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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

Gilles Bonnet
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abc
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    First, try conditionnaly to the position of the first point. The result must be independent of this position. – mookid Mar 01 '14 at 13:43
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    @mookid Please show me how. – abc Mar 01 '14 at 13:45
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    Is this what you're looking for? – ElThor Mar 01 '14 at 13:45
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    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 01 '14 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 01 '14 at 13:49
  • @ElThor Sort of. Their explanation seems however over-simplified. – abc Mar 01 '14 at 13:50
  • @abc Here is how I would start this: let $\Theta, \Phi$ be the two angles defining the points.

    We want to compute $ED$ where $D = \sqrt{(\cos\Theta -\cos \Phi)^2 + (\sin\Theta -\sin \Phi)^2}$ and we know that $\Theta, \Phi$ are independent with distribution uniform on $[0, 2\pi]$.

    Now my guess is that by symetry, $E[D|\Phi]$ does not depend on $\Phi$. This makes one less integral to compute.

     – mookid Mar 01 '14 at 13:52
  • @GlenO changed it... What could be wrong more? – abc Mar 01 '14 at 13:57
  • @mookid Looks similar to my approach. Could you maybe post your answer? – abc Mar 01 '14 at 13:58

1 Answers1

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

Christian Blatter
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