Statistics – Mean Absolute Deviation of Normal Distribution
normal distribution
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?
Best Answer
Let $X\sim\mbox{N}\left(\mu,\sigma^{2}\right)$. So as usual the PDF
is given by $$f_{X}(a)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\left(\frac{a-\mu}{\sigma\sqrt{2}}\right)^{2}}.$$
Standard deviation and mean give a probability mass distribution, which is continuous. That means that you can't talk about discrete things (like saying "there are exactly $x$ elements greater than some cutoff $n$) just on the basis of mean and standard deviation. As such, you can't say with certainty that "there are no elements less than the lower bound $a$ or greater than the upper bound $b$", so you can't give discrete values to the bound $[a,b]$ over which your range would be defined.
What the mean and standard deviation of a normal distribution does let you do is make claims about the relative probability of an element existing at any spot along the distribution, so you could pick some set cutoff deviation (say, perhaps $3 \sigma$) and say that $99.7\%$ of the probability mass lies within a $\pm3\sigma$ bound.
From the Wikipedia entry, we know that the $p$-th central absolute moment of a $N(\mu,\sigma^2)$ random variable $X$ is
$$E[|X-\mu|^p]=\sigma^p\frac{2^{p/2}\Gamma(\frac{p+1}2)}{\sqrt\pi}.$$
If we know this number, and know $p$, we can determine $\tau=\sigma^p$
and then determine $\sigma=\tau^{1/p}$.
It might seem paradoxical, but even when $p=1$ this is sufficient to yield $\sigma$.
Best Answer
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, $a=\mu$ $\iff$ $b=0$.)