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

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Mean and relative second moment of independent bridge ratios

Statement

Let $X$ be a finite nonempty set, let $K \geq 1$ , and let $w_{0}, \dots, w_{K} : X \to (0, \infty)$ be finite positive weights. For each $k$ , set $Z_{k} = \sum_{x \in X} w_{k} (x)$ and $π_{k} (x) = w_{k} (x) / Z_{k}$ . For strictly positive probability mass functions $ν, μ : T \to (0, 1]$ on any finite nonempty set $T$ (so $\sum_{x \in T} ν (x) = \sum_{x \in T} μ (x) = 1$ ), define $χ^{2} (ν ∥ μ) = \sum_{x \in T} μ (x) (ν (x) / μ (x) - 1)^{2}$ . On a common probability space $(Ω, F, P)$ , let $X_{k, j} : Ω \to (X, P (X))$ for $0 \leq k < K$ and $1 \leq j \leq n_{k}$ be mutually independent random variables with $P (X_{k, j} = x) = π_{k} (x)$ for each $x \in X$ , where $n_{k} \geq 1$ ; these are the supplied stage samples in alg-csm-frontier-independent-bridge-ratios. Then

$E [\hat{R}] = \frac{Z_{K}}{Z_{0}}, \frac{E [{\hat{R}}^{2}]}{(Z_{K} / Z_{0})^{2}} = \prod_{k = 0}^{K - 1} (1 + \frac{χ^{2} (π_{k + 1} ∥ π_{k})}{n_{k}}) .$

Consequently, $Var (\hat{R}) / (Z_{K} / Z_{0})^{2}$ is the displayed product minus $1$ . This is an exact finite-sample result under independent exact draws; it gives no claim for reused trajectories, approximate Gibbs draws, or adaptively selected bridges.

Statement references

Independent finite bridge-ratio estimator

Finite partition-ratio estimator and overlap variance

Facts & Assumptions

Given: The finite positive weights, stage laws, sample counts, and mutually independent stage samples in the Statement.

[F1]

Finite partition-ratio estimator and overlap variance: for a finite positive pair of weights and $n_{k}$ independent draws with common law $π_{k}$ , the adjacent raw-ratio sample mean has mean $Z_{k + 1} / Z_{k}$ and relative variance $χ^{2} (π_{k + 1} ∥ π_{k}) / n_{k}$ .

[F2]

Independent finite bridge-ratio estimator: the procedure returns the product of those adjacent sample means and requires all draws across stages to be mutually independent.

Proof

Given: [F1], [F2], and the finite positive weights and mutually independent stage samples.

1.1F1givenalgebra

For each $k$ , apply [F1] to $(w_{k}, w_{k + 1})$ and the $n_{k}$ independent samples with common law $π_{k}$ . With $r_{k} = Z_{k + 1} / Z_{k}$ , the stage mean ${\overline{R}}_{k}$ satisfies $E [{\overline{R}}_{k}] = r_{k}$ and $Var ({\overline{R}}_{k}) / r_{k}^{2} = χ^{2} (π_{k + 1} ∥ π_{k}) / n_{k}$ . Since $E [{\overline{R}}_{k}^{2}] = Var ({\overline{R}}_{k}) + (E [{\overline{R}}_{k}])^{2}$ , it follows that $E [{\overline{R}}_{k}^{2}] / r_{k}^{2} = 1 + χ^{2} (π_{k + 1} ∥ π_{k}) / n_{k}$ . All moments are finite because $X$ is finite and the weights are positive and finite.

2.1F1F2givenstep 1.1algebra∎

Write $B_{k} = (X_{k, 1}, \dots, X_{k, n_{k}}) \in X^{n_{k}}$ . The blocks are independent finite-valued random elements, and ${\overline{R}}_{k}$ is a function of $B_{k}$ . If $μ_{k} (b) = P (B_{k} = b)$ , then for any functions $G_{k}$ of the respective blocks, independence and finite summation give $E [\prod_{k = 0}^{K - 1} G_{k} (B_{k})] = \sum_{b_{0} \in X^{n_{0}}, \dots, b_{K - 1} \in X^{n_{K - 1}}} \prod_{k = 0}^{K - 1} μ_{k} (b_{k}) G_{k} (b_{k}) = \prod_{k = 0}^{K - 1} E [G_{k} (B_{k})] .$ Apply this first with $G_{k} = {\overline{R}}_{k}$ and then with $G_{k} = {\overline{R}}_{k}^{2}$ . Thus $E [\hat{R}] = \prod_{k} E [{\overline{R}}_{k}] = \prod_{k} (Z_{k + 1} / Z_{k}) = Z_{K} / Z_{0}$ , while $E [{\hat{R}}^{2}] / (Z_{K} / Z_{0})^{2} = \prod_{k} E [{\overline{R}}_{k}^{2}] / r_{k}^{2}$ . Substitution from step 1.1 gives the stated product, and subtracting one yields the relative variance.

Mean and relative second moment of independent bridge ratios

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources