CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Two visits can alias random-slope and residual variance components

Statement refuted

It is false that two visits at times $0$ and $1$ , together with a full-rank fixed design, always identify the random-intercept/random-slope covariance matrix and the residual variance in a Gaussian longitudinal mixed model.

Facts & Assumptions

Given: One subject has observations at $t = 0, 1$ , with fixed and random design $X = Z = (\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}),$ and the Gaussian branch of Finite-dimensional linear mixed model with explicit longitudinal roles.

[F1]

The marginal covariance is $Z G Z^{T} + R$ . Marginal covariance of the linear mixed model.

[F2]

In the Gaussian model, covariance-parameter identifiability is equivalent to injectivity of $θ \mapsto V (θ)$ . Gaussian variance-component identifiability is covariance-map injectivity.

[F3]

Under AC, the Gaussian branch gives jointly normal latent and residual vectors; scalar normal characteristic functions, the affine transformation rule, and uniqueness of characteristic functions identify the resulting multivariate-normal response law from its mean and covariance. Gaussian branch of the linear mixed model, Multivariate normal law, including singular covariance, Characteristic function of a normal law, Characteristic functions under affine maps and independent sums, Characteristic function of a multivariate normal law, Uniqueness of a law from its characteristic function.

[F4]

A real square matrix is invertible exactly when its determinant is nonzero. A finite square real matrix is invertible if and only if its determinant is nonzero.

[A1]

AC holds. The Axiom of Choice.

Construction

Set $R_{1} = I_{2}$ and $G_{1} = I_{2}$ . For a second parameter tuple set $G_{2} = (\begin{array}{cc} 4 / 5 & 1 / 5 \\ 1 / 5 & 3 / 5 \end{array}), R_{2} = \frac{6}{5} I_{2} .$ Use the same fixed parameter $β$ for both models.

Verification

Given: The parameter tuples and design in the Counterexample.

1.1F4givenalgebra

The matrix $G_{2}$ is positive definite because for every $(x, y) \neq (0, 0)$ , $(\begin{array}{cc} x & y \end{array}) G_{2} (\begin{array}{c} x \\ y \end{array}) = \frac{4}{5} {(x + \frac{1}{4} y)}^{2} + \frac{11}{20} y^{2} > 0.$ Also $G_{1} = I_{2}$ and $R_{1} = I_{2}$ are positive definite, while $R_{2} = (6 / 5) I_{2}$ is positive definite since $6 / 5 > 0$ . The design has determinant $1$ , so [F4] makes it invertible and therefore full rank.

1.2F1givenalgebra

For the first tuple, $V_{1} = Z G_{1} Z^{T} + R_{1} = (\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array}) + I_{2} = (\begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array}) .$

2.1F1givenstep 1.1algebra

Direct multiplication gives $Z G_{2} Z^{T} = (\begin{array}{cc} 4 / 5 & 1 \\ 1 & 9 / 5 \end{array}), V_{2} = Z G_{2} Z^{T} + R_{2} = (\begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array}) = V_{1} .$ The covariance parameter tuples are distinct, so the covariance map is not injective.

3.1A1F2F3givenstep 2.1∎

Under either Gaussian tuple, $Y - X β$ has every scalar projection normal: in direction $u$ , the coefficient vector on $(b, ε)$ is $(Z^{T} u, u)$ and its variance is $u^{T} V_{i} u$ . Since $V_{1} = V_{2}$ , these projections have identical characteristic functions, so [F3] gives equal response laws. The covariance-component tuples are distinct, so the parameter is not identifiable.

Remarks

The witness keeps $G_{2}$ and both residual covariance matrices positive definite. It shows an exact covariance alias at two visits; more replication of the same two-visit design preserves the same alias.

Depends on

Used by

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Sources

Fan Li, STA610 Lecture 6: Linear Mixed-Effects Models, PDF pp. 13–15 (standard reference, not scraped)
Fan Li, STA610 Lecture 7: The Longitudinal Linear Mixed Model, PDF pp. 6–8 (standard reference, not scraped)