Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 40 / 1 / a

Write $H={\int }_{\mathbb{R}}h(x)dx$, and assume $0<H<\infty$; a uniform distribution on the region requires positive area. Set $x=V/U$, $s=U$, so the inverse transformation is $U=s$, $V=sx$. Its Jacobian determinant has absolute value $s$ for $s>0$. The region becomes $x\in \mathbb{R}$, $0<s\le \sqrt{h(x)}$, and its area isThe original joint probability density function is therefore $2/H$ on $C_h$. The change of variables formula givesThe boundary $U=0$ has probability zero, so division by $U$ causes no ambiguity for the resulting random variable.

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.

Here $h(x)$ attains its maximum one at $x=1$, so $a=1$. For $x\le 0$, put $t=-x\ge 0$. Then $x^{2}h(x)=e^{-1/2}{t}^2{e}^{-t/2}$. Its derivative is $e^{-1/2}{e}^{-t/2}t(2-t/2)$, so its maximum is at $t=4$, with value $16e^{-5/2}$.

For $0\le x\le 1$, $x^{2}h(x)=x^2{e}^{(x-1)/2}$ is increasing and has maximum one. For $x\ge 1$, $x^{2}h(x)=x^2{e}^{-(x-1)/2}$ has logarithmic derivative $2/x-1/2$, so its maximum is at $x=4$, with value $16e^{-3/2}>1$. Therefore the Laplace ratio-of-uniforms envelope has the finite boundsThese calculations establish finiteness and identify the tight axis-aligned bounds, rather than relying only on exponential tail decay.

Independently generate $U$ uniform on $(0,1)$ and $V$ uniform on $(b_1,b_2)$, using the bounds above. Accept the proposal ifotherwise generate a fresh pair. On acceptance return $X=V/U$. The bounding rectangle contains the entire ratio region: $U\le \sqrt{h(X)}\le a$, and $V=XU$ lies between the extrema of $x\sqrt{h(x)}$. Conditioning a uniform rectangle point on acceptance makes it uniform on $C_h$, so the preceding marginal distribution calculation proves that the returned value has the desired Laplace distribution.

For this target $H=4$, so the region has area two. The rejection sampling acceptance probability isEndpoints have zero probability and can be excluded to avoid divisions by zero in implementation.