Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 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.

Mildly ill-posed analytic q=2 detection rate

Example

Assume α,β>0, q=2, aj=eαj, and σj=jβ in the Gaussian sequence detection experiment. The minimax separation rate over Θ2(r) has order rε∗≍ε(log⁡ε−1)β+1/4.

Facts & Assumptions

Given: α,β>0, q=2, aj=eαj, σj=jβ, and a nonempty Θ2(r) with a finite extremal profile for sufficiently small r. For this rate comparison, Xε≍Yε means the ratio is bounded above and away from zero by constants independent of ε.

[F1]

q=2 Gaussian-sequence ellipsoid detection boundary: r2=AJ1/J2, uε2=(r/ε)4J0/(2J12), and the uε boundary determines minimax distinguishability.

[F2]

Verification

1.1F1F2givenconstructalgebra

Let m be the largest active index. Then Ae2αm<1≤Ae2α(m+1), so t=Ae2αm∈[e−2α,1). For j=m−ℓ with 0≤ℓ<m, the profile factor is ρℓ=1−te−2αℓ. For ℓ≥1, 1−e−2α≤ρℓ≤1. Hence the terms with 1≤ℓ≤m/2 give a lower bound for J1 and J0, while the full sums are bounded above by ∑j≤mj4β≤m4β+1. Thus J1≍J0≍m4β+1. In J2, the summand at ℓ is te−2αℓ(m−ℓ)4βρℓ. Its sum is at most m4β∑ℓ≥0e−2αℓ≤Cαm4β, and the ℓ=1 term is at least cα,βm4β for m≥2. Therefore J2≍m4β. Formula [F1] gives r2≍e−2αmm,uε2≍ε−4r4m−(4β+1). Taking logarithms of the first relation shows m≍log⁡(1/r).

2.1F1step 1.1givenalgebra∎

Set L=log⁡ε−1 and rε∗=εLβ+1/4. At this radius, log⁡(1/rε∗)∼L, so [F1] and step 1.1 give uε2≍1. If rε/rε∗→0, then log⁡(1/rε)≥log⁡(1/rε∗)∼L and uε2≍(rε/rε∗)4(log⁡(1/rε∗)log⁡(1/rε))4β+1→0. If rε/rε∗→∞ and rε→0, the logarithm ratio is at least one, so uε→∞. If the radii do not tend to zero, use rˉε=rεrε∗. Nonempty alternatives satisfy rε2≤a1−2 because ∑jσj2ηj2≤a1−2∑jaj2σj2ηj2≤a1−2. Hence rˉε→0, rˉε/rε∗→∞, and Θ2(rε)⊆Θ2(rˉε); consistent tests for the larger class work on the original class. The two risk implications in [F1] prove the claimed rate.

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