ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Shared random bins do not factor across encoders

Example

Take independent uniform binary sources $X, Y$ and two encoders with $M \geq 2$ bins each. Each distinct random map assigns independent uniform labels in $[M]$ to its two input symbols. For one source symbol, compare the possible source pairs $(0, 0)$ and $(1, 1)$ . If the two encoders use independent random maps, the probability that both pairs have the same message vector is $M^{- 2}$ . If both use one shared random map, that probability is $M^{- 1}$ instead. Thus the product collision probability used in independent random binning depends on independence across encoders.

Facts & Assumptions

Given: The independent fair binary source, integer $M \geq 2$ , and the two random-map ensembles in the Example.

[F1]

Finite-alphabet IID multisource: the independent fair pair has all four binary pairs in its support.

[F2]

Finite fixed-length multiterminal lossless code: each terminal sends its own finite message index to the common decoder.

[F3]

Finite-IID k-source Slepian–Wolf rate region: the achievability proof assigns uniform bin labels to source words independently across words and encoders.

Verification

1.1F1F2F3algebra

Under [F1], each of the pairs $(0, 0)$ and $(1, 1)$ has source probability $1 / 4$ . Under independent maps $F_{X}, F_{Y} : {0, 1} \to [M]$ , each equality $F_{X} (0) = F_{X} (1)$ and $F_{Y} (0) = F_{Y} (1)$ has probability $1 / M$ , because the two labels in each map are independent and uniform. The two equalities are independent across encoders, so both competing pairs collide in both message coordinates with probability $M^{- 2}$ .

1.2F2algebra

Under a shared map $F_{X} = F_{Y} = F$ , both message-coordinate collision conditions are the same event $F (0) = F (1)$ . Its probability is $1 / M$ , not $1 / M^{2}$ . At $M = 2$ , the two probabilities are $1 / 4$ and $1 / 2$ .

2.1F3step 1.1step 1.2∎

The calculation identifies why the independent-map hypothesis in [F3] permits multiplication of per-encoder collision probabilities. It does not claim that every shared-map code is poor; it shows that this particular product bound cannot be used for the shared ensemble.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

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