8

Choose, at random, three points on the unit circle. Interpret them as cuts that divide the circle into three arcs. Compute the expected length of the arc that contains the point (1, 0).

Generalise for N points.

alexandreC
  • 457
  • 7
    Not a proof, but an intuitive guess: the problem is equivalent to the total length of the first and last segment of a line of length $2\pi$ that has been randomly cut into four pieces. By symmetry each piece is expected to be $\frac{\pi}{2}$ long, for an answer of $\pi$ ($\frac{4\pi}{n+1}$ in the general case.) – user7530 Jan 08 '13 at 17:23
  • 1
    Argue by symmetry. – André Nicolas Jan 08 '13 at 17:31

3 Answers3

13

The length of the arc from $(1,0)$ to the next point is the minimum $M$ of three (or $n$) independent uniform random numbers $X_1,\ldots, X_n$ from $[0,2\pi)$. Since $P(M>x)=\prod P(X_i>x)=\left(1-\frac x{2\pi}\right)^n$, we find $$E(M) = \int_0^{2\pi}P(M>x)dx =\left. -\frac{2\pi}{n+1}\left(1-\frac x{2\pi}\right)^{n+1}\right|_0^{2\pi}=\frac{2\pi}{n+1}.$$ The same holds for the arc length to the the other arc end, hence the expected length of the full arc containing $(0,1)$ is $$\frac{4\pi}{n+1}.$$

SBF
  • 36,568
Hagen von Eitzen
  • 382,479
  • 1
    For 3 cuts, this reasoning gives you an expected length of $\pi$. However, with just 2 cuts, the expected length of one of the resulting 2 pieces chosen at random is $\pi$. For 3 cuts, it must be less. I don't think you can treat $(1,0)$ like a cut point that you then stitch back together. It really just signifies random selection of a segment after the fact. – Zimul8r Jan 08 '13 at 18:08
  • 1
    That's not true. If the circle is cut into two pieces, one much bigger than the other, $(1,0)$ is more likely to be in the bigger piece. – user7530 Jan 08 '13 at 19:44
  • 1
    Right. OK, so from http://math.stackexchange.com/questions/14190/average-length-of-the-longest-segment?rq=1 and http://math.stackexchange.com/questions/13959/if-a-1-meter-rope-is-cut-at-two-uniformly-randomly-chosen-points-to-give-three?rq=1, in the 2 cut case, the pieces, being the longest and the shortest by default, have expected length $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ with prob of containing $(1,0)$ at $25%$ and $75%$, resp. So the expected length would be $\left(\frac{1}{4}\right) \left(\frac{\pi}{2}\right) + \left(\frac{3}{4}\right) \left(\frac{3\pi}{2}\right) = \frac{5\pi}{4}$? – Zimul8r Jan 08 '13 at 20:16
  • 1
    Shouldn't the random variables lie from (0, Pi) as after pi the length again start decreasing? – Rishabh Jain Sep 14 '18 at 18:43
  • 1
    @RishabhJain To make it clear, I first look for the anti-clockwise next, not the overall closest cut point; therefore up to $2\pi$ is possible. – Hagen von Eitzen Sep 15 '18 at 08:55
  • @HagenvonEitzen That makes sense. Can you also tell me how is E(X) = P(M>x) integrated over all x. I can't seem to find any reference to finding expectation this way. All i can get is E(M) = MP(X) integrated over the range. – Rishabh Jain Sep 15 '18 at 12:37
  • @RishabhJain It's a commonly used result. https://math.stackexchange.com/questions/64186/intuition-behind-using-complementary-cdf-to-compute-expectation-for-nonnegative – Funaizhang Aug 22 '22 at 12:13
4

You can cut the circle at (1,0) point and make it a line. The both side of the line corresponds to point (1,0) in the unit circle.

We also know that the answer for the circle is $2\pi$ times the answer of the uniform[0,1] case.

For uniform[0,1] case with the cutting method mentioned above, The expectation is $E(min(X_1,...,X_n)+1-max(X_1,...,X_n))$

We know that in this case $E(min(X_1,...,X_n))=\dfrac{1}{n+1}$ and $E(max(X_1,...,X_n)) = \dfrac{n}{n+1}$, so we have the final answer as

$2\pi (1+\dfrac{1}{n+1}-\dfrac{n}{n+1})=\dfrac{4\pi}{n+1}$

wei1210516514
  • 41
1

A bit late to the party here, I have a less clever solution but I believe it may be illustrative so I am including it.

Imagine starting at the point $(1,0)$ and walking along the unit circle in both the CW and CCW directions until you first encounter points in both. Say the total angle traveled in both directions was $2 \pi - \theta$. This implies that all $n$ points lie in a wedge with an angle exactly $\theta$ and that the desired arc length in this instance is $2 \pi - \theta$.

We can use conditional expectation to find the expected length by conditioning on the event that all of the point lie in some wedge of angle $\theta$, that does not contain the origin, and that wedge cannot be reduced in size and still contain all of the points.

What is the probability that $n$ points define and lie in a given wedge of angle $\theta$? Well, there are $\binom{n}{2}$ ways to pick two points to be on the boundary and $2!$ ways to order them. They land on the boundary locations with probability $1/2\pi$ each. The remaining points lie in/on the wedge with probability $\left(\frac{\theta}{2 \pi}\right)^{n - 2}$. Putting these together, for a given wedge of with width $\theta$, the probability that all points are on/in it is $$2 \binom{n}{2}\left(\frac{\theta}{2 \pi}\right)^{n - 2}.$$ We still need to account for all the wedges of this size, let the CW point on the wedge be at angle $\phi$ and then the proability of all points being in\on some wedge of width exactly $\theta$ is $$\int_0^{2 \pi - \theta} 2 \binom{n}{2}\left(\frac{\theta}{2 \pi}\right)^{n - 2} d \phi.$$ Note $\phi$ only goes to $2 \pi - \theta$ so that the wedge does not contain the point $(1,0)$.

Now we apply our conditional expectation formula. For a given $\theta$ then length of the desired arc is $2 \pi - \theta$, taking the expectation over all values of $\theta \in [0, 2\pi)$ yields the resulting integral

$$ \begin{align} \int_0^{\pi}(2 \pi - \theta)\left(\int_0^{2 \pi - \theta} 2 \binom{n}{2}\left(\frac{\theta}{2 \pi}\right)^{n - 2} d \phi\right) d\theta & = \frac{2}{(2 \pi)^n} \binom{n}{2} \int_0^{\pi}(2 \pi - \theta)^2 \theta^{n-2}\; d \theta\\ & = \frac{4\pi}{n + 1} \end{align} $$

Connor Colombe
  • 11