Mean of positive only normal distribution

Question

Take the normal distribution. Get rid of the negative side (density=0). Double the positives to normalize it.

Alternatively, if we have some normally distributed $X$ (which gives us the normal distribution), we can get this distribution from $|X|$.

Looks like this.

What is the mean for this distribution?

Answer 1

The desired answer is $$\frac 2 {\sqrt{2\pi}}\int_0^\infty x e^{-x^2/2}\,dx = \frac 2 {\sqrt{2\pi}} \int_0^\infty e^{-t}\,dt = \frac 2 {\sqrt{2\pi}},$$ with the change of variable $t = x^2/2$ so $dt = x\,dx$.

Answer 2

You have to compute

$$\mathbb E(|X|)=\sqrt{\frac{2}{\pi}}\int_0^{\infty}xe^{-x^2/2}=\sqrt{\frac{2}{\pi}} \left[-e^{-x^2/2}\right]_0^{\infty}=\sqrt{\frac{2}{\pi}}$$

Indeed $\dfrac{d}{dx}e^{-x^2/2}=-xe^{-x^2/2}$

Answer 3

It's called the folded normal distribution. I reproduce the results for completeness of the answer below.

If $Y = \mathcal{N}(\mu,\sigma)$ then:

$$E[|Y|] = \sigma\sqrt{\frac{2}{\pi}}\exp\left(\frac{-\mu^2}{2\sigma^2}\right) + \mu\left(1-2\Phi\left(\frac{-\mu}{\sigma}\right)\right)$$

$$Var[|Y|] = \mu^2 + \sigma^2 - E[|Y|]^2$$