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CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Expected information gain under a wrong operator does not identify the operator-parameter pair

Statement refuted

The following implication is false: in every finite dominated inverse experiment, if expected information gain for a parameter is positive under a fixed nominal forward operator, then an unknown truth-operator pair is identifiable from the observation law, where identifiable means that distinct pairs induce distinct conditional observation laws.

Example

Take one scalar observation Y=aθ+ϵ on Y={1,2,3} with counting measure and independent noise ϵ uniform on {0,1}. In the actual model the unknown pair H=(θ,a) is uniform on {(1,1),(2,12)}. In the nominal model the operator is fixed at a=1 and θ is uniform on {1,2}. Under the actual model both pairs give the same law, uniform on {1,2}; under the nominal model the expected information gain for θ is 12log⁡2>0.

Facts & Assumptions

Given: AC; the finite actual and nominal experiments in the Counterexample; the parameter priors and the counting measure on Y.

[A1]

AC is inherited from the library KL convention. The Axiom of Choice

[F1]

Expected information gain equals D(Pθ,Y∥Pθ⊗PY) for the specified dominated experiment. Expected information gain is joint relative entropy

[F2]

The expected information gain is the prior-predictive integral of posterior-to-prior KL divergence. Expected information gain

[F3]

KL divergence is nonnegative and vanishes exactly when the two probability measures agree. Kullback–Leibler divergence is nonnegative

[F4]

The natural logarithm is strictly increasing and log⁡1=0. Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm

[F5]

On a finite probability space, D(P∥Q)=∑xP({x})log⁡(P({x})/Q({x})), with zero-mass terms equal to zero. Kullback–Leibler divergence

Verification

Given: The finite models in the Counterexample.

1.1A1F1F3givenalgebra

In the actual model, aθ=1 for both possible pairs. Hence P(Y∈{1,2}∣H=(1,1))=P(Y∈{1,2}∣H=(2,12))=1, and both conditional laws assign probability 12 to each of 1 and 2. Thus H and Y are independent, so [F1] and [F3] give actual-pair EIG D(PH,Y∥PH⊗PY)=0. The two distinct pairs have identical observation laws and are not identifiable.

1.2givenalgebra

In the nominal model, θ=1 gives Y∈{1,2} equiprobably and θ=2 gives Y∈{2,3} equiprobably. Therefore the predictive probabilities are P(Y=1)=14, P(Y=2)=12, and P(Y=3)=14.

2.1A1F2F4F5step 1.2algebra

The joint probabilities are P(θ=1,Y=1)=14, P(θ=1,Y=2)=14, P(θ=2,Y=2)=14, and P(θ=2,Y=3)=14. Dividing by the marginals in step 1.2 gives posterior δ1 at Y=1, the original uniform prior at Y=2, and δ2 at Y=3. By [F5], the corresponding KL values are log⁡2, 0, and log⁡2. Thus [F2] gives nominal EIG 14log⁡2+12⋅0+14log⁡2=12log⁡2>0 by [F4].

3.1F1F3step 1.1step 2.1algebra∎

By [F1] and [F3] as applied in step 1.1, the actual-pair EIG is zero although the pairs are distinct and observationally identical; by step 2.1 the nominal fixed-operator EIG is positive. Thus the claimed implication fails.

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