How does the joint distribution of $X$ and $Y$ is the standard bivariate normal distribution using the change of variables We have $$ X = \sqrt{-2 \log(U)} \cos(2 \pi V)$$ and $$Y = \sqrt{-2 \log(U)} \sin(2 \pi V)$$
where $U$ and $V$ are independent uniform random variables over $[0,1]$.
I started solving it using the change of variables, so
$U = e^{\frac{-1}{2}(X^2 +Y^2)}$ and $ V = \frac{1}{2 \pi}\operatorname{atan2}(Y,\,X)$.
then using formula of the joint distribution $$f_{X,Y}(x,y)=|J| f(u(x,y),v(x,y))$$ I should obtain the joint as $\frac{1}{2 \pi} e^{\frac{-1}{2} (x^2 +y^2)}$
while what I found is totally different. Could anyone help me please in this last step.
What I found 
$ f(u,v)= 1$ since both variables are uniform and independent.
then the the determinant of Jacobian is $$ - \frac{x^2}{2 \pi(x^2+y^2)} e^{\frac{-1}{2}(x^2+y^2)} + \frac{y^2}{2 \pi(x^2+y^2)} e^{\frac{-1}{2}(x^2+y^2)}$$ which can't be simplified to the normal.
 A: $$f(U,V)=1$$
$$
f(X,Y)=|\partial(U,V)/\partial(X,Y)|f(U,V)=|\partial(X,Y)/\partial(U,V)|^{-1}f(U,V)
=\frac{U}{2\pi}=\frac{1}{2\pi}e^{-\frac{1}{2}(X^2+Y^2)}
$$
A: The correct calculation is$$J=\left\|\begin{array}{cc}
\partial_{X}U & \partial_{Y}U\\
\partial_{X}V & \partial_{Y}V
\end{array}\right\|=\left\|\begin{array}{cc}
-XU & -YU\\
-\frac{Y}{2\pi}\frac{1}{X^2+Y^2} & \frac{X}{2\pi}\frac{1}{X^2+Y^2}
\end{array}\right\|=\left|\frac{-U}{2\pi}\right|=\frac{U}{2\pi}.$$You've had a sign error somewhere, probably in $\partial_XV$ or $\partial_YV$.
A: \begin{align}
x & = \sqrt{-2\log u}\, \cos(2\pi v) \\[8pt]
y & = \sqrt{-2\log u}\, \sin(2\pi v)
\end{align}
\begin{align}
& \frac{\partial x}{\partial u} = \frac{-\cos(2\pi v)}{u\sqrt{-2\log u}} & & \frac{\partial y}{\partial u} = \frac{-\sin(2\pi v)}{u\sqrt{-2\log u}} \\[12pt]
& \frac{\partial x}{\partial v} = \sqrt{-2\log u}\,\sin(2\pi v) & & \frac{\partial y}{\partial v} = \sqrt{-2\log u}\,\cos(2\pi v)
\end{align}
Therefore
$$
\left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \frac 1 u.
$$
So $ dx\,dy = \dfrac{du\,dv} u.$
$$
x^2 + y^2 = -2\log u
$$
$$
- \frac{x^2+y^2} 2 = \log u
$$
$$
e^{-(x^2+y^2)/2} = u
$$
$$
e^{-(x^2+y^2)/2}\,dx\,dy = u\cdot\frac{du\,dv} u = du\,dv
$$
