Suppose I have $N$ points on a unit circle. I am aware that for a uniform distribution the average angle of one point is not defined, but I want to calculate the average arc length between nearest-neighbors. I disregard arc lengths larger than $\pi$.

For $N = 2$ I am formulating the problem as: \begin{equation} \langle \Delta x \rangle = \frac 1 {\pi^2}\int\limits_0^\pi \int\limits_0^\pi dx_1 dx_2 \, |x_1-x_2| = \frac \pi 3. \end{equation}

This answer does not make sense to me. Did I make a mistake constructing this integral?

What would happen to $N = 3$, etc?


1 Answer 1


I do not completely understand the integral that you provided. Let the point $O$ is placed somewhere on the sircle, and then $X_1,X_2\in [0,2\pi]$ are chosen independently and uniformly at random on the circle. There are two arcs arises: one containes $O$ and the other does not. You want to find expected value of minimum lengths of these two arcs?

To compute this value, it is better to reduce the problem to uniform samples on $[0,2\pi]$. Suppose we select on a circle $N$ points. If we take the first selected point as zero and open the circle into an interval, the rest randomly selected $N-1$ points will divide the interval into $N$ parts. The joint distribution of this $N$ parts is the same as joint distribution of $N$ uniform spacings (distances between order statistics of uniform distribution). We can deal with interval $[0,1]$ instead of $[0,2\pi]$ and scale results afterwards.

Let $X_1,\ldots,X_{N-1}$ are independent r.v.'s from $U(0,1)$ distribution. Let $$X_{1:N-1}\leq X_{2:N-1}\leq\ldots \leq X_{N-1:N-1}$$ are order statistics. Define spacings $$ Y_{1:N-1}=X_{1:N-1} - 0,\ Y_{2:N-1}=X_{2:N-1}-X_{1:N-1}, \ldots, \ Y_{N:N-1}=1-X_{N-1:N-1}. $$ Let $$Z_{1:N-1}=\min\{Y_{1:N-1}, Y_{2:N-1}, \ldots, Y_{N:N-1}\}$$ is the smallest distance between randomly chosen $N-1$ poins (including distances to endpoints of the interval). This is the same as the length of arc between nearest neighbors among $N$ points chosen randomly on the circle with unit circumference.

The distribution of $Z_{1:N-1}$ is well-known: $$ \mathbb P(Z_{1:N-1} > x) = (1-Nx)^{N-1}, 0\leq x\leq \frac1N. $$ Look, e.g., Limit results for ordered uniform spacings by Ismihan Bairamov, Alexandre Berred & Alexei Stepanov, Statistical Papers volume 51, Article number: 227 (2008). This is formula (5) at page 230.

From this distribution you can get expected values directly: $$ \mathbb E[Z_{1:N-1}] = \int_0^\frac1N (1-Nx)^{N-1} \,dx = \frac1{N^2}. $$
So, for $N=2$, $\mathbb E[Z_{1:1}]=\frac14$, for $N=3$, $\mathbb E[Z_{1:2}]=\frac19$ and so on.

Return to the circle, multiply the results by $2\pi$. So for two randomly chosen points, mean length of the shortest arc is $\frac{\pi}{2}$, for three points $\frac{2\pi}{9}$ and so on. For $N$ points it is $\frac{2\pi}{N^2}$.


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