Random points on a circle

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

• 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?

Solution 1:

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.

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 :)

Solution 2:

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.

Solution 3:

• 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.