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Given a small number $n$ (in my case, $n=2$) and a large number $N$, generate a set $S_N$ with $N$ points uniformly at random on $[0, 1)^n$. I'm looking for an asymptotic for the average largest gap between neighbors $$ \max_{s\in S}\min_{t\in S}\|s-t\| $$ as $N\to\infty.$ I assume the ratio of this to the average spacing diverges to $\infty$ for all $n\ge1$, is that right?

Actually, even better (a more direct answer to my underlying question) would be: what is the largest $r$ such that there is a point $x\in[0,1)^n$ such that the ball of radius $r$ centered at $x$ does not contain any point in $S$: $$ B_r(x)\cap S=\emptyset. $$ Of course I'm looking for some kind of an average over all random choices of $S$ and taking $N\to\infty$ as before.

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Charles
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1 Answers1

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Assuming we are in the unit square, for each such point $x$, where $r$ is less than the distance from the boundary of the unit square*, the probability that there is no point in the interior of $B_r(x)$ is given by $(1-\pi r^2)^N$. As $N$ goes to infinity, the probability goes to zero for all positive $r$. So the largest such $r$ is $0$. This answer holds regardless of dimension.

*For points closer to the boundary, you can throw in a factor of $1/4$ to bound the result and get the same answer.

Paul
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  • I'm interested in the asymptotics, especially the ratio of this distance to the average distance. It's not surprising that the maximum distance, like the average distance, goes to 0, but I'm interested in how quickly. – Charles Nov 04 '16 at 19:13
  • @Charles, then the question is not clear. Asymptotics of what in terms of $r$ as $N$ goes to infinity? Could you specify what you mean by "average," because it is not clear what exactly you mean here. – Paul Nov 04 '16 at 20:15
  • I disagree with @Paul. See for example (https://en.wikipedia.org/wiki/Nearest_neighbour_distribution). Because of an underlying Poisson distribution with a certain intensity $\lambda$, the probability that no event has occured in a circle with radius $r$ with center a given point is the P(N=0) in a Poisson distribution (Poisson process more exactly). Thus the probability that the distance from this given point to its closest neigbour is larger than $r$ is : $e^{-\lambda \pi r^2}$ ($\pi r^2$=area of the disk). (ctd...) – Jean Marie Nov 04 '16 at 20:32
  • (ctd...) But what we have done defines a cdf (cumulative density function)$F(r)=1-e^{-\lambda \pi r^2}$. Deriving this cdf, you obtain the pdf (prob. density function) $f(r)=2 \pi \lambda re^{-\pi \lambda r^2}$ which is the tool (Rayleigh density) to work with, for example to obtain its mean, variance... – Jean Marie Nov 04 '16 at 20:32
  • A reference in Math SE : (http://math.stackexchange.com/q/1429831) – Jean Marie Nov 04 '16 at 20:37
  • Asymptotics as $N$ goes to infinity, as in the question. Average with respect to the random selection of $S$; once $S$ is chosen $r$ is defined uniquely. – Charles Nov 04 '16 at 22:33