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Suppose we have $n$ iid random variables, $U_1,...,U_n\sim\text{Unif}(0,1)$. What is the expected minimum distance between any two of these random points?

Now, I know this has already been asked and answered here (see for example this link on MO, this link on MSE, or this link on stats.SE), and the result is $\frac1{n^2-1}=\frac1{n+1}\frac1{n-1}$. I'm wondering if there is a way to derive this by using symmetry.

In particular, we can let $U_0:=0$ and $U_{n+1}:=1$, and set $X_j=U_{(j+1)}-U_{(j)}$ for $0\leq j \leq n$. We are interested in $E \min\limits_{1\leq j<n}X_j$, and all the $X_j$ are identically $\text{Beta}(1,n)$ distributed, but they are not independent (if one is large, the others must be small). Yet, $(X_0,...,X_n)$ is exchangeable and $\sum_{j=0}^n X_j \equiv 1$, so $E X_j \equiv \frac1{n+1}$, for example. Similarly, if we set $I_j$ as the indicator of the event that $X_j=\min\limits_{1\leq k<n}X_{k}$, then again by symmetry we find $E I_j \equiv \frac1{n-1}$.

Knowing the answer already, we can write $E (\min\limits_{1\leq j<n}X_j) = E(X_1)E(I_1)$, which is why I wonder if there is a probabilistic interpretation of this equality. That is, a way to prove this directly (using e.g. symmetry of some sort), and so to deduce the result above.

I tried for a while, but couldn't figure it out. Is there something here? Or, is there a convincing argument why proving it this way shouldn't work out?

aaaaaaaaaaaaaaaaaanon
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1 Answers1

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I don't think there is an answer in terms of symmetry, per se. But I can explain the proof given by MikeEarnest in an intuitive way that I think may help explain roughly why the whole "product of expectations" thing works out, even though it shouldn't in general.

A simpler question is to find the expected length of the shortest segment of a circle, after it has been subdivided by $n$ randomly placed cuts. This value is $1/n^2$, as a fraction of the circle's circumference. Again, one can "compute" this by noting that the expected size of each particular segment is $1/n$, and the chance that a given segment is the shortest is also $1/n$.

But this is not how probability theory works, and for an actual proof, we need a different technique. It turns out to be nice to think about transforming our space of outcomes in some way.

Let's say we cut the circle and find that segment $j$ is the minimum, and has size $l_\text{min}$. Well, we can subtract $l_\text{min}$ from all the other segments, and give all that length to segment $j$, so its new size is $nl_\text{min}$. Since $l_\text{min}$ was the shortest length, all the other segments still have non-negative length. And since the minimum segment can have length at most $1/n$, the new length satisfies $nl_\text{min} \leq 1$. Such a transformation is invertible. We can start from a completely random division of the circle. We divide segment $j$ into $n$ equal pieces. It keeps one of them, while the rest are given to the other segments. Segment $j$ is now guaranteed to be the minimum, since its length $l_\text{min}$ was added to all the other segments. If it's clearer in equations:

$$ f_j(l_1, l_2, \dots l_n) = (l_1 - l_j, l_2 - l_j \dots nl_j \dots l_{n-1} - l_j, l_n - l_j) $$

$$ f_j^{-1}(l_1, l_2, \dots l_n) = (l_1 + \frac{l_j}{n}, l_2 + \frac{l_j}{n} \dots \frac{l_j}{n} \dots l_{n-1} + \frac{l_j}{n}, l_n + \frac{l_j}{n}) $$

So far, I've described a transformation $f_j$ and its inverse. If $C$ is the set of all possible cuts of the circle, and $J_j$ is the subset of cuts where the shortest segment is segment $j$, then we have:

$$ f_j: J_j \hookrightarrow C $$

$$ f_j^{-1}: C \hookrightarrow J_j $$

From those given facts, we now actually know that $f_j$ is a bijection between these sets. (also, $f_j$ is linear, so it doesn't do anything bad to the measure)

$$ f_j: J_j \leftrightarrow C $$

So we can sample from $C$ according to the original distribution by picking a random $j$ to be the minimum, sampling $c\in C$, and then taking our output to be $f^{-1}_j(c)$. The virtue of this is that we now actually know that segment $j$ is definitely the minimal one. So we can compute: $\langle l_\text{min}\rangle$ as follows:

  • When sampling $c$, we have $\langle l_\text{j} \rangle = \frac{1}{n}$, just like any other segment.
  • After applying $f^{-1}$, we have $l_\text{min} = \frac{1}{n}l_j$
  • So $\langle l_\text{min}\rangle = \frac{1}{n^2}$

The first factor of $1/n$ comes from the expectation of the length of a segment, just as you say. The second doesn't come directly from the probability of a segment being minimal, though. It comes from the fact that our transformation $f^{-1}$ splits length between $n$ different segments.

Generalization to your particular problem should be possible now. Now the expected segment length is $1/(n+1)$, while a transformation analogous to $f^{-1}$ splits one segment into $n-1$ pieces.

pb1729
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