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

Statement

Let $X \in R^{n \times p}$ have full column rank with $p < n$ , and let $K, K^{'} \in R^{n \times (n - p)}$ be two error-contrast matrices for $X$ . For every covariance parameter $θ$ in Restricted maximum likelihood from error contrasts, there is an invertible matrix $A \in R^{(n - p) \times (n - p)}$ , independent of $θ$ , such that $K^{'} = K A$ and $L_{R}^{K^{'}} (θ; y) = ∣ \det A ∣^{- 1} L_{R}^{K} (θ; y)$ 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 $K^{T} y$ and covariance $K^{T} V_{θ} K$ . Restricted maximum likelihood from error contrasts.

[F3]

Determinants are multiplicative and invariant under transpose. For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , For every square matrix over a commutative ring, $\det (A^{T}) = \det (A)$ .

[F4]

For a linear map on a finite-dimensional space, dimension of kernel plus rank equals the dimension of the domain. Rank-nullity: $\dim_{F} V = nullity T + rank T$ .

[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 $K^{'} = K A$ . If $A c = 0$ , 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 $\det A \neq 0$ .

2.1F2F3step 1.1algebra

Put $q = K^{T} y$ and $S_{θ} = K^{T} V_{θ} K$ . Then $K^{' T} y = A^{T} q$ and $K^{' T} V_{θ} K^{'} = A^{T} S_{θ} A$ . Multiplication verifies $(A^{T} S_{θ} A)^{- 1} = A^{- 1} S_{θ}^{- 1} A^{- T}$ , so the two Gaussian quadratic forms are equal. By [F3], $\det (A^{T} S_{θ} A) = (\det A)^{2} \det (S_{θ})$ .

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.

Depends on

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Sources

Fan Li, STA610 Lecture 4: Maximum Likelihood Estimation and Inference for the Hierarchical Normal Model, PDF pp. 4–5 (standard reference, not scraped)
Duchateau, Janssen, and Rowlands, Linear Mixed Models, Chapter 2 §2.4.2 (standard reference, not scraped)