Let $U$ be distributed uniformly on $[0,a]$. Now suppose we generate $n$ independent points according to $U$.

What is the average minimum distance between these $n$ points? That is \begin{align} E\left[ \min_{i,j\in [1,n]: i\neq j} |U_i-U_j|\right] \end{align}

Is this a correct formulation of average minimum distance between?


1 Answer 1


Assume $a=1$ for simplicity. Let $M$ be the minimum distance among these $n$ points. Note $M$ is always at most $\frac1{n-1}$, which occurs when the points are evenly spaced. To calculate $EM$, we first calculate $P(M>m)$, for $0\le m \le 1/(n-1)$.

By symmetry, $P(M>m)=n!\times P(M>m\text{ and } U_1<U_2<\dots<U_n).$ And the latter is equal to the volume of the below subset $S$ of the unit hypercube $Q=[0,1]^n$: $$ S = \{(x_1,\dots,x_n)\in Q:x_i+m\le x_{i+1}\text{ for }i=1,\dots,n-1\} $$ Now, consider the transformation $f:S\to [0,1]^n$, given by $$ (x_1,\dots,x_n)\mapsto (x_1,x_2-m,x_3-2m,\dots,x_n-(n-1)m) $$ A little thought shows that $f$ is a volume-preserving bijection from $S$ to the region $S'$ of the smaller hypercube $Q'=[0,1-(n-1)m]^n$ given by $$ S'=\{(y_1,\dots,y_n)\in Q':y_i\le y_{i+1}\text{ for }i=1,\dots,n-1\} $$ By symmetry of permuting the $y_i$, we have $\text{Vol}(S')=1/n! \ \times\text{Vol}(Q')$. But $\text{Vol}(Q')=(1-(n-1)m)^n$. Putting this all together, $$ \begin{align} P(M>m)&=n!\cdot \text{Vol}(S)\\ &=n!\cdot \frac1{n!}(1-(n-1)m)^n\\ &=(1-(n-1)m)^n \end{align} $$ and therefore $$ \begin{align} EM &=\int_0^{1/(n-1)}P(M>m)\,dm\\ &=\int_0^{1/(n-1)}(1-(n-1)m)^n\,dm\\ &=\left.\frac{1}{n-1}\frac{(1-(n-1)m)^{n+1}}{n+1}\right|_{0}^{1/(n-1)}\\ &=\boxed{\frac{1}{n^2-1}} \end{align} $$ To get the answer for the interval $[0,a]$, simply multiply this result by $a$.

  • $\begingroup$ Do you think your proof can be adapted to the case when we put $n$ points uniformly on shall of the two-dimensional ball? $\endgroup$
    – Boby
    Nov 7, 2016 at 16:46
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    $\begingroup$ @Boby Probably not. This question might be related. $\endgroup$ Nov 7, 2016 at 18:29
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    $\begingroup$ Why is $E[M]=\int_0^{1/(n-1)} P[M>m]dm$ in the first place? I don't see how to obtain this expression from the usual expression of the expected value of a function of random variables. $\endgroup$ Nov 18, 2016 at 12:10
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    $\begingroup$ @LoveTooNap29 See "continuous distribution taking nonnegative values" in en.m.wikipedia.org/wiki/… $\endgroup$ Nov 18, 2016 at 15:57
  • $\begingroup$ @MikeEarnest How do you come up with this idea. In which course i can learn proof like this. $\endgroup$ Apr 27, 2020 at 3:49

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