CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.
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A balanced binary type class is not exactly exponential in entropy

Statement refuted

For every finite alphabet, every $n \geq 1$ , and every $n$ -type $q$ , the type-class size is exactly $∣ T_{q} ∣ = e^{n H (q)}$ .

Facts & Assumptions

Given: The binary alphabet $A = {0, 1}$ and block length $n = 2$ .

[F1]

Empirical types, n-types, and type classes: $T_{q}$ consists of the words whose empirical vector is $q$ .

[F2]

Finite distributions, entropy, divergence, and stochastic matrices: $H (q) = - \sum_{a : q (a) > 0} q (a) \ln q (a)$ .

[F3]

Polynomially sharp bounds for type-class cardinality: $(n + 1)^{- (k - 1)} e^{n H (q)} \leq ∣ T_{q} ∣ \leq e^{n H (q)}$ for an alphabet of size $k$ .

Counterexample

Given: $n = 2$ and the binary type $q (0) = q (1) = 1 / 2$ .

1.1F1givenconstruct

The words with type $q$ are $01$ and $10$ , so $∣ T_{q} ∣ = 2$ .

2.1F2step 1.1algebra

By [F2], $H (q) = \ln 2$ , and therefore $e^{2 H (q)} = 4 \neq 2 = ∣ T_{q} ∣$ . This single attainable type refutes the universal exact-equality claim.

3.1F1F2F3algebra∎

At the boundary type $q = (1, 0)$ , the only word is $00$ , and both $∣ T_{q} ∣$ and $e^{2 H (q)}$ equal $1$ ; this boundary equality does not rescue the refuted universal claim and is consistent with [F3].

Depends on

Used by

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Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, author-hosted book draft (standard reference, not scraped)
Thomas M. Cover and Joy A. Thomas, Elements of Information Theory, second edition, Chapter 11 (standard reference, not scraped)