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Let $U \in \mathbb{R}^3$ be distributed uniformly in the Ball in $\mathbb{R}^3$ centered at zero. That is $U \sim f_U(u)= \frac{1}{ \frac{4}{3} \pi R^3}$ for all $\|u\|\le R$ where $R$ is the radius of the ball.

Now suppose we generate $n$ points i.i.d. according distribution of $U$.

My questions is: Can we compute the expected minimum distance between the generated points, that is \begin{align} E\left[ \min_{i,j\in \{1,2,,,n\}} \| U_i-U_j\| \right], \end{align} where $\| U_i-U_j\|$ is Euclidean distance.

This question is related to a number of other questions. For example, Average distance between two random points in a square
Average minimum distance between $n$ points generate i.i.d. with uniform dist.

I feel that this question should have been addressed before but not sure where to look.

There is a conjecture that the minimum distance behaves as $\frac{1}{n^{\frac{2}{3}}}$ but I am not sure how to show this?

Update See a recently add proof of this statement for the case when 'border' effects are negligible. That is the answer is asymptotic. The question know is how to take into account the border effects?

Thank you very much.

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Boby
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1 Answers1

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An approximation for large $n$.

Let $m=\frac12 n(n-1) \approx \frac12 n^2 $ , and let $j = 1, 2 \cdots m$ index all the pairs of points. Let $S_j(x)$ be the event that the distance between the points is larger than $x$.

Neglecting border effects (reasonable if $n$ is large) we can write $$P(S_j(x)) \approx 1- \frac{x^3}{R^3} \tag{1}$$

Now, the joint event (all pairs are separated by more than $x$) is approximately

$$P(\cap S_j(x)) \approx \prod P(S_j(x)) \approx \left(1-\frac{x^3}{R^3}\right)^m \tag{2}$$

Of course, this is not strictly true, because the events $S_j$ are not independent. But we can expect that for large $n$ this error turn negligible.

Then, letting $t$ be the minimum distance among the points

$$\begin{align} E(t) &\approx \int_0^{R}\left(1-\frac{x^3}{R^3}\right)^m dx \\ &= \, \Gamma\left(\frac43\right) \frac{\Gamma(m+1)}{\Gamma(m+\frac43)} R \\ &\approx \, \Gamma\left(\frac43\right) m^{-1/3} R\\ &\approx \frac{1.12508368}{n^{2/3}}R \tag{3} \end{align} $$

The border effect can be included by computing the probability exactly, as in here (lets assume $R=1$ to save notation, it's just a scale factor), so $(1)$ can be expressed exactly as

$$1-x^3+\frac{9}{16}x^4 -\frac{1}{32}x^6 \tag{4}$$

The integral gets more complicated, but the (first order) asymptotic result $(3)$ is not altered.

leonbloy
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  • Could you give more explanation how you arrive at eq (1). Also where does your definition of $D_1$ comes from? – Boby Nov 14 '16 at 14:16
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    @Boby for eq (1) suppose WLOG (neglecting border effects) that the reference ball is at the origin; then the probability that a new random ball overlaps it (both of diameter $D$) is the volume of the "excluded" space (a ball of radius $D$) divided over the total available volume (the ball of radius $R$) – leonbloy Nov 14 '16 at 14:54
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    @Boby $D_1$ ($D_2$ in the improved version) is the point where the function $g(D,N)$ reaches zero. – leonbloy Nov 14 '16 at 14:56
  • When you say $S_i$ are asymptotically independent you mean when $M$ is large or $R$ is large? – Boby Nov 14 '16 at 17:01
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    @Boby $M$ is large. $R$ is fixed here. Notice that I'm not at all sure if that asymptotical independence assumption is justified. – leonbloy Nov 14 '16 at 17:19
  • Can you also give me an example of what a 'border effect' is? Is it when the ball escapes from the big sphere? – Boby Nov 14 '16 at 18:34
  • Simply that eq $(1)$ or $(5)$ are not valid for points near the border. This is typical for this kind of problems. You just hope that, as $N$ grows (but the dimension and volume are fixed!), the proportion of balls near the border diminishes, and hence you can ignore them. – leonbloy Nov 14 '16 at 19:28
  • Is there a way of writing (5) by including border effect? I want to see how the asymptotic result behaves. For example, can we write $P(S_j)=1-\frac{D^3}{R^3}+\Delta$ where $\Delta$ takes into account border effects? – Boby Nov 15 '16 at 13:34
  • This is getting too long... You should simply replace $v(D)$ for the volumen of the intersection of the small ball with the big ball, over the volume of the big ball. – leonbloy Nov 15 '16 at 14:37
  • Do you mean it should be $P(S_i)= 1-\frac{Vol (B(r) \cap B(R) )}{B(R)}$?? Is this correct? – Boby Nov 15 '16 at 14:53
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    In (5), instead of $D^3/R^3$ use this http://math.stackexchange.com/a/1413112/312 – leonbloy Nov 15 '16 at 16:03
  • Final question, about the computation of (5) ? If the small balls have diameter $D$ and the big ball has radius $R$ should the answer be $P(S_j)=1-\frac{D^3}{2^3R^3}$ instead of $P(S_j)=1-\frac{D^3}{R^3}$? Am I missing something? – Boby Nov 16 '16 at 21:16
  • See a similar question that I asked here: http://math.stackexchange.com/questions/2018453/average-minimum-distance-between-npoints-generate-i-i-d-uniformly-on-the-shel Maybe you will be interested. – Boby Nov 17 '16 at 14:09
  • Any idea how to solve this? https://math.stackexchange.com/questions/2250734/uniformly-distributed-points-distance-question – Susan_Math123 Apr 25 '17 at 03:07