Past exam of the options II course of the University of Oxford / 2010 / M / M4

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 densityA value at $u=0$ may be assigned arbitrarily without changing the distribution.

Integrate the joint density over $u$:Since both original densities are even, substitution $u\mapsto -u$ shows that the negative and positive halves contribute equally. Hence

For positive standard deviations $a,b$, substitute the two normal densities:The integral is $1/(a^{-2}+v^2{b}^{-2})$, by substituting the exponent's quadratic argument. Thus, with $c=b/a$,This is a centered Cauchy distribution of scale $c$. Since $P(X=0)=0$, the desired probability is

Take $U=Y$ and $V=X{Y}^2$. The inverse is $y=u$, $x=v/u^2$, with absolute Jacobian $u^{-2}$. The original unit-square support becomes $0<u<1$ and $0<v<u^2$, so the joint density is $u^{-2}$ there. For a fixed $0<v<1$, the admissible range is $\sqrt{v}<u<1$. ThereforeIt is nonnegative and integrates to ${\int }_0^1(v^{-1/2}-1)dv=1$, as required.