Suppose that I have a sequence of $n$ i.i.d. chi-squared random variables with $k$ degrees of freedom $X_1, X_2, \ldots, X_n$, and denote $X_{\max}=\max(X_1, X_2, \ldots, X_n)$. Let $k$ be increasing but not as fast as $n$, i.e. $k=\omega(1)$ and $n=\omega(k)$.

I am wondering how far away $X_{\max}$ is from the rest of the sequence as $k\rightarrow\infty$. In particular, what is known about the asymptotics of the expected value $E[\Delta]$ where $\Delta$ is the fraction of the instances of the random variable in the sequence that are at least $\sqrt{k/\log(n)}$ less than $X_{\max}$ (formally, $\Delta=|\{X_i:X_i\leq X_{\max}-\sqrt{k/\log(n)}\}|/n$)? Is $\lim_{n\rightarrow\infty}E[\Delta]=0$ or $\lim_{n\rightarrow\infty}E[\Delta]=1$ (I doubt it's between zero and one)?

  • $\begingroup$ 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 $1$. $\endgroup$ Sep 21, 2013 at 12:07
  • $\begingroup$ 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... $\endgroup$
    – Bullmoose
    Sep 21, 2013 at 22:23
  • $\begingroup$ 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 $\endgroup$ Sep 22, 2013 at 2:59
  • $\begingroup$ Oops; I meant to write $E(\Delta)$ converges to $1$ - the fraction near the max converges to 0. Sorry. $\endgroup$ Sep 30, 2013 at 10:39


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