Let $X_1, X_2 \sim N(0, 1)$ be independently and identically distributed, and $Y_1 = X_1^2+X_2^2$, $Y_2 = X_1/X_2$. Are $Y_1$ and $Y_2$ independent?

$Y_1 \sim \chi^2$ and $Y_2$ is Cauchy, but is there any way to come up with the joint distribution other than the expensive integration?


Let us find joint CDF: for $t>0$, $u\in\mathbb R$ $$ \mathbb P\left(Y_1\leq t, Y_2\leq u\right)=\mathbb P\left(X_1^2+X_2^2\leq t,\, X_1\leq uX_2,\, X_2> 0\right)+\mathbb P\left(X_1^2+X_2^2\leq t,\, X_1\geq uX_2,\, X_2< 0\right). $$ The second probability is equal to the first one, since $X_i$ can be replaced by $-X_i$, and then the second event coincides with the first one, and the probability will not change. Therefore $$ \mathbb P(Y_1\leq t, Y_2\leq u)=2\mathbb P(X_1^2+X_2^2\leq t,\, X_1\leq uX_2,\, X_2> 0)=2\iint\limits_D \dfrac{1}{2\pi}e^{-\tfrac{x^2+y^2}{2}}\,dx\,dy. $$ Here $D=\left\{(x,y): x^2+y^2\leq t,\, y\leq ux,\, x>0\right\}$

Polar change of variables $$x=r\cos\varphi$$ $$y=r\sin\varphi$$ $$dx\,dy=r\, dr\,d\varphi$$ $$x^2+y^2=r^2$$ $$D=\{(r,\varphi): r\leq \sqrt{t},\, \tan(\varphi)<u\} $$ leads to $$\mathbb P(Y_1\leq t, Y_2\leq u)=\frac{1}{\pi}\int\limits_{-\pi/2}^{\arctan(u)}\int\limits_0^\sqrt{t}r e^{-r^2/2}\,dr\,d\varphi=\dfrac{\arctan(u)+\pi/2}{\pi}\times \left(1-e^{-t/2}\right). $$ This is the product of CDF's of $Y_1$ and $Y_2$: Cauchy distribution and $\chi_2^2$.

No expensive integration is required.

You can also express $X_1$ and $X_2$ as $X_2=r\cos\varphi$, $X_1=r\sin\varphi$, $$X_1^2+X_2^2=r^2 \text{ and } X_1/X_2=\tan\varphi$$ and use that the joint distribution of $(X_1,X_2)$ is spherically symmetric distribution, which should imply the independence of these values.

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