Let $P\_1, P\_2, ..., P\_n$ be points randomly placed on a unit circle from a uniform distribution. Consider the product $D$ of all pairwise distances:


$D=\displaystyle \prod\_{1\leq i < j \leq n} \overline{P\_iP\_j}$ 


I wonder...


1) What is the probability density function for $D$? What is the expected value of $D$?


2) When the $P\_i$ are equally spaced, we know $D=n^{n/2}$. Is this an absolute maximum for $D$?



Product of random diagonals on the unit circle
To 2, the answer is yes. This is easy. Re-enumerate points according to their cyclic order. For every fixed $k$ ($1\le k<n$), the product
$$
 D\_k:= \prod\_i |P\_i P\_{i+k}|
$$
(where the indices are taken modulo $n$), equals
$$
 D\_k=\prod\_i 2\sin (t\_i/2) = 2^n\exp \left(\sum\_i \log\sin(t\_i/2) \right)
$$
where $t\_i$ is the (oriented) angle from $P\_i$ to $P\_{i+k}$. Since $t\_i\in(0,2\pi)$, $\sum t\_i=2\pi k$, and the function $t\mapsto \log\sin(t/2)$ is concave on $(0,2\pi)$, by Jensen's inequality the maximum is attained when all $t\_i$ are equal, i.e., when the points are equally spaced. Since $D=\prod\_k D\_k^{1/2}$, the result follows.


