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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?

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

BCLC
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    You can normalize so that $A$ is fixed. Then using the metric bijection with the unit interval (possibly with a factor of $2\pi$ somewhere), $B$ and $C$ are independently sampled uniformly from $[-\frac12,\frac12]$. The minimum distance in question is then $\min{|B|,|C|,|B-C|,|B-C-1|,|B-C+1|}$, which won't be hard to integrate over $[-\frac12,\frac12]^2$. (One can simplify this a little more even by assuming $B$ is positive by symmetry.) – Greg Martin Mar 12 '21 at 07:23
  • Thanks @GregMartin may I just let $B,C$ be iid unif(0,1)? (or unif[0,1] i guess) – BCLC Mar 12 '21 at 07:26
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    You can apply the same idea as in previous cases to obtain the answer $6(1-3m)$ for $m<1/3$ and $0$ otherwise. – user Mar 12 '21 at 08:24
  • @user Thanks user. You can post as answer. But 1 - Where did you get the 3 in 3m and the factor of 6? 2 - with the substitution $v=3m$, why isn't this just the same thing as in part 1 (Random points on a circle)? (and even part 2 (More random points on a circle) ) – BCLC Mar 12 '21 at 10:39
  • @GregMartin is it that $M=\min{B-\min{A,C},|C-A|-(B-\min{A,C}),|C-A|}$ ? I ask about this here: Modelling random points on a circle – BCLC Mar 14 '21 at 08:25
  • I should have said explicitly that $A$ is fixed at $0$. – Greg Martin Mar 14 '21 at 08:39
  • @GregMartin sooo $M = \min {B,|C|-B,|C|}$ ? – BCLC Mar 14 '21 at 08:42
  • Not quite: you don't want to take $|C|-B$ because of examples like $B=\frac14$, $C=-\frac14$; and you need to take $|B-C-1|$ because of examples like $B=0.49$, $C=-0.49$. – Greg Martin Mar 14 '21 at 08:46
  • @GregMartin A,B,C are supposed to be in $(0,1)$...? well still i guess it's possible $|C|-B$ could be negative in which case...oh hell ummm....perhaps change $|C|-B$ to $\max{0,|C|-B}$ ? Or really has to be $|B-C-1|$? – BCLC Mar 14 '21 at 08:49

3 Answers3

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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 $a=6$.

BCLC
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user
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  • Thanks user, but I don't quite find this ansatz to be, like, legit or something. I understand the 6 cases: it's 2 cases of A<B<C, C<B<A. and then in each case there are 3 choices fo minimum. And then all 6 cases are symmetric because iid. And then in each case the probability is given by $1-3m$ for $m < \frac13$. How are you able to say that the pdf is some constant times $1-3m$? (ruling out the aforementioned cases thing) – BCLC Mar 14 '21 at 07:07
  • I can do this because the integral have to be equal to 1 (it is the probability that one of the three distances is the shortest one). – user Mar 14 '21 at 07:55
  • 1 - you know what actually, thanks, but can i just do the cases thing instead? i'm hoping to just avoid this ansatz – BCLC Mar 14 '21 at 08:02
  • 2 - Btw is it that $M= \min{Z,|C-A|-Z,|C-A|}$, where $Z$ is the one from the previous question? I ask here: Modelling random points on a circle – BCLC Mar 14 '21 at 08:04
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    You can do the cases instead. – user Mar 14 '21 at 08:04
  • Right thanks. as for (2)..? – BCLC Mar 14 '21 at 08:05
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    I would not try to reduce the problem to the previous one. In my view it would make it more complicated. – user Mar 14 '21 at 08:07
  • ah thanks. i just thought that was kind of the point of the sequence of the questions. anyway, can it be that $M=\min{B-\min{A,C},|C-A|-(B-\min{A,C}),|C-A|}$ ? – BCLC Mar 14 '21 at 08:14
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    To split this into cases, in the case where $A,B$ are the closest pair you have distance $m \leq \frac13$ between $A$ and $B$ while $C$ is somewhere on an arc of length $1-3m$ within the longer arc between $A$ and $B.$ The distribution of points is still uniform. The other cases are the same except for the names of the points. This gives us a distribution function on $m$ that is proportional to $\int_0^m (1-3m),dm,$ and a density proportional to $1-3m.$ I agree with this answer. – David K Mar 14 '21 at 17:00
  • Thanks @DavidK do we perhaps have $M=\min{X,1-X,Y,1-Y,g(X),h(Y)}$ where $X$ and $Y$ are from your answer here, for some $g$ and $h$? This would be much like how $W=\min{X,1-X,Y,1-Y}$. – BCLC Mar 15 '21 at 10:31
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    @JohnSmithKyon I think what you're looking for is $M=\min{X,1-X,Y,1-Y,|X-Y|,1-|X-Y|}.$ I don't think you can do it with just functions of $X$ and $Y$ individually. This seems to be a difficult way to approach the problem. – David K Mar 15 '21 at 12:59
  • @DavidK Oh nice thanks a lot!!! So $P(M > m)$ = area of the region ${X>m,1-X>m,Y>m,1-Y>m, |X-Y| > m, 1 - |X-Y| > m}$ ? – BCLC Mar 17 '21 at 06:10
  • @DavidK I asked about distribution of $M$ here – BCLC Mar 17 '21 at 07:07
  • @DavidK re $M$, i posted answer. please let me know what you think. – BCLC Mar 17 '21 at 23:30
  • @DavidK btw you can post that as an answer – BCLC Mar 17 '21 at 23:32
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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 $$p(x)=6(1-3x)^2.$$

mjqxxxx
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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 $M$.

all the intervals

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

peng yu
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