easy derivation is in order.
The variance of any random variable $X$ can be written as 
$$
V[X] = E[X^2] - (E[X])^2
$$
Solving for the needed quantity gives
$$
E[X^2] = V[X] + (E[X])^2
$$
But for our case, $E[X] = 0$, so the answer of $\sigma^2$ is immediate.