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A support-losing candidate has infinite directional KL

Example

Let Θ=[0,1] with its Borel sigma-algebra, let the observation space be the singleton {x0} with counting measure, and let the latent space be {0,1} with counting measure. Set pθ(x0,1)=θ and pθ(x0,0)=1−θ. At current value θt=1/2, the latent law is (1/2,1/2); candidate θ=0 has latent law (1,0). The observed marginal is one at both parameters, but the current-to-candidate KL divergence is infinite. The extended diagnostic integral of log⁡p0(x0,Z) under the current latent law is −∞ when log⁡0=−∞; the algorithm's Q(0∣1/2) is undefined because its Definition assigns Q only when the integral is finite.

Facts & Assumptions

Given: The displayed finite complete-data model, fixed observation x0, current value θt=1/2, candidate value θ=0, and the convention log⁡0=−∞.

[F1]

Latent-variable joint and observed-data factorization: a complete-data density is normalized over the observation and latent coordinates for every parameter.

[F2]

Kullback–Leibler divergence: if the first probability is not absolutely continuous with respect to the second, the directional divergence is +∞; this branch uses no Radon–Nikodym representative or Axiom of Choice.

[F3]

Expectation-maximization: Q is defined only when its integral is finite; admissibility requires finite directional KL and integrability of the candidate log density, and the qt≪̸qθ rejection branch is choice-free while the qt≪qθ finite-KL branch assumes AC.

Verification

1.1givenF1algebra

For each θ∈[0,1], the complete-data masses sum to θ+(1−θ)=1, and the observed marginal is mθ(x0)=1. The current and candidate latent laws are therefore qt=(1/2,1/2) and q0=(1,0).

2.1givenF2F3step 1.1algebra

The singleton set {1} has q0-mass zero and qt-mass 1/2, so qt≪̸q0 and D(qt∥q0)=+∞ by [F2]. The extended diagnostic integral is 12log⁡(1−0)+12log⁡0=−∞; it is not a finite value of Q(0∣1/2) under [F3].

3.1F3step 2.1∎

The observed log likelihood is zero at both parameters, but the candidate violates the support-compatible admissibility condition of [F3]. It is not an update allowed by the defined procedure.

Depends on

Used by

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