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Let $X_1,X_2,\dots$ be continuous random variables with full support (I need the result when they follow AR time series $X_i=\alpha X_{i-1}+\varepsilon_i$ for iid epsilons. But if you will consider iid case, it may also help with the time series case) with their stationary distribution function $F$. Notation $X_{n; (k)}$ represents the $k-$th order statistic, i.e. $X_{n; (1)}=\min_{i\leq n} X_i$.

Let $k_n\in\mathbb{N}$ fulfill $$k_n\to\infty, \frac{k_n}{n}\to 0 \text{, for } n\to\infty.$$

Let $$u_n=quantile(X_1, 1-\frac{k_n}{n}), \text{ i.e. } P(X_1>u_n)=\frac{k_n}{n}.$$

Is the following true $$\frac{n}{k_n}P(X_{n; (n-k_n)}>X_1>u_n)\overset{n\to\infty}{\to}0?$$

Maybe a helpful observation is that this is equal to $$ \frac{n}{k_n}P(X_{n; (n-k_n)}>X_1>u_n)=P(X_{n; (n-k_n)}>X_1\mid X_1>u_n)=1-P(\hat{F}(X_1)>1-\frac{k_n}{n}\mid F(X_1)>1-\frac{k_n}{n}). $$

RobPratt
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Albert Paradek
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  • Thanks for the thought. Firstly, in your way of thinking, it doesnt really depend on u_n. It shoild though, think if u_n is reeeeeally large, then if X_1 is even largen than that, it will be surely one of the top few values of the Sample. Secondly, X_{(n-kn)} goes to infinity almost surely, as forgottenarrow proved here https://math.stackexchange.com/q/4118680/615430 – Albert Paradek May 15 '21 at 11:46

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