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Let $U_1, U_2, \ldots, U_n$ be a set of random variable uniformly distributed over some box in $\mathbb{R}^3$ and let \begin{equation} R = \frac{1}{2 n(n-1)}\sum\limits_{i,j< i} |U_i - U_j| \end{equation} be the random variable corresponding to the average distance between the random points. What is the distribution of $R$? If no answer to that is available, then what is $E(R)$?

I have found somewhat similar questions, but couldn't related them to this one.

Thank you very much in advance.


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E(R) is rather clumsy to compute:… – leonbloy Jan 14 '13 at 15:30
up vote 2 down vote accepted

I take it that the $U_i$ are independent. As $\binom{n}{2}=n(n-1)/2$ I assume that you mean to investigate $$ R=\frac{2}{n(n-1)}\sum_{j<i}|U_i-U_j|. $$ The distribution of $R$ is not any of the well-known distributions. By linearity, $E[R]=E[|U_i-U_j|]$ which you can find by integration over $\text{box}\times\text{box}$, i.e. $$ E[R]=\frac{1}{\operatorname{vol}(\text{box})^2}\int_{\text{box}\times\text{box}}|x-y|\,d^3x\,d^3y. $$

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I am still confused about where the concentration of points enter. I mean, if $n$ is large, than the average distance between them should be smaller, but I don't see how that enters the calculation. Anyway, thank you for the help. – Gabriel Landi Jan 14 '13 at 15:30
"I mean, if n is large, than the average distance between them should be smaller" No. YOu are perhaps thinking of the minimum distance -the average distance to the nearest neighbour- which is another thing. – leonbloy Jan 14 '13 at 15:32
Oh! Yeah, I think that is it. The nearest-neighbour distance. I got everything confused. – Gabriel Landi Jan 14 '13 at 15:33

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