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 $\max\_{1 \le k \le n}|X\_k-Y\_k|$.



On a random partition
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.


