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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.

Use the ratio-of-uniforms method onWriting $x=v/u$ gives $u\le e^{-x^2/(4{\sigma }^2)}$ and $v=ux$. Consequentlybecause the derivative vanishes at $x=\sigma \sqrt{2}$. Thus the normal ratio-of-uniforms envelope is the rectangle $0<u<1$, $|v|<\sigma \sqrt{2/e}$.

Generate independent $U_1,U_2$ with uniform distribution on $(0,1)$, set $u=U_1$ and $v=(2U_2-1)\sigma \sqrt{2/e}$, and accept precisely whenReject and repeat otherwise. On acceptance return $X=v/u$. One can compare logarithms instead of exponentials to avoid underflow.

For a proof, the change of variables $(u,x)\mapsto (u,v=ux)$ has Jacobian determinant $u$. A uniform point in $\mathcal{R}$ therefore gives an $x$-density proportional toNormalizing yields the required normal distribution with mean zero and variance ${\sigma }^2$. The region area is $\frac{1}{2}\int h(x)dx=\sigma \sqrt{\pi /2}$; dividing it by the rectangle area $2\sigma \sqrt{2/e}$ gives acceptance probability $\sqrt{\pi e}/4$.

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.