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.
A low-error random code can be pruned to satisfy a strict average cost constraint
Statement
Fix a blocklength , a message count , a codebook ensemble with exchangeable message codewords, and a decoder for each sampled codebook. Let be the random fraction of codewords whose normalized cost exceeds , and let be the ensemble codebook's average error. If and satisfy , then some deterministic subcode has every retained codeword of normalized cost at most , average error at most , and at least messages.
If this construction is available for each sufficiently large , with , then the retained code's rate differs from by a quantity tending to zero.
Facts & Assumptions
Given: A random length- codebook with exchangeable message codewords, nonnegative normalized codeword cost, and a decoder whose per-message errors define .
Chebyshev-Markov inequality for the integral: for a nonnegative measurable random variable and , .
The finite union bound: the probability of a union of two events is at most the sum of their probabilities.
Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: the natural logarithm is continuous and increasing on , with and for .
A nonnegative measurable function has integral exactly when it vanishes almost everywhere: a nonnegative measurable random variable with expectation zero vanishes almost surely.
Proof
Put , including when . If , both nonnegative variables and have expectation zero, so [F4] makes each zero almost surely. Their zero events intersect by [F2], giving a realization with both variables zero. If , then . By [F1], and . The union bound [F2] makes the probability of either failure at most , so a realization exists with both quantities at most .
Retain exactly the codewords in that realization whose normalized cost is at most , and let be their number. Since , . Change the decoder only by mapping decisions for removed messages to one fixed retained message. An originally correct decision for a retained message is unchanged, so its error probability cannot increase; hence the retained average error is at most . Each retained word satisfies the cost constraint by construction.
If along blocklengths, then and . Since , the order and product laws in [F3] give . Continuity of at and in [F3] yield ; hence the rate loss tends to zero. The error bound in step 2.1 also tends to zero. The conclusion is per blocklength, so no sequence of witnesses is selected in this lemma.
Depends on
Used by
Dependency tree · two levels
21 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
- Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, Chapter 20 §20.1 (standard reference, not scraped)
- Robert Gallager, Principles of Digital Communication I, Chapter 8 §8.7.4, MIT OpenCourseWare (standard reference, not scraped)