## Mean Absolute Deviation/Standard Deviation Ratio

I ran across an interesting statistic today while doing some research, but it was stated as a matter of fact without explanation and there appears to be a dearth of material on it. It was stated that the Mean Absolute Deviation ("MAD") of a Normal (Gaussian) Distribution is .7979 of a Normal Distribution's Standard Deviation ("SD"). The simple equation offered was MAD:SD=SQRT (2/pi).

Question 1: Assuming this statement is true, why is it true? That is, what is it about the Normal Distribution that would cause a MAD to be .7979 of the SD?

Question 2: Again, assuming this statment is true, how would you reconcile two samples, one of which has a more favorable Jarque-Bera Test Statistic than another, but a less favorable MAD/SD Ratio?

Kimberley
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Mentor
 Quote by kimberley The simple equation offered was MAD:SD=SQRT (2/pi). Question 1: Assuming this statement is true, why is it true? That is, what is it about the Normal Distribution that would cause a MAD to be .7979 of the SD?
It's an easy calculation. The mean absolute deviation is the expected value of the absolute value of the random variable:

$$E(|x|) = \int_{-\infty}^{\infty}|x|\, p(x) dx$$

As both absolute value and the standard Gaussian distribution are even functions,

$$E(|x|) = 2\int_0^{\infty}x \frac 1 {\sigma\sqrt{2\pi}} e^{-\,\frac {x^2} {2\sigma^2} } dx$$

A simple u-substitution does the trick here, $u=\exp(-x^2/(2\sigma^2))$:

$$E(|x|) = \left.-\,\sigma \sqrt{\frac 2 {\pi}} e^{-\,\frac {x^2} {2\sigma^2} }\right|_0^{\infty} = \sigma \sqrt{\frac 2 {\pi}}$$

Recognitions:
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## Mean Absolute Deviation/Standard Deviation Ratio

Or MAD = $E(|x-\mu|) = \int_{-\infty}^{\infty}|x-\mu|\, p(x) dx$, if $\mu\ne 0$.

http://en.wikipedia.org/wiki/Mean_absolute_deviation
 i have a vaguely related question... consider random variables X and Y with E[X]=E[Y]=0 E[X^2]=E[Y^2] (ie the same standard deviation) and for n>=3 E[X^n]>=E[Y^n]>=0 is there a way to conclude that E[|X|]>=E[|Y|] ?

 Quote by judoudo i have a vaguely related question... consider random variables X and Y with E[X]=E[Y]=0 E[X^2]=E[Y^2] (ie the same standard deviation) and for n>=3 E[X^n]>=E[Y^n]>=0 is there a way to conclude that E[|X|]>=E[|Y|] ?

I assume you mean that the higher moments (3, 4, etc.) of |X| are always higher for X than Y. If that is what you mean, then I think I can provide a counter example. Consider X to be the absolute value of a rescaled version (standard deviation 1) of a central T distribution on 9 degrees of freedom. Suppose X is the absolute value of a Standard mean zero normal variable. Then I believe (from simulation) that higher moments (2, 3, ...) of the absolute value of the rescaled t variable are all greater than the corresponding higher moments of the absolute normal. Yet the expected value of the absolute T-variable is LESS than that of the absolute normal. Indeed you would not intuitively expect the relationship to be "higher, equal, higher, higher..." as the exponent goes 1, 2, 3, 4, ... rather it goes "less, equal, higher, higher, ..."

 Quote by judoudo i have a vaguely related question... consider random variables X and Y with E[X]=E[Y]=0 E[X^2]=E[Y^2] (ie the same standard deviation) and for n>=3 E[X^n]>=E[Y^n]>=0 is there a way to conclude that E[|X|]>=E[|Y|] ?

After I sent my post to the question you did NOT ask (with absolute value signs), I realized you might have meant it just the way you posed it! Here is a counterexample to the question you ACTUALLY posed.

Let X and Y both have 2-point distributions.

X is 9 with probability 0.1 and -1 with probability 0.9. E(X) = 0, Var(X) = 9

Y is 4.5 with probability 4/13 and -2 with probability 9/13. E(Y) = 0, Var(Y) = 9

E( |X| ) = 1.8 and E ( |Y| ) is clearly greater than 1.8, since it has to be between 2 and 4.5. Yet the higher moments of X are clearly all greater than those of Y.

I apologize for assuming you did not mean to leave off the absolute value signs.

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