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.
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.