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Let $I=(0,1) \subset \mathbb{R}$. We denote by $m$ the Lebesgue measure on $I$. For $n \in \mathbb{N}$, we set $n^{-1}\mathbb{N}=\{k/n \mid k \in \mathbb{N}\}$ and $I_n=I \cap n^{-1}\mathbb{N}$. Let $\{X_k\}_{k=1}^n$ be i.i.d random variables on a probability space $(\Omega,\mathcal{F},P)$. We assume that \begin{align*} P[X_1 \in A]=m(A),\quad A \in \mathcal{B}((0,1)). \end{align*}

Question. For $n \in \mathbb{N}$ and $1 \le k \le n$, we define $Y_k \in I_n$ in such a way that \begin{align*} \min_{x \in I_n}|X_k-x|=|X_k-Y_k|. \end{align*} Then, can we find positive numbers $\{\alpha_n\}_{n=1}^\infty$ with the following properties ?

  1. $\lim_{n \to \infty} n\alpha_n=0$.
  2. For large sufficiently $n$, we have $ P\left[\max_{1 \le k \le n}|X_k-Y_k| \le \alpha_n \right]=1.$

This is a question of how uniform the partition determined by $\{X_k\}_{k=1}^n$ is.

For the requirement 2, it seems better to use the Borel--Cantelli lemma. However, I feel that it is a little difficult to calculate $\max_{1 \le k \le n}|X_k-Y_k|$.

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1 Answer 1

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For fixed $n$ and $k \le n$ the distance $|X_k-Y_k|$ is uniformly distributed in $[0,1/(2n)]$. Thus if $\alpha_n<1/(2n)$, then $$P\left[\max_{1 \le k \le n}|X_k-Y_k| \le \alpha_n \right]=(1-2n\alpha_n)^n \,. $$ On the other hand, if $\alpha_n \ge 1/(2n)$, then this probability is clearly 1. So this gives a negative answer to the question.

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  • $\begingroup$ Thank you for your comment. $|X_n-Y_n|$ is also uniformly distributed, right? $\endgroup$ Commented Dec 25, 2020 at 19:44
  • $\begingroup$ Yes, indeed. This is fixed now. $\endgroup$ Commented Dec 25, 2020 at 19:48
  • $\begingroup$ Thank you. I understood. $\endgroup$ Commented Dec 25, 2020 at 19:50

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