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

The inverse change of variables is $x=u$, $y=uv$, whose Jacobian determinant is $u$. Since $X,Y$ are independent, the original joint density is $f(x)g(y)$. ThusThe transformation is nonsingular away from $u=0$; this line has probability zero because $X$ has a density, so no extra mass is omitted.

Integrate the joint density in $u$. Symmetry gives $f(-u)g(-uv)=f(u)g(uv)$, so the integrand $|u|f(u)g(uv)$ is even. HenceTonelli's theorem justifies marginalization for these nonnegative densities.

Assume $a,b>0$, the standard deviations. Substitute the two normal densities into (ii):The last step uses ${\int }_0^{\infty }u{e}^{-A{u}^2/2}du=1/A$. With $c=b/a$ this is $\boxed{f_V(v)=c/[\pi (c^2+v^2)]}$, the centered Cauchy density of scale $c$.

The desired event is $|V|<1$, outside the null event $X=0$. Integrating the density gives

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.