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✓ 12 results · all verified · 11 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. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Mixed effects and longitudinal models

1 · Prerequisites

2 · Summary

Objective

Develop the finite-dimensional linear mixed model and its longitudinal interpretation with explicit subject, visit, and observation-pattern roles. The general covariance identity uses only finite second moments and independence. Separate results add known-covariance GLS, a Gaussian conditional-law branch, moment-only BLUP, error-contrast REML, and a precise covariance-map criterion for Gaussian variance-component identifiability.

Scope

The Gaussian branch explicitly carries the Axiom of Choice required by the current multivariate-normal supplier. Local results prove the Gaussian law of an error contrast and supply the nondegenerate Gaussian density formula. The conditional formula allows singular latent covariance but requires positive-definite marginal observation covariance. The known-covariance GLS and BLUP results are moment-only; plug-in variance-component estimates do not inherit their exact covariance guarantees. An observation pattern is modeled separately from the complete-data response law: the mixed-model covariance equations imply no missingness or ignorability assumption. Error contrasts use columns of a matrix K∈Rn×(n−p) so KTY is defined and annihilates the fixed design; a basis change multiplies the likelihood by a parameter-independent factor.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Jointly Gaussian blocks with zero cross-covariance are independent, including singular blocks

Statement

Assume AC. Let (U1,U2) be a jointly multivariate-normal vector with mean (m1,m2) and covariance (S00T), where S and T are allowed to be singular positive-semidefinite matrices. Then the random vectors U1 and U2 are independent.

Facts & Assumptions

Given: AC and a finite-dimensional jointly Gaussian vector with the block covariance in the Statement.

[F1]

A multivariate-normal law with positive-semidefinite covariance exists, may be realized as its mean plus the covariance square root times independent standard-normal coordinates, and allows singular covariance. Multivariate normal law, including singular covariance.

[F2]

The nonnegative square root of a finite-dimensional nonnegative operator exists and is unique. A non-negative operator has a unique non-negative square root.

[F3]

Disjoint groups of an independent sigma-algebra family remain independent. Disjoint groups of an independent sigma-algebra family remain independent.

[F4]

Measurable functions of independent random elements remain independent. Measurable coordinatewise functions preserve independence.

[F5]

The multivariate-normal characteristic function uniquely determines its law, including for singular covariance. Characteristic function of a multivariate normal law.

[A1]

AC holds for the multivariate-normal construction. The Axiom of Choice.

Proof

Given: The hypotheses in the Statement.

1.1F2algebra

Let S1/2 and T1/2 be the nonnegative square roots from [F2]. The block matrix A=diag⁡(S1/2,T1/2) is nonnegative and satisfies A2=diag⁡(S,T), so uniqueness in [F2] makes A the nonnegative square root of the joint covariance. This argument also applies when either block is singular.

2.1A1F1F3F4step 1.1

By [A1] and [F1], the joint law is the law of (m1,m2)+AZ for a vector Z with independent standard-normal coordinates. Partition those coordinates into vectors Z1,Z2 of the matching dimensions. Their sigma-algebras are independent by [F3], and the separate affine maps m1+S1/2Z1 and m2+T1/2Z2 remain independent by [F4]. Hence the coordinate blocks of the constructed vector are independent.

3.1A1F5step 2.1∎

Any jointly multivariate-normal vector with the stated mean and covariance has this constructed law by the uniqueness clause of [F5]. Independence is a property of the joint law, so U1 and U2 are independent.

Remarks

No inverse of S or T is used. Positive definiteness of the full covariance is not required for this independence result.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A positive-definite multivariate normal law has its Gaussian density

Statement

Assume AC. Let d≥1, m∈Rd, and let Σ be a real symmetric positive-definite d×d matrix. If X∼Nd(m,Σ), then its law has Lebesgue density fm,Σ(x)=(2π)−d/2det⁡(Σ)−1/2exp⁡ ⁣(−12(x−m)TΣ−1(x−m)),x∈Rd.

Facts & Assumptions

Given: AC, a finite dimension d≥1, a mean vector m, and a positive-definite covariance matrix Σ.

[F1]

The standard normal law has density (2π)−1/2e−z2/2, and the multivariate-normal construction permits singular covariance. Standard normal and normal laws, Multivariate normal law, including singular covariance.

[F2]

Finite independent random elements have product joint law. Independent random elements have product joint law.

[F3]

In finite Euclidean dimension, product Lebesgue measure agrees with Euclidean Lebesgue measure on Borel sets; nonnegative product integrals obey Tonelli. On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product.

[F4]

Under countable choice, a C1 diffeomorphism satisfies the change-of-variables formula for every nonnegative Lebesgue-measurable integrand; continuous maps have Borel preimages of Borel sets. A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions, A continuous map has Borel preimages of Borel sets.

[F5]

Positive-definite Σ has a Cholesky factorization Σ=LLT with L lower triangular and positive diagonal. A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique.

[F6]

Characteristic functions of finite independent sums and affine images follow the stated product and affine rules. Characteristic functions under affine maps and independent sums.

[F7]

Scalar normal characteristic functions determine their laws. Characteristic function of a normal law, Uniqueness of a law from its characteristic function.

[F8]

A multivariate-normal law is specified by its scalar projection laws and is uniquely determined by its multivariate characteristic function. Multivariate normal law, including singular covariance, Characteristic function of a multivariate normal law.

[A1]

AC holds; the cited bridge supplies countable and dependent choice (CC and DC), as defined by The Axiom of Countable Choice (ACω) and The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain, for the Gaussian construction and linear change-of-variables theorem. The Axiom of Choice, AC supplies countable selections and prescribed serial paths.

Proof

Given: The hypotheses in the Statement.

1.1F1F2F3A1

By [A1], take d independent standard-normal coordinates Z1,…,Zd. Their joint law is the product of the coordinate laws by [F2]. Thus, using [F1] and [F3], the vector Z=(Z1,…,Zd)T has Lebesgue density fZ(z)=(2π)−d/2exp⁡(−zTz/2).

2.1F5F6F7F8step 1.1algebra

Let Σ=LLT be the factorization from [F5]. For any u∈Rd, the scalar uTLZ is a finite linear combination of the independent standard normals. By [F6] its characteristic function is ∏k=1dexp⁡ ⁣(−s2(LTu)k2/2)=exp⁡ ⁣(−s2uTΣu/2), so [F7] gives uTLZ∼N(0,uTΣu). Therefore m+LZ has all the projection laws defining Nd(m,Σ), and [F8] identifies its law with that of X.

3.1A1F4F5step 1.1step 2.1

Let T(z)=m+Lz. It is a C1 diffeomorphism with Jacobian determinant det⁡L>0. For any Borel set E⊆Rd, define the nonnegative Borel function g(x)=1E(x)fZ(L−1(x−m))/det⁡L. By [F4] and the countable choice supplied by [A1], ∫EfZ(L−1(x−m))(det⁡L)−1 dx=∫T−1(E)fZ(z) dz=P(TZ∈E). Thus the density in step 2.1 is fZ(L−1(x−m))(det⁡L)−1. The determinant is positive because L is triangular with positive diagonal.

4.1F9step 1.1step 3.1algebra∎

Since det⁡Σ=det⁡(LLT)=(det⁡L)2 by [F9], and (L−1)TL−1=Σ−1, substituting the expression for fZ from step 1.1 into step 3.1 gives exactly fm,Σ(x).

Remarks

Positive definiteness is essential for this Lebesgue-density formula. Singular Gaussian laws remain defined, but are supported on lower-dimensional affine subspaces and are not covered by this theorem.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Error-contrast matrix for a full-rank fixed design

Definition

Let X∈Rn×p have full column rank, with p<n, and let col⁡(X)⊥ denote the orthogonal complement of its column space in Euclidean Rn (The orthogonal complement W⊥={v:⟨v,w⟩=0 for all w∈W}). An error-contrast matrix for X is a matrix K∈Rn×(n−p) whose columns form a basis of col⁡(X)⊥. Thus K has full column rank and XTK=0, so KTY is an (n−p)-dimensional vector of linear contrasts that annihilate Xβ. Such a matrix exists by A full-rank error-contrast matrix exists when n exceeds p ↗.

This column-basis convention makes the product KTY dimensionally defined. The Gaussian restricted-likelihood definition below fixes one such K independently of the covariance parameter.

Remarks

The condition p<n is needed for a nonzero error-contrast space. When p=n, the orthogonal complement has dimension zero and this page makes no variance-likelihood claim from contrasts.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Generalized least-squares estimator for known covariance

Definition

Let X∈Rn×p have full column rank in a fixed-design first-moment regression model, so EβY=Xβ under each law Pβ (Fixed-design linear regression model). Suppose the same known symmetric positive-definite matrix V is the covariance of Y under every Pβ. Put C=(XTV−1X)−1andβ^GLS:=CXTV−1Y. This is a statistic and estimator of the identity target β under Statistics and estimators and Estimands on a statistical model: for fixed X and V, the displayed expression is a linear, hence Borel-measurable, function of the response. The positive definiteness of XTV−1X and existence of its inverse are established by The GLS information matrix is positive definite ↗.

This definition uses the true covariance as known. Replacing V by an estimate defines a plug-in procedure; the theorem below does not assign that procedure the same exact covariance or BLUE guarantee.

Remarks

The marginal-covariance lemma later identifies V=ZGZT+R for the mixed model. The GLS definition itself applies to any fixed-design first-moment model with the stated common covariance.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite-dimensional linear mixed model with explicit longitudinal roles

Definition

Let n,p≥1 and q≥0 be finite integers. A finite-dimensional linear mixed model on one probability space consists of fixed matrices X∈Rn×p and Z∈Rn×q, a fixed parameter β∈Rp, and square-integrable random vectors b∈Rq and ε∈Rn such that Y=Xβ+Zb+ε,Eb=0,Eε=0,Cov⁡(b)=G,Cov⁡(ε)=R, where b and ε are independent random elements. Here G and R are the resulting symmetric positive-semidefinite covariance matrices. The fixed-design convention for X and the first-moment regression model is that of Fixed-design linear regression model. The explicit scientific record of units, quantities, observation mechanism, and adequacy assumptions is required by Scientific statistical model contracts.

For longitudinal data, let j index subjects and let i∈Sj index the visits scheduled for subject j. A row of Y, X, and Z corresponds to one observed subject–visit pair (j,i); the time tji is measured from a stated origin. Blocks of b may represent subject-specific latent effects, while R records the residual covariance across observed rows. If a visit pattern is fixed or conditioned on, the corresponding rows are selected from the scheduled vectors and matrices. If the pattern is random, its observation mechanism is specified separately; the displayed model does not assert that this mechanism is independent of the full outcomes or ignorable for likelihood inference.

The definition imposes no Gaussianity, no identical visit schedule, and no claim that a scientific application satisfies the proposed mean or covariance structure.

Remarks

The vector b is a latent random effect in the stated probability model; this definition gives it no Bayesian interpretation. Singular G and singular R are allowed here. In later results, positive definiteness of the marginal covariance V=ZGZT+R is an additional hypothesis when an inverse is used.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Gaussian branch of the linear mixed model

Definition

The Gaussian branch of Finite-dimensional linear mixed model with explicit longitudinal roles adds the following assumption: under the Axiom of Choice, the concatenated vector W=(bε) has the multivariate-normal law Nq+n(0,diag⁡(G,R)) of Multivariate normal law, including singular covariance. Thus the full latent-effect and residual vector is jointly Gaussian. This is an additional branch; the moment-only definition and the covariance identity do not assume Gaussianity.

Remarks

This definition carries AC because the current multivariate-normal supplier does. Singular G or R is permitted by that supplier; any later inverse or density requires a separate positive-definiteness hypothesis.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Longitudinal observation patterns and missingness mechanisms

Definition

Let m=∑j∣Sj∣ and let Y∗∈Rm be the complete scheduled response vector for subjects j and visits i. Let the declared observed covariates take values in a measurable space (X,A). The finite set of observation patterns is R={0,1}m; rji=1 means visit (j,i) is observed. A missingness mechanism is a function g:R×Rm×X→[0,1] such that, for every (y∗,x), ∑r∈Rg(r∣y∗,x)=1, and, for each fixed r, (y∗,x)↦g(r∣y∗,x) is measurable for B(Rm)⊗A. Thus g(⋅∣y∗,x) is a probability mass function on R, and its dependence on the full data state is the finite-output kernel specified by the observation-mechanism field in Scientific statistical model contracts, which requires a probability measure at each state and measurable state dependence. For a pattern r, write Yobs(r) for the subvector of Y∗ at coordinates with rji=1.

The mechanism is missing completely at random (MCAR) if, for each pattern r, g(r∣y∗,x) does not depend on y∗. It is missing at random (MAR) if, for each r, it depends on the full response vector only through Yobs(r) and x. It is missing not at random (MNAR) relative to this partition if the MAR condition fails for at least one pattern on the declared support. Monotone dropout means that, within each subject, observed visits form an initial segment of the scheduled visits, possibly empty or complete.

These are properties of the declared observation kernel. A mixed-model mean or covariance equation does not specify the kernel or imply MCAR, MAR, or MNAR.

Remarks

For an LMM written on observed rows, Oj={i:rji=1} determines which coordinates and design rows are retained. If r depends on the full scheduled outcomes, the distribution of the selected rows conditional on that pattern need not equal the corresponding unselected marginal subvector law.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A full-rank error-contrast matrix exists when n exceeds p

Statement

For every full-column-rank X∈Rn×p with p<n, there exists an error-contrast matrix K∈Rn×(n−p) as defined in Error-contrast matrix for a full-rank fixed design.

Facts & Assumptions

Given: X∈Rn×p has full column rank and p<n.

[F1]

For every subspace W of finite-dimensional Rn, dim⁡W+dim⁡W⊥=n. In finite dimension, W⊥⊥=W and dim⁡W+dim⁡W⊥=dim⁡V.

[F3]

An error-contrast matrix has full-column-rank columns forming col⁡(X)⊥. Error-contrast matrix for a full-rank fixed design.

[A1]

The columns of X are linearly independent.

Proof

Given: A full-column-rank X∈Rn×p with p<n.

1.1A1F1F2

Set W=col⁡(X). By [A1] and [F2], the columns of X form a basis of W, so dim⁡W=p. Applying [F1] gives dim⁡W⊥=n−p; by [F2], this means W⊥ has a basis with n−p elements.

2.1F2step 1.1choose

Since W⊥ has dimension n−p, [F2] gives a basis k1,…,kn−p of W⊥. Put these vectors into the columns of K. The columns are independent and span W⊥, so K has full column rank.

3.1F3step 2.1∎

Every column kℓ is orthogonal to each column of X, hence XTkℓ=0. Thus XTK=0 and K satisfies the error-contrast definition [F3]. This proves existence.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The GLS information matrix is positive definite

Statement

If X∈Rn×p has full column rank and V∈Rn×n is symmetric positive definite, then XTV−1X is symmetric positive definite. In particular it is invertible.

Facts & Assumptions

Given: The full-column-rank design X and known positive-definite covariance V in Generalized least-squares estimator for known covariance.

[F1]

Every real symmetric positive-definite matrix has a Cholesky factorization V=LLT with L invertible. A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique.

[A1]

X has full column rank, and V is symmetric positive definite.

Proof

Given: X∈Rn×p has full column rank and V is positive definite.

1.1A1F1algebra

By [A1] and [F1], write V=LLT with L invertible. Hence V−1=L−TL−1 and V−1 is symmetric.

2.1A1step 1.1algebra∎

For any nonzero c∈Rp, full column rank gives Xc≠0, so L−1Xc≠0. Therefore cTXTV−1Xc=(L−1Xc)T(L−1Xc)>0. The matrix is symmetric by step 1.1, proving positive definiteness and invertibility.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Marginal covariance of the linear mixed model

Statement

For the moment model Y=Xβ+Zb+ε of Finite-dimensional linear mixed model with explicit longitudinal roles, the marginal covariance is Cov⁡(Y)=V:=ZGZT+R. This identity uses only the stated means, covariances, and independence; it does not require Gaussianity.

Facts & Assumptions

Given: The finite-dimensional linear mixed model and its finite second moments.

[F1]

Y=Xβ+Zb+ε, Eb=Eε=0, Cov⁡(b)=G, Cov⁡(ε)=R, and b and ε are independent. Finite-dimensional linear mixed model with explicit longitudinal roles.

[F2]

Covariance is bilinear in finite linear combinations of square-integrable real random variables. Covariance is symmetric and bilinear in finite linear combinations.

[F3]

Measurable coordinate functions of independent random elements remain independent. Measurable coordinatewise functions preserve independence.

[F4]

Expectations of products of independent integrable real random variables factor. Expectations factor over finite products of independent random variables.

[F5]

Expectation is linear on finite sums of integrable real random variables. The Lebesgue integral is linear on L1(μ).

Proof

Given: A fixed β, finite matrices X,Z, and random vectors b,ε satisfying [F1].

1.1F1F5algebra

Since Eb=Eε=0, linearity [F5] gives EY=Xβ and Y−EY=Zb+ε.

2.1F1F2step 1.1algebra

By [F2], each covariance entry of Zb+ε is the sum of the entries from Zb and ε and the two cross-covariance entries. The first two matrices are ZGZT and R by [F1] and finite matrix multiplication.

3.1F1F2F3F4step 2.1algebra∎

For each coordinate pair bk,εℓ, [F3] gives independence and [F4] gives E[bkεℓ]=E[bk]E[εℓ]=0. Both means vanish by [F1], so every cross-covariance entry is zero. The remaining covariance matrix is therefore ZGZT+R.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

The REML error-contrast covariance is positive definite

Statement

Let X∈Rn×p have full column rank with p<n, let K∈Rn×(n−p) be an error-contrast matrix for X, and let V∈Rn×n be symmetric positive definite. Then KTVK is symmetric positive definite. No Gaussian assumption is needed for this matrix claim.

Facts & Assumptions

Given: V∈Rn×n is symmetric positive definite and K∈Rn×(n−p) has full column rank.

[F1]

An error-contrast matrix has full column rank and columns spanning col⁡(X)⊥. Error-contrast matrix for a full-rank fixed design.

[A1]

The matrix V is symmetric positive definite.

Proof

Given: The full-rank design, its error-contrast matrix, and a symmetric positive-definite V.

1.1A1algebra

The matrix KTVK is symmetric because V is symmetric.

1.2F1A1algebra

Let a∈Rn−p be nonzero. By [F1], K has full column rank, so Ka≠0. Therefore aTKTVKa=(Ka)TV(Ka)>0 by positive definiteness of V. Thus KTVK is positive definite.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove symmetry and strict positivity of the quadratic form for every nonzero a, which is exactly positive definiteness of KTVK.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A Gaussian error contrast has the induced multivariate-normal law

Statement

Assume AC and the Gaussian branch of the finite-dimensional mixed model. Let X∈Rn×p have full column rank with p<n, let K∈Rn×(n−p) be an error-contrast matrix, and let V=ZGZT+R=Cov⁡(Y). Then KTY∼Nn−p(0,KTVK). The contrast law does not depend on β. The matrix KTVK may be singular; positive definiteness is a separate hypothesis when a density is needed.

Facts & Assumptions

Given: The centered Gaussian mixed model, AC, a full-rank fixed design, and an error-contrast matrix K.

[F1]

Y=Xβ+Zb+ε with centered b,ε, covariance matrices G,R, and b independent of ε. Finite-dimensional linear mixed model with explicit longitudinal roles.

[F2]

Under AC, the concatenated vector (b,ε) has a multivariate-normal law. Gaussian branch of the linear mixed model.

[F3]

The response covariance is V=ZGZT+R. Marginal covariance of the linear mixed model.

[F4]

An error-contrast matrix has shape n×(n−p) and satisfies XTK=0. Error-contrast matrix for a full-rank fixed design.

[F5]

Every scalar linear projection of a multivariate-normal vector has the specified univariate normal law, including variance zero. Multivariate normal law, including singular covariance.

[F6]

Expectation is linear on finite sums of integrable random variables. The Lebesgue integral is linear on L1(μ).

[F7]

Covariance is a centered second moment and is bilinear in finite linear combinations. Moments, variance, and covariance on a probability space, Covariance is symmetric and bilinear in finite linear combinations.

[F8]

A fixed finite-dimensional linear map is continuous and Borel measurable, so its composition with a Borel random vector is a Borel random vector. Every Euclidean linear map has a unique matrix and satisfies ∥Lh∥2≤K∥h∥2 for some K≥0, A continuous map has Borel preimages of Borel sets.

[F9]

A univariate normal law has characteristic function exp⁡(itm−t2σ2/2). Characteristic function of a normal law.

[F10]

The characteristic function of Nd(m,Σ) is exp⁡(itTm−tTΣt/2). Characteristic function of a multivariate normal law.

[F11]

The characteristic function of a multivariate-normal law uniquely determines its Borel probability law, including singular covariance. Characteristic function of a multivariate normal law.

[A1]

Proof

Given: The hypotheses in the Statement.

1.1F1F3F6

By [F1], Y−Xβ=Zb+ε is centered. Linearity [F6] gives EY=Xβ, and [F3] gives Cov⁡(Y)=V.

2.1A1F1F2F4F5F6F7step 1.1

Fix u∈Rn−p. Since XTK=0 by [F4], uTKTY=uTKT(Zb+ε), a scalar linear projection of the jointly Gaussian vector in [F2]; it is normal by [F5]. Its mean is zero by [F1] and [F6], and its variance is uTKTVKu by step 1.1 and covariance bilinearity [F7].

3.1F8F9F10F11step 2.1∎

By [F8], Q=KTY is a Borel random vector. For each u, [F9] and step 2.1 give EeiuTQ=exp⁡(−uTKTVKu/2). This is the characteristic function of Nn−p(0,KTVK) by [F10]; uniqueness [F11] gives the stated law. Its mean and covariance do not depend on β, so neither does its law.

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Conditional Gaussian law of random effects

Statement

Assume AC and the Gaussian branch of the finite-dimensional mixed model. Let V=ZGZT+R be positive definite. With conditional expectation taken componentwise given σ(Y), and conditional covariance defined entrywise by conditional expectations of centered products, one has E[b∣Y]=GZTV−1(Y−Xβ),Cov⁡(b∣Y)=G−GZTV−1ZG. The latent covariance G may be singular.

Facts & Assumptions

Given: The Gaussian mixed model, AC, and V≻0.

[F1]

Y=Xβ+Zb+ε, with centered independent b,ε and covariance matrices G,R; these are finite-second-moment covariances. Finite-dimensional linear mixed model with explicit longitudinal roles, Moments, variance, and covariance on a probability space.

[F2]

The marginal covariance is V=ZGZT+R. Marginal covariance of the linear mixed model.

[F3]

Under AC, (b,ε) is jointly multivariate normal with covariance diag⁡(G,R). Gaussian branch of the linear mixed model, Multivariate normal law, including singular covariance.

[F4]

Jointly Gaussian coordinate blocks with zero cross-covariance are independent, including when either covariance is singular. Jointly Gaussian blocks with zero cross-covariance are independent, including singular blocks.

[F5]

Independence of random elements means independence of their generated sigma-algebras. Independent random elements.

[F6]

A conditional-expectation version is characterized by its integrals on every event in the conditioning sigma-algebra. Conditional expectation given a sigma algebra.

[F7]

Measurable functions of independent random elements remain independent, and integrable products of independent variables factor. Measurable coordinatewise functions preserve independence, Expectations factor over finite products of independent random variables.

[F8]

Products of square-integrable variables are integrable by Cauchy–Schwarz. Cauchy-Schwarz inequality for L2.

[F9]

Covariance is bilinear in finite linear combinations. Covariance is symmetric and bilinear in finite linear combinations.

[F10]

Expectation is linear on finite sums of integrable random variables. The Lebesgue integral is linear on L1(μ).

[F11]

Finite-dimensional linear maps are continuous, and continuous maps between Euclidean spaces have Borel preimages of Borel sets. Every Euclidean linear map has a unique matrix and satisfies ∥Lh∥2≤K∥h∥2 for some K≥0, A continuous map has Borel preimages of Borel sets.

[A1]

AC holds, V is symmetric positive definite, and all variables have the finite second moments stated in [F1]. The Axiom of Choice.

Proof

Given: The Gaussian branch with V≻0 and AC.

1.1A1F1F2F5F7F9F10algebra

Put Yc=Y−Xβ. Linearity [F10] and [F1] give EY=Xβ, so centering by this deterministic vector leaves covariance unchanged; [F2] gives Cov⁡(Yc)=V. Coordinatewise independence of b and ε gives Cov⁡(b,ε)=0 by [F5] and [F7], so bilinearity [F9] gives Cov⁡(b,Yc)=Cov⁡(b,Zb)=GZT. Since V is invertible by [A1], set H=GZTV−1 and U=b−HYc. Bilinearity yields Cov⁡(U,Yc)=GZT−HV=0, and expectation linearity [F10] gives EU=0.

2.1A1F3F4F11step 1.1

The vectors U and Yc are Borel random vectors by [F11]. For any coefficient vectors s,t, the scalar sTU+tTYc is a linear combination of the coordinates of (b,ε), so it is normal by the projection definition in [F3]. Thus (U,Yc) is jointly multivariate normal. Its covariance is block diagonal by step 1.1, so [F4] implies that U and Yc are independent.

3.1A1F5F6F7F10step 1.1step 2.1

By step 2.1 and [F5], U and Yc generate independent sigma-algebras. For every event E∈σ(Yc), [F7] applied to Uk and 1E gives E[Uk1E]=E[Uk]P(E)=0. Since HYc is square-integrable and σ(Yc)-measurable, [F10] gives E[bk1E]=E[(HYc)k1E] for every such event; [F6] identifies HYc as a version of E[b∣Yc]. Finally, σ(Yc)=σ(Y), so the stated conditional-mean formula follows.

4.1F6F7F8step 1.1step 2.1step 3.1algebra

For each coordinate pair k,ℓ, UkUℓ is integrable by [F8]. As a measurable function of U, it is independent of Yc by [F7] and step 2.1. For every E∈σ(Yc), factorization gives E[UkUℓ1E]=E[UkUℓ]P(E), so [F6] yields E[UkUℓ∣Yc]=E[UkUℓ]. Together with the conditional mean from step 3.1, step 1.1, and σ(Yc)=σ(Y), the defined entrywise conditional covariance of b=HYc+U is Cov⁡(U).

5.1F1F2F9step 1.1step 3.1step 4.1algebra∎

Expanding Cov⁡(U)=Cov⁡(b−HYc) by [F9] and [F1]–[F2] yields Cov⁡(U)=G−GZTV−1ZG−GZTV−1ZG+GZTV−1VV−1ZG=G−GZTV−1ZG. Together with steps 3.1–4.1 this proves both formulas. No inverse of G is used, so singular G is allowed.

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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↦(Gθ,Rθ) be a declared covariance-component parameterization of the Gaussian mixed model, with Vθ=ZGθZT+Rθ positive semidefinite for every θ∈Θ. Consider the parametric family Pβ,θ=Nn(Xβ,Vθ),(β,θ)∈Rp×Θ, and the estimand ψ(β,θ)=θ. Then ψ is identifiable if and only if the covariance map θ↦Vθ 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↦Vθ is injective. If Pβ,θ=Pβ′,θ′, equality of laws and [F3] give Vθ=Vθ′. 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≠θ′ with Vθ=Vθ′. For any fixed β, the two Gaussian laws Nn(Xβ,Vθ) and Nn(Xβ,Vθ′) 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.

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Restricted maximum likelihood from error contrasts

Definition

Assume AC and a Gaussian family of the form in Gaussian branch of the linear mixed model, indexed by a nonempty parameter set Θ. For each θ∈Θ, let Gθ and Rθ be the latent and residual covariance matrices in Vθ=ZGθZT+Rθ≻0. Let the fixed design X∈Rn×p have full column rank with p<n. Fix an error-contrast matrix K∈Rn×(n−p) as in Error-contrast matrix for a full-rank fixed design, independently of θ. By A Gaussian error contrast has the induced multivariate-normal law, the contrast vector KTY has law Nn−p(0,KTVθK) and its law does not depend on the fixed-effect parameter β. The local result The REML error-contrast covariance is positive definite shows that this covariance is positive definite, and A positive-definite multivariate normal law has its Gaussian density gives its Gaussian density. The restricted likelihood for this fixed contrast basis is LR(θ;y)=(2π)−(n−p)/2det⁡(KTVθK)−1/2exp⁡ ⁣(−12(KTy)T(KTVθK)−1(KTy)). This item defines the likelihood for a fixed, parameter-independent choice of contrast coordinates; it makes no claim about likelihood comparisons across different fixed-effect designs.

Remarks

The local column-basis convention is K∈Rn×(n−p) with columns spanning col⁡(X)⊥, so KTY is defined. The restriction p<n supplies a nonzero contrast space; when p=n, this definition makes no variance-likelihood claim.

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Known-covariance GLS is BLUE

Statement

Let the linear mixed model have fixed design X∈Rn×p of full column rank and known marginal covariance V=Cov⁡(Y)≻0. Then β^GLS=(XTV−1X)−1XTV−1Y is unbiased for β, has covariance (XTV−1X)−1, and is BLUE: for every linear unbiased estimator β~ and every c∈Rp, Var⁡(cTβ^GLS)≤Var⁡(cTβ~). Only the fixed-design mean and covariance are used; Gaussianity is not assumed.

Facts & Assumptions

Given: The fixed-design mixed model, full-column-rank X, known V≻0, and the estimator defined in Generalized least-squares estimator for known covariance.

[F1]

The fixed-design mean is EβY=Xβ. Fixed-design linear regression model.

[F2]

The marginal covariance in the moment model is V=ZGZT+R=Cov⁡(Y). Marginal covariance of the linear mixed model.

[F4]

In a full-rank fixed-design model with covariance σ2I for σ2>0, OLS is unbiased and BLUE for every estimable contrast, without Gaussianity. Gauss–Markov theorem for estimable contrasts.

[F5]

Covariance is bilinear in finite linear combinations. Covariance is symmetric and bilinear in finite linear combinations.

[F6]

Expectation is linear on finite sums of integrable random variables. The Lebesgue integral is linear on L1(μ).

[A1]

X has full column rank, and V is the same known symmetric positive-definite covariance under each Pβ.

Proof

Given: The hypotheses in the Statement.

1.1A1F1F2F3F5F6F7algebra

Choose L from [F3] and set Y~=L−1Y and X~=L−1X. The transformed response is Borel by [F7]. Linearity [F6] and [F1] give EβY~=X~β, and covariance bilinearity [F5] with [F2] gives Cov⁡β(Y~)=L−1VL−T=In. Since L is invertible, X~ has full column rank.

2.1A1F3F4step 1.1algebra

OLS on the transformed model is (X~TX~)−1X~TY~=(XTV−1X)−1XTV−1Y=β^GLS, because L−TL−1=V−1. The transformed model meets every hypothesis of [F4], so this estimator is unbiased and each of its linear contrasts has minimum variance among linear unbiased contrasts.

2.2F1F2F5F6step 1.1algebra

Put C=(XTV−1X)−1 and A=CXTV−1. Then AX=Ip, so linearity [F6] and [F1] give Eβ[AY]=β. By [F5] and [F2], Cov⁡β(AY)=AVAT=CXTV−1VV−1XC=C. Here C is symmetric because XTV−1X is symmetric positive definite.

3.1F4step 1.1step 2.1step 2.2∎

Let β~ be any linear unbiased estimator. For each c, cTβ~ is a linear unbiased estimator of cTβ in the transformed model of step 1.1. By [F4] and step 2.1 its variance is at least that of cTβ^GLS. This holds for every c∈Rp, so the covariance difference is positive semidefinite and GLS is BLUE.

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Changing an error-contrast basis changes the REML likelihood by a constant

Statement

Let X∈Rn×p have full column rank with p<n, and let K,K′∈Rn×(n−p) be two error-contrast matrices for X. For every covariance parameter θ in Restricted maximum likelihood from error contrasts, there is an invertible matrix A∈R(n−p)×(n−p), independent of θ, such that K′=KA and LRK′(θ;y)=∣det⁡A∣−1LRK(θ;y) for every response vector y. Thus changing the fixed contrast basis multiplies the likelihood by a positive factor independent of the covariance parameter and leaves its maximizers unchanged.

Facts & Assumptions

Given: The full-rank fixed design, two error-contrast bases for the same design, and the Gaussian REML likelihoods defined from them.

[F1]

An error-contrast matrix has columns forming a basis of col⁡(X)⊥. Error-contrast matrix for a full-rank fixed design.

[F2]

For a fixed error-contrast matrix K, the restricted likelihood is the Gaussian density with argument KTy and covariance KTVθK. Restricted maximum likelihood from error contrasts.

[F4]

For a linear map on a finite-dimensional space, dimension of kernel plus rank equals the dimension of the domain. Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T.

[F5]

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.

Proof

Given: K and K′ are fixed bases of the same (n−p)-dimensional contrast space.

1.1F1F4F5algebra

Each column of K′ lies in the span of the columns of K, so there is a matrix A with K′=KA. If Ac=0, then K′c=0; the full column rank of K′ gives c=0. Thus A is injective. By [F4], A has full rank and is surjective, so it is invertible; [F5] gives det⁡A≠0.

2.1F2F3step 1.1algebra

Put q=KTy and Sθ=KTVθK. Then K′Ty=ATq and K′TVθK′=ATSθA. Multiplication verifies (ATSθA)−1=A−1Sθ−1A−T, so the two Gaussian quadratic forms are equal. By [F3], det⁡(ATSθA)=(det⁡A)2det⁡(Sθ).

3.1F2step 1.1step 2.1algebra∎

Substitution into the two densities in [F2] leaves the exponential terms equal and changes the determinant prefactor by ∣det⁡A∣−1. The matrix A depends only on the two fixed bases, not on θ, so this factor is positive and parameter-independent. Therefore the likelihood maximizers are the same.

Remarks

The numerical likelihood value depends on the chosen contrast coordinates, while the maximizing covariance parameters do not. The result does not compare different fixed-effect designs.

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Known-covariance BLUP and its prediction-error covariance

Statement

In the moment-only linear mixed model, suppose X has full column rank and the known marginal covariance V=ZGZT+R is positive definite. Define b^=GZTV−1(Y−Xβ^GLS),β^GLS=(XTV−1X)−1XTV−1Y. Here best linear unbiased predictor (BLUP) means that b^ is unbiased for Eb=0 for every fixed β and its prediction-error covariance is minimal in positive-semidefinite order among all affine linear predictors a+BY with that same unbiasedness property. Then b^ is a BLUP and Cov⁡(b−b^)=G−GZTV−1ZG+GZTV−1X(XTV−1X)−1XTV−1ZG. No Gaussianity or Axiom of Choice is assumed. The covariance formula treats G and R as known; it does not cover the extra uncertainty from estimating variance components.

Facts & Assumptions

Given: The moment-only mixed model, full-column-rank X, known V≻0, and the GLS estimator.

[F1]

Y=Xβ+Zb+ε, with centered independent b,ε, covariance matrices G,R, and finite second moments. Finite-dimensional linear mixed model with explicit longitudinal roles.

[F2]

The marginal covariance is V=ZGZT+R. Marginal covariance of the linear mixed model.

[F3]

The best affine L2 predictor has an intercept fixed by means and coefficients satisfying covariance normal equations; singular predictor covariance is allowed. Normal equations for best affine L2 prediction.

[F4]

The GLS estimator is β^GLS=(XTV−1X)−1XTV−1Y, is unbiased, and has covariance C=(XTV−1X)−1. Generalized least-squares estimator for known covariance, Known-covariance GLS is BLUE.

[F5]

Covariance is bilinear in finite linear combinations. Covariance is symmetric and bilinear in finite linear combinations.

[F6]

Variance and covariance are defined by centered second moments. Moments, variance, and covariance on a probability space.

[F7]

Measurable coordinate functions preserve independence, and integrable products of independent coordinates factor. Measurable coordinatewise functions preserve independence, Expectations factor over finite products of independent random variables.

[F8]

Expectation is linear on finite sums of integrable random variables. The Lebesgue integral is linear on L1(μ).

[A1]

X has full column rank, the random vectors are square-integrable, b is independent of ε, and the known covariance V is positive definite.

Proof

Given: The hypotheses in the Statement.

1.1A1F1F2F3F5F7F8F9algebra

Put Yc=Y−Xβ. By [F1], each pair of coordinates bk,εℓ is independent; since both are centered, factorization in [F7] gives Cov⁡(b,ε)=0. Bilinearity [F5] then gives Cov⁡(b,Yc)=GZT, while [F2] gives Cov⁡(Yc)=V. The affine predictor is a Borel random vector by [F9]. For fixed β, apply [F3] coordinatewise to predict b affinely from Y: the covariance normal equations have the unique solution H=GZTV−1 because V is invertible by [A1]. The intercept is Eb−HEY=−HXβ by [F8] and [F1]. Thus H(Y−Xβ) is the best affine L2 predictor when β is treated as known.

2.1A1F1F8F9step 1.1algebra

Write C=(XTV−1X)−1 and P=H(I−XCXTV−1). Then b^=PY and PX=0, because CXTV−1X=Ip. By [F9], b^ is Borel; by [F8] and [F1], Eβb^=PXβ=0=Eb for every β. Every affine linear predictor a+BY unbiased for b for every β must have a=0 and BX=0.

3.1F1F2F4F5F6step 1.1step 2.1algebra

Let Yc=Y−Xβ and u=b−HYc. By step 1.1, bilinearity gives Cov⁡(u,Yc)=GZT−HV=0 and Cov⁡(u)=G−GZTV−1ZG−GZTV−1ZG+GZTV−1VV−1ZG=G−GZTV−1ZG. Also β^GLS−β=CXTV−1Yc, so its covariance is C by [F4], and its covariance with u is zero because step 1.1 gives Cov⁡(u,Yc)=0. Since b−b^=u+HX(β^GLS−β), the error covariance is G−GZTV−1ZG+HXCXTHT, which is the displayed matrix after substituting H and C.

4.1A1F1F2F4F5step 1.1step 2.1step 3.1givenalgebra∎

Let BY be any other affine linear predictor unbiased for b for every β. By step 2.1 its intercept is zero and BX=0. Put D=B−P, so DX=0 and the competing prediction error is (b−PY)−DY. Since PV=GZT−HXCXT, one has Cov⁡(b−PY,Y)=GZT−PV=HXCXT. Therefore Cov⁡(b−PY,DY)=HXCXTDT=HXC(DX)T=0. Expanding the competing error covariance now gives Cov⁡(b−BY)=Cov⁡(b−PY)+DVDT⪰Cov⁡(b−PY). This is exactly the stated BLUP criterion.

Remarks

If variance components are estimated and substituted into this expression, the result is an empirical BLUP (eBLUP). This theorem supplies no exact prediction-error covariance for that plug-in estimator.

5 · Examples, counterexamples and false statements

None yet.

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