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If I choose $n$ random independent points on a circle's circumference and use them to divide a circle into $n$ sectors, expected size of each sector in radians is $2\pi/n$.

If I change the problem to finding the expected size of a sector with additional condition that it contains a specific given polar coordinate angle $\theta$, I still expected it to be $2\pi/n$. But when I do a computer simulation it seems to be $4\pi/(n+1)$.

Am I doing something wrong? If not, what's the intuition behind this?

Simplest simulation is to break the circle using positive x axis and convert it to a line segment, and generate $n$ points to define sectors, then calculate the average size of sector containing $\theta=0$:

import random

C = 10000
N = 4
sum = 0.0
for i in range(C):
  arr = [random.random() for j in range(N)]
  sum += 1 - max(arr) + min(arr)
print(sum / C)


outputs approximately 0.4, which means sector size of 2pi * 0.4.

If I change the code (which makes the code a bit more complicated) to calculate expected size of sector containing any specific $\theta$ other than 0, I still get 0.4.

Thinking in terms of a line segment this seems intuitive, but not when I think in terms of a circle.

HMA
  • 121
  • I did not calculate the actual answer, but I would expect the average segment containing a given point to be larger than an average segment with no conditions, simply because the larger a segment is, the more likely it is to contain a given point. So fixing a point basically "chooses" a random segment, and is more likely to "choose" a larger one. – SVG Aug 31 '23 at 06:35

2 Answers2

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It's a known result (see here for example) that the expected minimum of $n$ random uniform $(0,1)$ variables is $E(m) = 1/(n+1)$, while the expected maximum is $E(M) = 1 - 1/(n+1)$. From your simulation, you need $1-E(M) + E(m)$ and so your guess is correct.

nicola
  • 813
2

The easiest way to see this is to turn the circumference of the circle into a straight line segment $[0,2\pi]$ and then turn it back into a circle a different way.

With your original circle, you could choose one of the points to split the circumference and you in effect have the other $n-1$ points distributed uniformly at random in that interval, splitting it into $n$ sub-intervals (including the one starting at $0$ and the one ending at $2\pi$) each with expected length $\frac{2\pi}{n}$.

If instead you had used $n$ points distributed uniformly at random then you would have $n+1$ sub-intervals each with expected length $\frac{2\pi}{n+1}$.

If you now turn these $n+1$ subintervals back into a circle, identifying $0\equiv 2\pi$ now as your special point (you can rotate the new circle by $\theta$ to fit the question) rather than as a randomly selected point, you want the expected length of the arc back to the previous random point plus the expected length of the arc forward to the next random point, which is $$\frac{2\pi}{n+1}+\frac{2\pi}{n+1}=\frac{4\pi}{n+1}$$ as your simulation suggests.

Henry
  • 169,616