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...

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...

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

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...

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 $n=\omega(k)$ increasing faster than $k$ but not too much fast, i.e. $\log(n)=o(t)$. For example, $n=k^2$. This is the same scenario as in my previous question.

Now, let

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

I am wondering about the limiting behavior of $S(n)$ as $n\rightarrow\infty$. My conjecture (based on the fact that the maximum isn't too far away from the rest of the sequence of random variables with exponential tail, as well as the answer and comment to the two related questions) is that $S(n)$ converges to unity in distribution, however, I am having hard time proving this formally, as I get bogged down in a nasty convolution when I try to do it (but, maybe I am approaching this wrong and perhaps I am incorrect). Any suggestions?

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