Solution. The distribution function is \[F(x) = \int _1^{x}\theta t^{-(\theta +1)}\,dt = 1 - x^{-\theta },\qquad x>1 .\] Setting \(u = 1-x^{-\theta }\) and solving, \[x^{-\theta } = 1-u \quad \Longrightarrow \quad x = (1-u)^{-1/\theta }.\] So generate \(U\sim \) UNIF\((0,1)\) and return \(X = (1-U)^{-1/\theta }\), or equivalently \(X = U^{-1/\theta }\) since \(1-U\) is uniform as well.

Solution. Form the cumulative probabilities \(0.2,\ 0.5,\ 0.9,\ 1.0\) and return the smallest value whose cumulative probability is at least \(U\): \[X = \begin {cases} 1, & 0<U\leq 0.2,\\ 2, & 0.2<U\leq 0.5,\\ 3, & 0.5<U\leq 0.9,\\ 4, & 0.9<U<1 . \end {cases}\] For \(U=0.63\) we have \(0.5<0.63\leq 0.9\), so \(X=3\).

Ordering the values by decreasing probability first — here \(3,2,1,4\) — reduces the expected number of comparisons, which is worth doing when the routine is called many times.

Solution. For \(x>0\), \[P\left (-\log U > x\right ) = P\left (U < e^{-x}\right ) = e^{-x},\] so \(-\log U\) has the exponential survival function with rate \(1\).

A GAM\((n,1)\) variable is the sum of \(n\) independent EXP\((1)\) variables — which is the closure result of the convolutions chapter. So generate \(U_1,\dots ,U_n\) independently and return \[X = -\sum _{i=1}^{n}\log U_i = -\log \left (\prod _{i=1}^{n}U_i\right ),\] the product form saving \(n-1\) logarithms.

Solution. Here \(g(x)=1\) on \((0,1)\), so \(c\) must satisfy \(f(x)\leq c\) for all \(x\), that is \(c \geq \max _{0<x<1} 6x(1-x)\). Differentiating, the maximum is at \(x=\tfrac 12\) with value \(6\cdot \tfrac 12\cdot \tfrac 12 = 1.5\). Hence \(c = 1.5\).

The acceptance probability is \(1/c = 2/3\), so the expected number of draws per accepted value is \(c = 1.5\). Two thirds of the proposals are kept, which for a target this close in shape to its envelope is about as efficient as rejection sampling gets.

Solution. The inverse transform requires \(F^{-1}\) in closed form. The standard normal distribution function \[\Phi (x) = \int _{-\infty }^{x}\frac {1}{\sqrt {2\pi }}e^{-t^{2}/2}\,dt\] has no elementary antiderivative, so \(\Phi ^{-1}\) cannot be written in elementary functions; it can only be evaluated numerically, which defeats the purpose of a cheap generator.

Two methods avoid this. Box–Muller converts two uniforms into two independent normals through the polar representation, using only a logarithm, a square root and two trigonometric calls. Rejection sampling with a heavier-tailed envelope — the Cauchy or a double exponential — also works, since both dominate the normal density after scaling. Numerical inversion of \(\Phi \) is a third route and is what most software actually uses, but it is an approximation rather than an exact transformation.