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For the unnormalized Laplace distribution probability density function $h(x)=e^{-|x-1|/2}$, the ratio-of-uniforms method uses $0<u\le \sqrt{h(v/u)}$. The maximum of $h$ is one. On $x\le 0$, the maximum of $x^{2}h(x)$ occurs at $x=-4$ and equals $16e^{-5/2}$. On $x\ge 0$, it occurs at $x=4$ and equals $16e^{-3/2}$. Thus the displayed rectangle contains the region. Its area is $4(e^{-3/4}+e^{-5/4})$, while the target region has area $\frac{1}{2}\int h=2$. Uniform proposals in the rectangle, accepted when $u^2\le h(v/u)$, therefore return the correct Laplace distribution with acceptance probability $[2(e^{-3/4}+e^{-5/4}){]}^{-1}$.

Integrating out the second coordinate gives the probability density function of the marginal distributionThus the ratio has the normalized target probability density function. The factor $s$ from the Jacobian determinant is essential; merely transforming the rectangular coordinates as if they remained uniform would give a different probability density function. This is the basis of the ratio-of-uniforms method.

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 $\sqrt{\pi e}/4$.

Let $h\ge 0$ be proportional to the target probability density function, with $0<Z=\int h(x)dx<\infty$. Define the ratio-of-uniforms method through the regionDraw $(U,V)$ uniformly with respect to area on $A_h$ and return $X=V/U$. To prove its law, change variables from $(u,x)$ to $(u,v)=(u,ux)$. The Jacobian determinant is $u$, and the allowed interval for $u$ is $0<u\le \sqrt{h(x)}$. ConsequentlyThe joint density of $(U,X)$ is therefore $2u/Z$ on that interval. Integrating out $u$ givesFor an already normalized density, $Z=1$ and the region has area $1/2$. The name refers to the two coordinates of a uniformly sampled region; drawing two unrelated uniforms on arbitrary intervals and taking their ratio would not give the target law.