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Small entropy density without homogeneous local convergence

Statement refuted

Assume AC. Fix $0 < θ < 1$ and $ϵ > 0$ , and let $P_{N}, H_{N}$ be the reference and energy of Energy concentration without local equivalence. The refuted implication is: for every sequence of laws $Q_{N} ≪ P_{N}$ , if $N^{- 1} D (Q_{N} ∥ P_{N}) \to 0$ , $H_{N} / N \to ϵ θ$ in $Q_{N}$ -probability, and $N^{- 1} H_{S h} (Q_{N}) \to h (θ)$ , then the first-coordinate marginal of $Q_{N}$ converges in event-supremum total variation to Bernoulli $(θ)$ . Here $H_{S h} (Q_{N}) = - \sum_{x} Q_{N} (x) \log Q_{N} (x)$ and $h (θ) = - θ \log θ - (1 - θ) \log (1 - θ)$ use natural logarithms.

Facts & Assumptions

Given: AC, the Bernoulli model and parameters above.

[A1]

AC is the choice hypothesis of the conditional entropy chain supplier (The Axiom of Choice).

[F1]

Conditional entropy chain and disjoint-block bounds: the finite product KL chain rule separates the first coordinate from its conditional remainder.

Counterexample

Given: The finite product laws and energy.

1.1A1F1givenalgebra

Set $A_{N} : = δ_{1} \otimes Bernoulli (θ)^{\otimes (N - 1)}$ , using the empty product for $N = 1$ . Under $A_{N}$ , $X_{1} = 1$ surely and the remaining coordinates retain their independent Bernoulli- $θ$ law. The likelihood $d A_{N} / d P_{N}$ is $1 / θ$ on ${X_{1} = 1}$ and zero elsewhere, so integration gives $D (A_{N} ∥ P_{N}) = \log (1 / θ)$ . Equivalently, [F1] decomposes the divergence into $D (δ_{1} ∥ B e r n o u l l i (θ)) = \log (1 / θ)$ and a zero conditional remainder. Dividing by $N$ gives vanishing entropy density.

2.1step 1.1algebra∎

Under $A_{N}$ , $H_{N} = ϵ (1 + Y_{N})$ with $Y_{N}$ binomial $(N - 1, θ)$ . Since $Y_{N}$ is the sum of $N - 1$ independent Bernoulli coordinates, $E Y_{N} = (N - 1) θ$ and $Var (Y_{N}) = (N - 1) θ (1 - θ)$ , by expanding the mean and variance and using zero cross-covariances. Thus $E (H_{N} / N) = ϵ [1 + (N - 1) θ] / N \to ϵ θ$ and $Var (H_{N} / N) = ϵ^{2} (N - 1) θ (1 - θ) / N^{2} \to 0$ . For any $η > 0$ , $1_{{∣ H_{N} / N - E (H_{N} / N) ∣ \geq η}} \leq (H_{N} / N - E (H_{N} / N))^{2} / η^{2}$ ; integrating gives the Chebyshev bound and convergence in probability. Its Shannon entropy is $H_{S h} (A_{N}) = (N - 1) h (θ)$ by adding the logarithms of the independent-coordinate probabilities, so $H_{S h} (A_{N}) / N = (N - 1) h (θ) / N \to h (θ)$ , where $h (θ) = - θ \log θ - (1 - θ) \log (1 - θ)$ . Yet the first coordinate remains one and is at total-variation distance $1 - θ$ from Bernoulli $(θ)$ . This proves all claims and refutes the statement.

Small entropy density without homogeneous local convergence

Statement refuted

Facts & Assumptions

Counterexample

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