Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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.

Independent random elements are characterized by finite rectangle probabilities

Statement

Let (Xi)i∈I be random elements Xi:(Ω,F,P)→(Si,Σi). Then the following are equivalent:

  1. the family (Xi)i∈I is independent;
  2. for every natural number n≥1, every choice of distinct indices i0,…,in−1∈I, and every choice of measurable sets Bk∈Σik, P(Xi0∈B0,…,Xin−1∈Bn−1)=∏k<nP(Xik∈Bk).

Facts & Assumptions

Given: Random elements Xi:(Ω,F,P)→(Si,Σi).

[L1]

A family of random elements is independent exactly when the sigma-algebras σ(Xi) are independent. (Independent random elements)

[L2]

Independent pi-systems containing the whole space generate independent sigma-algebras. (Independent pi-systems generate independent sigma-algebras)

Proof

technique · direct
1.1L1

If (Xi)i∈I is independent, then by [L1] the sigma-algebras σ(Xi) are independent. Since Xi−1(B)∈σ(Xi) for every B∈Σi, the displayed rectangle identity follows immediately.

1.2givenL2

Conversely, for each i let Πi:={Xi−1(B):B∈Σi}⊆F. Preimages preserve finite intersections, so each Πi is a pi-system containing Ω. The hypothesis in clause 2 says exactly that the family (Πi)i∈I is independent. Since σ(Πi)=σ(Xi) by definition, [L2] implies that the sigma-algebras σ(Xi) are independent.

2.1step 1.1step 1.2∎

Step 1.2 proves clause 2 implies clause 1, and step 1.1 proves the reverse implication. Therefore the two conditions are equivalent.

Depends on

Used by

Dependency tree · two levels

7 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