Box-Muller Transform Normality I'm trying to show that for uniformly distributed variables $X_1$ and $X_2$ that the vector
$$
(\sqrt{-2\log X_1} \cos(2\pi X_2), \sqrt{-2 \log X_1} \sin(2\pi X_2))
$$
is two-dimensional normal.  Letting $Y_1$ be the first component and $Y_2$ be the second component I see that $X_1 = e^{-(Y_1^2 +Y_2^2)/2}$ and $X_2 = \frac{1}{2\pi} \arctan(Y_2/Y_1)$.  Is there a way to proceed from here to show that $(Y_1, Y_2)$ is normal?
 A: Remember the substitution rule (if we can characterize the CDF, then
we are done). In particular, consider the absolute value of the determinant
of the Jacobian matrix
\begin{align*}
\left|\det\left[\begin{array}{cc}
\frac{\partial X_{1}}{\partial Y_{1}} & \frac{\partial X_{1}}{\partial Y_{2}}\\
\frac{\partial X_{2}}{\partial Y_{1}} & \frac{\partial X_{2}}{\partial Y_{2}}
\end{array}\right]\right| & =\left|\det\left[\begin{array}{cc}
-Y_{1}\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right) & -Y_{2}\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right)\\
-Y_{2}/\left[2\pi\left(Y_{1}^{2}+Y_{2}^{2}\right)\right] & Y_{1}/\left[2\pi\left(Y_{1}^{2}+Y_{2}^{2}\right)\right]
\end{array}\right]\right|\\
 & =\left|-Y_{1}^{2}\frac{\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right)}{2\pi\left(Y_{1}^{2}+Y_{2}^{2}\right)}-Y_{2}^{2}\frac{\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right)}{2\pi\left(Y_{1}^{2}+Y_{2}^{2}\right)}\right|\\
 & =\left|-\frac{\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right)}{2\pi}\right|\\
 & =\frac{\exp\left(Y_{1}^{2}/2\right)}{\sqrt{2\pi}}\frac{\exp\left(Y_{2}^{2}/2\right)}{\sqrt{2\pi}}.
\end{align*}
