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Let $\Omega=[0,L]$ and $X_1,\dotsc, X_n$ be uniformly distributed i.i.d.r.v on $\Omega$. It is known that (see here) $$\mathbb{E}\left[\min_{1\leq i \neq j \leq n} |X_i-X_j|\right]=\frac{L}{n^2-1}.$$

We wish to choose $\epsilon >0$ with the intention of ensuring $N_{\epsilon}(X_i)\cap N_{\epsilon}(X_j)=\emptyset$ almost surely. If this is even possible, then $\epsilon$ obviously needs to be less than the above average minimum distance, but how small? What would be an upperbound to ensure the result?

Is this a well-formulated problem? Can it be done? How would one approach solving this problem?

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Nap D. Lover
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  • I am not sure about your question - since the infimum of the minimum distance is $0$, or in other words $\Pr{\min|X_i - X_j| < 2\epsilon} > 0$ for any $\epsilon > 0$, so you may need to refine your question? – BGM Nov 20 '16 at 19:37
  • I see so the issue you raise is that given any radius $\epsilon$ of a neighborhood, I can always find $n$ (uniformly distributed) points whose average minimum distance is less than $2\epsilon$ with non zero probability and thus the intersection would probably be nonempty? If I understand correctly, this is definitely a problem and I'll think of ways how to refine the question. If you have suggestions let me know. – Nap D. Lover Nov 20 '16 at 19:49

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