Prove that $\sqrt{X^2+Y^2}$ and $\frac{Y}{X}$ are independent Given $X,Y$~$N(0,1)$independent so (X,Y) has density $\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}$I need to prove that $S=\sqrt{X^2+Y^2}$ and $R=\frac{Y}{X}$ are independent. I wanted to calculate distribution of $S$ and $R$ and to show that its product is in fact (S,R). Using polar cordinates:
$$P(S<t)=P(X^2+Y^2<t^2)=\int_{0}^{2\pi}\int_{0}^{t}re^{-\frac{r^2}{2}}drd\alpha$$ So we can clearly calculate that one. But I have a problem with $R$.
$$P(R<t)=P(\frac{Y}{X}<t)=\int_{-\infty}^{\infty}\int_{-\infty}^{tx}\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}dxdy$$ How should I parametrize it? Using polar cordinates doesn't seems to work.
 A: The pdfs for $R$ and $S$ are given by:
$$f_R(r) = {1 \over \pi} \frac{1}{1+r^2}, \quad r \in \mathbb{R}$$
$$f_S(s) = se^{-{1 \over 2} s^2}, \quad s \geq 0.$$
The joint pdf is given by
$$f_{R,S}(r,s) = \frac{1}{2\pi} \iint_\mathbb{R^2} dxdy \ e^{-(x^2+y^2)/2} \delta(y/x -r )\delta(\sqrt{x^2+y^2}-s) $$
Go to polar coordinates:
$$...=\frac{1}{2\pi} \int_0^{2\pi}d\theta \int_0^\infty d\rho \ \rho e^{-\rho^2/2}  \delta(\tan \theta -r) \delta(\rho-s).$$
These seperate:
$$= se^{-s^2/2} \ \left[ \frac{1}{2\pi}\int_0^{2\pi} d\theta \delta (\tan \theta -r) \right].$$
It remains to show that the bracket is the pdf for $R$. Using delta function properties:
$$\delta (\tan \theta -r) =\sum_i \frac{\delta(\theta - \theta_i)}{|\cos^2 \theta_i|}$$
with $\tan \theta_i =r$. There are two such roots on the interval $[0,2\pi]$ with the same weight, namely
$$\frac{1}{|\cos^2 \theta_i|} = \frac{1}{1+r^2},$$
therefore the result is 
$$f_{R,S}(r,s)  = \left[ se^{-s^2/2} \right] \ \left[{1 \over \pi} \frac{1}{1+r^2}\right] = f_S(s) f_R(r), \quad s\geq 0, r\in \mathbb{R}.$$
Therefore $R$ and $S$ are independent.
