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My question is about the asymptotic behavior of the ratio between the largest and second largest values of $n$ independent chi-square random variables.

Let $X_1, \ldots, X_n$ be $n$ independent and identically distributed random variables with distribution $\chi_1^2$. Let $X_{(n)}$ be the largest and $X_{(n-1)}$ be the second largest of these $n$ random variables. I was wondering what is the asymptotic order of $X_{(n)}/ X_{(n-1)}$. My conjecture is that this quantity has the order of $1 + O(1/ \log n)$.

Can we show that \begin{equation} \frac{X_{(n)}} {X_{(n-1)}} - 1 \asymp \frac{C}{\log n} \end{equation} with high probability, where $C$ is a constant?

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  • $\begingroup$ Set $\Phi(x)= \mathbb P(X_1 >x)$. I would say the largest values, once shifted by $\Phi^{-1}(1/n) \sim 2 \log n$, converge to some Poisson point measure, the intensity of which I have yet to compute. I would therefore go for a convergence in distribution of $X_{(n)}- X_{(n-1)}$ towards some non degenerate random variable (that is explicit once you have the intensity). This random variable would take the role of $2C$ in your expression. I'll write this down tomorrow if nobody has. $\endgroup$ – Olivier Oct 25 '16 at 20:49
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    $\begingroup$ On this kind of convergence of extreme values, you may check the lecture notes by Anton Bovier wt.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Anton_Bovier/… $\endgroup$ – Olivier Oct 25 '16 at 21:01
  • $\begingroup$ if you replace $\chi_1^2$ by $\chi_2^2$ you have exponentials and as you have the inverse cumulant it should be easy to decide, then you have the issue of getting a good grid on the inverse cumulant of the $\chi_1^2$. $\endgroup$ – user83457 Oct 26 '16 at 6:19
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    $\begingroup$ @michael Good remark, the ordered statistics of the exponential random variable are indeed very simple :-) With that choice it is quick to realize your $C$ should be random. @ Steve: $X_{(n)}/2\log(n)$ converges in probability to $1$ and (to be checked) $X_{(n)}-X_{(n-1)}$ weakly converges to an exponential with parameter $1/2$. I can give more details if you need. $\endgroup$ – Olivier Oct 26 '16 at 8:38
  • $\begingroup$ Thanks @micheal and @Olivier! Your remarks are very helpful. I think I can solve that using the fact that $X_{(n)} - X_{(n-1)} $ converges to an exponential distribution. $\endgroup$ – Steve Oct 30 '16 at 9:49

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