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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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Finite-Blocklength Bounds and Reliability — Examples

1 · Prerequisites

2 · Summary

These examples evaluate the one-shot dependence-testing expression and meta-converse on a binary symmetric channel, then give a finite noiseless-channel counterexample separating small average error from maximal error. The calculations retain the null/alternative convention for βϵ and distinguish average-error bounds from per-message guarantees.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Small expected average code error need not control maximal error

Statement refuted

Expected average error tending to zero for singleton random-code ensembles on finite DMCs forces the corresponding maximal errors to tend to zero.

Facts & Assumptions

Given: M≥2, input/output alphabet {1,…,M−1}, and the noiseless DMC W(y∣x)=1{y=x}.

[F1]

Fixed-length codes for a finite DMC: a fixed-length code consists of an encoder from messages to input words and a deterministic decoder from outputs to messages.

[F2]

Average and maximal error of a fixed-length code: average error is the uniform-message mean, and maximal error is the largest individual error.

Counterexample

For every integer M≥2, there is a singleton random-code ensemble on a finite noiseless DMC whose expected average error is 1/M while the code's maximal error is 1. Thus a small expected average error alone does not make every message's error small.

1.1F1givenconstruct

Use blocklength one, encode message m as x=m for 1≤m≤M−1 and encode message M as x=1. Decode output y as message y. These maps are defined on the stated finite alphabets and form a code by [F1].

2.1F2step 1.1∎

On the noiseless channel, messages 1,…,M−1 are decoded correctly, while message M is decoded as 1 and is always wrong. Hence [F2] gives average error 1/M and maximal error 1. Take the random-code ensemble to put probability one on this single code; its expected average error remains 1/M, while its maximal error remains 1. Letting M→∞ proves the claimed separation.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Exact evaluation of a one-use BSC dependence-testing bound

Example

For a one-use BSC with crossover p=0.1, uniform input, and M=2, the dependence-testing upper bound on achievable average error is exactly 0.35.

Facts & Assumptions

Given: The BSC transition probabilities W(y∣x)=0.9 for y=x and 0.1 for y≠x, uniform input, and M=2.

[F1]

Finite-DMC dependence-testing achievability bound: a deterministic code exists whose average error is at most Eexp⁡(−[ı(X;Y)−ln⁡((M−1)/2)]+).

[F2]

Finite discrete memoryless channels: a BSC row is a probability vector, and its output law under an input vector is the matrix product PW.

Verification

1.1F2givenalgebra

Under the uniform input, the output is uniform. Therefore ı(X;Y)=ln⁡(0.9/0.5)=ln⁡1.8 with probability 0.9 and ı(X;Y)=ln⁡(0.1/0.5)=ln⁡0.2 with probability 0.1.

2.1F1step 1.1casesalgebra∎

The threshold is ln⁡((M−1)/2)=ln⁡(1/2). On the correct-output event, the integrand is exp⁡[−(ln⁡1.8−ln⁡(1/2))]=1/3.6=5/18, contributing 0.9(5/18)=1/4. On the crossover event, ln⁡0.2<ln⁡(1/2), so the positive part is zero and the integrand is one, contributing 0.1. Thus the expectation is 1/4+1/10=7/20=0.35, and [F1] supplies a code with average error at most this value.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

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 PMY and the uniform-message/uniform-output auxiliary law Q~MY, the oriented testing value is exactly βp(PMY,Q~MY)=1/2=1/M.

Facts & Assumptions

Given: Messages m∈{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: βϵ(P,Q) minimizes EQ(1−ϕ) over tests ϕ∈[0,1] with EPϕ≤ϵ.

[F2]

Finite-DMC hypothesis-testing meta-converse: every M-message code with average error ϵ satisfies βϵ(PMY,Q~MY)≤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≠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 EPψ≥1−p, and the objective is EQ~ψ. Let s be the sum of ψ on the two correct pairs and t its sum on the two crossover pairs. Then 0≤s,t≤2, the constraint is (1−p)s/2+pt/2≥1−p, and the objective is (s+t)/4.

3.1F1F2step 2.1casesalgebra∎

If s≥2, the objective is at least 1/2. If s<2, the constraint forces t≥(1−p)(2−s)/p, so s+t≥s+1−pp(2−s)≥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 βp=1/2=1/M, as also bounded by [F2].

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