Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 2 / 11F (Probability)

All the following distributions are conditional on the point lying in the unit disk. Initially the joint probability density of $(S,T)$ is $1/4$ on the square. The disk has area $\pi$, so the conditioning event has probability $\pi /4$. Dividing by this probability gives the constant conditional density $1/\pi$ on the disk, proving its uniform distribution there.

The polar coordinates $S=r\cos \theta ,T=r\sin \theta$ have Jacobian determinant $r$. ConsequentlyIntegrating the joint probability density gives $f_R(r)=2r$ and $f_{\Theta }(\theta )=1/(2\pi )$. Since the joint density equals the product of these marginal densities on a product support, $R$ and $\Theta$ are independent. In particular $U=R^2$ has the uniform distribution on $(0,1)$.

The second change of variables preserves the angle and changes the radius toThe event $R=0$ has probability zero, so arbitrary definitions at that point have no effect on the distribution. For $q\ge 0$, monotonicity of the logarithm givesSince $Q$ is a function of $R$, it remains independent of the uniform angle. Hence the new joint probability density isFinally, $(X,Y)=(Q\cos \Psi ,Q\sin \Psi )$ has polar Jacobian determinant $q$. Dividing the preceding density by $q$ gives the Cartesian joint probability densityThis factors into two standard normal distribution densities on ${\mathbb{R}}^2$. Thus $X$ and $Y$ are independent, each with mean zero and variance one. This derivation is the radial form of the Box-Muller transform.