Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck pass
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.

Empirical laws of a finite valued iid sample

Example

For IID samples with law μ=∑j=1mpjδaj on distinct points a1,…,am∈Rd, where pj≥0 and ∑jpj=1, the empirical laws converge weakly almost surely. On outcomes whose samples all lie in this finite set, weak convergence is equivalent to convergence of all atom frequencies to pj.

Facts & Assumptions

[F1]

Empirical measures of iid euclidean samples converge weakly: For IID Rd-valued samples (Xi) with common law μ and finite d≥1, the empirical probabilities μ^n=n−1∑i=1nδXi converge weakly to μ almost surely on one common event.

[F2]

Finite and countable subadditivity of measures: Let μ be a measure and let (Ek)k∈N be measurable. Then

μ(⋃k∈NEk)≤∑k=0∞μ(Ek).

For every m∈N one also has

μ(⋃k<mEk)≤∑k<mμ(Ek),

including m=0, where both sides are 0.

Verification

Given: The objects, hypotheses and definitions in the statement. Its conclusions are to be established below.

1.1F1F2

Apply F2 to the union of the sample-outside-support events. By F1, the empirical laws converge weakly almost surely. Also all samples lie in the finite set on a conull event, since each outside event has probability zero and there are countably many coordinates.

1.2givenalgebra

On such an outcome put qn,j=n−1#{i≤n:Xi=aj}. If each qn,j→pj, then for bounded continuous f, ∫f dμ^n=∑jqn,jf(aj)→∑jpjf(aj)=∫f dμ, proving weak convergence.

2.1givenalgebra∎

Conversely, for m≥2 put rj=12min⁡l≠j∥aj−al∥>0 and fj(x)=max⁡(0,1−∥x−aj∥/rj). This bounded continuous test is one at aj and zero at every other al. Its empirical integral is qn,j and its μ integral is pj, so weak convergence implies qn,j→pj. If m=1, both frequencies are identically one and the constant test suffices.

Depends on

Used by

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Dependency tree · two levels

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Sources