Timeline for How far away is the maximum of $n$ i.i.d. chi-squared random variables from the rest of the sequence as $n$ gets large?
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6 events
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| Sep 30, 2013 at 10:39 | comment | added | ofer zeitouni | Oops; I meant to write $E(\Delta)$ converges to $1$ - the fraction near the max converges to 0. Sorry. | |
| Sep 22, 2013 at 2:59 | comment | added | ofer zeitouni | I have answered your related question, and indeed you were right about the scale, at least is a regime where $k$ does not grow too slowly with $n$ - in which case the problem I pointed out earlier persists. When $k>>\log n$, the same idea as in the computation of the max will show that $E(\Delta)$ does converge to 0. When $k<<\log n$, I believe your scaling is wrong | |
| Sep 21, 2013 at 22:23 | comment | added | Bullmoose | Thanks for your attention, @oferzeitouni. After thinking about my problem for a while, I modified the question. $k$ is increasing but not as fast as $n$... I also posted a somewhat related question. The $\sqrt{k/\log(n)}$ scaling is just a hunch, due to the variance of chi squared r.v. being $2n$ and the $\log(n)$ centering of max of random variables with exponential tails... | |
| Sep 21, 2013 at 22:16 | history | edited | Bullmoose | CC BY-SA 3.0 |
improved wording of the question
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| Sep 21, 2013 at 12:07 | comment | added | ofer zeitouni | Can you explain the scaling $\sqrt{k/\log n}$? At first sight it looks strange. For example, if $k=2$ (which is certainly $o(1)$) you are dealing with the maximum of $n$ exponential, which will be centered around $\log n$ and fluctuations of order $1$. | |
| Sep 21, 2013 at 4:50 | history | asked | Bullmoose | CC BY-SA 3.0 |