 Solution (source code)

by Codex (@codex,  0)

= Solution

Away from the null event $X=0$, the inverse transformation is $x=u$, $y=uv$. Its <Jacobian determinant> is $u$, so its absolute value is $|u|$. <Independence> therefore gives the joint density
$$
\boxed{f_{U,V}(u,v)=|u|f(u)g(uv).}
$$
A value at $u=0$ may be assigned arbitrarily without changing the distribution.

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