Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)
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.

Relative Entropy Variational, Chain, and Conditioning Rules

Statement

Assume AC. Let P,Q be probability measures on one measurable space and use D(P∥Q) as defined in Relative Entropy of Probability Measures. Then D(P∥Q)=sup⁡ϕ∈Bb(P(ϕ)−log⁡Q(eϕ)), where Bb is the set of bounded real measurable functions. In particular D≥0. If T is measurable, then D(P∘T−1∥Q∘T−1)≤D(P∥Q).

For standard-Borel spaces X,Y, probability reference measures QX,QY, and a joint probability PXY≪QX⊗QY with finite D(PXY∥QX⊗QY), let PY∣X=x be the conditional law supplied by Conditional density formula. Then D(PXY∥QX⊗QY)=D(PX∥QX)+∫XD(PY∣X=x∥QY) PX(dx). More generally, for random variables R,S,Y taking values in standard-Borel spaces R,S,Y, with R=r(S) for a measurable r:S→R, and a probability reference QY on Y, the conditional increments G(R):=EP ⁣[D(PY∣R∥QY)],G(S):=EP ⁣[D(PY∣S∥QY)] satisfy 0≤G(R)≤G(S) whenever D(PS,Y∥PS⊗QY)<∞.

Facts & Assumptions

Given: AC, probability measures, standard-Borel observation spaces where conditioning is used, and the exact finite-region relative entropy definition.

[A1]

AC supplies Radon–Nikodym densities on these finite measure spaces (The Axiom of Choice, A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density). The kernel construction and event identities of Conditional density formula require no AC when the joint density and filler probability are supplied; only its conditional-expectation class convention assumes AC. Here the filler for a conditional law on Y is the already supplied QY.

[F1]

For P≪Q, h=dP/dQ is nonnegative, Q(h)=1, unique almost everywhere, and D(P∥Q)=Q(hlog⁡h) (Relative Entropy of Probability Measures).

[F2]

A joint density on a product of sigma-finite probability spaces produces an everywhere conditional probability kernel by dividing by its finite positive marginal density; zero and infinite marginal fibers are filled without division (Conditional density formula).

[F3]

Nonnegative product-measurable functions have measurable section integrals and Tonelli iterated-integral equality (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F4]

Dominated and monotone convergence justify the bounded truncation limits in the entropy integrals (Dominated convergence, Monotone convergence for the integral).

Proof

Given: The hypotheses and exact definitions in Facts & Assumptions.

1.1A1F1algebra

For P≪Q with density h, the negative part of hlog⁡h is integrable because −ulog⁡u≤1/e for 0≤u≤1; the density definition is therefore unambiguous. Also ulog⁡u≥u−1 for u≥0, so D(P∥Q)=∫hlog⁡h dQ≥∫(h−1) dQ=0. If P≪̸Q, the defined value is +∞≥0.

2.1step 1.1F1F5construct

For bounded ϕ, let Qϕ have density eϕ/Q(eϕ) relative to Q. When P≪Q and D(P∥Q)<∞, the density logarithm identity gives D(P∥Qϕ)=D(P∥Q)−P(ϕ)+log⁡Q(eϕ)≥0 by step 1.1. Thus every variational test is at most D(P∥Q); the same upper bound is immediate if D=+∞. If P≪̸Q, choose a measurable Q-null set N with P(N)>0 and use ϕ=t1N: then Q(eϕ)=1 and P(ϕ)=tP(N)→∞.

3.1step 1.1step 2.1F1F4F5

In the remaining case P≪Q, let h=dP/dQ and set ϕm=max⁡(−m,min⁡(m,log⁡h)), taking ϕm=−m on {h=0}. Then eϕm→h and eϕm≤1+h, so [F4] gives Q(eϕm)→1. The negative part of hϕm is bounded by 1/e, and its positive part increases to (hlog⁡h)+; hence [F4] gives P(ϕm)=∫hϕm dQ→D(P∥Q), including when the positive entropy part is infinite. Combining this with step 2.1 proves the variational identity.

4.1step 3.1F5

If T is measurable, every bounded test ψ on the target pulls back to the bounded test ψ∘T on the source, and Q(eψ∘T)=(Q∘T−1)(eψ) while P(ψ∘T)=(P∘T−1)(ψ). Taking suprema in step 3.1 proves the restriction inequality.

5.1A1F1F2F3F4step 1.1step 2.1step 3.1step 4.1

Finiteness of D(PS,Y∥PS⊗QY) gives a joint density and its conditional kernel by [F2]. Pushing forward by (s,y)↦(r(s),y) and using step 4.1 gives finite D(PR,Y∥PR⊗QY), so [F2] supplies the corresponding conditional kernel for R as well. Thus all conditional densities and references in this clause are defined. For a conditioning observation R, applying the variational identity pointwise to the conditional density kR(r,y) shows G(R)=sup⁡ϕ∈Bb(EPϕ(R,Y)−EP ⁣[log⁡∫eϕ(R,y)QY(dy)]), where the tests range over bounded measurable functions of (R,Y). The upper bound follows from step 2.1 for each conditional law; the reverse bound follows by using the jointly measurable bounded truncations of log⁡kR from step 3.1 and then [F3]–[F4]. When R is a measurable function of S, every test depending on (R,Y) is an admissible test depending on (S,Y) with the same objective value, since the QY normalizer still depends only on R. Thus G(R)≤G(S); nonnegativity is step 1.1 applied conditionally.

5.2A1F1F2F3step 4.1F5

Suppose PXY has density h(x,y) relative to QX⊗QY. Put hX(x)=∫h(x,y)QY(dy). By [F3], hX is a density for PX. Where 0<hX(x)<∞, [F2] supplies the conditional density k(x,y)=h(x,y)/hX(x) relative to QY; elsewhere use its supplied filler. On the full PX-measure set of finite positive hX, h=hXk and log⁡h=log⁡hX+log⁡k PXY-almost everywhere. Integrating this identity, with nonnegative conditional entropies and finite total entropy, gives the displayed chain rule. If a restriction entropy is infinite, step 4.1 makes the joint entropy infinite as well, so no subtraction of two infinite terms is used.

6.1step 1.1step 2.1step 3.1step 4.1step 5.2step 5.1∎

Steps 1.1, 2.1 and 3.1 prove nonnegativity and the variational identity, step 4.1 proves restriction monotonicity, step 5.2 proves the product chain rule, and step 5.1 proves monotonicity under additional conditioning.

Depends on

Dependency tree · 2 levels

Sources