Transformation approaches of linear random-effects models

Abstract

Assume that a linear random-effects model $\mathbf{y}= \mathbf{X}\varvec{\beta }+ \varvec{\varepsilon }= \mathbf{X}(\mathbf{A}\varvec{\alpha }+ \varvec{\gamma }) + \varvec{\varepsilon }$ is transformed as $\mathbf{T}\mathbf{y}= \mathbf{T}\mathbf{X}\varvec{\beta }+ \mathbf{T}\varvec{\varepsilon }= \mathbf{T}\mathbf{X}(\mathbf{A}\varvec{\alpha }+ \varvec{\gamma }) + \mathbf{T}\varvec{\varepsilon }$ by pre-multiplying a given matrix $\mathbf{T}$ of arbitrary rank. These two models are not necessarily equivalent unless $\mathbf{T}$ is of full column rank, and we have to work with this derived model in many situations. Because predictors/estimators of the parameter spaces under the two models are not necessarily the same, it is primary work to compare predictors/estimators in the two models and to establish possible links between the inference results obtained from two models. This paper presents a general algebraic approach to the problem of comparing best linear unbiased predictors (BLUPs) of parameter spaces in an original linear random-effects model and its transformations, and provides a group of fundamental and comprehensive results on mathematical and statistical properties of the BLUPs. In particular, we construct many equalities for the BLUPs under an original linear random-effects model and its transformations, and obtain necessary and sufficient conditions for the equalities to hold.

Appendix

Statistical methods in many areas of application often involve mathematical computations with vectors and matrices. In particular, formulas and algebraic tricks for handling matrices in linear algebra and matrix theory play important roles in the derivation of predictors/estimators and characterizations of their properties under linear regression models. Recall that the rank of matrix is a conceptual foundation in linear algebra and matrix theory, which is the most significant finite nonnegative integer in reflecting intrinsic properties of matrices. The mathematical prerequisites for understanding the rank of matrix are minimal and do not go beyond elementary linear algebra. It has long history to establish rank formulas for block matrices and use the formulas in statistical inference, and a pioneer work in this aspect can be found in Guttman (1944). The intriguing connections between generalized inverses of matrices and rank formulas of matrices were recognized in 1970s, and a seminal work on rank formulas for matrices and their generalized inverses was presented in Marsaglia and Styan (1974). In order to establish and characterize various possible equalities for predictors/estimators under LRMs, and to simplify various matrix equalities composed by the Moore–Penrose inverses of matrices, we need a variety of matrix rank formulas for matrices and their Moore–Penrose generalized inverses as simple and effective tools. The results in the following lemma were established in Marsaglia and Styan (1974) and Tian (2004).

Lemma 2

Let $\mathbf{A}\in \mathbb R^{m \times n},$ $\mathbf{B}\in \mathbb R^{m \times k},$ and $\mathbf{C}\in \mathbb R^{l \times n}.$ Then

$$\begin{aligned} r[\, \mathbf{A}, \, \mathbf{B}\, ]&= r(\mathbf{A})+ r(\mathbf{E}_{\mathbf{A}}\mathbf{B})= r(\mathbf{B}) + r(\mathbf{E}_{\mathbf{B}}\mathbf{A}), \end{aligned}$$

(A. 1)

$$\begin{aligned} r\begin{bmatrix} \mathbf{A}\\ \mathbf{C}\end{bmatrix}&= r(\mathbf{A}) + r(\mathbf{C}\mathbf{F}_{\mathbf{A}}) = r(\mathbf{C}) + r(\mathbf{A}\mathbf{F}_{\mathbf{C}}), \end{aligned}$$

(A. 2)

$$\begin{aligned} r\begin{bmatrix} \mathbf{A}\mathbf{A}^{\prime }&\mathbf{B}\\ \mathbf{B}^{\prime }&\mathbf{0}\end{bmatrix}&= r[\, \mathbf{A}, \, \mathbf{B}\,] + r(\mathbf{B}). \end{aligned}$$

(A. 3)

If ${\mathscr {R}}(\mathbf{A}_1^{\prime }) \subseteq {\mathscr {R}}(\mathbf{B}_1^{\prime })$, ${\mathscr {R}}(\mathbf{A}_2) \subseteq {\mathscr {R}}(\mathbf{B}_1)$, ${\mathscr {R}}(\mathbf{A}_2^{\prime }) \subseteq {\mathscr {R}}(\mathbf{B}_2^{\prime })$ and ${\mathscr {R}}(\mathbf{A}_3) \subseteq {\mathscr {R}}(\mathbf{B}_2),$ then

$$\begin{aligned} r(\mathbf{A}_1\mathbf{B}_1^{+}\mathbf{A}_2)&= r\begin{bmatrix} \mathbf{B}_1&\mathbf{A}_2\\ \mathbf{A}_1&\mathbf{0}\end{bmatrix} - r(\mathbf{B}_1), \end{aligned}$$

(A. 4)

$$\begin{aligned} r(\mathbf{A}_1\mathbf{B}_1^{+}\mathbf{A}_2\mathbf{B}_2^{+}\mathbf{A}_3)&= r\begin{bmatrix} \mathbf{0}&\mathbf{B}_2&\mathbf{A}_3\\ \mathbf{B}_1&\mathbf{A}_2&\mathbf{0}\\ \mathbf{A}_1&\mathbf{0}&\mathbf{0}\end{bmatrix} - r(\mathbf{B}_1) - r(\mathbf{B}_2). \end{aligned}$$

(A. 5)

In addition,  the following results hold.

• (a) $r[\, \mathbf{A}, \, \mathbf{B}\,] = r(\mathbf{A}) \Leftrightarrow {\mathscr {R}}(\mathbf{B}) \subseteq {\mathscr {R}}(\mathbf{A}) \Leftrightarrow \mathbf{A}\mathbf{A}^{+}\mathbf{B}= \mathbf{B}\Leftrightarrow \mathbf{E}_{\mathbf{A}}\mathbf{B}= \mathbf{0}.$

• (b) $r\begin{bmatrix} \mathbf{A}\\ \mathbf{C}\end{bmatrix} = r(\mathbf{A}) \Leftrightarrow {\mathscr {R}}(\mathbf{C}^{\prime }) \subseteq {\mathscr {R}}(\mathbf{A}^{\prime }) \Leftrightarrow \mathbf{C}\mathbf{A}^{+}\mathbf{A}= \mathbf{C}\Leftrightarrow \mathbf{C}\mathbf{F}_{\mathbf{A}} = \mathbf{0}.$

• (c) $r[\, \mathbf{A}, \, \mathbf{B}\,] = r(\mathbf{A}) + r(\mathbf{B}) \Leftrightarrow {\mathscr {R}}(\mathbf{A}) \cap {\mathscr {R}}(\mathbf{B}) = \{ \mathbf{0}\} \Leftrightarrow {\mathscr {R}}[(\mathbf{E}_\mathbf{A}\mathbf{B})^{\prime }] = {\mathscr {R}}(\mathbf{B}^{\prime }) \Leftrightarrow {\mathscr {R}}[(\mathbf{E}_\mathbf{B}\mathbf{A})^{\prime }] = {\mathscr {R}}(\mathbf{A}^{\prime }).$

• (d) $r \begin{bmatrix} \mathbf{A}\\ \mathbf{C}\end{bmatrix} = r( \mathbf{A}) + r(\mathbf{C}) \Leftrightarrow {\mathscr {R}}(\mathbf{A}^{\prime }) \cap {\mathscr {R}}(\mathbf{C}^{\prime }) = \{ \mathbf{0}\} \Leftrightarrow {\mathscr {R}}(\mathbf{C}\mathbf{F}_{\mathbf{A}}) = {\mathscr {R}}(\mathbf{C}) \Leftrightarrow {\mathscr {R}}(\mathbf{A}\mathbf{F}_{\mathbf{C}}) = {\mathscr {R}}(\mathbf{A}).$

Lemma 3

(Penrose 1955) The linear matrix equation $\mathbf{A}\mathbf{X}= \mathbf{B}$ is consistent if and only if $r[\mathbf{A}, \, \mathbf{B}\,] = r(\mathbf{A}),$ or equivalently,  $\mathbf{A}\mathbf{A}^{+}\mathbf{B}= \mathbf{B}.$ In this case,  the general solution of the equation can be written in the following parametric form $\mathbf{X}= \mathbf{A}^{+} \mathbf{B}+ ( \mathbf{I}- \mathbf{A}^{+}\mathbf{A}) \mathbf{U},$ where $\mathbf{U}$ is an arbitrary matrix.

Once a statistical model is formulated, its parameters can be estimated and predicted by various optimization methods. A brief survey on modern optimization methods in statistical analysis can be found in Lange et al. (2014). In any case, we expect that the optimization problems occurred in predictions/stimations of parameter spaces in linear models have analytical solutions so that we can use the analytical solutions to establish precise theory in the corresponding statistical inferences of the models. In order to directly solve the matrix minimization problem in (2.4), we need the following result on constrained quadratic matrix-valued function minimization problem, which have been recently proved in Tian (2015a) and have been applied in Gan et al. (2017), Lu et al. (2016), Tian (2015a, b) and Tian and Jiang (2016, 2017) to establish exact expressions of BLUPs under linear models with fixed and random effects.

Lemma 4

Let

$$\begin{aligned} f(\mathbf{L}) = \left( \mathbf{L}\mathbf{C}+ \mathbf{D}\right) \mathbf{M}\left( \mathbf{L}\mathbf{C}+ \mathbf{D}\right) ^{\prime } \ \ s.t. \ \ \mathbf{L}\mathbf{A}= \mathbf{B}, \end{aligned}$$

where $\mathbf{A}\in {\mathbb R}^{p \times q}$, $\mathbf{B}\in {\mathbb R}^{n \times q}$, $\mathbf{C}\in {\mathbb R}^{p\times m}$ and $\mathbf{D}\in {\mathbb R}^{n\times m}$ are given,  $\mathbf{M}\in {\mathbb R}^{m \times m}$ is positive semi-definite, and the matrix equation $\mathbf{L}\mathbf{A}= \mathbf{B}$ is consistent. Then there always exists a solution $\mathbf{L}_0$ of $\mathbf{L}_0\mathbf{A}= \mathbf{B}$ such that

$$\begin{aligned} f(\mathbf{L}) \succcurlyeq f(\mathbf{L}_0) \end{aligned}$$

holds for all solutions of $\mathbf{L}\mathbf{A}= \mathbf{B}.$ In this case,  the matrix $\mathbf{L}_0$ satisfying the above inequality is determined by the following consistent matrix equation

$$\begin{aligned} \mathbf{L}_0\left[ \mathbf{A}, \, \mathbf{C}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp } \right] = \left[ \mathbf{B}, \, -\mathbf{D}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp } \right] . \end{aligned}$$

In this case,  the general expression of $\mathbf{L}_0$ and the corresponding $f(\mathbf{L}_0)$ and $f(\mathbf{L})$ are given by

$$\begin{aligned} \mathbf{L}_0&= \mathop {{{\mathrm{argmin}}}}\limits _{\mathbf{L}\mathbf{A}= \mathbf{B}} f(\mathbf{L}) = \left[ \mathbf{B}, \, -\mathbf{D}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp }\right] \left[ \mathbf{A}, \, \mathbf{C}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp } \right] ^{+} + \mathbf{U}\left[ \mathbf{A}, \, \mathbf{C}\mathbf{M}\mathbf{C}^{\prime }\right] ^{\perp },\\ f(\mathbf{L}_0)&= \min _{\mathbf{L}\mathbf{A}= \mathbf{B}} f(\mathbf{L}) = \mathbf{K}\mathbf{M}\mathbf{K}^{\prime } - \mathbf{K}\mathbf{M}\mathbf{C}^{\prime }\mathbf{T}\mathbf{C}\mathbf{M}\mathbf{K}^{\prime },\\ f(\mathbf{L})&= f(\mathbf{L}_0) + \left( \mathbf{L}\mathbf{C}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp } + \mathbf{D}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp }\right) \mathbf{T}\left( \mathbf{L}\mathbf{C}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp } + \mathbf{D}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp }\right) ^{\prime }, \end{aligned}$$

where $\mathbf{K}= \mathbf{B}\mathbf{A}^{+}\mathbf{C}+ \mathbf{D},$ $\mathbf{T}= \left( \mathbf{A}^{\perp }\mathbf{C}\mathbf{M}\mathbf{C}^{\prime }\mathbf{A}^{\perp }\right) ^{+},$ and $\mathbf{U}\in {\mathbb R}^{n\times p}$ is arbitrary.

With the supports of the formulas in Lemmas 2–4, we are capable of proving all the assertions in Sects. 2 and 3.

Proof of Lemma 1

It is obvious that $\mathrm{E}\left( \mathbf{L}{Ty} - \varvec{\phi }\,\right) = \mathbf{0}\Leftrightarrow \mathbf{L}\widehat{\mathbf{X}}\varvec{\alpha }- \mathbf{K}\varvec{\alpha }=\mathbf{0}\ \hbox {for all } \varvec{\alpha }\Leftrightarrow \mathbf{L}\widehat{\mathbf{X}}= \mathbf{K}$. From Lemma 3, the matrix equation is consistent if and only if (2.10) holds. $\square$

Proof of Theorem 1

Under (2.9), (2.4) is equivalent to finding a solution $\mathbf{L}_0$ of $\mathbf{L}_0\mathbf{T}\widehat{\mathbf{X}} = \mathbf{K}$ such that

$$\begin{aligned} f(\mathbf{L}) \succcurlyeq f(\mathbf{L}_0) \ \ \hbox {s.t.} \ \ \mathbf{L}\mathbf{T}\widehat{\mathbf{X}} = \mathbf{K} \end{aligned}$$

(A. 6)

holds in the Löwner partial ordering. From Lemma 4, there always exists a solution $\mathbf{L}_0$ of $\mathbf{L}_0\mathbf{T}\widehat{\mathbf{X}} = \mathbf{K}$ such that $f(\mathbf{L}) \succcurlyeq f(\mathbf{L}_0)$ holds for all solutions of $\mathbf{L}\mathbf{T}\widehat{\mathbf{X}} = \mathbf{K}$, and the $\mathbf{L}_0$ is determined by the matrix equation $\mathbf{L}_0[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] = [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]$, establishing the matrix equation in (2.11). The general solution of the equation is given in (2.13) by Lemma 3.

Results (a)–(d) are routine consequences of (2.13).

Taking dispersion operation of (2.13) yields (2.14) and (2.16). Also from (1.33) and (2.13), the covariance matrix between ${\mathrm{BLUP}}_{{\mathscr {N}}}(\varvec{\phi })$ and $\varvec{\phi }$ is

$$\begin{aligned} \mathrm{Cov}\{\,{\mathrm{BLUP}}_{{\mathscr {N}}}(\varvec{\phi }), \, \varvec{\phi }\,\}&= \mathbf{L}\mathbf{T}\mathrm{Cov}\{\,\mathbf{y}, \, \varvec{\phi }\}\\&= [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}\mathbf{W}^{\prime }, \end{aligned}$$

thus establishing (2.15). Further by (2.13),

$$\begin{aligned} \mathbf{L}\mathbf{T}\widetilde{\mathbf{X}} = \left( [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+} + \mathbf{U}[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp } \right) \mathbf{T}\widetilde{\mathbf{X}}. \end{aligned}$$

Substituting it into (2.9) yields

$$\begin{aligned} \mathrm{Cov}(\varvec{\phi }- \mathbf{L}\mathbf{T}\mathbf{y})&= \left( [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}\widetilde{\mathbf{X}} - \mathbf{J}\right) \varvec{\Sigma }\\&\ \ \ \times \left( [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}\widetilde{\mathbf{X}} - \mathbf{J}\right) ^{\prime }, \end{aligned}$$

thus establishing (2.17).

Note that the arbitrary matrix $\mathbf{U}$ in (2.13) can be rewritten as $\mathbf{U}= \mathbf{U}_1 + \mathbf{U}_2 + \mathbf{U}_3$, while $[\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}$ in (2.13) can be decomposed as the sum of three terms

$$\begin{aligned}&[\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}\\&= [\,\mathbf{K}, \, \mathbf{0}\,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}\\&\quad + [\,\mathbf{0}, \, \mathbf{G}[\mathbf{I}_p, \, \mathbf{0}]\varvec{\Sigma }(\mathbf{T}\widetilde{\mathbf{X}})^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,][\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}\\&\quad + [\,\mathbf{0}, \, \mathbf{H}[\mathbf{0}, \, \mathbf{I}_n]\varvec{\Sigma }(\mathbf{T}\widetilde{\mathbf{X}})^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,][\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}\\&: = \mathbf{Q}_1 + \mathbf{Q}_2 + \mathbf{Q}_3. \end{aligned}$$

Hence

$$\begin{aligned} {\mathrm{BLUP}}_{{\mathscr {N}}}(\varvec{\phi })&= \left( \mathbf{Q}_1 + \mathbf{U}_1[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp }\mathbf{T}\right) \mathbf{y}\\&\ \ \ \ + \left( \mathbf{Q}_2 + \mathbf{U}_2[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp }\mathbf{T}\right) \mathbf{y}\\&\ \ \ \ + \left( \mathbf{Q}_3 + \mathbf{U}_3[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp }\mathbf{T}\right) \mathbf{y}\\&= \mathrm{BLUE}_{{\mathscr {N}}}(\mathbf{K}\varvec{\alpha }) + {\mathrm{BLUP}}_{{\mathscr {N}}}(\mathbf{G}\varvec{\gamma }) + {\mathrm{BLUP}}_{{\mathscr {N}}}(\mathbf{H}\varvec{\varepsilon }), \end{aligned}$$

thus establishing (2.18).

From (2.13), the covariance matrix between $\mathrm{BLUE}_{{\mathscr {N}}}(\mathbf{K}\varvec{\alpha })$ and ${\mathrm{BLUP}}_{{\mathscr {N}}}(\mathbf{G}\varvec{\gamma }+ \mathbf{H}\varvec{\varepsilon })$ is

$$\begin{aligned}&{\mathrm{Cov}}\{\, {\mathrm{BLUE}}_{{\mathscr {N}}}(\mathbf{K}\varvec{\alpha }), \, {\mathrm{BLUP}}_{{\mathscr {N}}}(\mathbf{G}\varvec{\gamma }+ \mathbf{H}\varvec{\varepsilon }) \,\} \nonumber \\&= [\,\mathbf{K}, \, \mathbf{0}\,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}\mathbf{V}\mathbf{T}^{\prime }\left( [\,\mathbf{0}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+} \right) ^{\prime }. \end{aligned}$$

(A. 7)

Applying (A. 5) to (A. 7) and simplifying, we obtain

$$\begin{aligned}&r({\mathrm{Cov}}\{\,{\mathrm{BLUE}}_{{\mathscr {N}}}(\mathbf{K}\varvec{\alpha }), \, {\mathrm{BLUP}}_{{\mathscr {N}}}(\mathbf{G}\varvec{\gamma }+ \mathbf{H}\varvec{\varepsilon })\,\}) \\&\quad = r\left( [\,\mathbf{K}, \, \mathbf{0}\,][\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}{\widehat{\mathbf{X}}})^{\perp } \,]^{+}\mathbf{T}\mathbf{V}\mathbf{T}^{\prime }\left( [\,\mathbf{0}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}{\widehat{\mathbf{X}}})^{\perp } \,] [\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}{\widehat{\mathbf{X}}})^{\perp } \,]^{+}\right) ^{\prime } \right) \\&\quad = r\begin{bmatrix} \mathbf{0}&\begin{bmatrix} (\mathbf{T}{\widehat{\mathbf{X}}})^{\prime }\\ (\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }\mathbf{T}\mathbf{V}\mathbf{T}^{\prime } \end{bmatrix}&\begin{bmatrix} \mathbf{0}\\ (\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }\mathbf{T}\mathbf{W}^{\prime } \end{bmatrix} \\ [\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }\,]&\mathbf{T}\mathbf{V}\mathbf{T}^{\prime }&\mathbf{0}\\ [\,\mathbf{K}, \, \mathbf{0}\,]&\mathbf{0}&\mathbf{0}\end{bmatrix} - 2r[\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }]\\&\quad = r\begin{bmatrix} \begin{bmatrix} \mathbf{0}&\mathbf{0}\\ \mathbf{0}&- (\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }\mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}{\widehat{\mathbf{X}}})^{\perp } \end{bmatrix}&\begin{bmatrix} (\mathbf{T}{\widehat{\mathbf{X}}})^{\prime }\\ \mathbf{0}\end{bmatrix}&\begin{bmatrix} \mathbf{0}\\ (\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }\mathbf{T}\mathbf{W}^{\prime } \end{bmatrix} \\ [\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{0}\,]&\mathbf{T}\mathbf{V}\mathbf{T}^{\prime }&\mathbf{0}\\ [\,\mathbf{K}, \, \mathbf{0}\,]&\mathbf{0}&\mathbf{0}\end{bmatrix} - 2r[\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }] \\&\quad = r\begin{bmatrix} \mathbf{0}&(\mathbf{T}{\widehat{\mathbf{X}}})^{\prime } \\ \mathbf{T}{\widehat{\mathbf{X}}}&\mathbf{T}\mathbf{V}\mathbf{T}^{\prime } \\ \mathbf{K}&\mathbf{0}\end{bmatrix} + r\left[ \, (\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }\mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }, \, (\mathbf{T}{\widehat{\mathbf{X}}})^{\perp }\mathbf{T}\mathbf{W}^{\prime } \right] - 2r\left[ \,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}{\widetilde{\mathbf{X}}}\varvec{\Sigma }\,\right] \\&\quad = r\begin{bmatrix} \mathbf{T}{\widehat{\mathbf{X}}}\\ \mathbf{K}\end{bmatrix} + r\begin{bmatrix} (\mathbf{T}{\widehat{\mathbf{X}}})^{\prime }\\ \mathbf{T}\mathbf{V}\mathbf{T}^{\prime } \end{bmatrix} + r\left[ \,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}{\widetilde{\mathbf{X}}}\varvec{\Sigma }, \, \mathbf{T}\mathbf{W}^{\prime }\right] \\&\qquad - r(\mathbf{T}{\widehat{\mathbf{X}}}) - 2r[\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}{\widetilde{\mathbf{X}}}\varvec{\Sigma }\,] \quad (by (A.1) \hbox { and } (A.3)) \\&\quad = r(\mathbf{T}{\widehat{\mathbf{X}}}) + r\begin{bmatrix} (\mathbf{T}{\widehat{\mathbf{X}}})^{\prime }\\ \varvec{\Sigma }(\mathbf{T}{\widetilde{\mathbf{X}}})^{\prime } \end{bmatrix} + r[\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}{\widetilde{\mathbf{X}}}\varvec{\Sigma }\,] - r(\mathbf{T}{\widehat{\mathbf{X}}}) - 2r[\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}{\widetilde{\mathbf{X}}}\varvec{\Sigma }\,] \\&\quad = 0, \end{aligned}$$

which implies that ${\mathrm{Cov}}\{\,{\mathrm{BLUE}}_{{\mathscr {N}}}(\mathbf{K}\varvec{\alpha }), \, {\mathrm{BLUP}}_{{\mathscr {N}}}(\mathbf{G}\varvec{\gamma }+ \mathbf{H}\varvec{\varepsilon })\,\}$ is a zero matrix, establishing (2.19). Equation (2.20) follows from (2.18) and (2.19). Result (g) follows from (2.13). $\square$

Proof of Theorem 2

From Definition 3(a), ${\mathrm{BLUP}}_{{\mathscr {M}}}(\varvec{\phi }) = \mathbf{G}\mathbf{T}\mathbf{y}$ holds for some matrix $\mathbf{G}$ if and only if the coefficient matrix $\mathbf{G}\mathbf{T}$ of $\mathbf{y}$ satisfies the matrix equation in (2.22), i.e.,

$$\begin{aligned} \mathbf{G}\mathbf{T}[\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } ] = [\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } ]. \end{aligned}$$

(A. 8)

From Lemma 3, the equation in (A. 8) is solvable for $\mathbf{G}$ if and only if

$$\begin{aligned} r\begin{bmatrix} \mathbf{T}{\widehat{\mathbf{X}}}&\quad \mathbf{T}\mathbf{V}{\widehat{\mathbf{X}}}^{\perp } \\ \mathbf{K}&\quad \mathbf{W}{\widehat{\mathbf{X}}}^{\perp } \end{bmatrix} = r[\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}{\widehat{\mathbf{X}}}^{\perp } \,], \end{aligned}$$

(A. 9)

where by (1.5), (1.33), and (A. 2),

$$\begin{aligned}&r\begin{bmatrix} \mathbf{T}\widehat{\mathbf{X}}&\quad \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \\ \mathbf{K}&\quad \mathbf{W}\widehat{\mathbf{X}}^{\perp } \end{bmatrix} = r\begin{bmatrix} \mathbf{T}\widehat{\mathbf{X}}&\quad {\mathrm{Cov}}\{\mathbf{T}\mathbf{y},\, \widehat{\mathbf{X}}^{\perp }\mathbf{y}\}\\ \mathbf{K}&\quad {\mathrm{Cov}}\{\varvec{\phi },\, \widehat{\mathbf{X}}^{\perp }\mathbf{y}\} \end{bmatrix} = r\begin{bmatrix} \mathbf{T}{\widehat{\mathbf{X}}}&\quad {\mathrm{Cov}}\{\mathbf{T}\mathbf{y},\, \mathbf{y}\}\\ \mathbf{0}&\quad \widehat{\mathbf{X}}^{\prime }\\ \mathbf{K}&\quad {\mathrm{Cov}}\{\varvec{\phi },\, \mathbf{y}\} \end{bmatrix} - r(\widehat{\mathbf{X}}),\\&r[\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}{\widehat{\mathbf{X}}}^{\perp } \,] = r[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathrm{Cov}\{\mathbf{T}\mathbf{y},\, {\widehat{\mathbf{X}}}^{\perp }\mathbf{y}\}\,] = r\begin{bmatrix} \mathbf{T}{\widehat{\mathbf{X}}}&\quad {\mathrm{Cov}}\{\mathbf{T}\mathbf{y},\, \mathbf{y}\}\\ \mathbf{0}&\quad {\widehat{\mathbf{X}}}^{\prime } \end{bmatrix} - r({\widehat{\mathbf{X}}}). \end{aligned}$$

Hence (A. 9) is equivalent to (d) and (e), respectively. Thus, (a), (d), and (e) are equivalent. The equivalence of (e) and (f) follows from Lemma 2(b).

From Definition 3(a), ${\mathrm{BLUP}}_{{\mathscr {M}}}(\varvec{\phi })= {\mathrm{BLUP}}_{{\mathscr {N}}}(\varvec{\phi })$ holds definitely if and only if the coefficient matrix of $\mathbf{y}$ in (2.13) satisfies the matrix equation in (2.22), i.e.,

$$\begin{aligned}&\left( [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}+ \mathbf{U}[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp }\mathbf{T}\right) [\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]\nonumber \\&\quad = [\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } \,]. \end{aligned}$$

(A. 10)

From Lemma 3, this equation is solvable for $\mathbf{U}$ if and only if

$$\begin{aligned}&r\begin{bmatrix} [\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } \,] - [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \,] \\ [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp }[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \,] \end{bmatrix} \nonumber \\&\quad = r([\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp }[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]). \end{aligned}$$

(A. 11)

Simplifying both sides by elementary block matrix operations, we obtain

$$\begin{aligned}&r\begin{bmatrix} [\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } \,] \quad -[\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \,] \\ [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp }[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \,] \end{bmatrix}\\&\quad =r\begin{bmatrix} [\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } \,] - [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]&\mathbf{0}\\ [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]&[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] \end{bmatrix}\\&\qquad \,- \ r[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]\\&\quad =r\begin{bmatrix} [\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } \,]&[\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] \\ [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]&[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] \end{bmatrix} - r[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\widetilde{\mathbf{X}}\varvec{\Sigma }\,]\\&\quad =r\begin{bmatrix} \mathbf{K}&\quad \mathbf{W}\widehat{\mathbf{X}}^{\perp } \\ \mathbf{T}\widehat{\mathbf{X}}&\quad \mathbf{T}\mathbf{V}\widehat{\mathbf{X}}^{\perp } \end{bmatrix} - r[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\widetilde{\mathbf{X}}\varvec{\Sigma }\,]\\&\quad = r\begin{bmatrix} \mathbf{T}\widehat{\mathbf{X}}&\quad {\mathrm{Cov}}\{\mathbf{T}\mathbf{y},\, \widehat{\mathbf{X}}^{\perp }\mathbf{y}\}\\ \mathbf{K}&\quad {\mathrm{Cov}}\{\varvec{\phi },\, \widehat{\mathbf{X}}^{\perp }\mathbf{y}\} \end{bmatrix} - r[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\widetilde{\mathbf{X}}\varvec{\Sigma }\,]\\&\quad = r\begin{bmatrix} \mathbf{T}\widehat{\mathbf{X}}&\quad {\mathrm{Cov}}\{\mathbf{T}\mathbf{y},\, \mathbf{y}\}\\ \mathbf{0}&\quad \widehat{\mathbf{X}}^{\prime }\\ \mathbf{K}&\quad {\mathrm{Cov}}\{\varvec{\phi },\, \mathbf{y}\} \end{bmatrix} - r(\widehat{\mathbf{X}}) - r[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\widetilde{\mathbf{X}}\varvec{\Sigma }\,], \end{aligned}$$

and

$$\begin{aligned}&r([\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}{\widehat{\mathbf{X}}})^{\perp } \,]^{\perp }[\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\mathbf{V}{\widehat{\mathbf{X}}}^{\perp } \,]) = r[\, \mathbf{T}\widehat{\mathbf{X}}, \, {\mathrm{Cov}}\{\mathbf{T}\mathbf{y},\, \widehat{\mathbf{X}}^{\perp }\mathbf{y}\}\,] - r[\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\widetilde{\mathbf{X}}\varvec{\Sigma }\,]\\&\quad = r\begin{bmatrix} \mathbf{T}{\widehat{\mathbf{X}}}&\quad {\mathrm{Cov}}\{\mathbf{T}\mathbf{y},\, \mathbf{y}\}\\ \mathbf{0}&\quad {\widehat{\mathbf{X}}}^{\prime } \end{bmatrix} - r({\widehat{\mathbf{X}}}) - r[\,\mathbf{T}{\widehat{\mathbf{X}}}, \, \mathbf{T}\widetilde{\mathbf{X}}\varvec{\Sigma }\,]. \end{aligned}$$

Substituting the two equalities into (A. 11) leads to the equivalences of (b), (d), and (e).

From Definition 3(b), ${\mathrm{BLUP}}_{{\mathscr {M}}}(\varvec{\phi })= {\mathrm{BLUP}}_{{\mathscr {N}}}(\varvec{\phi })$ holds with probability 1 if and only if the coefficient matrices in (2.13) and (2.23) satisfy

$$\begin{aligned}&\left( [\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } \,] [\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]^{+} + \mathbf{U}_1[\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]^{\perp } \right) [\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]\\&\quad = \left( [\,\mathbf{K}, \, \mathbf{W}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,] [\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{+}\mathbf{T}+ \mathbf{U}[\,\mathbf{T}\widehat{\mathbf{X}}, \, \mathbf{T}\mathbf{V}\mathbf{T}^{\prime }(\mathbf{T}\widehat{\mathbf{X}})^{\perp } \,]^{\perp }\mathbf{T}\right) \\&\qquad \quad \times \,[\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,], \end{aligned}$$

which can further reduce to (A. 10) from

$$\begin{aligned} {[}\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } \,][\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]^{+}[\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,] = [\,\mathbf{K}, \, \mathbf{W}\widehat{\mathbf{X}}^{\perp } \,] \ \ {\mathrm{and}} \ \ [\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,]^{\perp }[\,\widehat{\mathbf{X}}, \, \mathbf{V}\widehat{\mathbf{X}}^{\perp } \,] = \mathbf{0}. \end{aligned}$$

Hence, (b) and (c) are equivalent. $\square$

Proof of Theorem 3

Equations (3.1) and (3.2) follow from (2.18) and (2.28) by setting $\varvec{\phi }= \mathbf{T}\mathbf{y}= \mathbf{T}\widehat{\mathbf{X}}\varvec{\alpha }+ \mathbf{T}\mathbf{X}\varvec{\gamma }+ \mathbf{T}\varvec{\varepsilon }$. $\square$

Proof of Theorem 4

Results (a)–(c) follow from Theorem 2 by setting $\varvec{\phi }=\mathbf{T}\mathbf{X}\varvec{\beta }$, $\mathbf{T}\widehat{\mathbf{X}}\varvec{\alpha }$, $\mathbf{T}\mathbf{X}\varvec{\gamma }$, and $\mathbf{T}\varvec{\varepsilon }$, respectively. $\square$