# Convergence rate of the sum of squares of inverse distances of random points which become dense in a region

$$n$$ points $$\{X_i\}$$ are drawn at random from a uniform distribution over a domain $$\Omega\subset \mathbb{R}^m$$ with a Lipschitz boundary. $$D_n$$ is defined as $$D_n = \sqrt{\frac{1}{\sum\limits_{1\le i,j \le n,i \ne j} \frac{1}{\|X_i-X_j\|_2^2}}}$$

I want help in finding the convergence rate of $$D_n$$ as $$n\to\infty$$, in terms of $$n$$ and $$m$$.

• $D_n$ is not well-defined as written as whenever $i = j$ you have zero in the denominator. Maybe you wanted to sum only over $i < j$? Commented Aug 1 at 9:32
• @MartinModrák : Thank you. I have edited to make $i$ not equal to $j$. Commented Aug 1 at 13:00

This is a partial answer, not sure a complete proof covering the case $$m=2$$ would fit within an MO answer. Let's look at $$A_n = 1/(2D_n^2) = \sum_{i, so the question is how fast does $$A_n$$ diverge.
When $$m > 2$$, the function $$X \mapsto |X|^{-2}$$ is integrable so that, by the law of large numbers, $$A_n$$ is of order $$n^2$$ with a prefactor you can easily guess. When $$m \le 2$$ on the other hand, $$A_n$$ has infinite expectation and the sum will be influenced by extreme events. In dimension $$1$$, points are ordered and distances between consecutive points are well approximated by i.i.d. $$\exp(1/n)$$ random variables. The minimum of $$n$$ such random variables is of order $$1/n^2$$, so one would expect to have $$A_n \approx n^4$$ in that case. (Not in the sense of having expectation of that size, but in the sense that the laws of $$A_n/n^4$$ are tight.)
Dimension $$2$$ is the most delicate case. By performing again a Poisson approximation and chopping up $$\Omega$$ into about $$n^2$$ disjoint boxes of area about $$1/n^2$$, one finds that with order $$1$$ probability each box contains only one of the $$X_i$$'s. On larger domains however a law of large numbers kicks in, so the expectation would be that $$A_n$$ is of order of its expectation, provided that we force the $$X_i$$'s to be at least at distance $$1/n$$ from each other, which yields $$A_n \approx n^2\log n$$.