CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Identifiability and KL support do not replace testability

Statement refuted

Under the library AC convention for KL divergence and density representatives, refute the claim that in every well-specified dominated IID model an injective parameter-to-law mapping and positive prior mass in every KL neighborhood of the truth suffice for posterior consistency in every metric on the model laws, even without a uniform testing condition.

Facts & Assumptions

Given: Refute the universal claim with a well-specified dominated IID model that is identifiable and has KL support at its truth, but has no uniformly consistent tests for a metric-separated complement and has posterior mass one on that complement at every sample size.

[A1]

The Axiom of Choice is the premise in the general published definitions of KL divergence and likelihood-density representatives; the Bernoulli densities below are explicitly supplied. The Axiom of Choice

[F1]

A parametric model is identifiable exactly when its parameter-to-law map is injective. Identifiability of an estimand

[F2]

KL divergence is directional: $D (P ∥ Q) = + \infty$ if $P ≪̸ Q$ , and otherwise it is the $P$ -expectation of $\log (d P / d Q)$ . Kullback–Leibler divergence

[F3]

Under the root-L2 Hellinger and library event-supremum conventions, $∥ P - Q ∥_{T V} \leq h (P, Q)$ and $h (P, Q)^{2} \leq 2 ∥ P - Q ∥_{T V}$ . Hellinger invariance and comparison with library total variation

[F4]

Library total variation is $\sup_{A} ∣ P (A) - Q (A) ∣$ . Total variation distance for probability laws

[F5]

For a proper prior and a jointly measurable dominated probability kernel, a finite-positive evidence fibre has posterior proportional to its likelihood times the prior. Bayes formula for dominated kernels

[F6]

An IID family is mutually independent and identically distributed. Identical distribution and IID families

[F7]

A finite vector of independent random elements has the product of its marginal laws. Independent random elements have product joint law

[F8]

A dominated experiment specifies a sigma-finite measure dominating each law, and the likelihood-density convention supplies finite-valued nonnegative density representatives. Dominated statistical experiments, Likelihood densities

[F9]

A parameter-measurable statistical model is a probability kernel whose event-probability sections are measurable. Parameter-measurable statistical models as kernels

[F10]

For $∣ r ∣ < 1$ , $\sum_{k = 1}^{\infty} r^{k} = r / (1 - r)$ ; in particular $\sum_{k = 1}^{\infty} 2^{- k} = 1$ . For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges

[F11]

For $x > 0$ , $\log x = \int_{1}^{x} d t / t$ . The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t

[F12]

A probability measure has total mass one. Probability measures and probability spaces

Counterexample

1.1F1F8F9givenconstruct

Let $X = {0, 1}$ with counting measure, let $Θ = {0, 1, 2, \dots}$ with its power-set sigma-algebra, and put $δ_{k} = (k + 2)^{- 2}$ for $k \geq 1$ . Define $P_{0} = Bernoulli (1 / 2)$ and $P_{k} = Bernoulli (1 / 2 + δ_{k})$ . Every mass is strictly between zero and one, and the map $k \mapsto P_{k}$ is injective because the $δ_{k}$ are distinct. Thus the model is identifiable by [F1]. The parameter and observation spaces are discrete, so every event-probability section is measurable by [F9] and every function on their countable product, including the pointwise Bernoulli density, is jointly measurable. The counting measure is sigma-finite and the displayed densities are normalized, so this is a dominated model by [F8].

2.1A1F2F10F11F12step 1.1algebra

Give the true parameter $0$ prior mass zero and set $Π ({k}) = 2^{- k}$ for $k \geq 1$ . By [F10] these masses sum to one, so [F12] makes $Π$ a probability. For $δ = δ_{k}$ , the explicit densities and [F2] give $D (P_{0} ∥ P_{k}) = \frac{1}{2} \log \frac{1 / 2}{1 / 2 + δ} + \frac{1}{2} \log \frac{1 / 2}{1 / 2 - δ} = - \frac{1}{2} \log (1 - 4 δ^{2})$ . For $0 < x < 1$ , [F11] gives $0 \leq - \log (1 - x) = \int_{1 - x}^{1} d t / t \leq x / (1 - x)$ ; hence $0 \leq D (P_{0} ∥ P_{k}) \leq 2 δ_{k}^{2} / (1 - 4 δ_{k}^{2}) \to 0$ . Every KL neighborhood of $P_{0}$ therefore contains some $P_{k}$ with prior mass $2^{- k} > 0$ .

2.2F4F5F10step 1.1algebragiven

Put $d (P_{j}, P_{k}) = 1$ for distinct model laws and $d (P_{j}, P_{j}) = 0$ . This is a metric: if the endpoints in a triangle inequality differ, at least one adjacent pair differs and the sum of its two distances is at least one; if they agree, the left side is zero. Thus the parameter complement is $V_{1 / 2} = {k \geq 1}$ . For every finite data vector, all $P_{k}^{n} (x_{1 : n}) > 0$ and are at most one, so the evidence $\sum_{k \geq 1} 2^{- k} P_{k}^{n} (x_{1 : n})$ is finite and positive. By [F5], the posterior is supported on ${k \geq 1}$ ; consequently $Π_{n} (V_{1 / 2} ∣ x_{1 : n}) = 1$ for every data vector and every $n$ .

2.3A1F3F4F6F7step 1.1algebra

Let $ρ_{k} = \sum_{x \in {0, 1}} \sqrt{P_{0} ({x}) P_{k} ({x})}$ . Then $0 \leq ρ_{k} \leq 1$ , since $h (P_{0}, P_{k})^{2} = 2 - 2 ρ_{k}$ is an integral of a square. The IID product law [F6, F7] and distributivity give product affinity $ρ_{k}^{n}$ , so $h (P_{0}^{n}, P_{k}^{n})^{2} = 2 (1 - ρ_{k}^{n}) \leq n 2 (1 - ρ_{k}) = n h (P_{0}, P_{k})^{2}$ , since $1 - r^{n} = (1 - r) \sum_{j = 0}^{n - 1} r^{j} \leq n (1 - r)$ for $0 \leq r \leq 1$ . Also $∥ P_{0} - P_{k} ∥_{T V} = δ_{k}$ : the four events in ${0, 1}$ have probability differences $0, δ_{k}, - δ_{k}, 0$ . Thus [F3] gives $∥ P_{0}^{n} - P_{n}^{n} ∥_{T V} \leq h (P_{0}^{n}, P_{n}^{n}) \leq \sqrt{2 n δ_{n}} \to 0$ .

3.1F4step 2.3algebra

Write $Δ_{n} (x) = P_{0}^{n} ({x}) - P_{n}^{n} ({x})$ on the finite sample space. Since $\sum_{x} Δ_{n} (x) = 0$ , the positive and negative sums each have magnitude $∥ P_{0}^{n} - P_{n}^{n} ∥_{T V}$ : the event ${x : Δ_{n} (x) > 0}$ attains the positive sum, and every event sum lies between the negative and positive sums. For any randomized test $0 \leq ϕ_{n} \leq 1$ , it follows term-by-term that $∣ P_{0}^{n} ϕ_{n} - P_{n}^{n} ϕ_{n} ∣ = ∣ \sum_{x} ϕ_{n} (x) Δ_{n} (x) ∣ \leq ∥ P_{0}^{n} - P_{n}^{n} ∥_{T V}$ . Hence $\sup_{k \geq 1} P_{k}^{n} (1 - ϕ_{n}) \geq P_{n}^{n} (1 - ϕ_{n}) \geq 1 - P_{0}^{n} ϕ_{n} - ∥ P_{0}^{n} - P_{n}^{n} ∥_{T V}$ . If the type-I error tends to zero, step 2.3 makes this uniform type-II lower bound tend to one. No tests with both errors tending to zero exist for this complement, although the model is identifiable and the prior has KL support.

4.1step 1.1step 2.1step 2.2step 2.3step 3.1∎

Steps 1.1 and 2.1 establish identifiability and KL support, while steps 2.2, 2.3, and 3.1 establish posterior inconsistency and absence of uniformly consistent tests. This refutes the statement.

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