Expected value of the distance between 2 uniformly distributed points on circle

Question

I have the following problem (related to Bertrand):

Given a circle of radius $a=1$. Choose 2 points randomly on the circle circumference. Then connect these points using a line with length $b$. What is the expected length of this line? ($\mathbb{E}[b]$=..?)

I have tried this:

$x_i=\cos(\theta_i), y_i=\sin(\theta_i), \quad i=1,2$, where $\theta_i$ is uniformly distributed on $[0,2\pi]$

Then I tried to compute the squared distance. The squared distance between two points in the Eucledian space is:

$$d^2=(\cos(\theta_1)-\cos(\theta_2))^2+(\sin(\theta_1)-\sin(\theta_2))^2 $$

Now taking expectations I got:

$$E(d^2)=2-2 \ ( \ E(\cos(\theta_1)\cos(\theta_2) + E(\sin(\theta_1)\sin(\theta_2) \ )$$ (as $E(\cos^2(\theta_i))=E(\sin^2(\theta_j))$

Then $$E(\cos(\theta_1)\cos(\theta_2))\overset{uniform}=\int_0^{2\pi}\int_0^{2\pi}\theta_1 \theta_2\cos^2(\frac{1}{2\pi})\ \mathrm{d}\theta_1 \ \mathrm{d}\theta_2 = 4\pi^4 \cos^2(\frac{1}{2\pi})$$

and

$$E(\sin(\theta_1)\sin(\theta_2))\overset{uniform}=\int_0^{2\pi}\int_0^{2\pi} \theta_1 \theta_2\sin^2(\frac{1}{2\pi})\ \mathrm{d}\theta_1 \ \mathrm{d}\theta_2 = 4\pi^4 \sin^2(\frac{1}{2\pi})$$

so that $$d^2=2-4 \pi^2 \left(\cos^2(\frac{1}{2 \pi}) + \sin^2(\frac{1}{2\pi})\right)=2-4 \pi^2$$

But that doesn't make sense since it is negative. Any help would be appreciated

Answer 1

You may assume the first point $A$ at $(1,0)$ and the second point $B=(\cos\phi,\sin\phi)$ being uniformly distributed on the circle. The probability measure is then given by ${1\over2\pi}{\rm d}\phi$. The distance $D:=|AB|$ computes to $2\left|\sin{\phi\over2}\right|$, and we obtain $${\mathbb E}(D)={1\over 2\pi}\int_{-\pi}^\pi 2\left|\sin{\phi\over2}\right|\ d\phi={1\over \pi}\int_0^\pi 2\sin{\phi\over2}\ d\phi={4\over\pi}\ .$$

Answer 2

Fix the first point at $(1,0)$ and consider the angle $\theta\in[0,\pi/2]$ (chosen uniformly). Then the distance between the point $(1,0)$ and the point determined by $\theta$ is equal to $2\cos\theta$. The probability measure is given by $2\theta/\pi$. This gives us $$ \mathbb{E}[D] = \int_{0}^{\pi/2}2\cos(\theta)\cdot 2d\theta/\pi = \frac{4}{\pi}\underbrace{\int_{0}^{\pi/2}\cos(\theta)d\theta}_{=1} = \frac{4}{\pi}. $$

Answer 3

A key mistake the OP made is: Knowing $\mathbb E(d^2)$ doesn't really tell us $\mathbb E(d)$. We should compute $\mathbb E(d)$ instead.

We can in fact continue with the OP's approach as follows: $d^2=2-2(\cos \theta_1\cos\theta_2+\sin\theta_1\sin\theta_2)=2(1-\cos(\theta_1-\theta_2))=4\sin^2\frac{\theta_1-\theta_2}2$. So, \begin{align}\mathbb E[d]&=\int_0^{2\pi}\left(\int_{0}^{2\pi} 2\left\vert \sin\frac{\theta_1-\theta_2}2\right\vert\frac1{2\pi}d\theta_1\right)\frac1{2\pi}d\theta_2\\ &\overset{(a)}{=}\int_0^{2\pi}\left(\int_{0}^{2\pi} 2\left\vert \sin\frac{\theta_1}2\right\vert\frac1{2\pi}d\theta_1\right)\frac1{2\pi}d\theta_2 =\int_{0}^{2\pi} 2\left\vert \sin\frac{\theta_1}2\right\vert\frac1{2\pi}d\theta_1,\end{align} which evaluates to $\pi/4$, as in Christian Blatter's answer. Note that (a) is because $\vert \sin(x/2)\vert$ is periodic with period of $2\pi$.