LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Fixed-quality net gain coefficients are uniformly bounded

Statement

Assume Countable Choice. Let $a_{1}, \dots, a_{n}$ be a $(λ, t, m, d)$ -net in base $b$ , with $n = λ b^{m}$ and $1 \leq λ < b$ . For every nonempty $u \subseteq {1, \dots, d}$ and $κ \in N^{∣ u ∣}$ , the gain from ANOVA variance identity for nestedly scrambled point sets satisfies $Γ_{u, κ} = 0 if ∣ κ ∣ + ∣ u ∣ \leq m - t,$ and otherwise $∣ Γ_{u, κ} ∣ \leq λ b^{t + ∣ u ∣ - 1} \leq λ b^{t + d - 1} .$

Facts & Assumptions

Given: The net, its parameters, the scramble, and $u, κ$ .

[F1]

$(λ, t, m, d)$ -nets in base $b$ : Every base- $b$ elementary interval of total digit depth $m - t$ contains exactly $λ b^{t}$ points.

[F2]

Pair-count formula for scrambled-net gain coefficients: The gain is the alternating finite sum of pair counts $C_{u} (κ + 1_{V})$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is inherited through [F2].

Proof

technique · direct

Given: Countable Choice, the net, and the parameters in the Statement.

1.1F1givenalgebra

If an elementary interval has total digit depth $L \leq m - t$ , it is partitioned into $b^{m - t - L}$ intervals of depth $m - t$ . By [F1] it therefore contains $λ b^{t} b^{m - t - L} = λ b^{m - L} = n b^{- L}$ points. Thus the number of ordered point pairs agreeing at any coordinate-depth vector $ℓ$ with $∣ ℓ ∣ \leq m - t$ is $C_{u} (ℓ) = b^{∣ ℓ ∣} (n b^{- ∣ ℓ ∣})^{2} = n^{2} b^{- ∣ ℓ ∣}$ .

2.1A1F2step 1.1algebra

If $∣ κ ∣ + ∣ u ∣ \leq m - t$ , every depth vector $κ + 1_{V}$ in [F2] has total depth at most $m - t$ . Step 1.1 then makes $b^{∣ V ∣} C_{u} (κ + 1_{V}) = n^{2} b^{- ∣ κ ∣}$ independent of $V$ . The alternating sum over all $V \subseteq u$ is $(1 - 1)^{∣ u ∣} = 0$ , so $Γ_{u, κ} = 0$ .

2.2F1F2step 1.1algebra

Otherwise, a nonzero product of $Υ$ factors requires the second point to share with the first its coarse prefixes of total depth $∣ κ ∣$ . If $∣ κ ∣ \leq m - t$ , step 1.1 and $∣ κ ∣ \geq m - t - ∣ u ∣ + 1$ bound the number of such points by $n b^{- ∣ κ ∣} \leq λ b^{t + ∣ u ∣ - 1}$ . If $∣ κ ∣ > m - t$ , the coarse cell lies inside a depth- $(m - t)$ cell containing $λ b^{t} \leq λ b^{t + ∣ u ∣ - 1}$ points. Each factor $Υ$ has absolute value at most $1$ , so for each first point the absolute row sum is at most this cell count. Summing over $n$ first points and dividing by $n$ gives $∣ Γ_{u, κ} ∣ \leq λ b^{t + ∣ u ∣ - 1}$ .

3.1step 2.2algebra∎

Since $∣ u ∣ \leq d$ , the first bound in step 2.2 is at most $λ b^{t + d - 1}$ , proving the remaining inequality.

Depends on

Used by

Smooth-integrand variance rate for nestedly scrambled (λ,t,m,d)-nets Theorem

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, Local Antithetic Sampling with Scrambled Nets, §5.1, Theorem 3 (standard reference, not scraped)
Art B. Owen, Monte Carlo Variance of Scrambled Net Quadrature, §7 (standard reference, not scraped)