Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 40 / 3 / c

For $\sigma >0$, use the unnormalized normal density $h(x)=e^{-x^2/(2{\sigma }^2)}$. Its ratio region isThe bound on $u$ is one because $h$ is maximized at zero. To bound $|v|$, maximize $x\sqrt{h(x)}=x{e}^{-x^2/(4{\sigma }^2)}$ for $x\ge 0$. Its derivative vanishes at $x=\sqrt{2}\sigma$, giving $b=\sigma \sqrt{2/e}$. Thus the normal ratio-of-uniforms envelope is $0<u<1$, $-b<v<b$.

Draw independent $U_1,U_2\sim \mathrm{Unif}(0,1)$, set $u=U_1$ and $v=b(2U_2-1)$, and accept precisely whenreturning $X=v/u$. Equivalently compare logarithms to avoid underflow. Since $Z=\sigma \sqrt{2\pi }$, the region area is $\sigma \sqrt{\pi /2}$ and the rectangle area is $2\sigma \sqrt{2/e}$. ThereforeIt is independent of the normal scale. In the diagram the vertical coordinate is normalized by $\sigma$, so the same shape works for every positive scale.