Most random points on a circle

Question

Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $M$ be the minimum distance between any 2 of the points.

Find the pdf of $M$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_M(m)= P(M \le m)$ 1st and then hope the cdf is absolutely continuous.)

My model: The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

Question: Well, I hope to find the pdf of $M$ via its cdf, which I think I'm able to compute if I know what $M$ is. What is $M$?

I think $M=\min\{X,Y,1-Y,1-X$ $,|X-Y|,1-|X-Y|\}$, where $X$ and $Y$ are the anti-clockwise distances from $B$ to, resp, $A$ and $C$. Somehow $X$ and $Y$ are iid Unif(0,1).

How would I compute $M$'s distribution?

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These questions are all related, but I hope I made each self-contained

• Random points on a circle
• More random points on a circle
• Most random points on a circle
• Modelling random points on a circle
• Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

Answer 1

Three points on a circle divide it in three arcs. One from the arcs is the shortest one. Let its length be $m$. Where can we place the third point? It cannot be closer than $m$ to any of both points. It means it cannot be on the shorter arc connecting the points, and it cannot be on the parts of the larger arc which are closer than $m$ to one of the two first points. Hence in total the arc of the length $3m$ is not accessible for the third point and there remains only the arc of $1-3m$ for it. Therefore the pdf for the shortest distance is: $$\rho(m)= \begin{cases} a(1-3m),&m<\frac13\\ 0,&m>\frac13 \end{cases} $$ The simplest way to compute the constant $a$ is to use the condition: $$ \int_0^1\rho(m)\,dm=1 $$ which results in the value $a=6$.

Answer 2

If the minimum distance is at least $x$, then the three points, encountered in clockwise order and with the first taken to be at $0$, are at positions $(0, y, y+z)$, where $y \ge x$, $z \ge x$, and $y+z\le 1-x$. The volume of this in configuration space is $$ V(x)=\int_{y=x}^{1-2x}dy\int_{z=x}^{1-x-y}dz=\int_{y=x}^{1-2x}(1-2x-y)dy=(1-2x)y-\frac{1}{2}y^2\bigg\vert_{x}^{1-2x}=(1-2x)^2-\frac{1}{2}(1-2x)^2-(1-2x)x+\frac{1}{2}x^2=\frac{1}{2}(1-2x)^2-(1-2x)x+\frac{1}{2}x^2=\frac{1}{2}(1-3x)^2. $$ There are two equally probable cases, corresponding to the orderings $ABC$ and $ACB$, so the full probability that the minimum distance is at least $x$ is $(1-3x)^2$; the probability that it is $\le x$ (the cdf of $M$) is $1-(1-3x)^2$; and the pdf of $M$ is the derivative of this, which is $$p(x)=6(1-3x)^2.$$

Answer 3

I'm not exactly sure what is $X, Y$. But i assume it's $BA$, $BC$ and $1-X$ is $AB$. $1-Y$ is $CB$, similarly, $|X-Y|$ is $AC$, $1-|X-Y|$ is $CA$.

Given those, the goal is to find a circle with radius $r$ and three points $A, B, C$ on the circle, the minimum interval length is the random variable $M$.

using the same arc length idea. we can simplify this minimum interval problem into the minimum angle.

First, it should be easy to prove. the minimum can't be the sum of two intervals. Namely $M \neq BC+CA, CA+AB, AB+BC$, since some of the $1-?$ is the sum of two other intervals, E.g $1-BA = AC+CB$

We can simplify $M = r*min(\alpha, \beta, \gamma)$, Since $r$ is a constant, we can remove it for easy of analysis.

Also, as $\alpha, \beta, \gamma \sim U(0, 2\pi)$.

$P(min(\alpha, \beta, \gamma) \geq z) = P(\alpha \geq z)P(\beta \geq z)P(\gamma \geq z) = (2\pi-z)^3$

The CDF would be $P(min(\alpha, \beta, \gamma) \leq z) = 1 - P(min(\alpha, \beta, \gamma) \geq z)$