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A singular pair has no full likelihood ratio

Statement refuted

On a finite sample space with counting measure, for any probability laws $P_{0}, P_{1}$ , the pointwise quotient $r (x) = p_{1} (x) / p_{0} (x)$ is finite at every point and represents $P_{1}$ as a density with respect to $P_{0}$ .

Facts & Assumptions

Given: AC (for the linked likelihood-ratio definition and theorem), and probability measures on $X = {0, 1}$ with $A = 2^{X}$ .

[A1]

The Axiom of Choice: AC states that every family of nonempty sets has a choice function.

[F1]

Sigma-algebras and Measurable spaces and measurable sets: the full power set of a finite set is a sigma-algebra and gives a measurable space.

[F2]

Probability measures and probability spaces: a probability measure is a measure of total mass $1$ .

[F3]

Counting measure on an arbitrary set, Counting measure is a measure, and Measures on sigma-algebras: counting measure assigns unit mass to each singleton and is a measure on the finite power-set space.

[F4]

Absolute continuity of a signed or complex measure with respect to a positive measure: a measure is absolutely continuous when it vanishes on every null set of the reference measure.

[F5]

Mutual singularity for signed or complex measures: singularity is witnessed by disjoint measurable carriers on which each measure vanishes off its own carrier.

[F6]

Likelihood ratio as a Radon–Nikodym derivative: the likelihood-ratio class for a general pair is the derivative of the absolutely continuous component; the singular component remains separate.

[F9]

The Lebesgue decomposition of a sigma-finite signed measure is unique: a decomposition into an absolutely continuous and singular part is unique.

[F7]

Likelihood-ratio change of measure: the change-of-measure identity assumes $P_{1} ≪ P_{0}$ .

[F8]

A nonnegative integral over a null set vanishes: the integral of a nonnegative measurable function over a null set is zero.

Counterexample

Take $P_{0} = δ_{0}$ and $P_{1} = δ_{1}$ , where $δ_{i} (A) = 1$ if $i \in A$ and $0$ otherwise. With respect to counting measure their densities are $p_{0} (0) = 1, p_{0} (1) = 0$ and $p_{1} (0) = 0, p_{1} (1) = 1$ .

Given: $P_{0} = δ_{0}$ and $P_{1} = δ_{1}$ on the finite measurable space above.

1.1F1F2F4F5givenalgebra

The full power set is a sigma-algebra by [F1]. Each $δ_{i}$ has total mass $1$ and is countably additive: in a disjoint sequence, at most one set contains $i$ , so the sum of its values equals its value on the union. Thus $P_{0}, P_{1}$ are probability measures by [F2]. The set ${1}$ is $P_{0}$ -null and has $P_{1}$ -mass $1$ , so $P_{1} ≪̸ P_{0}$ by [F4]. The singleton carriers show $P_{0} ⊥ P_{1}$ by [F5].

2.1A1F3F5F6F9step 1.1given

The decomposition relative to $P_{0}$ is $P_{1} = 0 + P_{1}$ : the zero measure is absolutely continuous with respect to $P_{0}$ , and $P_{1}$ is singular to $P_{0}$ by step 1.1. By uniqueness [F9], this is the decomposition in [F6], so the likelihood-ratio class is zero $P_{0}$ -almost everywhere. At $x = 1$ , the point-mass quotient has denominator $p_{0} (1) = 0$ and is undefined.

3.1A1F7F8step 1.1givencontradiction: algebra∎

If a nonnegative density $q$ represented all of $P_{1}$ with respect to $P_{0}$ , then $P_{1} ({1}) = \int_{{1}} q d P_{0} = 0$ by [F8], contradicting $P_{1} ({1}) = 1$ . Thus no full density exists. The hypothesis of [F7] fails, so this does not contradict the dominated change-of-measure theorem.

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