ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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.

Exact beta value for a one-use BSC meta-converse test

Example

For the one-use BSC $(p)$ with $0 < p < 1 / 2$ , the identity encoder/decoder with two equiprobable messages has average error $p$ . With its true joint law $P_{M Y}$ and the uniform-message/uniform-output auxiliary law ${\tilde{Q}}_{M Y}$ , the oriented testing value is exactly $β_{p} (P_{M Y}, {\tilde{Q}}_{M Y}) = 1 / 2 = 1 / M$ .

Facts & Assumptions

Given: Messages $m \in {0, 1}$ are uniform, the input is $x = m$ , the decoder outputs $y$ , and the BSC crossover probability is $0 < p < 1 / 2$ .

[F1]

Finite type-II testing function with explicit orientation: $β_{ϵ} (P, Q)$ minimizes $E_{Q} (1 - ϕ)$ over tests $ϕ \in [0, 1]$ with $E_{P} ϕ \leq ϵ$ .

[F2]

Finite-DMC hypothesis-testing meta-converse: every $M$ -message code with average error $ϵ$ satisfies $β_{ϵ} (P_{M Y}, {\tilde{Q}}_{M Y}) \leq 1 / M$ for any auxiliary output law.

[F3]

Finite discrete memoryless channels: the BSC has transition probability $1 - p$ on $y = x$ and $p$ on $y \neq x$ .

Verification

1.1F3given

The four true joint masses are $(1 - p) / 2$ on $(0, 0), (1, 1)$ and $p / 2$ on $(0, 1), (1, 0)$ . The auxiliary law assigns $1 / 4$ to each pair. The identity decoder is wrong with total true mass $p$ , so its average error is $p$ .

2.1F1step 1.1algebra

Write $ψ = 1 - ϕ$ for the acceptance probability. The constraint in [F1] is $E_{P} ψ \geq 1 - p$ , and the objective is $E_{\tilde{Q}} ψ$ . Let $s$ be the sum of $ψ$ on the two correct pairs and $t$ its sum on the two crossover pairs. Then $0 \leq s, t \leq 2$ , the constraint is $(1 - p) s / 2 + p t / 2 \geq 1 - p$ , and the objective is $(s + t) / 4$ .

3.1F1F2step 2.1casesalgebra∎

If $s \geq 2$ , the objective is at least $1 / 2$ . If $s < 2$ , the constraint forces $t \geq (1 - p) (2 - s) / p$ , so $s + t \geq s + \frac{1 - p}{p} (2 - s) \geq 2,$ where the last inequality follows from $(1 - p) / p > 1$ and $s < 2$ . Thus every randomized feasible test has objective at least $1 / 2$ . Taking $ψ = 1$ on the two correct pairs and $0$ on the crossover pairs attains constraint equality and objective $1 / 2$ . Hence $β_{p} = 1 / 2 = 1 / M$ , as also bounded by [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, Chapter 22 §22.3 (standard reference, not scraped)
Yury Polyanskiy, H. Vincent Poor, and Sergio Verdú, Channel Coding Rate in the Finite Blocklength Regime (standard reference, not scraped)