Distributions, transformations and marginals Suppose $X$ and $Y$ are two independent random variables each distributed as $\mathrm{Uniform}(0,1)$.
1) Find the joint distribution of $X$ and $Y$.
2) Let $U = \cos(2\pi Y)\sqrt{-2\ln(X)}$  and $V = \sin(2\pi Y)\sqrt{-2\ln(X)}$. Find the joint distribution
of $U$ and $V$ assuming that the transformation is one-to-one?
3) Find the marginal distributions of $U$ and $V$?
For 1, I got that my distribution is 1 by multiplying the two distributions together.  For 2, I have been getting stuck trying to get my equations in terms of $X$ and $Y$ to perform the transformation.  I simplified and got $Y = \frac{1}{2}\tan^{-1}\left(\frac{V}{U}\right)$ but something seems wrong about this and am having trouble trying to get $X$.  For 3, I am obviously stuck and even not totally sure on the support.  Any help is appreciated.
 A: Consider the transformation $T:[0,1]^2\to \mathbb R^2$ defined by
$$
T(x,y)=(\cos(2\pi y)\sqrt{-2\log x},\sin(2\pi y)\sqrt{-2\log x}).
$$
Then, for every $(u,v)$ in $\mathbb R^2\setminus\{(0,0)\}$,
$$
f_{U,V}(u,v)=f_{X,Y}(T^{-1}(u,v))\cdot|\det J_T(T^{-1}(u,v))|^{-1},
$$
provided $T$ is bijective and the Jacobian matrix $J_T$ is nonsingular on $[0,1]^2$. This remark allows to skip the computation of $T^{-1}$. Your next step is to compute the $2\times2$ matrix $J_T$, to deduce its determinant, and to plug this value into the expression of $f_{U,V}$.
A shortcut the physicists (and others) use is to rely on the formalism of differential forms. That is, one starts from
$$
u=\cos(2\pi y)\sqrt{-2\log x},\quad v=\sin(2\pi y)\sqrt{-2\log x},
$$
and one differentiates $u$ and $v$. This yields
$$
\mathrm du=-2\pi\sin(2\pi y)\sqrt{-2\log x}\mathrm dy-\frac{\cos(2\pi y)}{x\sqrt{-2\log x}}\mathrm dx,
$$
and
$$
\mathrm dv=+2\pi\cos(2\pi y)\sqrt{-2\log x}\mathrm dy-\frac{\sin(2\pi y)}{x\sqrt{-2\log x}}\mathrm dx.
$$
Now, $\mathrm du\mathrm dv$ is the product of these, with the convention that $\mathrm du\mathrm du=\mathrm dv\mathrm dv=0$ and that $\mathrm dv\mathrm du=-\mathrm du\mathrm dv$. In other words, if $\mathrm du=\alpha\mathrm dx+\beta\mathrm dy$ and $\mathrm du=\gamma\mathrm dx+\delta\mathrm dy$, then
$$
\mathrm du\mathrm dv=|\alpha\delta-\beta\gamma|\,\mathrm dx\mathrm dy.
$$
Using either the mathematician's way or the physicist's way, in the present case one gets
$$
\mathrm du\mathrm dv=\frac{2\pi}x\,\mathrm dx\mathrm dy,
$$
hence
$$
|\det J_T(x,y)|^{-1}=\frac{x}{2\pi}.
$$
A last remark is that, if $T(x,y)=(u,v)$, then $x=\exp(-(u^2+v^2)/2)$, hence
$$
f_{U,V}(u,v)=\frac{\mathrm e^{-(u^2+v^2)/2}}{2\pi}.
$$
From here, the other questions you were asked should be direct.
