Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck pending
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Conditional joint-typicality packing bound for finite alphabets

Statement

Let (V,U,Z) have any joint law PVUZ on finite alphabets. Fix η>0. For every sufficiently small strong-typicality tolerance, every sufficiently large n, and every fixed (vn,zn) typical for the marginal PVZ, an independent candidate word drawn from the product law PU∣V⊗n(⋅∣vn) satisfies Pr⁡{(vn,Un,zn) is jointly typical for PVUZ}≤e−n(IP(U;Z∣V)−η). The conclusion includes the case that V is constant, giving the exponent IP(U;Z).

Facts & Assumptions

Given: A finite joint law PVUZ, η>0, and fixed typical (vn,zn).

[F1]

Finite strongly typical sets with support restriction: typicality restricts empirical distributions to a finite-alphabet neighborhood of the target law and excludes unsupported atoms.

[F2]

Finite conditional mutual information: I(U;Z∣V) is the finite conditional-KL average in nats.

Proof

1.1F1F2givenalgebra

For any jointly typical supported candidate un, ln⁡PU∣V⊗n(un∣vn)PU∣V,Z⊗n(un∣vn,zn)=n∑v,u,zP^vnunzn(v,u,z)ln⁡P(u∣v)P(u∣v,z). The logarithms are finite on the positive target support. At PVUZ the finite sum equals −IP(U;Z∣V) by [F2], and it is linear in the empirical probabilities. Choose the typicality tolerance small enough that every jointly typical triple has this sum at most −IP(U;Z∣V)+η.

2.1F1step 1.1algebra∎

Sum the likelihood-ratio expression of step 1.1 over the jointly typical candidates. The target conditional law is a probability vector, so its mass on that set is at most one; the exponential likelihood ratio is at most e−n(IP(U;Z∣V)−η). This gives the claimed bound. Joint typicality excludes unsupported triples; on each supported target triple both conditional probabilities in the ratio are positive.

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