Timeline for Concentration for sum of order statistics
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 24 at 23:50 | vote | accept | user139952 | ||
| Feb 5 at 5:14 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Feb 4 at 0:39 | history | edited | kodlu | CC BY-SA 4.0 |
edited title
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| Feb 3 at 15:44 | comment | added | Christian Remling | More specifically, this gives $P(x_1+\ldots +x_k\le s)= \binom{n}{k}\int\ldots\int (1-t)^{n-k}\, dt_1\ldots dt_k$, where the integration is over $t_1+\ldots +t_k\le s$, and $t=\max t_j$. | |
| Feb 3 at 15:34 | comment | added | Christian Remling | Can't you just do this as a calculus problem? The joint density of the order statistics is $n!$ where it is non-zero, so the probability that $x_1+\ldots +x_k\le b$ equals $n!$ times the volume of the region defined by this condition and $0\le x_1\le x_2\le \ldots \le x_n\le 1$. | |
| Feb 3 at 15:06 | comment | added | James Martin | If $k$ and $\alpha$ are fixed and $n\to\infty$, the statement isn't true in fact. For example, the probability that $x_1$ exceeds $\alpha(k+1)/(2(n+1))$ converges to $e^{-\alpha(k+1)/2}$ -- in that case the sum $x_1+\dots+x_k$ certainly exceeds the given threshold. For $k\to\infty$, a useful approach may be that $x_{k+1}$ concentrates, and then conditional on $x_{k+1}$, the sum $x_1+ \dots+x_k$ has the same distribution as $x_{k+1}(U_1+\dots+U_k)$ where $U_i$ are i.i.d. $U[0,1]$, to which you can indeed apply Chernoff or whatever. | |
| Feb 3 at 12:24 | comment | added | user139952 | It could be fixed, but could also be a function of $n$. | |
| Feb 3 at 8:27 | comment | added | mathworker21 | Do you view $k$ as fixed? | |
| Feb 3 at 7:35 | history | asked | user139952 | CC BY-SA 4.0 |