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A generalized EM step can increase Q without maximizing it

Example

Let Θ=(0,1) with its Borel sigma-algebra, and let X=Z={0,1} with the power-set sigma-algebra and counting measure. Under parameter θ, take X,Z independent Bernoulli(θ) variables, so pθ(x,z)=θx+z(1−θ)2−x−z. Observe x=1 at current value θt=1/2. The candidate θ=7/10 strictly increases Q but is not its exact maximizer; the observed log likelihood also increases.

Facts & Assumptions

Given: AC for the finite-KL admissibility branch, the displayed two-coordinate Bernoulli model, observation x=1, current parameter θt=1/2, and candidate θ=7/10.

[A1]

The Axiom of Choice: AC supplies the Radon–Nikodym branch needed to evaluate finite KL when the current law is absolutely continuous with respect to the candidate law.

[F1]

Latent-variable joint and observed-data factorization: each parameter fiber of a complete-data density integrates to one over observed and latent coordinates.

[F2]

The observed density is the latent marginal under Tonelli: integrating out the latent coordinate gives the observed-data density, and positive finite fibers yield a normalized conditional density.

[F3]

Kullback–Leibler divergence: under AC, D(P∥Q) is directional and uses the Radon–Nikodym derivative when P≪Q.

[F4]

Expectation-maximization: Q(θ∣θt) averages the complete-data log density under the current latent law; exact mode selects an attained maximizer over an admissible candidate set containing the current iterate.

Verification

1.1givenA1F1F2F3F4algebra

For every θ∈(0,1), summing pθ(x,z) over (x,z)∈{0,1}2 gives (θ+(1−θ))2=1. At x=1, mθ(1)=θ and the conditional law of Z is Bernoulli(θ), so the current law is Bernoulli(1/2). Every candidate has positive finite marginal, the same two-point latent support, and D(qt∥qθ)=12log⁡(1/(2θ))+12log⁡(1/(2(1−θ))) is finite by [A1, F3]. Their complete-data log values are finite, so every θ∈(0,1) is admissible by [F4].

2.1givenF2step 1.1algebra

Averaging log⁡pθ(1,Z) under that current conditional law gives Q(θ∣1/2)=32log⁡θ+12log⁡(1−θ)=12log⁡(θ3(1−θ)). At θ=1/2, this is log⁡(1/4); at θ=7/10, it is 12log⁡(1029/10000). Their difference is 12log⁡(1029/625)>0, so the candidate strictly raises Q.

3.1givenF4step 2.1algebra∎

For 0<θ<1, 27256−θ3(1−θ)=(θ−3/4)2((θ+1/4)2+1/8)≥0, with equality only at θ=3/4. Thus the exact maximizer of Q is 3/4, so 7/10 is a genuinely nonmaximizing generalized step. The observed log-likelihood difference is log⁡(7/10)−log⁡(1/2)=log⁡(7/5)>0.

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