Mean Absolute Deviation of normal distribution

Question

The Mean Absolute Deviation of the normal distribution is simply $$\sqrt{\frac{2}{\pi}}\sigma,$$ where $\sigma$ is the standard deviation of the normal distribution. (Wikipedia, Mathworld.)

How do I prove this?

Answer 1

Let $X\sim\mbox{N}\left(\mu,\sigma^{2}\right)$. So as usual the PDF is given by $$f_{X}(a)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\left(\frac{a-\mu}{\sigma\sqrt{2}}\right)^{2}}.$$

The mean absolute deviation is

\begin{alignat*}{1} \mathbf{E}\left[\left|X-\mu\right|\right] & =\int_{-\infty}^{\infty}\left|a-\mu\right|f_{X}(a)da\\ & =\int_{-\infty}^{\mu}\left(\mu-a\right)f_{X}(a)da+\int_{\mu}^{\infty}\left(a-\mu\right)f_{X}(a)da\\ & \overset{1}{=}2\int_{\mu}^{\infty}\left(a-\mu\right)f_{X}(a)da\\ & =2\int_{\mu}^{\infty}\frac{a-\mu}{\sigma\sqrt{2\pi}}e^{-\left(\frac{a-\mu}{\sigma\sqrt{2}}\right)^{2}}da\\ & \overset{2}{=}\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}be^{-b^{2}}\sigma\sqrt{2}db\\ & =2\sqrt{\frac{2}{\pi}}\sigma\int_{0}^{\infty}be^{-b^{2}}db\\ & =2\sqrt{\frac{2}{\pi}}\sigma\left[\frac{e^{-b^{2}}}{-2}\right]_{0}^{\infty}\\ & =\sqrt{\frac{2}{\pi}}\sigma\left[e^{0}-e^{-\infty}\right]\\ & =\sqrt{\frac{2}{\pi}}\sigma. \end{alignat*}

$\overset{1}{=}$ uses: Normal distribution is symmetric about the mean $\mu$.

$\overset{2}{=}$ uses the substitution: $b=\frac{a-\mu}{\sigma\sqrt{2}}$. (Thus, $\sigma\sqrt{2}db=da$. Also, $a=\mu$ $\iff$ $b=0$.)

Answer 2

Hint:

Rewrite $$\displaystyle \int_{-\infty}^{\infty} |x-\mu| \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}\,dx $$ as $$\displaystyle 2 \int_{\mu}^{\infty} \frac{1}{\sqrt{\pi}}\left(\frac{x-\mu}{\sqrt{2}\sigma}\right) e^{-((x-\mu)/(\sqrt{2}\sigma))^2}\,dx $$ and use fairly simple calculus to show this is $\sqrt{\dfrac{2}{\pi}}\sigma$