TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.

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Gaussian variance-component identifiability is covariance-map injectivity

Statement

Assume AC. Let $Θ$ be a nonempty parameter set and let $θ \mapsto (G_{θ}, R_{θ})$ be a declared covariance-component parameterization of the Gaussian mixed model, with $V_{θ} = Z G_{θ} Z^{T} + R_{θ}$ positive semidefinite for every $θ \in Θ$ . Consider the parametric family $P_{β, θ} = N_{n} (X β, V_{θ}), (β, θ) \in R^{p} \times Θ,$ and the estimand $ψ (β, θ) = θ$ . Then $ψ$ is identifiable if and only if the covariance map $θ \mapsto V_{θ}$ is injective on $Θ$ .

Facts & Assumptions

Given: The parameterized Gaussian model and target in the Statement.

[F1]

An estimand is identifiable exactly when equal sampling laws force equal target values. Identifiability of an estimand.

[F2]

Under AC, a multivariate-normal law is determined by its mean and covariance, including in singular cases. Multivariate normal law, including singular covariance, Characteristic function of a multivariate normal law.

[F3]

Variance and covariance are moments of the sampling law, so equal laws with finite second moments have equal covariance matrices. Moments, variance, and covariance on a probability space.

[A1]

AC holds, and every covariance parameter value in $Θ$ yields the stated Gaussian law. The Axiom of Choice.

Proof

Given: The Gaussian family and target $ψ (β, θ) = θ$ .

1.1A1F1F3algebra

Suppose $θ \mapsto V_{θ}$ is injective. If $P_{β, θ} = P_{β^{'}, θ^{'}}$ , equality of laws and [F3] give $V_{θ} = V_{θ^{'}}$ . Injectivity then gives $θ = θ^{'}$ , so the target values agree. By [F1], $ψ$ is identifiable.

1.2A1F1F2choose

Conversely, if the covariance map is not injective, choose $θ \neq θ^{'}$ with $V_{θ} = V_{θ^{'}}$ . For any fixed $β$ , the two Gaussian laws $N_{n} (X β, V_{θ})$ and $N_{n} (X β, V_{θ^{'}})$ are equal by [F2], while the target values are different. Thus $ψ$ is not identifiable by [F1].

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove both implications.

Depends on

Used by

Two visits can alias random-slope and residual variance components Counterexample

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Fan Li, STA610 Lecture 6: Linear Mixed-Effects Models, PDF pp. 10–11 (standard reference, not scraped)
Duchateau, Janssen, and Rowlands, Linear Mixed Models, Chapter 2 §§2.3.3–2.4.2 (standard reference, not scraped)