Timeline for Is mean absolute deviation smaller than standard deviation for $n\ge 3$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2019 at 19:28 | comment | added | Sextus Empiricus | The comment "(n-1) isn't enough to make up for..." sounded a bit difficult to me. In some cases it can be enough. | |
| Oct 2, 2019 at 19:18 | comment | added | meh | @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. | |
| Oct 2, 2019 at 19:00 | comment | added | Sextus Empiricus | 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. | |
| Oct 2, 2019 at 18:53 | comment | added | Sextus Empiricus | 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 $$n^2+2n+1 = (n+1)^2 \> (n+3)(n-1) =n^2+2n-3$$ | |
| Oct 2, 2019 at 18:32 | comment | added | Sextus Empiricus | 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 $. | |
| Oct 2, 2019 at 16:10 | history | answered | meh | CC BY-SA 4.0 |