= Solution
Assume $a,b>0$, the <standard deviations>. Substitute the two <normal probability density>[normal densities] into (ii):
$$
f_V(v)=\frac1{\pi ab}\int_0^\infty u\exp\left[-\frac{u^2}2\left(\frac1{a^2}+\frac{v^2}{b^2}\right)\right]du=\frac{ab}{\pi(b^2+a^2v^2)}.
$$
The last step uses $\int_0^\infty ue^{-Au^2/2}du=1/A$. With $c=b/a$ this is $\boxed{f_V(v)=c/[\pi(c^2+v^2)]}$, the centered <Cauchy sequence>[Cauchy] <probability density function>[density] of scale $c$.
The desired event is $|V|<1$, outside the null event $X=0$. Integrating the <probability density function>[density] gives
$$
\boxed{P(|Y|<|X|)=\frac2\pi\arctan(1/c)=\frac2\pi\arctan(a/b).}
$$
Back to article page
