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Ordinary importance sampling is unbiased

Statement

Let $(X, X, μ)$ be a measure space, let $\tilde{p} \geq 0$ and $q \geq 0$ be measurable densities with $Z_{p} = \int \tilde{p} d μ \in (0, \infty)$ and $\int q d μ = 1$ , and set $π (d x) = \tilde{p} (x) μ (d x) / Z_{p}$ and $Q (d x) = q (x) μ (d x)$ . Let $f : X \to R$ be measurable with $\int ∣ f ∣ \tilde{p} d μ < \infty$ . Suppose $q > 0$ $μ$ -almost everywhere on ${f \tilde{p} \neq 0}$ and $X_{1}, \dots, X_{n}$ are IID with law $Q$ . For the ordinary importance sampling estimator in Ordinary importance sampling, $E_{Q} [{\hat{μ}}_{I S}] = \int f d π .$

Facts & Assumptions

Given: The densities, integrability, support, and IID sampling hypotheses in the Statement.

[F1]

Ordinary importance sampling: With $Y_{i} = f (X_{i}) \tilde{p} (X_{i}) / (Z_{p} q (X_{i}))$ when $q (X_{i}) > 0$ and $Y_{i} = 0$ otherwise, the estimator is ${\hat{μ}}_{I S} = n^{- 1} \sum_{i = 1}^{n} Y_{i}$ .

[F2]

Change of variables for expectation: For a random element $X$ with law $Q$ , a nonnegative measurable $g$ satisfies $E [g (X)] = \int g d Q$ , possibly with value $+ \infty$ ; if $g (X)$ is integrable, the same identity holds as a finite integral.

[F3]

The Lebesgue integral is linear on $L^{1} (μ)$ : The Lebesgue integral is linear on integrable functions; in particular, the expectation of a finite average of integrable random variables is the average of their expectations.

[F4]

Expectation of a nonnegative or integrable random variable: The expectation of an integrable real random variable is its finite Lebesgue integral.

[F5]

Integrating against a density agrees with integrating the product: for a nonnegative measurable density $r$ and nonnegative measurable $h$ , $\int h d (r d μ) = \int h r d μ$ , including infinite values. Integrable real and complex functions, and their integrals: an absolutely integrable real function has integral equal to the finite difference of the integrals of its positive and negative parts.

Proof

Given: Use the target expectation $θ : = \int f d π$ and the estimator specified in Ordinary importance sampling.

1.1givenF1F2F4algebraF5

Use [F5] with density $r = q$ for integration against $Q$ and with $r = \tilde{p} / Z_{p}$ for integration against $π$ . For every $i$ , apply the nonnegative case of [F2] to $∣ Y_{i} ∣$ . Since $X_{i}$ has law $Q$ , $E_{Q} ∣ Y_{i} ∣ = \int_{{q > 0}} \frac{∣ f (x) ∣ \tilde{p} (x)}{Z_{p} q (x)} q (x) d μ (x) = \frac{1}{Z_{p}} \int_{{q > 0}} ∣ f ∣ \tilde{p} d μ \leq \frac{1}{Z_{p}} \int_{X} ∣ f ∣ \tilde{p} d μ < \infty .$ Thus $Y_{i}$ is $Q$ -integrable. For any real measurable $h$ whose absolute integral is finite against $r d μ$ , apply [F5] separately to $h^{+} = \max (h, 0)$ and $h^{-} = \max (- h, 0)$ : both product integrals are finite because $h^{\pm} \leq ∣ h ∣$ . Subtracting their finite identities yields $\int h d (r d μ) = \int h r d μ$ , since $(h r)^{\pm} = h^{\pm} r$ for $r \geq 0$ (with the density convention $0 \cdot \infty = 0$ ). The integrable case of [F2] therefore gives $E_{Q} [Y_{i}] = \int_{{q > 0}} \frac{f (x) \tilde{p} (x)}{Z_{p} q (x)} q (x) d μ (x) = \frac{1}{Z_{p}} \int_{{q > 0}} f \tilde{p} d μ .$ The support condition makes $f \tilde{p} = 0$ $μ$ -almost everywhere on ${q = 0}$ ; hence this last integral equals $Z_{p}^{- 1} \int_{X} f \tilde{p} d μ = \int f d π = θ$ .

2.1F1F3F4step 1.1algebra∎

By [F1], ${\hat{μ}}_{I S} = n^{- 1} \sum_{i = 1}^{n} Y_{i}$ . Apply [F3] and step 1.1 to obtain $E_{Q} [{\hat{μ}}_{I S}] = n^{- 1} \sum_{i = 1}^{n} E_{Q} [Y_{i}] = θ$ .

Depends on

Used by

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Sources

Quan Zhou, STAT 689 Unit 1: Introduction, Rejection and Importance Sampling, §1.3, Theorem 1.2 (standard reference, not scraped)
Thomas B. Schön and Fredrik Lindsten, Learning of dynamical systems: Particle filters and Markov chain methods, Chapter 4 §§4.3.1, 4.3.3 (standard reference, not scraped)
Data Science and Statistical Modelling, Chapter 5 §5.3.1, Theorem 5.3 (standard reference, not scraped)
Tuan Anh Le, Importance Sampling, §Importance Sampling (standard reference, not scraped)