LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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Conditional entropy chain and disjoint-block bounds

Statement

Assume AC. Let $X_{1}, \dots, X_{m}$ be standard-Borel spaces, let $Q_{i}$ be probability measures on $X_{i}$ , and let $P$ be a probability on $\prod_{i = 1}^{m} X_{i}$ . Write $P_{i}$ for its $i$ th marginal and use the directional extended relative entropy $D (\cdot ∥ \cdot)$ of Relative Entropy of Probability Measures. Then $D (P ∥ Q_{1} \otimes \dots \otimes Q_{m}) \geq \sum_{i = 1}^{m} D (P_{i} ∥ Q_{i}) .$ If the left side is finite, the conditional chain rule gives the stronger equality $D (P ∥ ⨂_{i} Q_{i}) = D (P_{1} ∥ Q_{1}) + \sum_{i = 2}^{m} E_{P} D (P_{X_{i} ∣ X_{1}, \dots, X_{i - 1}} ∥ Q_{i}),$ and each conditional increment is at least $D (P_{i} ∥ Q_{i})$ . If the left side is infinite, the displayed inequality is understood in the extended order and is automatic. For an integer $d \geq 1$ and disjoint spatial blocks with fixed particle counts $n_{i} \in {0, 1, 2, \dots}$ and fixed positions, apply this result to their mark spaces $X_{i} = (R^{d})^{n_{i}}$ , with Borel sigma-algebras and probability mark references $Q_{i}$ . The zero-fold product $X_{i} = (R^{d})^{0}$ is the singleton consisting of the empty mark tuple. On that singleton both $P_{i}$ and $Q_{i}$ are its unique probability measure, so a zero-occupancy mark block contributes zero. No zero-divergence claim is made for an unconditioned occupancy law.

Facts & Assumptions

Given: Standard-Borel block spaces, probability reference laws, and the stated joint probability.

[A1]

AC supplies the conditional-law and entropy chain-rule hypotheses recorded in the current supplier.

[F1]

Relative Entropy Variational, Chain, and Conditioning Rules: $D (R ∥ Q) = \sup_{ϕ \in B_{b}} {R (ϕ) - \log Q (e^{ϕ})}$ , where $B_{b}$ consists of bounded real measurable functions; relative entropy is nonnegative and cannot increase under a measurable map; finite relative entropy against a product probability reference has the conditional chain rule.

[F2]

Relative Entropy of Probability Measures: $D (P ∥ Q)$ is the extended directional KL divergence, equal to $\int h \log h d Q$ for $P ≪ Q$ , with value $+ \infty$ otherwise.

Proof

Given: The hypotheses and exact entropy conventions above.

1.1A1F1F2

If $D (P ∥ ⨂_{i} Q_{i}) = + \infty$ , then the inequality holds in the extended order. Suppose instead the joint entropy is finite. For $1 \leq i \leq m$ , let $P_{1 : i}$ be the marginal on the first $i$ coordinates and set $D_{i} = D (P_{1 : i} ∥ ⨂_{j = 1}^{i} Q_{j})$ . Projection contraction in [F1] gives $D_{i} \leq D_{m} < \infty$ . For $i \geq 2$ , apply the finite-entropy chain rule to $P_{1 : i}$ with reference $(⨂_{j < i} Q_{j}) \otimes Q_{i}$ . It gives $D_{i} = D_{i - 1} + G_{i}$ , where $G_{i} = E_{P} D (P_{X_{i} ∣ X_{1}, \dots, X_{i - 1}} ∥ Q_{i})$ . Each $G_{i}$ is nonnegative and finite. Telescoping these equalities gives the stated chain equality without subtracting infinite terms.

2.1F1F2step 1.1∎

For $i \geq 2$ , write $S = (X_{1}, \dots, X_{i - 1})$ and $K_{i} (s, \cdot) = P_{X_{i} ∣ S = s}$ for the conditional probability kernel in step 1.1. For every bounded real measurable $ϕ : X_{i} \to R$ , the variational formula [F1] gives $D (K_{i} (s, \cdot) ∥ Q_{i}) \geq \int ϕ d K_{i} (s, \cdot) - \log Q_{i} (e^{ϕ})$ . Integrating against $P_{S}$ and using the conditional-law identity yields $G_{i} \geq P_{i} (ϕ) - \log Q_{i} (e^{ϕ})$ . Taking the supremum over these tests and applying [F1] again gives $G_{i} \geq D (P_{i} ∥ Q_{i})$ . Summing with the first chain term $D_{1} = D (P_{1} ∥ Q_{1})$ proves the inequality. For $m = 1$ , the assertion is equality. On a zero-fold mark space, the density of its unique probability relative to itself is $1$ , so its marginal and conditional divergences are $1 \log 1 = 0$ .

Conditional entropy chain and disjoint-block bounds

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources