How can I derive the variance of the half-normal distribution?

Question

Let $X$ follow a half-normal distribution with pdf $f(x|\sigma)=\frac{\sqrt{2}}{\sigma\sqrt{\pi}}exp(-\frac{x^2}{2\sigma^2}), x >0$. How can I derive $Var(X)$?

my work:

I know that $Var(X)=E(X^2)-(EX)^2$, but I cannot solve for $E(X^2)$:

$E(X^2)=\int^\infty_0\frac{\sqrt{2}x^2}{\sigma\sqrt{\pi}}exp(-\frac{x^2}{2\sigma^2})dx$.

How can I integrate this?

Answer 1

Hint: if $X \sim \text{HalfNormal}(\sigma^2)$ and $Y \sim \text{Normal}(0,\sigma^2)$, then for any symmetric and integrable $f$ $$ E[f(X)] = E[f(|Y|)] = E[f(Y)|Y \ge 0] = 2 E[f(Y) 1(Y \ge 0)]. $$

I always get tripped up a little on the difference between conditioning and multiplying by indicators.

Answer 2

See Wikipedia on half normal. Then consider the simulation in R below. You already have some of the key results.

set.seed(2020)     # for reproducibility
z = rnorm(10^7)    # standard normal
mean(z);  mean(z^2)
[1] -2.9034e-06    # aprx E(Z) = 0
[1] 0.9996958      # aprx E(Z^2) = 1
x = abs(z)
var(x);  mean(x^2);  mean(x)
[1] 0.3633301      # aprx Var(X)= 1-2/pi = 0.363 
[1] 0.9996958      # aprx E(x^2)=E(Z^2)=1
[1] 0.7977253      # aprx E(x)=sqrt(2/pi) = 0.798
sqrt(2/pi)
[1] 0.7978846
1- 2/pi
[1] 0.3633802
mean(x^2) - mean(x)^2
[1] 0.3633301      # aprx Var(X) again

In Comment, @Dilip was trying to get you to see that $E(Z^2) = E(X^2).$