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Mean Absolute Deviation/Standard Deviation Ratio

  1. Nov 21, 2007 #1
    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?

    Thank you in advance.

  2. jcsd
  3. Nov 21, 2007 #2


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    For any arbitrary probability distribution F, MAD(F) < SD(F) is always the case.

    As for Q.1., the normal dist. has the characteristic that as its "spread" increases one unit as measured by squared deviations (i.e., the variance), its spread increases 0.7979 of a unit as measured by absolute deviations. Remember that SD is "the sqrt of Var" = "sqrt of average squared error," and squaring inflates outliers.

    (This link shows a MAD of 0.681 for the Normal -- my guess is it is simulated data; which contain some error.) Contrast this with the Double Exponential, whose MAD is about half of its SD. Compared to SD, MAD gives less weight to outliers, so distributions with light tails tend to have a MAD/SD ratio closer to 1. This is confirmed by the observation that the Normal does indeed have lighter tails than the Double Exp'l.

    Q.2 is challenging. MAD and Skew measure different characteristics: MAD is dispersion, which is a second-order moment, so is SD. But the ratio MAD/SD is akin to a fourth-order moment (Kurtosis). Skew is third-order. It should be possible to devise a joint test of skewness and Kurtosis, which would be a Golden Key, but I don't have a ready formula for that.

    A practical approach may be to say "a third-order statistic is obviously more important than a fourth-order one," and devise an ad-hoc two-step test: If distribution F has excess skew closer to 0 than distribution G, then we conclude F is more normal than G. If distributions F and G have "practically" the same excess skew, then we compare their MAD/SD ratios; the one closer to 0.7979 is more normal.

    Again, the Golden Key would be to devise a joint test.
    Last edited: Nov 22, 2007
  4. Nov 22, 2007 #3

    D H

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    It's an easy calculation. The mean absolute deviation is the expected value of the absolute value of the random variable:

    [tex]E(|x|) = \int_{-\infty}^{\infty}|x|\, p(x) dx[/tex]

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

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

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

    [tex]E(|x|) =
    \left.-\,\sigma \sqrt{\frac 2 {\pi}} e^{-\,\frac {x^2} {2\sigma^2} }\right|_0^{\infty}
    = \sigma \sqrt{\frac 2 {\pi}}[/tex]
  5. Nov 22, 2007 #4


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    Last edited: Nov 23, 2007
  6. Nov 29, 2007 #5
    i have a vaguely related question...
    consider random variables X and Y with
    E[X^2]=E[Y^2] (ie the same standard deviation)
    and for n>=3

    is there a way to conclude that
  7. Mar 29, 2011 #6

    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, ..."
  8. Mar 29, 2011 #7

    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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