= Solution
At a common proposal budget, the two asymptotic variances are
$$
v_{\rm IS}=\int\phi^2\frac{f^2}{g}-\theta^2,\qquad v_{\rm rej}=M\left(\int\phi^2f-\theta^2\right).
$$
Thus \b[prefer the estimator with the smaller proposal-budget <variance>; the stated assumptions do not give a universal winner]. <Importance sampling> uses every proposal and avoids an empty accepted sample, but those facts alone do not establish <variance> dominance. The bound from part (a) only says $v_{\rm IS}\leq Mv_f+(M-1)\theta^2$. If $\theta=0$, it does imply that <importance sampling> is at least as efficient asymptotically.
For explicit nonconstant counterexamples in both directions, take $g(x)=1$ and $f(x)=2x$ on $0<x<1$, and zero outside, with the sharp envelope $M=2$. For $\phi(x)=x$, direct integration gives $\theta=2/3$, $v_f=1/18$ and
$$
\boxed{v_{\rm IS}=16/45>1/9=v_{\rm rej}.}
$$
The accepted-sample mean is better. For the same densities but $\phi(x)=x-1$, the target <variance> is unchanged, while $\theta=-1/3$ and
$$
\boxed{v_{\rm IS}=1/45<1/9=v_{\rm rej}.}
$$
<Importance sampling> is now better. This is the <variance reversal under additive shifts of an importance-sampling integrand>.
There is a useful distinction about the <Rao-Blackwell theorem>. The weighted importance estimator is the conditional expectation, given the proposals, of the fixed-denominator rejection estimator $(M/n)\sum_i I_i\phi(X_i)$. It is not the conditional expectation of the random-denominator mean $N^{-1}\sum_i I_i\phi(X_i)$. Rao-Blackwell <variance> reduction for the former therefore does not prove a comparison with the latter. This is the <Rao-Blackwell identity for a fixed-denominator rejection estimator>. The appropriate answer is the <proposal-budget variance comparison of importance and rejection sampling>, with the empty-output convention specified.
Back to article page
