 Solution

With $r$, $Z$, and $W$ as above, independence of the draws gives the variance of a sample mean:This formula allows infinite variance. To identify the optimum, write $\pi (\theta )=p(\theta \mid \mathcal{D})=r(\theta )/Z$. ThenThe last equality uses $\int \pi =\int q=1$. Its nonnegative integrand provesIndeed, all importance weights then equal $Z$. This is a formal optimum: evaluating the normalized posterior density generally already requires the unknown marginal likelihood.

Poor coverage of regions with substantial posterior mass produces large weights. In particular, excessively light proposal tails, a proposal that is too narrow, or a missed mode can make $\int r^2/q$ enormous or infinite. A proposal that is unnecessarily diffuse can also waste nearly all draws on negligible posterior probability. Omitting target support altogether violates unbiasedness, rather than merely increasing variance.