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Two points, $A$ and $B$, are independently and uniformly selected from a unit circle.

What is the distribution of the shortest arc length between them?

The answer is supposed to be that the shortest arc length is distributed as a uniform distribution, i.e., ${\cal U}[0, \pi]$, which seems intuitive to me, but I can't show why this should be the case.

Any thoughts would be greatly appreciated.

Thanks in advance!

(BTW, this is my first time posting, so forgive me if I've violated any etiquette.)

David G. Stork
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user418749
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1 Answers1

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Here's one way of thinking about it. Once $A$ and $B$ are put on the circle, rotate your head so that $A$ is positioned at the top of the circle. Now $B$ is clearly uniformly distributed anywhere along the circle, so the result pops right out!

Now, you don't actually need to rotate your head, as David G. Stork says in the comments. The key idea is to think about $B-A$ as a uniformly chosen independent variable of its own.

fractal1729
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  • That seems reasonable. But then it feels like we're conditioning on a fixed value b.

    Would a formal way of finding it be looking at P(|A-b|<d | B = b)*P(B=b). Then proceeding from there?

     – user418749 May 18 '17 at 01:04
  • No need to rotate the circle so that $A$ is positioned at the top. Simply measure distance (angle) from $A$'s position. – David G. Stork May 18 '17 at 01:06
  • @DavidG.Stork you are correct. I included the rotate the circle part as intuition that may help the problem-solver focus on the difference between the points as a single independent variable of its own. To the OP, this is the key idea. You could formalize it in the way your comment says but it's even easier to simply focus on $B-A$ and recognize that it is its own random variable. – fractal1729 May 18 '17 at 01:09
  • fractal1729 and @DavidG.Stork may you please help out in these related questions: More random points on a circle ? – BCLC Mar 15 '21 at 10:55