Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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.

The Markov property is past-future conditional independence

Statement

Assume Choice. Let X be adapted to (Fn) and put Tn+=σ(Xn,Xn+1,…). For every n, the following are equivalent:

  1. for every bounded Tn+-measurable random variable V, E[V∣Fn]=E[V∣σ(Xn)]a.s.; 2. Fn and Tn+ are conditionally independent given σ(Xn). Every homogeneous K-chain satisfies these conditions, with the first conditional expectation equal to h(Xn) for a measurable h. Conversely, if the equivalent conditions hold and a single kernel K satisfies E[g(Xn+1)∣σ(Xn)]=Kg(Xn)a.s. for every bounded measurable g and every n, then X is a homogeneous K-chain. Thus conditional independence characterizes the absence of extra past information; the additional displayed hypothesis identifies the same time-homogeneous kernel at every time.

Facts & Assumptions

Given: Choice and the adapted process in the statement. Adaptedness gives σ(Xn)⊆Fn.

[F1]

Conditional independence is the conditional product identity, and its equivalence proof identifies it with invariance of a conditional law after the other side is adjoined. (Conditional-independence equivalences and preservation)

[F2]

A K-chain satisfies the bounded future-functional identity with a measurable function of its present state. (Markov property for bounded future path functionals)

Proof

1.1F1

Assume (1), take bounded U measurable for Fn and bounded V [F1] measurable for Tn+, and put W=E[V∣σ(Xn)]. Conditioning first on Fn gives E[UV∣σ(Xn)]=E[UE(V∣Fn)∣σ(Xn)]=E[UW∣σ(Xn)]=E[U∣σ(Xn)]W. This is the sigma-algebra form of conditional independence in [F1], so (2) holds. Constants, zero, and one cause no exception.

1.2F1

Conversely assume (2) and keep V,W as above. For every [F1] A∈Fn, the conditional product identity gives E[1AV]=E[E(1AV∣σ(Xn))]=E[E(1A∣σ(Xn))W]=E[1AW]. Since W is Fn-measurable, this is exactly the defining event test for W=E[V∣Fn]. Hence (1). Empty and full A are included.

2.1F2step 1.1step 1.2

If X is a homogeneous K-chain, apply [F2] to every bounded measurable [F2, step 1.1, step 1.2] path functional H and V=H(Xn,Xn+1,…). Such variables generate the bounded Tn+-measurable variables by the event/simple-function argument in [F2], and [F2] gives a σ(Xn)-measurable version h(Xn). Thus (1), and hence (2), holds.

3.1step 1.1step 1.2∎

For the converse qualification, take V=g(Xn+1) in (1). Combining (1) [step 1.1, step 1.2] with the stated present-state kernel identity gives E[g(Xn+1)∣Fn]=Kg(Xn). Indicators recover the K-chain definition. Without the single-K hypothesis, conditional independence alone allows time-inhomogeneous present-state kernels, so it would not justify the stronger homogeneous conclusion. Choice is used by the conditional-expectation interfaces throughout.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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