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Suppose $X$ and $Y$ are independent standard normal random variables. Let $U = X^2 + Y^2$ and $V = \frac{X}{\sqrt{X^2 + Y^2}}$. (a) Find the joint pdf of $U$ and $V$. (b) Show that $U$ and $V$ are independent.

For this question, I calculated the modulus of the Jacobian of transformation to be $\frac{1}{2\sqrt{1-v^2}}$. Then, on multiplying with the joint pdf of $X$ and $Y$, I get the joint pdf to be $$f_{U,V}(u,v) = f_X(\sqrt{U}V)f_Y(\sqrt{U-UV^2})\frac{1}{\sqrt{1-v^2}}$$. However, upon substituting the normal pdfs and working through the calculations in part (b), I cannot seem to get $f_{U,V}(u,v) = f_{U}f_{V}$. On the right hand side I get $\frac{e^{-u^2/2}}{8\pi\sqrt{1-v^2}}$ - a factor of a half different to the left hand side. I used limits of $0 \leq u \leq \infty$ and $-1 \leq v\leq 1$ for the integrals on the right hand side. Clearly something is wrong, so can anyone give a full solution?

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I think it might just be a typo on your part, but the factor with $u$-dependence should be $e^{-u/2},$ not $e^{-u^2/2}.$

The Jacobian method gets the wrong answer here (or, rather, you are misapplying it). The reason is that the coordinate transformation is not one-to-one. Notice that $U$ and $V$ do not change when $Y\to -Y.$

It might help to consider the simpler problem that is extremely related. Let $A$ be uniform on $(0,2\pi)$ and then consider the distribution of $B = \cos(A).$ Note that the Jacobian gets the wrong answer here too. And a simpler but less related example would be for $A$ uniform on $(-1,1)$ and $B=A^2.$ Do you see how to use symmetry to fix the problem in each case?

The reason the first example is "extremely related" is that note that in polar coordinates $U=R^2$ and $V=\cos(\Theta).$ Note if you took your second variable to be $\Theta$ rather than $\cos(\Theta),$ the transformation would be one-to-one (except for the singularity at $R=0,$ but that doesn't cause an issue). And intuitively, don't we expect the angle to be uniformly distributed here?

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It is clear from the definition of $U$ and $V$ that a polar transformation would be useful.

Changing variables $(X,Y)\to (R,\Theta)$, you get the joint density of $(R,\Theta)$:

$$f_{R,\Theta}(r,\theta)=\frac{r}{2\pi}e^{-r^2/2}\mathbf1_{r>0\,,\,0<\theta<2\pi}$$

This was a one-to-one transformation.

But now if you consider the transformation $(R,\Theta)\to(U,V)$ such that $U=R^2$ and $V=\cos\Theta$, this is no longer one-to-one. Because for $v=\cos\theta$ and $0<\theta<2\pi$, you get two preimages for $\theta$, namely $\theta=\cos^{-1}v$ and $\theta=2\pi-\cos^{-1}v$, with $-1<v<1$. Let $J_1$ and $J_2$ be the two jacobians of transformation, one for each preimage.

Then we see that $$|J_1|=|J_2|=\frac{1}{2\sqrt{u}\sqrt{1-v^2}}$$

So by the transformation formula, density of $(U,V)$ takes the form

\begin{align} f_{U,V}(u,v)&=f_{R,\Theta}(\sqrt u,\cos^{-1}v)|J_1|+f_{R,\Theta}(\sqrt u,2\pi-\cos^{-1}v)||J_2| \\&=2\times \frac{\sqrt u}{2\pi}e^{-u/2}\frac{1}{2\sqrt{u}\sqrt{1-v^2}}\mathbf1_{u>0\,,\,|v|<1} \\&=\frac{1}{2}e^{-u/2}\mathbf1_{u>0}\frac{\mathbf1_{|v|<1}}{\pi\sqrt{1-v^2}} \end{align}

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