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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.

The support condition for importance sampling is missing from the printed assertion. Assume $M>1$, $E_f|h|<\infty$, and $Mg(z)>f(z)$ for $f$-almost every $z$. For a proposal $Z\sim g$, rejection has conditional probability $1-f(Z)/(Mg(Z))$. Its unconditional probability is $(M-1)/M$, so the conditional density of a rejected draw isThus the rejected-draw weight is precisely $f(z)/r(z)$, andConditionally on any positive rejection count $n$ among $N$ independent proposals, the rejected locations have this iid conditional density. Henceis an unbiased estimator conditionally on $n$, and therefore also conditionally on the event $0<n<N$. This is importance sampling from the rejected-proposal distribution.

The nonstrict assumption $f\le Mg$ alone does not suffice: the rejected distribution may completely miss part of the target support. For example, let $g$ be uniform on $[0,1]$, $f(z)=21_{[0,1/2]}(z)$, $M=2$, and $h=1$. All rejected proposals lie in $(1/2,1]$, where $f=0$, so the displayed estimator is always $0$, whereas $E_{f}h=1$. This remains a counterexample after conditioning on $0<n<N$. A strict envelope, such as choosing $M$ larger than the essential supremum of $f/g$, repairs the missing support condition. Near-equality can still give high-variance weights, even when it remains an unbiased estimator.