Independence of $X^2+Y^2$ and $\frac{X}{Y}$

Question

Let $X$ and $Y$ be two independent random variables with standard normal distributions. It is known that $X^2+Y^2$ is a random variable with $\chi^2(2)$-distribution, while $\frac{X}{Y}$ is a random variable with $F(1,1)$-distribution. Is there any easy way to show that $X^2+Y^2$ and $\frac{X}{Y}$ are independent? I think that the brutal computation may work, but there ought to be an easy way.

Answer 1

It is well-known that the pdf of the two-dimensional standard normal distribution $(X,Y)$ has the symmetries of a circle. This means that the pdf can be factored in polar coordinates as $f_R(r)f_\Theta(\theta)$ – with $f_\Theta(\theta)$ being uniform on the unit circle – so $R,\Theta$ are independent; since $R^2=X^2+Y^2$ and $\tan\Theta=\frac YX$, $X^2+Y^2$ and $\frac YX$ are also independent.

Answer 2

Geometrically $X^2+Y^2$ is the radius of $(X,Y)$, and $X/Y$ depends on its argument, so they are independent.

To prove it formally, this idea of radius and argument is well summarised by the Box-Muller transform: there exist independent random variables $R$ and $\Theta$ such that $(X,Y)$ has the same joint distribution as $(R\cos\Theta,R\sin\Theta)$. Then $(X^2+Y^2,X/Y)$ has the same distribution as $(R^2,1/\tan\Theta)$, hence we have independence.