 Solution

Put $A=\int |\theta (x)|f(x)dx$, with $0<A<\infty$. The Cauchy-Schwarz inequality applied to $|\theta |f/\sqrt{g}$ and $\sqrt{g}$ givesEquality holds exactly when $|\theta |f/\sqrt{g}$ is proportional to $\sqrt{g}$ on the contributing probability support. Therefore the minimum-variance importance distribution isIf $\theta$ has one sign, $A=|\mu |$ and the ideal proposal gives zero variance. Its normalization or exact sampling may nevertheless be impractical. If $A=0$, the integrand vanishes almost everywhere and the estimator is identically zero.

The minimizing probability density function may vanish where $\theta =0$ even though $f>0$ there. This is valid for the ordinary estimator, which needs probability support only for its integrand. If one insists on exactly the same positive probability support as $f$, the bound is instead an infimum whenever this vanishing occurs, approached by $g_{\delta }=(1-\delta )g_0+\delta f$ as $\delta \downarrow 0$: because $g_{\delta }\ge (1-\delta )g_0$, the second moment lies between $A^2$ and $A^2/(1-\delta )$, proving convergence to the bound.