Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / M / M4 / b / Solution

Use $U=Y$, $V=X{Y}^2$, with inverse $y=u$, $x=v/u^2$. The Jacobian is $u^{-2}$, and the transformed support is $0<u<1$, $0<v<u^2$. Thus, for $0<v<1$,Therefore $\boxed{f_V(v)=(v^{-1/2}-1)1_{(0,1)}(v)}$. The integral over its support is $2-1=1$, checking normalization; there are no endpoint atoms.