= Solution
Let $w(x)=f(x)/g(x)$ where $g(x)>0$, and set it to zero on the g-null set where $g=0$. The domination assumption makes $f=0$ there and gives $0\leq w\leq M$. Integrating it also gives $M\geq1$. For iid proposals $X_i\sim g$, the <importance sampling> estimator is
$$
\boxed{\widehat\theta_1=\frac1n\sum_{i=1}^n\phi(X_i)w(X_i).}
$$
<Cauchy-Schwarz inequality> under $f$ gives $|\theta|\leq(\int\phi^2f)^{1/2}<\infty$. Direct integration proves unbiasedness, $\mathbb E_g[\phi(X)w(X)]=\theta$. Moreover
$$
\mathbb E_g[\phi(X)^2w(X)^2]=\int_{g>0}\phi(x)^2\frac{f(x)^2}{g(x)}\,dx\leq M\int\phi(x)^2f(x)\,dx<\infty.
$$
Thus
$$
\boxed{\mathbb E\widehat\theta_1=\theta,\qquad\operatorname{Var}(\widehat\theta_1)=\frac{v_{\rm IS}}n,\quad
v_{\rm IS}=\int\phi^2\frac{f^2}{g}-\theta^2.}
$$
The iid <central limit theorem> applies to these finite-<variance> weighted observations:
$$
\boxed{\sqrt n(\widehat\theta_1-\theta)\ \Longrightarrow\ N(0,v_{\rm IS}).}
$$
If the <variance> is zero, this denotes the point mass at zero. This is the <bounded-weight importance-sampling moment bound>.
Back to article page
