Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 2 / 10F (Probability)

For a uniform random point in a disk, the event $R<x$ occupies a disk of area $\pi x^2$ when $0<x<1$. ThusThe radial probability density function is $f_R(r)=2r$ on $(0,1)$. Integrating givesChoose the polar angle $\Theta \in [0,2\pi )$. The Cartesian density is $1/\pi$ in the disk, and the polar coordinates Jacobian determinant is $r$. HenceThe product support and product density prove that $R$ and $\Theta$ are independent, with uniform angle.

Reflection symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$ and $\mathbb{E}(XY)=0$, so $\mathrm{Cov}(X,Y)=0$. They are nevertheless not independent. The events $X>3/4$ and $Y>3/4$ each have positive probability, but their intersection is empty inside the disk, because it would require $X^2+Y^2>18/16>1$. Independence would give a positive product probability for that intersection.

Except at the probability-zero center, $X/R=\cos \Theta$ and $Y/R=\sin \Theta$. ThereforeAn arbitrary definition at the center does not change these expected values. This example distinguishes uncorrelated Cartesian coordinates from independent polar coordinates.