How to apply the Borel-Cantelli lemma to the Hausdorff distance between a uniform sample on $[0,1]$ and the aforementioned interval

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Let $n\in\mathbb Z^+$ and $X_1\ldots X_n$ i.i.d. $\operatorname{Unif}[0,1]$, that is, an independent sample from a uniform distribution on the interval $[0,1]$.

I've spent way too much time trying to prove that, for any $\varepsilon\in\mathbb R^+$,

$\operatorname{P}\{d_H(\chi_n,[0,1])>\varepsilon\;\text{ i. o.}\}=0.$

The LHS on the previous inequality refers to the Hausdorff distance between the sets $[0, 1]$ and $\chi_n:=\{X_i\}_{i=1}^n$. In addition, by i .o. I mean infinitely often, as usual.

I've reduced this to the slightly simpler statement:

$\displaystyle\operatorname{P}\left\lbrace\sup_{x\in[0,1]}\min_{i=1\ldots n}|X_i-x|>\varepsilon\;\text{ i. o.}\right\rbrace=0,$

which I also know to be true.

So my main goal rn is proving this last claim, which is driving me crazy, although I do I know how to do it if I delete the supremum and use an arbitrary $x\in[0,1]$.

I've tried lots of things, which I don't want to disclose for the moment, in order not to "corrupt" your minds. Anyway, if anybody can give me some (explicit) hints, I'd me most grateful, for it's almost sure that I'm not going to figure it out by myself.

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Well, in the end I came up with a rather convoluted way to solve the problem, which basically consists in adapting the proof of theorem 3 in http://sci-hub.cc/10.1239/aap/1086957575, in case anyone's interested.

But I still feel that there must be some simple way to do it, probably in the same line of Nate Eldredge's hints.

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Hint 1: $d_H(\chi_n, [0,1])$ is monotone decreasing in $n$ (since $\chi_n$ just adds more points as $n$ increases). So the limit as $n \to \infty$ exists, and your claim is equivalent to showing it equals 0 almost surely.

Hint 2: If $\omega$ is such that $\lim d_H(\chi_n(\omega), [0,1]) > 1/k$, then there exists an interval $(a,b)$, having rational endpoints and of length $> 1/(2k)$, such that no $X_n$ is in $(a,b)$. What is the probability of that?