Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
How statement and proof provenance work

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.

Additive composition of finite sampling error budgets

Statement

Let θ,θT,m,m^,m~∈R be, respectively, a target quantity, a finite/truncated target, the exact mean under the specified finite sampling law, one realized sample statistic, and its rounded computed value. Let ϵtr,ϵmix,ϵstat,ϵarith≥0 have the same units as these quantities. If ∣θT−θ∣≤ϵtr,∣m−θT∣≤ϵmix,∣m^−m∣≤ϵstat,∣m~−m^∣≤ϵarith, then ∣m~−θ∣≤ϵtr+ϵmix+ϵstat+ϵarith. Here the statistical inequality is conditional on the explicitly supplied realized-error event; the statement assigns no probability to that event. The four terms respectively record truncation, finite-sampling bias or mixing, statistical deviation, and arithmetic rounding. If a corresponding discrepancy is separately proved to vanish, its budget term can be omitted. Using only the four assumed bounds, omitting any strictly positive budget term is not valid uniformly over the admissible quantities: the additive bound can be attained with all four discrepancies equal to their respective bounds and having the same sign.

Facts & Assumptions

Given: The five real quantities, four nonnegative error bounds, and the four inequalities in the Statement.

[A1]

The statistical deviation bound is an explicit event assumption for the realized sample; it is not a deductive consequence of the sampling algorithm.

Proof

Given: [A1] and the four stated error inequalities.

1.1givenalgebra

Insert the intermediate real quantities in the order m~,m^,m,θT,θ and apply the real triangle inequality repeatedly: ∣m~−θ∣≤∣m~−m^∣+∣m^−m∣+∣m−θT∣+∣θT−θ∣. This bounds the total error by the sum of its four intermediate discrepancies.

2.1givenA1step 1.1algebra

Substitute the four assumed bounds into step 1.1 to obtain ∣m~−θ∣≤ϵarith+ϵstat+ϵmix+ϵtr. The proof is conditional on the statistical event in [A1] and asserts no confidence level or probability for it.

3.1givenstep 1.1step 2.1algebra∎

A discrepancy known to vanish contributes zero in step 1.1, so its budget term can be omitted. For any prescribed nonnegative budgets, set θ=0, θT=ϵtr, m=ϵtr+ϵmix, m^=ϵtr+ϵmix+ϵstat, and m~=ϵtr+ϵmix+ϵstat+ϵarith. All four signed increments are their respective nonnegative budgets, all four hypotheses hold with equality, and the total error equals their sum. Omitting any strictly positive term would make the proposed right-hand side smaller than this attained error, proving the uniform non-omission claim. This does not exclude a smaller bound in a particular instance with cancellations or additional information.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

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