TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Finite partition-ratio estimator and overlap variance

Statement

Let $X$ be a finite nonempty set equipped with counting measure. For $a \in {0, 1}$ , let $w_{a} : X \to (0, \infty)$ be a finite positive weight, define $Z_{a} = \sum_{x \in X} w_{a} (x)$ , and define the probability mass function $π_{a} (x) = w_{a} (x) / Z_{a}$ . For an integer $n \geq 1$ , let $X_{1}, \dots, X_{n}$ be independent random variables with common law $π_{0}$ . Define $R (x) = w_{1} (x) / w_{0} (x)$ , ${\hat{r}}_{n} = n^{- 1} \sum_{j = 1}^{n} R (X_{j})$ , and $χ^{2} (π_{1} ∥ π_{0}) = \sum_{x \in X} π_{0} (x) (π_{1} (x) / π_{0} (x) - 1)^{2}$ . Then

$E_{π_{0}} [{\hat{r}}_{n}] = \frac{Z_{1}}{Z_{0}}, \frac{{Var}_{π_{0}} ({\hat{r}}_{n})}{(Z_{1} / Z_{0})^{2}} = \frac{χ^{2} (π_{1} ∥ π_{0})}{n} .$

The estimator uses only the finite raw-weight ratios $w_{1} (X_{j}) / w_{0} (X_{j})$ ; the partition sums define the target ratio and probability laws but are not inputs to its computation.

Statement references

Ordinary importance sampling

Ordinary importance sampling has finite variance and a CLT under a weighted second moment

Facts & Assumptions

Given: The finite nonempty state set, strictly positive finite weights $w_{0}, w_{1}$ , their finite normalizing sums $Z_{0}, Z_{1}$ , and independent samples $X_{j} \sim π_{0}$ .

[F1]

Ordinary importance sampling: for an unnormalized target $\tilde{p}$ , proposal density $q$ , integrand $f$ , and target normalizer $Z_{p}$ , the weighted value is $f \tilde{p} / (Z_{p} q)$ on $q > 0$ and zero otherwise. Its support and integrability conditions are explicit, and its normalizing constant $Z_{p}$ is a known input to that algorithm.

[F2]

Ordinary importance sampling has finite variance and a CLT under a weighted second moment: under the ordinary importance-sampling hypotheses and finite weighted second moment, the finite-sample variance is the variance of one weighted value divided by $n$ ; the finite-sample conclusion is choice-free.

Proof

Given: [F1], [F2], and the finite positive weights and IID $π_{0}$ sample in the Statement.

1.1F1givenalgebra

Put $R = w_{1} / w_{0}$ . Finite summation gives $E_{π_{0}} R = \sum_{x} (w_{0} (x) / Z_{0}) (w_{1} (x) / w_{0} (x)) = Z_{1} / Z_{0}$ . Since $w_{0} (x), w_{1} (x) > 0$ on every point of $X$ , the formal ordinary-importance-sampling setup with counting measure, $\tilde{p} = w_{1}$ , $q = π_{0}$ , $f \equiv 1$ , and $Z_{p} = Z_{1}$ has full support and finite weighted second moment. Its weighted value is $Y = (Z_{0} / Z_{1}) R = π_{1} / π_{0}$ , and finite summation gives $E_{π_{0}} Y = (Z_{0} / Z_{1}) E_{π_{0}} R = 1$ . Here $Z_{1}$ is used only to identify this mathematical random variable; it is not supplied to or computed by the raw-weight ratio estimator.

2.1F2givenstep 1.1algebra

Apply [F2] to this formal weighted variable with mean $1$ from step 1.1. It gives $Var ({\overline{Y}}_{n}) = n^{- 1} (E_{π_{0}} [Y^{2}] - 1)$ . Since ${\overline{Y}}_{n} = (Z_{0} / Z_{1}) {\hat{r}}_{n}$ , rescaling gives $Var ({\hat{r}}_{n}) / (Z_{1} / Z_{0})^{2} = n^{- 1} (\sum_{x} π_{0} (x) (π_{1} (x) / π_{0} (x))^{2} - 1)$ .

3.1givenstep 2.1algebra∎

The probability masses sum to one, so expanding the square gives $χ^{2} (π_{1} ∥ π_{0}) = \sum_{x} π_{0} (x) (π_{1} (x) / π_{0} (x))^{2} - 2 \sum_{x} π_{1} (x) + \sum_{x} π_{0} (x) = \sum_{x} π_{0} (x) (π_{1} (x) / π_{0} (x))^{2} - 1$ . Substitution into step 2.1 proves the variance identity.

Remarks

The proof uses the ordinary importance-sampling variance identity only after symbolically normalizing the target. It does not treat the ordinary importance-sampling algorithm, whose contract requires a known target normalizer, as a procedure for computing an unknown partition ratio.

Finite partition-ratio estimator and overlap variance

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources