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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.

For independent draws $X_i$ from a normalized proposal $g$, use the importance sampling identityThus the unbiased estimator and its variance areThe probability support condition is $g>0$ wherever $\theta f\ne 0$, and the variance formula requires a finite second moment. Since both probability densities are normalized, no self-normalization of weights is needed.

In the proposed example, normalize $g$ to $g(x)=3x^2$ on $(0,1)$. It is sampled by $X=U^{1/3}$ for uniform $U$. Taking $\theta (x)=x$ givesAs a check, ${\mu }_g=\frac{4}{\pi }{\int }_0^{1}x\sqrt{1-x^2}dx=4/(3\pi )$. The second moment of one weighted observation isso $\mathrm{Var}({\hat{\mu }}_g)=16/(9{\pi }^{2}n)$. The weight's singularity at zero does not make this variance infinite.

To prove the optimal proposal rule, apply the Cauchy-Schwarz inequality:Equality holds when the two factors are proportional almost everywhere, giving the minimum-variance importance distributionFor the present nonnegative integrand, normalization givesUnder this proposal, $xf(x)/g_0(x)=4/(3\pi )$ is constant, so the variance is exactly zero. If sampling is wanted, its distribution function is $1-(1-x^2{)}^{3/2}$, with inverse $X=\sqrt{1-(1-U{)}^{2/3}}$. In general the optimal proposal's normalizing constant can itself be the unknown integral, limiting this zero-variance construction's practical use.

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 integrand is positive, so the minimum-variance importance distribution is its target-weighted probability density function, normalized by $\theta =\sqrt{e}$:By completing the square,Thus $q^{\star }$ is the multivariate normal distribution $N(-e_1,I_d)$, where $e_1=(1,0,\dots ,0)$ is the first standard basis vector. Its importance weight and weighted integrand areA variance cannot be negative, and this proposal makes every summand constant, proving optimality directly. It is admissible under the arbitrary-proposal condition even though its mean vector is not zero. Although optimal proposals usually involve an unknown normalizing integral, here both that integral and the resulting normal distribution are explicit.