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Can someone show why the ratio is $\sqrt{\frac{2}{\pi}}$ ?

kjetil b halvorsen
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user34829
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  • Need to specify the conditions. Here https://en.wikipedia.org/wiki/Average_absolute_deviation says the condition is " X is a normally distributed random variable with expected value 0". Is it the same as yours? – user158565 Oct 13 '18 at 22:24
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    missed "with expected value 0". It says you can find the answer from Geary, R. C. (1935). The ratio of the mean deviation to the standard deviation as a test of normality. Biometrika, 27(3/4), 310–332. – user158565 Oct 13 '18 at 22:38
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    I don't see why X needs to have expectation 0. MAD and standard deviation will not change if the mean shifts. Yes wiki gives a reference where essentially there is simply the conclusion, which is why I am putting the question here. – user34829 Oct 14 '18 at 01:16
  • See small number correction and this. It is the same problem. – Carl Oct 14 '18 at 01:32
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    Try differentiating the normal pdf $\phi(x)$ w.r.t. $x$ and then stare at the result to figure out how it is relevant to filling in the missing steps in the following calculation: $$\operatorname{MAD}=\int_{-\infty}^\infty |x|\phi(x)dx=2\int_0^\infty x\phi(x)dx=~\cdots ~= \sqrt{\frac 2\pi}$$ for a standard normal random variable. Then extend the calculation to a $N(\mu,\sigma^2)$ random variable. – Dilip Sarwate Oct 14 '18 at 06:52
  • @Carl Your "this" link is irrelevant to the question being asked. – Dilip Sarwate Oct 14 '18 at 06:55
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    The comment by Dilip Sarwate is essentially the answer.

    For $X\sim N(\mu,\sigma^2)$, after using the obvious change of variables $\frac{x-\mu}{\sigma}=t$, we have $$E|X-\mu|=\sigma \int_{\mathbb R}|t|\phi(t),dt$$

     – StubbornAtom Oct 14 '18 at 07:24
  • @DilipSarwate I do not find fault with your answer. The commonality for $\sqrt{\dfrac{2}{\pi}}$ is from SD$_{n=2}=\dfrac{|x_2-x_1|}{\sqrt{2}}$. The problems are similar enough that they appear to have the same form. – Carl Oct 14 '18 at 15:00

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