Given two independent random variables $A$ uniform on $[0,1]$ and $B$ uniform on $[0,2\pi]$. Obtain the joint pdf, tranform to the disk, if necessary modify to obtain the uniform pdf over the disk.
The joint pdf factors since $A$ and $B$ are independent. Hence, $f_{A,B}(a,b) = \frac{1}{2\pi}$, right?
Transformation to the disk is done by changing to polar coordinates, i.e. set $X = A \cos B$ and $Y = A \sin B$, then $A = (X^2+Y^2)^{\frac{1}{2}}$ and $B = \tan^{-1} \frac{Y}{X}$. Changing the coordinates then yields
$f_{X,Y}(x,y) = \frac{1}{2\pi}(x^2+y^2)^{-\frac{1}{2}}$, right?
I do not know about the third part though. Any hints?
Best Answer
Indeed, $f_{A,B}(a,b)=(2\pi)^{-1}\mathbf 1_{0\lt a\lt1}\mathbf 1_{0\lt b\lt2\pi}$ and $(X,Y)=(A\cos B,A\sin B)$ has density $f_{X,Y}$ where, for every $(x,y)$, $f_{X,Y}(x,y)=(2\pi)^{-1}(x^2+y^2)^{-1/2}\mathbf 1_{0\lt x^2+y^2\lt1}$.
Starting from some independent $A$ and $B$ with densities $f_A$ on $[0,1]$ and $f_B$ respectively, where $f_B$ is still defined by $f_B(b)=(2\pi)^{-1}\mathbf 1_{0\lt b\lt2\pi}$, a similar computation yields $$f_{X,Y}(x,y)=(2\pi)^{-1}f_A((x^2+y^2)^{1/2})(x^2+y^2)^{-1/2}\mathbf 1_{0\lt x^2+y^2\lt1}.$$ This is the uniform distribution on the unit disk if $f_A:a\mapsto2a\mathbf 1_{0\lt a\lt1}$.
To achieve this, one can use the original uniform random variables $A$ and $B$ and consider $$(X,Y)=(\sqrt{A}\cos B,\sqrt{A}\sin B).$$