Joint density of $V_1=X^2+Y^2$ and $V_2=X^2-Y^2$

Question

Let $X,Y$ be two independent normally distributed random variables, with $V_1=X^2+Y^2$ and $V_2=X^2-Y^2$. I have to find $f_{V_1,V_2}(v_1,v_2)$.

I'm quite stuck… I tried to pass to polar coordinates with $\left\{\begin{matrix} R\in [0,+\infty)\\\theta \in [0,2\pi) \end{matrix}\right.$ and I obtained

$f_{V_1,V_2}(v_1,v_2)=f_X(R\cos \theta)f_Y(R\sin \theta)|J|=\frac{R}{2\pi}e^{-\frac{R^2}{2}}$

but I'm almost certain that it's wrong. Can you please give me any suggests?

Thanks in advance.

Answer 1

It is true that $V_1$ has a chi squared distribution with 2 degrees of freedom. Can you tell me how you arrived at the density $f_{V_1}$ from $f_{A,B}$? It's not clear in your comments.

In order to find the joint density you're looking for $f_{V_1,V_2}$ from $f_{A,B}$ you should first note that $(A,B)$ is supported on $\Omega:=[0,\infty)^2$. Set $T(a,b):=(a+b,a-b)=(v_1,v_2)$. Check that image of $\Omega$ under $T$ is $T(\Omega)=\{(v_1,v_2)\in\mathbb{R}^2:-v_1\leq{v_2}\leq{v_1}\}$. Moreover, for any $(v_1,v_2)$ in the interior of $T(\Omega)$, we have that

$f_{V_1,V_2}(v_1,v_2)=f_{A,B}(\frac{v_1+v_2}{2},\frac{v_1-v_2}{2})\left\vert\frac{\partial(a,b)}{\partial(v_1,v_2)}\right\rvert=\frac{e^{-v_1/2}}{2(\Gamma(1/2))^2\sqrt{v_1^2-v_2^2}}$

Clearly $f_{V_1,V_2}(v_1,v_2)=0$ outside of the purple region shown here.

If you want to get the density of $V_2$ you can simply integrate away $v_1$ in the density above to get the marginal $f_{V_2}$. Even if we had information about the density of $V_2$ beforehand we wouldn't necessarily be able to say $f_{V_1,V_2}(v_1,v_2)$ factors into $f_{V_1}(v_1)f_{V_2}(v_2)$ since $V_1$ and $V_2$ are not guaranteed to be independent.

Answer 2

Here's a hint: If $X$ and $Y$ are i.i.d. normal with mean 0 and variance 1, then $A:=X^2$ and $B:=Y^2$ are i.i.d. random variables that have a chi squared distribution with $1$ degree of freedom. Then $V_1=A+B$, $V_2=A-B$ and $(A,B)$ has a joint distribution of $f_{A,B}$ where $f_{A,B}(a,b):=\frac{e^{\frac{-(a+b)}{2}}}{2(\Gamma(1/2))^2\sqrt{ab}}$.