# Average minimum distance between $n$ points generate i.i.d. uniformly in the ball

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.

• Is it uniform if you use polar coordinates? I think you need to use Euclidean coordinates. – Sungjin Kim Nov 8 '16 at 22:37
• @i707107 I am not sure. Why do you say it matters? – Boby Nov 8 '16 at 22:46
• Because in polar coordinates, you are $7$ times more likely to have $\frac{R}{2} \lt r \lt R$ than $0 \lt r \lt \frac{R}{2}$ – Henry Nov 8 '16 at 22:50
• @Henry Done. See now. – Boby Nov 8 '16 at 22:54
• I guess it behaves as $n^{-2/3}$ ($n^2$ pairs of points, so the probability that it is of order $\varepsilon$ is roughly $n^2 \varepsilon^3$. Tune $\varepsilon$ so that the result has unit size). – D. Thomine Nov 8 '16 at 23:22

Here's an asymptotic bound - hopefully it's tight.

We throw $N$ little balls of diameter $D$ randomly (uniformly) inside the big sphere of radius $R$.

UPDATE: The new version (lower half) is better than what follows.

Neglecting border effects (reasonable if $N$ is large) the probability that the ball $i$ is "free" (no other overlaps with it) is

$$P(F_i)=\left(1-v(D)\right)^{N-1} \tag{1}$$

where $v(D) \triangleq D^3/R^3$

The probability that all balls are free can be bounded as

$$P(\cap F_i)=1- P(\cup F_i^c)\ge 1 - N(1-P(F_i)) \triangleq g(D,N) \tag{2}$$

For large $N$

$$g(D,N) \approx 1 - N^2 \, v(D) \tag{3}$$

in the range where this is positive, ie. $0\le D \le D_1 \triangleq R/N^{2/3}$

Now, let $t$ be the minimun distance between the sphere centers. Then

$$P(t \ge D) = P(\cap F_i) \ge g(D,N) \tag{4}$$

And then

$$E(t) = \int_0^{\infty} P(t \ge D) dD \ge\int_0^{D_1} g(D,N) \, dD \approx D_1- N^2 \frac{ D_1^4}{4 R^3} = \frac{3 }{ 4 } \frac{R}{N^{2/3}}$$

(Simulation data suggests that the order is right, and so is the bound, but the real coefficient is around $1.12$ - perhaps $9/8$)

Update: (Improved version)

A better approach can be obtained by considering instead of $F_i$ (free ball) the event $S_j\equiv$ "separated pair" (pair of balls are separated, they do not overlap) where $j$ indexes the $M=N(N-1)/2 \approx N^2/2$ pairs.

By the same reasoning:

$$P(S_j)=1-v(D) =1 - \frac{D^3}{R^3} \tag{5}$$

$$P(\cap S_i) \ge \max(1 - M(1-P(S_i)),0)= \max\left(1 - M \frac{D^3}{R^3},0\right) \triangleq h(D,M) \tag{6}$$

The range where $h(D,M)$ is positive, ie. $0\le D \le D_2 \triangleq R/M^{1/3}$

Now, let $t$ be the minimun distance between the sphere centers. Then

$$P(t \ge D) = P(\cap S_i) \ge h(D,M) \tag{7}$$

And then

$$E(t) = \int_0^{\infty} P(t \ge D) dD \ge\int_0^{D_2} h(D,M) \, dD =\\ = \frac{3}{4} \frac{R}{M^{1/3}} \approx 0.945 \frac{R}{\sqrt[3]{N(N-1)}} \approx 0.945 \frac{R}{N^{2/3}} \tag{8}$$

Update 2 : A simple heuristic which seems to produce the correct coefficient:

Following the approach above, we could assume that $S_i$ are asympotically independent, and then:

$$P(\cap S_i) \approx \left(1-\frac{D^3}{R^3}\right)^M \tag{9}$$

Then $$E(t) \approx \int_0^{R}\left(1-\frac{D^3}{R^3}\right)^M dD =\\= R \, \Gamma(4/3) \frac{\Gamma(M+1)}{\Gamma(M+4/3)} \approx R \, \Gamma(4/3) M^{-1/3} \approx 1.12508368 \frac{R}{N^{2/3}} \tag{10}$$

Update 3 : Regarding corrections for border effects.

(Lets assume $R=1$ to save notation, it's just a scale factor)

If we wished to include border effects we should replace $(5)$ (computing the balls intersection as here) by

$$1-D^3+\frac{9}{16}D^4 -\frac{1}{32}D^6 \hspace{1cm} 0\le D\le 2$$

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

Lemma: For any positive differentiable function $g(x)$ which , in $[0,+\infty)$, has global maximum at $g(0)=1$, and which has zero first and second derivates $g(x)=1-a x^3 + O(x^4)$ we have (variation of Laplace method, see eg here sec 2.1.3)

$$\int_0^\infty g(x)^M dx = \frac{\Gamma(1/3)}{3 a^{1/3}} M^{-1/3}+ o(M^{-1/3})$$

which again leads as to $(10)$.

• 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
• @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
• @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
• @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
• In (5), instead of $D^3/R^3$ use this math.stackexchange.com/a/1413112/312 – leonbloy Nov 15 '16 at 16:03