Can someone help me identify a flaw in the following argument:
Let $C$ be the unit circle, and let $p$ be some fixed point in its interior. I pick two random points $x$ and $y$ on the boundary of the circle such that $p$ lies on the chord $xy$ (so I choose $xy$ uniformly at random from the set of chords passing through $p$). What is the expected length of $xy$?
First, if $p$ is the center of the circle the answer is obviously $2$, since all chords are just diameters. So what's going wrong in the following argument?
The average length can be written as $\frac{1}{\pi}\int_{xy}\int_{c\in xy} 1$, where $x$ ranges over all points on half of the circumference (and $y$ is determined by $x$ and $p$, so we take an average by dividing by $\pi$). But this is just $\frac{1}{\pi}\int_{c\in int(C)} 1$, since $c$ ends up ranging over all points in $C$. This is then $\pi/\pi = 1$… I feel like maybe in rewriting the integral there's some issue like the distribution over points in $C$ obtained by first choosing a random chord and then choosing a point on the chord is not the same as just choosing a random point (but I can't see why this is true…)
Best Answer
You can't rewrite the integral $\int_x\int_{c\in \mathrm{line}_{xy}}1$ as area integral. Yes, point $c$ runs through all points of $C$, but different locations of $c$ contribute differently to the integral.
Let solve the problem for $p=(0,0)$:
$$ \left<L\right> = \frac{1}{2\pi} \int_0^{2\pi} d\varphi\ 2\int_0^1 dr = \left<L\right> = \frac{1}{\pi} \int_0^{2\pi}\!\!\!\!\int_0^1 drd\varphi $$
For each angle $\varphi$ we integrate length over the radius and multiply by $2$. Since $dA = rdrd\varphi$:
$$ \frac{1}{\pi} \int_0^{2\pi}\!\!\!\!\int_0^1 drd\varphi = \frac{1}{\pi} \iint_{C} \frac{dA}{r} $$
Thus we are integrating not $1$ over the circle, but the function $1/r$. This function shows the contribution of each point in the result.
P.S. For arbitrary $p$ inside the circle, we must remember that arc $d\varphi$ on a circle is visible with some other angle $d\varphi'$ from the point P. This dependency will enter the function of integration.