Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-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.

Markov typicality for independently selected finite codewords

Statement

Let PXYZ(x,y,z)=PXY(x,y)PZ∣X(z∣x) be a finite-alphabet Markov law. For every ϵ>0 there are δ>0 and n0 such that the following holds for n≥n0: if (xn,yn)∈Tδ(n)(PXY) and a random Zn is supported on Tδ(n)(PZ∣X∣xn) and, conditional on each joint type of (xn,Zn), is uniform on that conditional type class, then Pr⁡{(xn,yn,Zn)∈Tϵ(n)(PXYZ)}≥1−Cn, where C<∞ depends only on the fixed finite law, alphabets, and ϵ, not on the particular typical source block or conditional type.

Facts & Assumptions

Given: The finite Markov law, ϵ>0, a typical fixed source block (xn,yn), and the stated conditional-type-uniform law of Zn.

[A1]

PXYZ=PXYPZ∣X: conditional on X=x, the target law of Z does not depend on Y.

[F1]

Finite strongly typical sets with support restriction: relative typicality excludes every zero-probability atom and bounds each positive empirical frequency by its relative tolerance.

[F2]

Finite random-codebook covering and selected-codeword law: conditional on a successful iid marginal-codebook search, the selected codeword is supported on the conditional typical set and is uniform within each conditional type class.

Proof

1.1A1F1F2construct

Fix a conditional type qZ∣X that can occur under the stated law. For each x with nx:=∣{t:xt=x}∣>0, there are nxy positions with (xt,yt)=(x,y) and nxz positions with (xt,Zt)=(x,z). Conditional-type uniformity means that, among the nx positions with X=x, the z labels are assigned uniformly over all arrangements with these counts. Hence Nxyz:=∣{t:(xt,yt,Zt)=(x,y,z)}∣ is hypergeometric, with ENxyz=nxynxznx,Var⁡(Nxyz)≤nx. When nx=0, all these counts vanish.

2.1A1F1step 1.1algebra

By [F1], nxy/n and nxz/n are uniformly close to PXY(x,y) and PXZ(x,z), respectively. For PX(x)>0, division by nx/n shows that nxz/nx is close to PZ∣X(z∣x). Thus, after choosing δ sufficiently small, the mean frequency ENxyz/n=(nxy/n)(nxz/nx) is within half the required relative tolerance of PXYZ(x,y,z) for every positive-probability triple. If PXYZ(x,y,z)=0, either (x,y) is excluded by the source support condition or (x,z) is excluded by the conditional support condition, so Nxyz=0.

3.1F1step 1.1step 2.1algebra∎

For a positive-probability triple, Chebyshev's inequality and step 1.1 give Pr⁡{∣Nxyz−ENxyz∣>12nϵPXYZ(x,y,z)}≤4nxn2ϵ2PXYZ(x,y,z)2≤4nϵ2PXYZ(x,y,z)2. There are finitely many triples. A union bound makes the probability that any empirical triple frequency misses its required tolerance at most C/n for a finite constant C depending only on the law, alphabets, and ϵ. This bound is uniform in every admissible conditional type, so it remains valid after mixing over the conditional-type distribution of Zn. The complement event is exactly strong typicality of (xn,yn,Zn), proving the claim.

Depends on

Used by

Dependency tree · two levels

7 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