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Can the mean deviation about mean exceed the standard deviation for the Pareto distribution?

I just went through some books and found they are claiming that it cannot.

How can I prove that? What is the underlying logic that confirms this conclusion?

Glen_b
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arnab samanta
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  • It has nothing to do with the distribution the data are from; if both had the $n$ divisor, it would immediately follow from Jensen's inequality that standard deviation is always at least as large as the mean deviation. [If you mean for samples (rather than populations), you also have that since the usual form of standard deviation has the $n-1$ divisor (before taking square roots), that makes the standard deviation always strictly larger.] – Glen_b Sep 23 '13 at 07:01
  • Is 'proving it' related to some subject? For the purpose of private study? (By 'nothing to do with' above I mean it's a result that always holds, so you don't need to refer to the distribution.) – Glen_b Sep 23 '13 at 07:03
  • yea , it's enough ! i understand !and thanks a lot ! – arnab samanta Sep 23 '13 at 07:09

1 Answers1

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There's a well-known result, Jensen's inequality, which for our present purposes can be taken as:

if X is a random variable and $\varphi$ is a convex function, then $\varphi\left(\mathbb{E}\left[X\right]\right) \leq \mathbb{E}\left[\varphi(X)\right]$

  1. Population mean deviation and standard deviation

    Consider the original variable to be $Y$, and let $X = |Y-\mu_Y|$. Further, let $\varphi(X)=X^2$. Then immediately by the above inequality,

    $$\left(\mathbb{E}\left[|Y-\mu_Y|\right]\right)^2 \leq \mathbb{E}\left[|Y-\mu_Y|^2\right]\quad\text{,}$$

    and since both expectations are positive,

    $$\mathbb{E}\left[|Y-\mu_Y|\right] \leq \sqrt{\mathbb{E}\left[(Y-\mu_Y)^2\right]}\quad\text{,}$$

    which is the required result.

  2. Sample mean deviation (MD) and standard deviation, $s$

    The above result applies to discrete distributions just as well as continuous ones, and so in samples, if we were to use an $n$ denominator for the standard deviation (i.e. $s_n$ instead of $s_{n-1}$) the same result clearly holds simply by applying the above result to the ECDF as if it were a CDF.

    Immediately we have, if $\underline x = (x_1, x_2, \ldots, x_n)$,

    $$\text{MD}(\underline{x}) \leq s_n(\underline{x})$$.

    Since $s_{n-1} > s_n$, in that case it becomes a strict inequality.

    (Alternatively, we could use one of the other forms of the inequality at the above link to get to the same result.)

As you see, we don't need to know it's Pareto at all. However, proving it specifically for the Pareto may be easier than proving Jensen's inequality in general, but probably not by as much as all that.

Glen_b
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