$R^2$ is uniform on (0,1) Let (X,Y) be uniformly distributed in a circle of radius 1. Show that if R is the distance from the center of the circle to (X,Y) then $R^2$ is uniform on (0,1). 
This is question from the Simulation text of Prof. Sheldon Ross. Any hints? 
 A: There are lots of approaches one could take, but the simplest one I can think of is to consider the shape of the CDF of $R^2$. 
What is the probability that $R^{2}<t$? Well, this is clearly $0$ for $t<0$ and $1$ for $t>1$. For $t\in [0,1]$, the probability is the area of the circle of radius $R$ (divided by $\pi$, since $1/\pi$ is the joint density in the unit circle) where $R^2=t$.  Substituting $R=\sqrt{t}$, this is:
$$\frac{\pi R^{2}}{\pi} = \frac{\pi t}{\pi} = t.$$
And we're done.  This is the CDF of the uniform distribution on $[0,1]$, as required!
A: Here is a much more tedious way of doing the same thing:
Let  $A \subset (0,1)$ be an open set.
For later convenience, let $N = \{(x,0) | x \in (-1,0] \}$.
Let $C_A = \{ x | x_1^2+x_2^2 \in A \}$, and note that $C_A = \{ \sqrt{t} (\cos \theta, \sin \theta) | t  \in A, \theta \in (-\pi, \pi] \} \subset B(0,1)$. Note that if we let
$C_A' = C_A \setminus N =\{ \sqrt{t} (\cos \theta, \sin \theta) | t  \in A, \theta \in (-\pi, \pi) \}  $, the measure remains the same.
Define $\phi: (0,\infty) \times (-\pi, \pi) \to \mathbb{R}^2 \setminus N$ by   $\phi(\alpha) = (\sqrt{\alpha_1} \cos \alpha_2, \sqrt{\alpha_1} \sin \alpha_2)$,  note that 
$\phi$ is smooth, bijective and
$\det {\partial \phi(\alpha) \over \partial \alpha} = {1 \over 2} >0$, hence $\phi$ is a diffeomorphism.
The important point is that $\phi(A \times (-\pi, \pi)) = C_A'$.
The change of variables theorem gives
$\int_{C_A'} dx = \int_{\phi(A \times (-\pi, \pi))} dx = \int_{A \times (-\pi, \pi)} | \det {\partial \phi(\alpha) \over \partial \alpha} | d \alpha = \pi m(A)$, where $m$ is the Lebesgue measure of $A$.
If $\mu$ is the uniformly distributed measure on the unit circle, we have $\mu = {1 \over \pi} m$, and so we have $\mu(C_A) = m(C_A') = m(A)$. Since this is true for open $A$, it is true for $G_\delta$ sets and since $m$ is outer regular and complete, it follows that it is true for all measurable sets.
Hence $R^2$ is distributed uniformly on $(0,1)$.
