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I want to compare the mean absolute deviation with standard deviation in general case with this definition:

$$MAD = \frac{1}{n-1}\sum_1^n|x_i - \mu|, \qquad SD = \sqrt{\frac{\sum_1^n(x_i-\mu)^2}{n-1}}$$

where $\mu =\frac{1}{n}\sum_1^n x_i$.

Is it true that $MAD \le SD$ for every $\{x_i\}^n_1$?

It's false for $n=2$, becouse $x+y \ge \sqrt{x^2+y^2}$, for every $x, y \ge 0$.

It's easy to show that:

$$MAD \le \sqrt{\frac{n}{n-1}} \times SD$$

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2 Answers 2

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No, in general this is not true.

A simple way to look at this is to simulate. I typically hack together an infinite loop that stops if it finds a counterexample. If it runs for a long time, I start thinking about whether the claim might be true. In the present case, my R code looks like this:

while ( TRUE ) {
    xx <- runif(3)
    mad <- sum(abs(xx-mean(xx)))/(length(xx)-1)
    sd <- sqrt(sum((xx-mean(xx))^2)/(length(xx)-1))
    if ( mad > sd ) break
}
xx

It yields this counterexample:

[1] 0.7852480 0.0760231 0.8295893
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    $\begingroup$ That's a clever way of using simulation! Saved me from incorrectly answering that the result always holds due to Jensen's inequality... which is apparently not applicable when you divide by $n-1$ instead of $n$ $\endgroup$
    – CloseToC
    Commented Sep 25, 2019 at 7:46
  • $\begingroup$ However I think perhaps an answer that compares $s_n$ to the mean deviation with $n$ denominator would, I think, be useful, because it would give context to the counterexample. $\endgroup$
    – Glen_b
    Commented Sep 25, 2019 at 22:52
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Here is a more mathematical approach. Firstly, it's probably true that by a change of variables, one can assume that the mean is zero. Certainly from the point of view of finding a counter example, this is acceptable. So, setting $ \mu = 0$, squaring both sides of the proposed inequality and multiplying through by (n-1) one is left with the proposed inequality -

$ (\sum_{i=1}^{i=n}|x_i|)^2 \leq (n-1)(\sum_{i=1}^{i=n}|x_i|^2))$

This looks fishy. (n-1) isn't enough to make up for all the $|x_i| |x_j|$ terms . Particularly if all the $x_i$ are the same in absolute value. My first guess was n=4 and $ x_1 = x_2 =1, x_3=x_4 = -1$. This leads to $\frac{4}{3} \leq \sqrt{\frac{4}{3}}$ . I would think that this sort of thing is well known to people interested in inequalities.

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  • $\begingroup$ For all even $n $ you can use your construction (every $x_i = \pm 1$) and $$MAD = \frac {n}{n-1} > \sqrt{\frac {n}{n-1}} = SD$$ thus it can not be true that $MAD \leq SD $ for all $x_i $. $\endgroup$
    – Sextus Empiricus
    Commented Oct 2, 2019 at 18:32
  • $\begingroup$ For all odd $n $ you can use my construction ($x_0=-2$, $x_1=x_2=1$ and then every other $x_i = \pm1$ with alternating plus minus). Then you have $$MAD = \frac {n+1}{n-1} > \sqrt {\frac {n+3}{n-1}} = SD $$ where the inequality can be made clear by multiplying by $n-1$ and squaring such that it becomes $$n^2+2n+1 = (n+1)^2 \> (n+3)(n-1) =n^2+2n-3$$ $\endgroup$
    – Sextus Empiricus
    Commented Oct 2, 2019 at 18:53
  • $\begingroup$ But it is not true that $MAD > SD $ for all possible $x_i $. The terms $|x_i| |x_j|$ (there's $n^2$ of them) can be made up for by the $(n-1) $ term when sufficient number of the $x_i $ are small. $\endgroup$
    – Sextus Empiricus
    Commented Oct 2, 2019 at 19:00
  • $\begingroup$ @Martijn All I was saying was that doing a little algebra pointed the way to finding counter-examples. I by no means think, and I don't think I even gave the impression I thought, that the inequality was always false or true. $\endgroup$
    – meh
    Commented Oct 2, 2019 at 19:18
  • $\begingroup$ The comment "(n-1) isn't enough to make up for..." sounded a bit difficult to me. In some cases it can be enough. $\endgroup$
    – Sextus Empiricus
    Commented Oct 2, 2019 at 19:28

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