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Let $U_1,U_2,...U_n,$ be iid samples from uniform distribution $U(0,1)$. And the order Statistic: \begin{equation} U_{n,1}\leq U_{n,2}\leq ...\leq U_{n,n}, \end{equation}
Given $p\in(0,1)$, if $1\leq K_n\leq n$ and $\frac{K_n}{n}\rightarrow p$. Proof $U_{n,k_n}\rightarrow p$ a.s.

My ideas so far:

The density function of $U_{n,k_n}$ is \begin{equation} f_n(x)=\frac{n!}{(k_n-1)!(n-k_n)!}x^{k_n-1}(1-x)^{n-k_n}. \end{equation} Then I would like to proof: \begin{equation} \sum_n P(|U_{n,k_n}-p|>\epsilon)<\infty \end{equation} Then by Borel-Cantelli lemma: \begin{equation} P(|U_{n,k_n}-p|>\epsilon \quad\text{i.o.})=0 \end{equation} which means: \begin{equation} U_{n,k_n}\rightarrow p = 0. \quad a.s. \end{equation} However, the integral of $P(|U_{n,k_n}-p|>\epsilon)<\infty$ is hard to calculate. Do I miss something? Does there exist some easy way to proof this?

Thanks in advance for any help!

Davide Giraudo
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Knotnet
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1 Answers1

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Let $\mu_n:=\mathsf{E}U_{n,k_n}$. Using an exponential bound found here, i.e., $$ \mathsf{P}(|U_{n,k_n}-\mu_n|>\epsilon)\le 2\exp(-\sqrt{n}\epsilon/(5\sigma_n)), $$ where $\sigma_n:=\mu_n(1-\mu_n)$, one gets $$ \sum_{n\ge 1}\mathsf{P}(|U_{n,k_n}-p|>\epsilon)\le \sum_{n\ge 1}2\exp(-\sqrt{n}\epsilon/(10\sigma_n))+1\{|\mu_n-p|>\epsilon/2\}<\infty $$ because $\mu_n\to p$. (See also this question regarding series convergence.)