Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 37 / 4

The generalized quantile function isThe tail limits of a cumulative distribution function make the defining set nonempty and its infimum finite. Monotonicity and right continuity imply $F^{-1}(u)\le x$ exactly when $u\le F(x)$. To see the direction involving the infimum, let points in the defining set decrease towards it; right continuity gives $F(F^{-1}(u))\ge u$, so every larger $x$ also qualifies. Consequently, for a uniform distribution variable $U$,The probability-zero endpoints $U=0,1$ can be assigned arbitrary outputs. This proves inverse transform sampling even for distributions with atoms or flat portions; continuity or strict monotonicity of $F$ is not required.

Choose a proposal probability density function $q$ positive wherever $|h|f$ is nonzero, up to null sets, and draw independent and identically distributed random variables $Y_1,\dots ,Y_n$ from $q$. The ordinary importance sampling estimator isThe support condition for importance sampling and absolute integrability give ${\mathbb{E}}_q[h(Y)f(Y)/q(Y)]=\int hf=\mu$, so this is an unbiased estimator. Its variance, possibly infinite, isLet $A=\int |h|f$. If $A>0$, the Cauchy-Schwarz inequality yieldsEquality holds for $q$ proportional to $|h|f$. Thus the minimum-variance importance distribution and the minimum variance areFor a constant-sign integrand this is zero variance. If $A=0$, the integrand vanishes almost everywhere and the zero estimator already has zero variance. Requiring the proposal to cover all of the target's support, including where $h=0$, can exclude $q_{\ast }$; in that stricter class, $(1-ϵ)q_{\ast }+ϵf$ approaches the same infimum as $ϵ\downarrow 0$. This distinguishes coverage of the integral from coverage of every target event.

Integrating the Weibull distribution density gives $F(x)=1-\exp (-x^{\alpha }/\beta )$ for $x>0$. Inverse transform sampling therefore givesReplacing $1-U_1$ by $U_1$ produces the same probability distribution. Both uniforms belong to $(0,1)$ with probability one. Here the conventional Weibull scale is ${\beta }^{1/\alpha }$, rather than $\beta$.

The Box-Muller transform uses independent uniforms to setThe radius has the unit Rayleigh distribution, with density $r{e}^{-r^2/2}$ for $r>0$, and the angle is independently uniform on $[0,2\pi )$. Their joint density is $(2\pi {)}^{-1}r{e}^{-r^2/2}$. The Cartesian change of variables has absolute Jacobian determinant $r$, so the joint density of $(X_1,X_2)$ isThe factorization proves that the outputs are independent random variables, each with a standard normal distribution.

Recognize the integrand as a Gaussian scale mixture. If $R$ has the unit Rayleigh distribution and $Z$ is an independent standard normal distribution variable, then the conditional density of $RZ$ given $R=r$ is $(\sqrt{2\pi }r{)}^{-1}{e}^{-x^2/(2r^2)}$. Multiplying by the radius density givesThus use $U_1$ for the independent radius and $U_2,U_3$ for the normal output of the Box-Muller transform:The mixture argument also proves that $g$ integrates to one, by the Tonelli theorem. As a check, $R^2$ is exponential of rate $1/2$, so the characteristic function of $RZ$ is $\mathbb{E}{e}^{-t^2{R}^2/2}=1/(1+t^2)$. This is the Rayleigh-normal scale mixture, with Laplace distribution density $e^{-|x|}/2$.