LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Pair-count formula for scrambled-net gain coefficients

Statement

Assume Countable Choice and use the gains $Γ_{u, κ}$ of ANOVA variance identity for nestedly scrambled point sets. For a nonempty $u \subseteq {1, \dots, d}$ and a vector $ℓ = (ℓ_{q})_{q \in u}$ of nonnegative integers, define the ordered pair count $C_{u} (ℓ) = \sum_{i = 1}^{n} \sum_{j = 1}^{n} 1 {⌊ b^{ℓ_{q}} a_{i, q} ⌋ = ⌊ b^{ℓ_{q}} a_{j, q} ⌋ for every q \in u} .$ Let $1_{V}$ be the vector that is $1$ on $V \subseteq u$ and $0$ on $u ∖ V$ . Then $Γ_{u, κ} = \frac{1}{n (b - 1)^{∣ u ∣}} \sum_{V \subseteq u} (- 1)^{∣ u ∣ - ∣ V ∣} b^{∣ V ∣} C_{u} (κ + 1_{V}) .$

Facts & Assumptions

Given: Countable Choice, the fixed input point set, base $b$ , the subset $u$ , and scale vector $κ$ .

[F1]

ANOVA variance identity for nestedly scrambled point sets: $n Γ_{u, κ} = \sum_{i, j} \prod_{q \in u} Υ_{i j, q, k_{q}}$ .

[F2]

ANOVA variance identity for nestedly scrambled point sets: $Υ_{i j, q, k} = (b - 1)^{- 1} (b 1_{{same prefix of length k + 1}} - 1_{{same prefix of length k}})$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is inherited through the variance-identity supplier [F1].

Proof

technique · direct

Given: The objects and definitions in the Statement.

1.1A1F1F2givenalgebra

For each $q \in u$ , expand the numerator of [F2] as the difference of its two prefix-equality indicators. In the product over $q$ , choose the finer-prefix term on a subset $V \subseteq u$ and the coarser-prefix term on its complement. The resulting coefficient is $(- 1)^{∣ u ∣ - ∣ V ∣} b^{∣ V ∣}$ , and the pair $(i, j)$ satisfies exactly the prefix equalities at depths $κ + 1_{V}$ .

2.1F1step 1.1algebra∎

Summing the indicator for those equalities over all ordered pairs gives $C_{u} (κ + 1_{V})$ by its definition. Summing over every subset $V$ and dividing by $n (b - 1)^{∣ u ∣}$ yields the displayed formula.

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Art B. Owen, Monte Carlo Variance of Scrambled Net Quadrature, §7 (standard reference, not scraped)