Consider a function $f: \mathcal X \to \mathbb R$ and a probability distribution $p$ with the support on $\mathcal X$ which we can evaluate up to a normalizing constant, i.e. we can only evaluate $\tilde p(x) = Zp(x)$ where $Z = \int_{\mathcal X} \tilde p(x) \,\mathrm dx$. Given an integral $$I = \int_{\mathcal X} p(x)f(x) \,\mathrm dx$$ that we want to estimate, a proposal distribution $q(x)$ which we can sample from and whose density we can evaluate, the self-normalized importance sampling estimator of $I$ is $$\hat I = \frac{\sum_{n = 1}^N W_n f(X_n)}{\sum_{n = 1}^N W_n},$$ where $X_n \sim q$ and $W_n = \tilde p(X_n) / q(X_n)$.

The optimal proposal (that minimizes the variance of $\hat I$) has the form $q_{\text{opt}}(x) \propto |f(x) - I| p(x)$ according to page 9 of Chapter 9 of Art Owen's Monte Carlo book, which in turn refers to Chapter 2 of Tim Hesterberg's thesis. However, I haven't been able to track down any proof of this in the thesis.

How would one go about showing this?


1 Answer 1


It should be noted that $q_{opt}$ actually minimizes the approximate variance given by the Delta method. You can get this by solving

$$ q_{opt} = \arg\min_q\mathbb{E}_q[w^2(X)(f(X)-I)^2], \; \text{ s.t.} \int q(x)dx=1 $$

Now, since: $$ \mathbb{E}_q[w^2(X)(f(X)-I)^2] = \int\frac{p^2(x)}{q(x)}(f(x)-I)^2dx = \int L(x,q(x))dx $$

for $L(x,q(x)) = \frac{p^2(x)}{q(x)}(f(x)-I)^2$, using Lagrange multipliers for calculus of variations yields $$ \begin{aligned} 0&=\frac{\partial L}{\partial q} +\lambda \\ &= -\frac{p^2(x)}{q^2(x)}(f(x)-I)^2 + \lambda \end{aligned} $$

Thus $$ q^2(x) = \frac{p^2(x)}{\lambda}(f(x)-I)^2 \implies q_{opt}(x) \propto p(x)|f(x)-I| $$

  • $\begingroup$ Should the variance be $\mathbb{V}_q w(x)f(x) = \mathbb{E}_q w^2f^2 - I^2$. Why do you subtract the expectation $I$ within the square? $\endgroup$ Commented Feb 17, 2022 at 14:03

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