I've been working on a question and it seems I have obtained a paradoxical answer. Odds are I've just committed a mistake somewhere, however, I will elucidate the question and my solution just in case anyone is interested.

I want to know what is the average distance between two points on a circle of radius 1 where we consider only the boundary points.

My attempt is as follows:

Consider a segment of the diameter x which is uniformly distributed between 0 and 2. Then you can calculate the distance between the points (2,0) and the point determined by x just by elementary geometry as this picture shows:

enter image description here

Here in the picture, the green segment is the geometric mean and the orange one is the distance whose distribution we want to know. Just by calculating the expected value, we obtain:

$E\left(\sqrt{(4-2X)}\right) = \int_{0}^{2} \sqrt{(4-2x)}\cdot\frac{1}{2} dx = 1.333.... = \frac{4}{3}$

Where $\sqrt{(4-2x)}$ is the transformation of the random variable and $\frac{1}{2}$ is the pdf of a uniform distribution $[0,2]$.

Also, if we derive the pdf of the transformation we obtain the same result:

$y = \sqrt{(4-2x)} , x = 2- \frac{y^2}{2}, \mid\frac{d}{dy}x\mid = y$

$g(y)=f(x)\cdot\mid\frac{d}{dy}x\mid = \frac{1}{2}\cdot y$

$E(Y)= \int_{0}^{2}y\cdot\frac{1}{2}\cdot y dy = 1.333.... = \frac{4}{3} $

I have seen a different approach somewhere else where the distribution of the angle is considered as a uniform distribution between 0 and $\pi$ and the final result was:

$1.27... = \frac{4}{\pi}$

That's pretty much the problem I found. Maybe I just did it wrong in some step but it all makes sense to me. I know this is not exactly what we refer as Bertrand paradox but it just suggests something like that because both problems handle with segments in circumference and maybe my result is wrong because it does not hold for rotations of the circle or something like that (I read a little bit about the Bertrand's Paradox).

That's pretty much it. Also sorry for my bad English and maybe I'm also wrong in something pretty elemental since I've just started learning about probability theory. It's also my first post so I will try to improve my exposition and LateX use in the following ones.

  • $\begingroup$ welcome to MSE. I have attempted to edit your question. In case you find any discrepancy with your idea please check. Also, you can look up the edits I have made that will help you for your future questions. Thank you! $\endgroup$ – Mann Apr 19 at 21:57

This is a very nice question, well-written and researched before posting. You have clearly put a lot of careful into your question and it is very much appreciated on this site (and I thought your English reads perfectly naturally). Thanks so much for taking the time to ask a serious and well-considered question. I hope this answer comes close to meeting the high standard of quality set by your question.

I don’t see any calculation errors on your part, and I’ll wager you checked those thoroughly. The flaw is something much more subtle and is related to the initial framing of the problem.

You have correctly solved the problem of “what is the average distance between the rightmost point of a circle and another point projected upwards from a randomly chosen point on the diameter”. But this doesn’t quite capture the same probability distribution as “rightmost point and another randomly chosen point on the circle” (which is sufficient to capture the dynamics of “two random points on the circle”).

There is a hidden non-uniformity in the projection process, and that’s because the circle has varying slope so the projection hits the circle differently at different positions.

Here’s an experiment that should help convince you that uniform distribution on the diameter does not yield uniform distribution on the circle boundary. Fundamentally, a uniform distribution should treat all arcs of the same length equally: a random point has a $1/4$ chance of falling on a quarter-circular arc, regardless of which quarter circle that is.

Now compare two specific quarter-circles and their projections onto the diameter: 1) the quarter-circle adjacent to $(2,0)$, i.e. 12 o’clock to 3 o’clock, and 2) the quarter-circle centered around the top-most point $(1,1)$, i.e. 10:30 to 1:30.

These both project onto the diameter with no self-overlap (avoiding double counting). But the first one projects down to the interval $[0,1]$ which has length $1$, and the second one projects down to $[-\sqrt{2}/2,\sqrt{2}/2]$, which has length $\sqrt2$.

This should indicate to you that the two distributions are not equivalent (though they may intuitively seem so). Consequently, the random process of choosing uniformly from the diameter and projecting onto the circle results in disproportionately favoring the top and bottom sectors of the circle over the left and right.

  • $\begingroup$ Thanks! I really appreciate your words and yes, it seems the problem really deals with something subtle. I'm going to add some considerations in an answer. $\endgroup$ – Hyz Apr 19 at 19:49

Thanks, Erick Wong for your feedback. After your answer, I calculated the distribution of the arc length subject to the uniform distribution of the point on the diameter. In fact: if we want to express the arc length $l$ as a function of $x$, $l = f(x)$ we obtain:

$l = \arccos(1-x), x = 1-\cos{l}, |\frac{d}{dl}(x)| = \sin{l}$

$l_{pdf} = x_{pdf} \cdot |\frac{d}{dl}x| = \frac{1}{2}\cdot \sin{l}$.

So the arc length does not distribute uniformly, we have "lost it", we might say. That's what was wrong. For instance, if the arc length obeys a uniform distribution [0, $\pi$], then we can calculate the segment $s$ as a function of the arc length:

We know $l = f(x)$ and want to know $s = h(l)$. If we calculate $s = g(x)$ we're done:

From the image on the question I post, $s = g(x)=\sqrt{2x}$ (or the opposite segment $\sqrt{4-2x}$ ) then $h = g \circ f^{-1}$, $s = \sqrt{2(1-cosl)}$

$E(s) = \int_0^\pi{\sqrt{2(1-cosl)}\frac{1}{\pi}dl}=1.273... = \frac{4}{\pi}$

Also the pdf:

$s = h(l) = \sqrt{2(1-cosl)} , l=h^{-1}(s)= 2\cdot \arcsin(\frac{s}{2}), |\frac{d}{ds}h^{-1}|=\frac{2}{\sqrt{4-s^2}}$

$s_{pdf} = \frac{1}{\pi} \frac{2}{\sqrt{4-s^2}}$

$E(s) = \int_0^2{s\cdot \frac{1}{\pi} \frac{2}{\sqrt{4-s^2}}ds} = 1.273... = \frac{4}{\pi}$

So we're done. Also, the pdf of the segment suggests something related to the Cauchy distribution. Not exactly but certainly it has to do something with it. If we read the description of Cauchy distribution in Wolfram MathWorld:

"The Cauchy distribution, also called the Lorentzian distribution or Lorentz distribution, is a continuous distribution describing resonance behavior. It also describes the distribution of horizontal distances at which a line segment tilted at a random angle cuts the x-axis."

And that's it. A really fascinating problem that introduces some subtle ideas of probability theory. If anyone knows something else please give me feedback. I really think there's a nice connection with Cauchy distribution.


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