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Consider 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 $B_{circ}$.

  • Let $Z$ be the random variable denoting the distance of the point $B_{circ}$ to its neighbour in the anti-clockwise direction.
  • Let $W$ be the minimum distance of the point B to its nearest neighbour, either clockwise or anti-clockwise, whichever is smaller.
  • Let $M$ be the minimum distance between any 2 of the points.

Is the following model correct?

  1. Let the other points be $A_{circ}$ and $C_{circ}$. They are all necessarily distinct (almost surely, but there's no measure theory here) because the probability that any 2 or all 3 of them are the same is zero. Map them all into unit half-open interval $[0,1)$, which is bijective with the circle, and call them respectively $B,A,C$. These are iid uniform(0,1) random variables, much like how the original $B_{circ}$, $A_{circ}$ and $C_{circ}$ are iid uniform(over the circle) random variables.

  2. In terms of the order of $A,B,C$, there are exactly $2=2!=(3-1)!$ cases here because of circular permutation. The cases are $A<B<C$ and $C<B<A$.

  3. These cases may also be respectively described as $B-A > 0$ and $B-C > 0$.

  4. $Z=(B-A) 1_{B-A > 0} + (B-C) 1_{B-C>0}$

    • 4.1. Alternatively, $Z=\max\{B-A,B-C\}=B+\max\{-A,-C\}=B-\min\{A,C\}$. Basically, the idea behind '$\max\{B-A,B-C\}$' is that exactly 1 of these quantities is going to be positive while the other is negative.

  5. $W=\min\{Z,U\}$, where $U$ is similar to $Z$: $U=\max\{C,A\}-B=\max\{C-B,A-B\}$.

    • 5.1. We have that $Z+U=|C-A|$ and then $W=\min\{B-\min\{A,C\},|C-A|-(B-\min\{A,C\})\} = \min\{Z,|C-A|-Z\}$.

  6. $M=\min\{W,|C-A|\} = \min\{Z,|C-A|-Z,|C-A|\} = \min\{B-\min\{A,C\},|C-A|-(B-\min\{A,C\}),|C-A|\}$

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

BCLC
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1 Answers1

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I see a fundamental problem with this approach.

It is true that there are really only two sequences of three points on the circle: $A,B,C$ are encountered in that sequence either clockwise or counterclockwise. But when you assign numerical coordinates to the three points via the bijection between the circle and $[0,1)$, and then start dealing with arithmetic operations and relations among the three numbers, you change the topology of the problem.

Here is an example. In one case you could have $A = 0.4,$ $B = 0.6.$ In another case you could have $A = 0.9,$ $B = 0.1.$ The two cases are isometric on the circle: $B$ is $0.2$ units counterclockwise from $A$. But numerically they are very different: $B - A = 0.2$ and $A < B$ in one case, $B - A = -0.8$ and $B < A$ in the other case.

Perhaps the most obvious contradiction is, if $A,B,C$ truly are iid uniform on $[0,1)$, then there is a non-zero probability that $A < C < B,$ and likewise for three other cases that you have ruled out in the question.

It is possible to exclude the other four cases from consideration, but to do so you have to give up the iid assumption and restrict the support of $(A,B,C)$ to a subset of the cube $[0,1)^3.$ The distribution of $(A,B,C)$ is then uniform on that subset.

Then to compute probabilities correctly you have to understand the shape of that subset of the cube. I think this makes the problem much harder than it needs to be.

David K
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