Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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 n, a message count M≥2, a codebook ensemble with exchangeable message codewords, and a decoder for each sampled codebook. Let Bn be the random fraction of codewords whose normalized cost exceeds Γ, and let εˉn be the ensemble codebook's average error. If qn:=EBn and an:=Eεˉn satisfy qn+an<1, then some deterministic subcode has every retained codeword of normalized cost at most Γ, average error at most qn+an/(1−qn+an), and at least M(1−qn+an) messages.

If this construction is available for each sufficiently large n, with qn+an→0, then the retained code's rate differs from n−1ln⁡Mn by a quantity tending to zero.

Facts & Assumptions

Given: A random length-n codebook with M≥2 exchangeable message codewords, nonnegative normalized codeword cost, and a decoder whose per-message errors define εˉn.

[F1]

Chebyshev-Markov inequality for the integral: for a nonnegative measurable random variable V and a>0, Pr⁡[V≥a]≤E[V]/a.

[F2]

The finite union bound: the probability of a union of two events is at most the sum of their probabilities.

[F3]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: the natural logarithm is continuous and increasing on (0,∞), with ln⁡1=0 and ln⁡(ab)=ln⁡a+ln⁡b for a,b>0.

[F4]

A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere: a nonnegative measurable random variable with expectation zero vanishes almost surely.

Proof

1.1F1F2F4givencasesalgebra

Put sn=qn+an<1, including sn=0 when qn+an=0. If qn+an=0, both nonnegative variables Bn and εˉn have expectation zero, so [F4] makes each zero almost surely. Their zero events intersect by [F2], giving a realization with both variables zero. If qn+an>0, then sn>0. By [F1], Pr⁡[Bn>sn]≤qn/sn and Pr⁡[εˉn>sn]≤an/sn. The union bound [F2] makes the probability of either failure at most sn<1, so a realization exists with both quantities at most sn.

2.1step 1.1algebracases

Retain exactly the codewords in that realization whose normalized cost is at most Γ, and let M′ be their number. Since Bn≤sn, M′≥M(1−sn)>0. 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 Mεˉn/M′≤sn/(1−sn). Each retained word satisfies the cost constraint by construction.

3.1F3step 1.1step 2.1givenalgebra∎

If qn+an→0 along blocklengths, then sn→0 and 1−sn>0. Since M(1−sn)≤M′≤M, the order and product laws in [F3] give 0≤(ln⁡M−ln⁡M′)/n≤−ln⁡(1−sn)/n. Continuity of ln⁡ at 1 and ln⁡1=0 in [F3] yield ln⁡(1−sn)→0; 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