LemmaStatement: Literature-sourcedProof: 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.

Conditional block law factors across IID coordinates

Statement

Let $Z_{t} = (X_{1, t}, \dots, X_{k, t})$ be a finite-alphabet IID multisource with joint law $P$ . For nonempty $S \subseteq [k]$ , put $C = S^{c}$ . If $(x_{S}^{n}, x_{C}^{n})$ is supported under $P^{\otimes n}$ , then the conditional block probability obtained from its joint and complement marginal satisfies $P^{\otimes n} (x_{S}^{n} ∣ x_{C}^{n}) = \prod_{t = 1}^{n} P_{X_{S} ∣ X_{C}} (x_{S, t} ∣ x_{C, t}) .$

Facts & Assumptions

Given: The finite-alphabet IID multisource, a nonempty $S \subseteq [k]$ , and supported blocks $(x_{S}^{n}, x_{C}^{n})$ .

[F1]

Finite-alphabet IID multisource: the joint letter law is $P$ , and $P^{\otimes n} (z^{n}) = \prod_{t = 1}^{n} P (z_{t})$ .

[F2]

Finite conditional entropy: for $P_{X_{S}, X_{C}} (s, c) > 0$ , the conditional probability is $P_{X_{S} ∣ X_{C}} (s ∣ c) = P_{X_{S}, X_{C}} (s, c) / P_{X_{C}} (c)$ .

[F3]

Marginals and positive-mass conditionals of a finite joint law are probability vectors: finite joint-law marginals are probability vectors; positive-mass conditional vectors are well-defined probability vectors.

Proof

1.1F1F3algebra

For each coordinate $t$ , support gives $P_{X_{S}, X_{C}} (x_{S, t}, x_{C, t}) > 0$ , and the finite marginal sum $P_{X_{C}} (x_{C, t}) = \sum_{x_{S}} P_{X_{S}, X_{C}} (x_{S}, x_{C, t})$ is therefore positive by [F3]. Summing the finite product law over all $x_{S}^{n}$ and distributing the finite sums gives $P_{X_{C}^{n}}^{\otimes n} (x_{C}^{n}) = \sum_{x_{S}^{n}} \prod_{t = 1}^{n} P_{X_{S}, X_{C}} (x_{S, t}, x_{C, t}) = \prod_{t = 1}^{n} P_{X_{C}} (x_{C, t}) > 0$ . Thus the conditional block probability is defined by dividing its joint mass by this positive complement marginal.

2.1F1F2step 1.1algebra∎

Using [F1] for the numerator and step 1.1 for the denominator, that ratio is $\frac{\prod_{t = 1}^{n} P_{X_{S}, X_{C}} (x_{S, t}, x_{C, t})}{\prod_{t = 1}^{n} P_{X_{C}} (x_{C, t})} = \prod_{t = 1}^{n} \frac{P_{X_{S}, X_{C}} (x_{S, t}, x_{C, t})}{P_{X_{C}} (x_{C, t})} = \prod_{t = 1}^{n} P_{X_{S} ∣ X_{C}} (x_{S, t} ∣ x_{C, t})$ , where each final factor is the conditional probability in [F2]. This proves the factorization.

Depends on

Used by

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Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, author-hosted book draft (standard reference, not scraped)
Thomas M. Cover and Joy A. Thomas, Elements of Information Theory, second edition, Chapter 15 (standard reference, not scraped)