Prove $E(|X-\mu|) = \sqrt{\frac{2}{\pi}}\sigma$

Question

Prove $$E(|X-\mu|) = \sqrt{\frac{2}{\pi}}\sigma,$$ if $X$ is a normal random variable with mean $\mu$ and variance $\sigma^2$.

I don't know where to begin with this problem and would like help! Thanks

Answer 1

To simplify a bit, standardize (i.e. take $Z = (X - \mu)/\sigma$). Write the expected value as an integral. Use symmetry to get rid of the absolute value.

Answer 2

This is an elegant problem of symmetry of the normal probability distribution. First of all, let $Y = X - \mu$. So $D(Y) = D(X) = \sigma$. Notice that the figure of the normal probability distribution which Y follows is symmetrical with respect to its expectation $E(Y)=0$. The probability distribution of $|Y|$ can be viewed as the result of "adding" the probability of $+y$ and $-y$ together. Intuitively, the figure of probability distribution of $|Y|$ is made by folding the left half into the right and they plus together so that each point of the right half doubles its original value. $$E(|X-\mu|) = E(|Y|) = \int_{0}^{\infty}2{\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}}xdx $$ let $t=\frac{x}{\sigma}$. So we have $$E(|Y|)=\sigma\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}e^{-\frac{t^2}{2}}tdt \\ =\sigma\sqrt{\frac{2}{\pi}}$$