TITLE: Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$
QUESTION [7 upvotes]: It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$
I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$
Then I am getting stuck at ranges of $\theta$

REPLY [2 votes]: Setting $U=\dfrac{XY}{\sqrt{X^2+Y^2}}, V=\dfrac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ and using moment generating functions:
\begin{align}
M_{(U,V)}(u,v) & =\mathbb{E}\left[\mathrm{e}^{\langle\,(u,v)\,;\,(U,V)\,\rangle}\right] \\[10pt]
& =\iint_{\mathbb{R}^2}\exp\left(u\frac{xy}{\sqrt{x^2+y^2}}+v\frac{x^2-y^2}{2\sqrt{x^2+y^2}}\right)\cdot f_{(X,Y)}(x,y)\,\mathrm{d}x\,\mathrm{d}y.
\end{align}
You can see here that $$M_{(U,V)}(u,v)=\exp\left(\frac18(u^2+v^2)\right)=\exp\left(\frac12(u,v)
\begin{pmatrix}1/4 & 0\\ 0 & 1/4\end{pmatrix}
(u,v)^T\right)$$
so that $(U,V)$ is normal multivariate with $\mathbf{0}$ mean and diagonal covariance matrix, meaning $U, V$ are independant $\mathcal{N}(0,1/4)$.