distribution of n point on the unit circle
- From: Ralf Goertz <r_goertz@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 30 Dec 2006 21:24:51 +0100
Hi,
I've been working on this problem for quite a while now and although it
didn't seem to be too difficult, I got stuck.
Consider n points distributed uniformly and independently on the unit
circle. What is the probability density function or the distribution
function for the smallest angle alpha such that all points are on a
sector of the unit circle with that opening angle.
For n=2 this is easy. alpha cannot be bigger than pi. Therefore
p.d.f.(alpha)=1/pi and d.f.(alpha)=alpha/pi.
In general, alpha must be smaller than or equal to 2pi*(n-1)/n. Using
Monte Carlo simulations I found out that for n=3 the p.d.f. is piecewise
linear with
3x/(2pi^2) for 0<=x<=pi
-9x/(2pi^2)+6/pi for pi<x<=4*pi/3
but I have no clue how to derive this analytically let alone how to
extend it to arbitrary n. Does anyone have an idea?
Ralf
.
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