Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 2 results · all verified · 2 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 2 also cleared it.

Domination and Likelihood Ratios — Examples

1 · Prerequisites

2 · Summary

The two-point Bernoulli pair gives an everywhere positive likelihood ratio and verifies the change-of-measure identity by direct summation. Two disjoint point masses show why an arbitrary pair of laws need not have a full likelihood-ratio density: the denominator gives zero mass to the numerator's support.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Bernoulli likelihood ratio and change of measure

Example

Let $X = {0, 1}$ , $A = 2^{X}$ , and let $μ$ be counting measure. Set $Θ = {0, 1}$ and define the experiment's probability measures by

P_{0} ({1}) = \frac{1}{3}, P_{0} ({0}) = \frac{2}{3}, P_{1} ({1}) = \frac{2}{3}, P_{1} ({0}) = \frac{1}{3} .

Both are dominated by $μ$ . Their likelihood-ratio density is $r (0) = 1 / 2$ and $r (1) = 2$ .

Facts & Assumptions

Given: AC (for the linked general theorem), the two-point measurable space, counting measure, and probability mass assignments in the example.

[F1]

Sigma-algebras: a sigma-algebra is an algebra of subsets closed under countable unions; the full power set of ${0, 1}$ is such an algebra.

[F2]

Measurable spaces and measurable sets: a measurable space is a set with a sigma-algebra.

[F3]

Counting measure on an arbitrary set, Counting measure is a measure, and Finite, sigma-finite, and semifinite measures: counting measure assigns mass $1$ to each singleton and is a measure; here its total mass is $2$ , so it is finite and sigma-finite.

[F4]

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

[F5]

Statistical experiments: an experiment is a parameter-indexed family of probability measures on one measurable sample space.

[F6]

Dominated statistical experiments: domination means each sampling law is absolutely continuous with respect to the supplied sigma-finite positive measure.

[F7]

Likelihood densities: for a fixed observation $x$ , likelihood is the selected density value $p_{θ} (x)$ .

[F8]

Likelihood ratio as a Radon–Nikodym derivative: for $P_{1} ≪ P_{0}$ , the full likelihood-ratio class is $d P_{1} / d P_{0}$ .

[F9]

Likelihood-ratio change of measure: when $P_{1} ≪ P_{0}$ , a nonnegative likelihood-ratio representative satisfies $E_{P_{0}} [g r] = E_{P_{1}} [g]$ for every nonnegative measurable $g$ and every $P_{1}$ -integrable real $g$ .

[A1]

The Axiom of Choice: AC states that every family of nonempty sets has a choice function; the finite example itself uses explicit mass functions.

Verification

Given: $X = {0, 1}$ , the four masses above, and counting measure $μ$ .

1.1F1F2F3F4F5F6givenalgebra

The full power set is a sigma-algebra by [F1], so [F2] gives a measurable space. The counting measure is finite with $μ (X) = 2$ by [F3], hence sigma-finite. Each displayed mass is nonnegative and the two masses for each $P_{i}$ sum to $1$ . For any disjoint sequence of events, each point belongs to at most one event, so the mass formula is countably additive; each $P_{i}$ is therefore a probability measure by [F4]. Since counting measure assigns positive mass to every nonempty subset, $P_{i} ≪ μ$ , and the family is a dominated experiment by [F5]–[F6].

1.2A1F7F8givenalgebra

Put $p_{i} (x) : = P_{i} ({x})$ for $i \in {0, 1}$ . For every $A \subseteq X$ , $\int_{A} p_{i} d μ = \sum_{x \in A} p_{i} (x) = P_{i} (A)$ , so these are density representatives as in [F7]. Since $p_{0}$ is positive at both points, $P_{1} ≪ P_{0}$ ; define $r (x) = p_{1} (x) / p_{0} (x)$ . For every $A \subseteq X$ , $\int_{A} r d P_{0} = \sum_{x \in A} r (x) p_{0} (x) = \sum_{x \in A} p_{1} (x) = P_{1} (A)$ ; thus $r$ represents the likelihood-ratio derivative [F8], with $r (0) = (1 / 3) / (2 / 3) = 1 / 2$ and $r (1) = (2 / 3) / (1 / 3) = 2$ .

2.1A1F9givenalgebra∎

For any real-valued $g$ on $X$ , $E_{P_{0}} [g r] = \frac{2}{3} g (0) \frac{1}{2} + \frac{1}{3} g (1) 2 = \frac{1}{3} g (0) + \frac{2}{3} g (1) = E_{P_{1}} [g]$ . Thus the change-of-measure identity holds directly, in agreement with [A1, F9].

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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.