Minimum 1st-neghbors distance between N random points on a ring We have $N$ points randomly and uniformly distributed on a ring of length 1.
Let $d_i$ be the distance between point $i$ and its first neighbor. 
We want to know the expected value of the smallest $d_i$, $E([d_i]_{min})$,
being specially interested in the case of a quite small value of $N$ ($N<10$).
From numerical simulation it looks like $E([d_i]_{min}^{(N)})=N^{-2}$.
I faced problems in getting rid of correlations. As a first step, I tried to consider neighbors on the right side only, looking for the minimum of $N$ quantities $d_i^{(R)}$ that I suppose to be uncorrelated, but this is not a good approximation for small $N$.
I would be grateful for any help. Thank you
Luce
 A: You can generate a point-set with the correct joint distribution of gap-sizes by running a rate 1 Poisson point process on the positive real numbers until you have $N$ events. Then wrap around to get a circle by identifying the time of the $N$th event with the origin. 
By standard properties of the Poisson process, conditioning on the time of the $N$th point, the first $N-1$ points (forgetting the ordering) are distributed like $N-1$ independent uniform points in the interval from zero to the time of the $N$th point. Therefore, if we rescale the circle to perimeter 1, this is exactly what you want.
The smallest gap is the smallest of $N$ independent standard exponential variables, which has expectation $1/N$. And then there is a factor $1/N$ from the rescaling. So your guess from simulations is correct! 
A: PYKE, R. (1965). Spacings. J. R. Stat. Soc. Ser. B Stat. Methodol. 27 395–449.
A: The problem is related to the following. Assume you have $K$ i.i.d. Uniform$[0,1]$ points put in the increasing order: $X_1,X_2,.., X_{K-1}$. Let $X_0=0$ and $X_K=1$. Define $D=D_K$ as the minimum distance amongst these $K+1$ points. What is the distribution of $D$?
To answer this, Let $0<Y_1<Y_2<...<Y_K$ be the increasing sequence of points of a Poisson process with rate 1. Let $W_1=Y_1, W_2=Y_2-Y_1, ..., W_K=Y_K-Y_{K-1}$ be the inter-arrival distances; hence they are i.i.d. $Exp(1)$ distributed.
Then $[0,Y_1/Y_{K+1},Y_2/Y_{K+1},...,Y_{K-1}/Y_K,1]$ has the same distribution as $[X_0,X_1,...,X_K]$. For any $a>0$, you can get that the joint probability ${\mathbb P}(\min_i W_i>a,W_1+...+W_K \in [L,L+\delta]) = \delta  \frac{(L-Ka)^{K-1} e^{-L} }{(K-1)!}$ for infinitesimally small $\delta>0$.
This will imply that ${\mathbb P}(\min_i W_i/L >v, W_1+...+W_K \in [L,L+\delta]) = \delta  \frac{L^{K-1}(1-Kv)^{K-1} e^{-L} }{(K-1)!}  $.
Since $W_1+...+W_K$ has a Gamma(K,1) distribution with density $\frac{L^{K-1} e^{-L} }{(K-1)!}$, we get that the conditional distribution
${\mathbb P}(\min_i W_i/L >v | W_1+...+W_K=L) = (1-Kv)^{K-1}$
and thus ${\mathbb P}(D<v)=1-(1-Kv)^{K-1}$
and ${\mathbb E}(D)=\int_0^{1/K} {\mathbb P}(D>v) dv=\frac{1}{K^2}$.
