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

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