Expected value of half normal using gamma function

Question

Let $X$ be $N(0,1)$. I want to compute $E(|X|)$. This is what I tried:

$$E(|X|)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty|x|e^{\frac{-x^2}{2}}dx=\frac{2}{\sqrt{2\pi}}\int_{0}^\infty xe^{\frac{-x^2}{2}}dx$$

In order to aim for the gamma function, I do the change of variables $x^2/2=y.$ This results in

$$\frac{2}{\sqrt{2\pi}}\int_{0}^\infty xe^{\frac{-x^2}{2}}dx=\frac{2}{\sqrt{\pi}}\int_{0}^\infty y^{1/2}e^{-y}dy=\frac{2}{\sqrt{\pi}}\Gamma(1/2+1)=\frac{2}{\sqrt{\pi}}\frac{\sqrt \pi}{2}=1$$

This is obviously wrong, but I cannot find my mistake.

Answer 1

Its just a slight mistake within your substitution. In fact by replacing $x^2/2$ with $y$ you have to consider that your differential changes to $xdx=dy$ and not only to $dy$. Therefore the single $x$ is absorbed within the $dy$ instead of remaining as $y^{1/2}$. This finally leads to

$$\int_0^{\infty}e^{-\frac{x^2}2}(xdx)=\int_0^{\infty}e^{-y}dy$$

where the last integral turns out to equal $1$ instead of $\Gamma(1/2+1)$. Therefore your evaluation has to be finished as

$$E(|X|)=\frac{2}{\sqrt{2\pi}}\int_{0}^\infty xe^{\frac{-x^2}{2}}dx=\frac{2}{\sqrt{2\pi}}\int_{0}^\infty e^{-y}dy=\sqrt{\frac{2}{\pi}}$$

Answer 2

Change of variable is wrong, you should have $\int (\ldots) \frac{dx}{dy} \, dy$ and you have omitted the (jacobian) $\frac{dx}{dy}$.