Timeline for Lowest index giving half of the sum
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2016 at 7:23 | vote | accept | tsmi | ||
| Mar 15, 2016 at 3:36 | answer | added | Anthony Quas | timeline score: 1 | |
| Mar 15, 2016 at 0:27 | comment | added | Anthony Quas | right... sorry. I should have read the question more carefully. Of course knowing one, you know the other. | |
| Mar 14, 2016 at 23:43 | comment | added | Iosif Pinelis | Anthony:I think your reasoning is good for the ascending order, but the order is descending. So, the mean should be flipped to about $n-n/\sqrt2$. | |
| Mar 14, 2016 at 22:27 | comment | added | Anthony Quas | The variance is of order $n$: if you take the samples up to $1/\sqrt 2$, there are $n/\sqrt 2+O(\sqrt n)$ of them and their sum is $n/4+O(\sqrt n)$. The full sum is $n/2+O(\sqrt n)$. So the way to get the "scale balance point" is to start with the ones up to $1/\sqrt 2$ and then move $O(\sqrt n)$ terms of the biggest terms below $1/\sqrt 2$ or the smallest terms above $1/\sqrt 2$ to the other side to get the two sides to balance. | |
| Mar 14, 2016 at 22:21 | comment | added | Anthony Quas | The mean should be close to $n/\sqrt 2$ (if you look at the samples from the $x_i$'s that are below $1/\sqrt 2$, there should be roughly $n/\sqrt 2$ of them and their average value should be $1/(2\sqrt 2)$, so that the sum is roughly $n/4$. On the other hand, the total of all the samples should be close to $n/2$. | |
| Mar 14, 2016 at 22:16 | comment | added | Anthony Quas | Quite an interesting question. A somewhat related topic, in case you don't know the name already, is order statistics. | |
| Mar 14, 2016 at 21:36 | review | First posts | |||
| Mar 14, 2016 at 22:30 | |||||
| Mar 14, 2016 at 21:35 | history | asked | tsmi | CC BY-SA 3.0 |