If $X$ and $Y$ are independent $N(0,\sigma^2)$, then $X^2+Y^2$ and $X/Y$ are independent?

If $X$ and $Y$ are independent, then $X^2+Y^2$ and $X/Y$ are independent? I was solving the problem for the case that $X$ and $Y$ are independent $N(0,\sigma^2)$.

So i found that $X^2+Y^2$ is negative exponential, $X/Y$ is Cauchy distribution. How can i prove that $X^2+Y^2$ and $X/Y$ are independent?

Also, what it will be if the case for general random variable $X$, $Y$ with $X^2+Y^2$ and $X/Y$ are defined?

Solution 1:

Joint density of $(X,Y)$ is

$$f_{X,Y}(x,y)=\frac{1}{2\pi\sigma^2}\exp\left[-\frac{1}{2\sigma^2}\left(x^2+y^2\right)\right]\qquad,\,(x,y)\in\mathbb R^2$$

Transforming to polar coordinates (as mentioned in comments)

$$(X,Y)\to (R,\Theta)$$ such that $$X=R\cos\Theta\quad,\quad Y=R\sin\Theta$$

Then, $$R^2=X^2+Y^2\quad,\quad \cot\Theta=\frac{X}{Y}$$

Now, $$(x,y)\in\mathbb R^2\implies r>0\,,\,0<\theta<2\pi$$

Absolute value of jacobian of transformation is $$|J|=r$$

So joint density of $(R,\Theta)$ is

\begin{align} f_{R,\Theta}(r,\theta)&=\frac{r}{2\pi\sigma^2}e^{-r^2/2\sigma^2}\,\mathbf1_{r>0\,,\,0<\theta<2\pi} \\\\&=\frac{r}{\sigma^2}e^{-r^2/2\sigma^2}\mathbf1_{r>0}\,\frac{1}{2\pi}\mathbf1_{0<\theta<2\pi} \end{align}

Clearly, $R$ and $\Theta$ are independently distributed.

At this point, you can conclude that $R^2$ and $\cot\Theta$ are independently distributed because they are measurable functions of $R$ and $\Theta$ respectively. Or if you want to fully derive the joint density of $(R^2,\cot\Theta)$ and then conclude independence, you would have to use another change of variables. But note that this time around, the transformation would not be one-to-one.

The distribution of $X^2+Y^2$ would be a scaled $\chi^2_2$ distribution, and that of $X/Y$, as you say, would be standard Cauchy.

As for the 'general' case, we need to know the exact joint distribution of $(X,Y)$ to say anything about independence of $(g(X,Y),h(X,Y))$ for arbitrary function $g$ and $h$.