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

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

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First, try conditionnaly to the position of the first point. The result must be independent of this position. –  mookid Mar 1 at 13:43
@mookid Please show me how. –  abc Mar 1 at 13:45
Is this what you're looking for? –  ElThor Mar 1 at 13:45
Your main mistake is that $(\cos(\theta_1)-\cos(\theta_2))^2+(\sin(\theta_1)-\sin(\theta_2))^2$ is equal to $2-2\cos(\theta_1)\cos(\theta_2)-2\sin(\theta_1)\cos(\theta_2)$ - you forgot the $2$ at the start of it. –  Glen O Mar 1 at 13:48
Also note that $\cos(\theta_1)\cos(\theta_2)+\sin(\theta_1)\sin(\theta_2) = \cos(\theta_1-\theta_2)$ –  Glen O Mar 1 at 13:49

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 $${\mathbb E}(D)={1\over 2\pi}\int_{-\pi}^\pi 2\left|\sin{\phi\over2}\right|\ d\phi={1\over \pi}\int_0^\pi 2\sin{\phi\over2}\ d\phi={4\over\pi}\ .$$