# Behavior of the sum of the exponents of chi-squared random variables normalized by their maximum

Let $X_1,X_2,\ldots,X_n$ be a sequence of $n$ i.i.d. chi-squared random variables with $k$ degrees of freedom, and denote by $X_\max$ the maximum of this sequence. Furthermore, let $k=\omega(1)$ increase, and $n$ be an increasing function of $k$, $n=f(k)$, where $f(k)$ is not increasing too fast, i.e. $\log(n)=o(k)$. For example, $n=k^d$ for some $d>0$. This is very similar to a scenario in my previous question.

Now, let

$$S(n)=\sum_{i=1}^n\exp\left[g(n)\frac{\sqrt{\log n}}{\sqrt{k}}(X_i-X_\max)\right]$$

where $g(n)=o(1)$ is a positive but slowly decreasing function (which we can pick arbitrarily, as long as it's decreasing).

I am wondering about the asymptotic behavior of the sum $S(n)$ as $n\rightarrow\infty$. Specifically, I wonder how does $S(n)$ grow in terms $n$, $k$ and $g(n)$? I.e., how much do the terms other than maximum matter as $n$ gets large? I'll be happy with an in-distribution convergence...

What I've done

Let's denote the terms in the sum by $Y_i=\exp\left[g(n)\frac{\sqrt{\log n}}{\sqrt{k}}(X_i-X_\max)\right]$. Clearly, each $Y_i$ is bounded: $0\leq Y_i\leq 1$. Now, using the fact that $\frac{X_\max-k}{\sqrt{2k\log n}}\rightarrow 1$ almost surely from the answer to my previous question, one can show that each $Y_i$ individually has low probability of being close to unity (i.e. $Y_\max$). So that means that $S(n)$ is not growing linearly with $n$. However, one can also show that $P(Y_i\leq \delta)\rightarrow 0$ for any $\delta=o(1/n^c)$, $c>0$. So $S(n)$ is growing with $n$, but how? Any hints/tips/suggestions would be appreciated...

Note

I substantially revised this question since I figured out the actual question that I wanted to ask. @ofer zeitouni's comment refers to the previous version of this question...

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Well, it is the same question really as your related question mathoverflow.net/questions/142772/… (except that I misread it then and thus my answer was wrong - now corrected). Your $S(n)$ will converge to $0$ because actually most of the sequence is FAR from the maximum in the relevant scale). To see that, use the same estimate that by now you should understand on the probability that a single variable is near the expected location of the maximum, and apply asymptotics for a binomial variable with that $p$. –  ofer zeitouni Sep 30 '13 at 10:43
@oferzeitouni I substantially revised this question. While the work I did on it is based on your answer to one of my previous question, I think it's now dissimilar from this question. If you have any ideas about what do here, please share. Thanks! –  Bullmoose Oct 30 '13 at 2:54
Well, it is still similar. Just estimate how many points there are at distance $x$ behind the expected location of the maximum. The first moment estimate will give the correct answer. The details depend of course on the function $g$. –  ofer zeitouni Oct 30 '13 at 5:48