# Is mean absolute deviation smaller than standard deviation for $n\ge 3$?

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:

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

## 2 Answers

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

• 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 $n$ – CloseToC Sep 25 at 7:46
• 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. – Glen_b Sep 25 at 22:52

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.

• 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 $x_i$. – Sextus Empiricus Oct 2 at 18:32
• 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 $$n^2+2n+1 = (n+1)^2 \> (n+3)(n-1) =n^2+2n-3$$ – Sextus Empiricus Oct 2 at 18:53
• 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. – Sextus Empiricus Oct 2 at 19:00
• @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. – aginensky Oct 2 at 19:18
• The comment "(n-1) isn't enough to make up for..." sounded a bit difficult to me. In some cases it can be enough. – Sextus Empiricus Oct 2 at 19:28