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I have seen this question be touched on tangentially in many posts, but there does not seem to be one distinct place that asks about it directly.

The following statement seems to be true, but I have not found a formal proof anywhere:

$X=Rcos\theta$, $Y=Rsin\theta$ are independent standard uniform random variables if and only if $R^{2}$ chi square with two degrees of freedom and $\theta$ is uniform $(0,2\pi)$, where $R^{2}$ and $\theta$ are also independent.

It would be nice to have the proof of both directions in one place, though I am particularly concerned with the backward direction. I have seen people write $R^{2}=X^{2}+Y^{2}$ and $\theta=\arctan(x/y)$ and blindly finding the jacobian without giving regard to the fact that $tan(\theta)=x/y$ has multiple inverses on $\theta \in (0,2\pi)$.

What is shocking to me is that ignoring the multiple inverse problem still gives them the right answer, but this is going against the variable transform theorem, so this makes me feel like I have no idea what is going on.

Any clarification would be truly appreciated.

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    $\begingroup$ The detailed example in my answer shows $rdr\wedge d\theta=dx\wedge dy$ for the transformation $(x,y)\leftrightarrow(r,\theta)$. For $r\gt 0, \theta\in [0,2\pi)$ this transformation is one-to-one with the punctured plane. The result is immediate. Note how the use of differential forms eliminates the somewhat shaky use of inverse tangents. $\endgroup$
    – whuber ♦
    Commented Nov 30, 2017 at 15:32
  • $\begingroup$ Thanks would you be able to link me to your answer. I have seen a few answers that make use of the $rdrd\theta=dxdy$ however, I am not completely clear why it should still work if that equality comes from the change of variable theorem for double integrals as well which involves a jacobian and so should fall into the same issue. $\endgroup$
    – DanRoDuq
    Commented Nov 30, 2017 at 15:58
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    $\begingroup$ I have pasted a link into that comment twice, and twice it has vanished! Let's try here: stats.stackexchange.com/a/154298/919. The methods shown there don't use Jacobians directly--they are designed to avoid them altogether. $\endgroup$
    – whuber ♦
    Commented Nov 30, 2017 at 16:07
  • $\begingroup$ This seems great. I would still be happy to see a more traditional approach this question as it seems like a very basic thing to know. $\endgroup$
    – DanRoDuq
    Commented Nov 30, 2017 at 16:35
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    $\begingroup$ Some people think the approach in my link is "very basic." :-) It has been in wide use in mathematics for over 120 years. $\endgroup$
    – whuber ♦
    Commented Nov 30, 2017 at 17:36

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