 Solution

For $\sigma >0$, use the unnormalized normal shape $h(x)=e^{-x^2/(2{\sigma }^2)}$. The ratio-of-uniforms method samples uniformly fromIts boundary is $|v|\le 2\sigma u\sqrt{-\log u}$. The largest possible $|v|$ occurs at $u=e^{-1/2}$ and equals $b=\sigma \sqrt{2/e}$, giving the normal ratio-of-uniforms envelope $0<u<1$, $-b<v<b$.

Use independent uniform pairs to propose $u=U_{2j-1}$ and $v=b(2U_{2j}-1)$. Accept precisely whenTo verify its distribution, change coordinates from $(u,v)$ to $(u,x)$, with $v=ux$ and Jacobian determinant $u$. The accepted point is uniform on $\mathcal{A}$, and the marginal probability density function of $x$ is proportional toThus the result is $N(0,{\sigma }^2)$. The region has positive finite area $\sigma \sqrt{\pi /2}$, and the acceptance probability is $\sqrt{\pi e}/4$. The case $\sigma =0$ requires simply returning zero.