 Minimum-variance importance distribution

For $\mu =\int hf$ and $0<A=\int |h|f<\infty$, the proposal $q_{\ast }=|h|f/A$ minimizes the variance of the ordinary unbiased importance sampling estimator. The Cauchy-Schwarz inequality gives $\int h^2{f}^2/q\ge A^2$, so the minimum variance for $n$ independent samples is $(A^2-{\mu }^2)/n$. The proposal need only cover the nonzero integrand.