Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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.

Two-series null centering with a final block prefix

Example

Apply moving-mode Null-constrained block bootstrap for independent series in H=R to the series (0,2,4) and (10,12,14,16), with b1=b2=2. Their observed means are 2 and 13, but both bootstrap series have conditional mean YˉP=58/7.

Verification

Given: The two fixed series and block lengths in the Example.

[F1] Null-constrained block bootstrap for independent series subtracts the within-series mean, then the candidate average at each within-block position, and adds the pooled mean; it retains exactly ni observations.

[F2] Exact null centering of the K-sample block bootstrap gives the common conditional mean for every retained observation and hence every series mean.

1.1F1givenalgebra

Here M=7, YˉP=(6+52)/7=58/7, N1=2, N2=3, and k1=k2=2. The first residual series is (−2,0,2); its candidate blocks are (−2,0) and (0,2), with position averages (−1,1). After subtracting those averages the two candidates are (−1,−1) and (1,1). The second residual series is (−3,−1,1,3); its candidates are (−3,−1),(−1,1),(1,3), also with position averages (−1,1), and become (−2,−2),(0,0),(2,2).

2.1F1F2step 1.1algebra∎

Let V1,V2 be independent uniforms on {−1,1} and W1,W2 independent uniforms on {−2,0,2}, all four independent. The outputs are (58/7+V1,58/7+V1,58/7+V2) and (58/7+W1,58/7+W1,58/7+W2,58/7+W2). The first output retains only the first entry of its final block. Since all four offsets have mean zero, E∗Yˉ1∗=E∗Yˉ2∗=58/7, explicitly realizing [F2] despite the unequal observed means and unequal sample sizes.

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