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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 $Z$ be the random variable denoting the distance of the point $B$ to its neighbour in the anti-clockwise direction.

Find the pdf of $Z$. (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 $F_Z(z)= P(Z \le z)$ 1st and then hope the cdf is absolutely continuous.)

A. My model:

  • A.1. 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).

  • A.2. (Not sure if any measure theory needed here, but they are all distinct $\mathbb P$-almost surely.)

  • A.3. Then $Z$'s image is also $(0,1)$ (or $[0,1)$ or whatever). (more details in part D.)

B. Answer (that I found online):

  • Edit: it says $Z$ is uniform, but I think this is wrong.

C. My intuition for the answer (but I would like to know precisely please):

  • Edit: Nvm. I think the answer is wrong.

For parts D,E,F,G: Edit: nvm this is wrong.

H. New section: What exactly is the formula of $Z$ in relation to $A,B,C$? Supposedly $f_Z(z)=2(1-z)1_{(0,1)}(z)$. But what exactly is $Z$? $Z=|A-C|$? $Z=\min\{?,?\}$

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

BCLC
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    Yes, the distribution of $D$ is triangular. It is symmetric around $0$. The distribution of $Z$ is uniform on $[0,1)$. But $Z$ cannot be independent of $A,B,C$ since you compute it deterministically from $A$ and $B.$ – David K Mar 14 '21 at 16:36
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    I don't understand what you mean by $x$ and $y$ in part G. I would have expected them to be values of $A$ and $B$, but then it doesn't make sense to allow $-1 < y < 0.$ In any case you end up with a wrong answer, because $Z$ actually is uniform on $[0,1)$ and therefore $F_Z(z) = z$ for $0\leq z \leq 1.$ – David K Mar 14 '21 at 16:41
  • Thanks @DavidK I took into account your comments about the $-1$ in part G for my new question here: Remodelling random points on a circle: Arc length between points distributed on circle is uniform? – BCLC Mar 15 '21 at 10:28
  • Your comment B above is correct. WLG, I take the circumference of the circle to be 1. Snip the circle at one of the points, and you have the remaining points uniformly distributed. And the counterclockwise gap is distributed as the minimum of those. So for two points, the clockwise gap is uniform. But for three points the CDF is x^2, and generally, for n points the CDF is x^(n-1). My problem is a bit trickier than this: I am looking for the PDF of the MAXIMUM of n gaps between n random points on a circle and for the MINIMUM of the n gaps. – richard1941 Dec 21 '23 at 15:49

3 Answers3

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From symmetry we can fix the point $B$ and consider the distribution of the other two point with respect to it. Let assume that the next point is at distance $z$. Then the remaining configuration space for the other point is $1-z$.

Therefore the pdf in question is $$ 2(1-z) $$ where the factor $2$ accounts for the number of ways to choose the closest point.

BCLC
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user
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  • The three points are independently and uniformly distributed on the circle

  • So given the location of $B$, the anticlockwise arc distances from $B$ of each of the other two points are independently uniformly distributed on $[0,1)$.

  • If $Z$ is the minimum of the anticlockwise arc distances from $B$ of each of the other two points, and $0 \le z \lt 1$, then:

    • the probability a particular one of these arc distances is greater than $z$ is $(1-z)$, given its uniform distribution

    • the probability both of these arc distances are greater than $z$ is $\mathbb P(Z\gt z)=(1-z)^2$, since they are independent

    • and the probability the nearest is less than or equal to $z$ is $F_Z(z)=\mathbb P(Z\le z)=1-(1-z)^2=2z-z^2$, as the complementary probability

    • making the probability density for the nearest $f_Z(z)=2-2z$, by taking the derivative of the absolutely continuous cumulative distribution function

You are correct in saying $Z$ does not have a uniform distribution. It has a triangular distribution on $[0,1)$ with the mode at $0$.

$Z$ is not independent of the positions of $A$, $B$ and $C$ (for example it is exactly equal to the anticlockwise arc distance between two of them), though it is pairwise independent of the position of any one of them.

Henry
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We can introduce an angular representation for $A, B, C$.

Since $A, B, C$ is considered iid. we pick $B$ as the origin for angles.

enter image description here

The distance between $A,B$ would be arc length $r\beta$, where $r$ is 1. Similarly $C,B$ would have length $r \alpha$. As both A, C uniformly distributed. So should both $\alpha, \beta \sim U(0, 2\pi)$.

Now our question is simply asking for the distribution of this random variable $\gamma = min(\alpha, \beta)$.

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

In the end you can change the coordinator back if you want :)

peng yu
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  • THANK YOU SO SO MUCH. I'll analyse later. – BCLC Mar 21 '21 at 08:23
  • Please help here also: https://math.stackexchange.com/questions/4058750/most-random-points-on-a-circle – BCLC Mar 21 '21 at 08:24
  • Something funny about that formula: for z=0, the probability is (2 Pi)^2, which is somewhat greater than 1. In college 60 years ago my teacher said that there is a secret higher level meaning to probabilities that are negative or greater than 1, but I never got that far... – richard1941 Dec 21 '23 at 15:55