On efficiency of some restricted estimators in a multivariate regression model

Abstract

In this paper, we study a constrained estimation problem in a multivariate measurement error regression model. In particular, we derive the joint asymptotic normality of the unrestricted estimator (UE) and the restricted estimators (REs) of the matrix of the regression coefficients. The derived result holds under the hypothesized restriction as well as under the sequence of alternative restrictions. In addition, we establish Asymptotic Distributional Risk for the UE and the REs and compare their relative performance. It is established that near the restriction, the restricted estimators (REs) perform better than the UE. But the REs perform worse than the UE when one moves far away from the restriction. Further, we explore by simulation the performance of the shrinkage estimators (SEs). The numerical findings corroborate the established theoretical results about the relative risk dominance between the REs and the UE. The findings also show that near the restriction, the REs dominate SEs but as one moves far away from the restriction, REs perform poorly while SEs dominate always the UE.

Appendix: Some technical results

In this appendix, we give technical results and proofs which are underlying the established results. The following lemma are useful in establishing the asymptotic distributions. Let

$$\begin{aligned}&\text {Cov}\left( {\varvec{U}},{\varvec{V}}\right) =\text {E}\left[ \left( {\text {vec}}({\varvec{U}})-\text {E}\left( {\text {vec}}({\varvec{U}})\right) \right) \left( {\text {vec}}({\varvec{V}})-\text {E}\left( {\text {vec}}({\varvec{V}})\right) ^{\prime }\right) \right] \mathrm{{and}} \,\,\, \mathrm{{let}} \\&\text {Cov}\left( {\varvec{U}}\right) =\text {E}\left[ \left( {\text {vec}}({\varvec{U}})-\text {E}\left( {\text {vec}}({\varvec{U}})\right) \right) \left( {\text {vec}}({\varvec{U}})-\text {E}\left( {\text {vec}}({\varvec{U}})\right) ^{\prime }\right) \right] \mathrm{{i.e.}}\\&\text {Cov}\left( {\varvec{U}}\right) =\text {Cov}\left( {\varvec{U}},{\varvec{U}}\right) . \end{aligned}$$

Lemma A.1

Let ${\varvec{w}}_{1},{\varvec{w}}_{2},\dots ,{\varvec{w}}_{n}$ be iid p- column random vectors with $\text {E}\left( {\varvec{w}}_{1}\right) ={\varvec{0}}$ and

${\text {Cov}}\left( {\varvec{w}}_{1}\right) ={\varvec{\Omega }}_{0}$ and let $\left\{ {\varvec{U}}_{n}\right\} _{n=1}^{\infty }$ be a $p_{0}\times p$-sequence of nonrandom matrices such that $\displaystyle {\lim _{n\rightarrow \infty }}{\varvec{U}}_{n}$ exists and non null. Then $\displaystyle n^{-1/2}\sum _{i=1}^{n}{\varvec{U}}_{i}{\varvec{w}}_{i}\xrightarrow [n\rightarrow \infty ]{d}{\varvec{Z}}_{0}\sim {\mathcal {N}}_{p}\left( {{\varvec{0}}},{\varvec{\Omega }}\right)$ where

$$\begin{aligned} {\varvec{\Omega }}=\lim _{n\rightarrow \infty } {\text {Cov}}\left[ \left( n^{-1/2}\sum _{i=1}^{n}{\varvec{U}}_{i}{\varvec{w}}_{i}\right) \right] . \end{aligned}$$

Proof

Let ${\varvec{\alpha }}$ be a $p_{0}$-column vector (no zero vector) and let $s_{n}^{2}{=}\displaystyle \sum _{i=1}^{n}{{\text {Cov}}}\left[ {\varvec{\alpha }}^{\prime }n^{-1/2}{\varvec{U}}_{i}{\varvec{w}}_{i}\right]$. This gives $s_{n}^{2} =n^{-1}\displaystyle \sum _{i=1}^{n}{\varvec{\alpha }}^{\prime }{\varvec{U}}_{i}{\varvec{\Omega }}_{0}{\varvec{U}}^{\prime }_{i}{\varvec{\alpha }}$. By Cesàro mean theorem, we conclude that

$\displaystyle {\lim _{n\rightarrow \infty }} {\text {Cov}}\left[ \left( n^{-1/2}\sum _{i=1}^{n}{\varvec{U}}_{i}{\varvec{w}}_{i}\right) \right]$ exists. Letting ${\varvec{\Omega }}$ this limit, we get $\displaystyle {\lim _{n\rightarrow \infty }}s_{n}^{2}={\varvec{\alpha }}^{\prime }{\varvec{\Omega }}{\varvec{\alpha }}$ and

$$\begin{aligned}&\displaystyle \lim _{n\rightarrow \infty }\left( \max _{1\le k\le n}{\text {Cov}}\left[ {\varvec{\alpha }}^{\prime } n^{-1/2}{\varvec{U}}_{k}{\varvec{w}}_{k}\right] \Big /s_{n}^{2}\right) \\&\quad =\displaystyle \lim _{n\rightarrow \infty }\left( \max _{1\le k\leqslant n}{\text {Cov}}\left[ {\varvec{\alpha }}^{\prime } n^{-1/2}{\varvec{U}}_{k}{\varvec{\Omega }}_{0}{\varvec{U}}^{\prime }_{k}{\varvec{\alpha }}\right] \Big /s_{n}^{2}\right) =0, \end{aligned}$$

for all ${\varvec{\alpha }}\in {\mathbb {R}}^{p_{0}}\setminus \{0\}$. Hence, by Lindeberg central limit theorem, we get

$$\begin{aligned} n^{-1/2}{\varvec{\alpha }}^{\prime }\displaystyle {\sum _{i=1}^{n}}{\varvec{U}}_{i}{\varvec{w}}_{i}/s_{n}\xrightarrow [n\rightarrow \infty ]{d}Z_{0}^{*} \end{aligned}$$

where $Z_{0}^{*}\sim {\mathcal {N}}\left( 0,\,1\right)$. Then, by Slutsky’s theorem, $n^{-1/2}{\varvec{\alpha }}^{\prime }\displaystyle {\sum \nolimits _{i=1}^{n}}{\varvec{U}}_{i}{\varvec{w}}_{i}\xrightarrow [n\rightarrow \infty ]{d} \left( {\varvec{\alpha }}^{\prime }{\varvec{\Omega }}{\varvec{\alpha }}\right) ^{1/2}Z_{0}^{*}$ and note that $\left( {\varvec{\alpha }}^{\prime }{\varvec{\Omega }}{\varvec{\alpha }}\right) ^{1/2}Z_{0}^{*}\sim {\mathcal {N}}\left( 0,{\varvec{\alpha }}^{\prime }{\varvec{\Omega }}{\varvec{\alpha }}\right)$ this is distributed as ${\varvec{\alpha }}^{\prime }{\varvec{Z}}_{0}$, where ${\varvec{Z}}_{0}\sim {\mathcal {N}}_{p_{0}}\left( {\varvec{0}},{\varvec{\Omega }}\right)$. This completes the proof. $\square$

It should be noticed that for the special case where $p_{0}=p$ and ${\varvec{U}}_{i}={\varvec{I}}_{p_{0}}$, the above Lemma A.1 becomes the classical multivariate central limit theorem. From Lemma A.1, we establish the following lemma which is useful in establishing Theorem 3.2. Let ${\varvec{\Sigma }}_{E}$ be the variance-covariance matrix of each row of ${\varvec{E}}$ i.e ${\varvec{\Sigma }}_{E}={\text {Cov}}\left( {\varvec{e}}_{(i)}\right)$ with ${\varvec{E}}=\left( {{\varvec{e}}}_{(1)},{{\varvec{e}}}_{(2)},\dots ,{{\varvec{e}}}_{(n)}\right) ^{\prime }$,

${\varvec{\Lambda }}_{\psi }=\text {E}\left( ({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)})\otimes ({{\varvec{\psi }}}_{(1)} {{\varvec{\psi }}}^{\prime }_{(1)})\right)$ and let ${\varvec{\Lambda }}_{\delta }=\text {E}\left( ({{\varvec{\delta }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)})\otimes ({{\varvec{\delta }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)})\right)$.

Lemma A.2

Suppose that Assumption 1 holds. Then,

$$\begin{aligned} \begin{aligned}&(1)\quad n^{-\frac{1}{2}}{\varvec{\Psi }}^{\prime }{\varvec{\Psi }}-n^{\frac{1}{2}}\sigma _\psi ^2{\varvec{I}}_{p} \xrightarrow [n\rightarrow \infty ]{d} {\varvec{W}}_1\sim \mathcal {MN}_{p\times p} ({\varvec{0}}_{p\times p},{\varvec{\Lambda }}_\psi -\sigma _\psi ^4{\text {vec}} ({\varvec{I}}_{p})({\text {vec}}({\varvec{I}}_{p}))^{\prime });\\&(2)\quad n^{-\frac{1}{2}}{\varvec{\Delta }}^{\prime }{\varvec{\Delta }}-n^{\frac{1}{2}}\sigma _\delta ^2{\varvec{I}}_{p} \xrightarrow [n\rightarrow \infty ]{d} {\varvec{W}}_2\sim \mathcal {MN}_{p\times p} ({\varvec{0}}_{p\times p},{\varvec{\Lambda }}_\delta -\sigma _\delta ^4{\text {vec}}({\varvec{I}}_{p}) ({\text {vec}}({\varvec{I}}_{p}))^{\prime });\\&(3)\quad n^{-\frac{1}{2}}{\varvec{M}}^{\prime }{\varvec{E}} \xrightarrow [n\rightarrow \infty ]{d} {\varvec{W}}_3\sim \mathcal {MN}_{p\times q}\left( {\varvec{0}}_{p\times q},({\varvec{\sigma }}_{\scriptscriptstyle M} {\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M})\otimes {\varvec{\Sigma }}_{\scriptscriptstyle E}\right) ;\\&(4)\quad n^{-\frac{1}{2}}{\varvec{M}}^{\prime }{\varvec{\Psi }} \xrightarrow [n\rightarrow \infty ]{d} {\varvec{W}}_4\sim \mathcal {MN}_{p\times p}\left( {\varvec{0}}_{p\times p},\sigma _\psi ^2 (({\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M})\otimes {\varvec{I}}_{p})\right) ;\\&(5)\quad n^{-\frac{1}{2}}{\varvec{M}}^{\prime }{\varvec{\Delta }} \xrightarrow [n\rightarrow \infty ]{d} {\varvec{W}}_5\sim \mathcal {MN}_{p\times p}\left( {\varvec{0}}_{p\times p},\sigma _\delta ^2 (({\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M})\otimes {\varvec{I}}_{p})\right) ;\\&(6)\quad n^{-\frac{1}{2}}{\varvec{\Psi }}^{\prime }{\varvec{\Delta }} \xrightarrow [n\rightarrow \infty ]{d} {\varvec{W}}_6\sim \mathcal {MN}_{p\times p}\left( {\varvec{0}}_{p\times p},\sigma _\psi ^2\sigma _\delta ^2 {\varvec{I}}_{p^2}\right) ;\\&(7)\quad n^{-\frac{1}{2}}{\varvec{\Psi }}^{\prime }{\varvec{E}} \xrightarrow [n\rightarrow \infty ]{d} {\varvec{W}}_7\sim \mathcal {MN}_{p\times q}\left( {\varvec{0}}_{p\times q},\sigma _\psi ^2({\varvec{I}}_{p}\otimes {\varvec{\Sigma }}_{\scriptscriptstyle E})\right) ;\\&(8)\quad n^{-\frac{1}{2}}{\varvec{\Delta }}^{\prime }{\varvec{E}} \xrightarrow [n\rightarrow \infty ]{d} {\varvec{W}}_8\sim \mathcal {MN}_{p\times q}\left( {\varvec{0}}_{p\times q},\sigma _\delta ^2({\varvec{I}}_{p}\otimes {\varvec{\Sigma }}_{\scriptscriptstyle E})\right) . \end{aligned} \end{aligned}$$

Proof

Parts (1) and (2): We have

$n^{-\frac{1}{2}}{\varvec{\Psi }}^{\prime }{\varvec{\Psi }}-\sigma _\psi ^2{\text {vec}}({\varvec{I}}_{p})= n^{-\frac{1}{2}}\sum \limits _{i=1}^n({{\varvec{\psi }}}_{(i)}{{\varvec{\psi }}^{\prime }}_{(i)} -\sigma _\psi ^2{\varvec{I}}_{p})$ with $\text {E}\left( {{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)}-\sigma _\psi ^2{\varvec{I}}_{p} \right) =0$ and

$$\begin{aligned}&\text {E}\left( ({\text {vec}}({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)}) -\sigma _\psi ^2{\text {vec}}({\varvec{I}}_{p})) \left( {\text {vec}}({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)}) -\sigma _\psi ^2{\text {vec}}({\varvec{I}}_{p})\right) ^{\prime }\right) \\&=\text {E}\left( {\text {vec}}({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)}) \left( {\text {vec}}({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)})\right) ^{\prime }\right) -\sigma _{\psi }^{4}{\text {vec}}({\varvec{I}}_{p})\left( {\text {vec}}({\varvec{I}}_{p}) \right) ^{\prime }\\&=\text {E}\left( ({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)}) \otimes ({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)})\right) -\sigma _{\psi }^{4} {\text {vec}}({\varvec{I}}_{p})\left( {\text {vec}}({\varvec{I}}_{p})\right) ^{\prime } ={\varvec{\Lambda }}_{\psi }-\sigma _{\psi }^{4}{\text {vec}}({\varvec{I}}_{p}) \left( {\text {vec}}({\varvec{I}}_{p})\right) ^{\prime } \end{aligned}$$

where ${\varvec{\Lambda }}_{\psi }=\text {E}\left( ({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)}) \otimes ({{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)})\right)$. Then, Part (1) follows from Lemma A.1. The proof of Part (2) is similar.

Parts (3)–(5): We have $n^{-\frac{1}{2}}{\varvec{M}}^{\prime }{\varvec{E}}=n^{-\frac{1}{2}}\sum \limits _{i=1}^n{{\varvec{m}}}_{(i)} {\varvec{e}}^{\prime }_{(i)}$ where ${\varvec{E}}=\left[ {{\varvec{e}}}_{(1)}, {{\varvec{e}}}_{(2)}, \cdots , {{\varvec{e}}}_{(n)}\right] ^{\prime }$ with ${{\varvec{e}}}_{(i)}=(\epsilon _{i1},\epsilon _{i2},\cdots ,\epsilon _{iq})^{\prime }$. Then, the proof of Part (3)follows by combining the conditions $({\mathcal {A}}_{1})$ and $({\mathcal {A}}_{4})$ of Assumption 1 along with Lemma A.1. The proof of Parts (4)-(5) are established in the similar way.

Parts (6)-(8): We have $n^{-\frac{1}{2}}{\varvec{\Psi }}^{\prime }{\varvec{\Delta }}=n^{-\frac{1}{2}}\sum \limits _{i=1}^n{{\varvec{\psi }}}_{(i)}{{\varvec{\delta }}^{\prime }}_{(i)}$ where ${{\varvec{\psi }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}$, ${{\varvec{\psi }}}_{(2)}{{\varvec{\delta }}}^{\prime }_{(2)}$, ...,${{\varvec{\psi }}}_{(n)}{{\varvec{\delta }}}^{\prime }_{(n)}$ are iid with $\text {E}\left( {{\varvec{\psi }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}\right) ={\varvec{0}}$ and ${\text {Cov}}\left( {{\varvec{\psi }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}\right) =\text {E}\left[ {\text {vec}}\left( {{\varvec{\psi }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)} \right) \left( {\text {vec}}\left( {{\varvec{\psi }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}\right) \right) ^{\prime } \right]$. Then

$$\begin{aligned} {\text {Cov}}\left( {{\varvec{\psi }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}\right)&=\text {E}\left[ \left( {{\varvec{\psi }}}_{(1)}\otimes {\varvec{I}}_{p}\right) {{\varvec{\delta }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}\left( {{\varvec{\psi }}}^{\prime }_{(1)}\otimes {\varvec{I}}_{p}\right) \right] \\ \quad { }&=\text {E}\left\{ \left( {{\varvec{\psi }}}_{(1)}\otimes {\varvec{I}}_{p}\right) \text {E}\left[ {{\varvec{\delta }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}\Big |{{\varvec{\psi }}}_{(1)}\right] \left( {{\varvec{\psi }}}^{\prime }_{(1)}\otimes {\varvec{I}}_{p}\right) \right\} . \end{aligned}$$

Then, since ${{\varvec{\delta }}}_{(1)}$ and ${{\varvec{\psi }}}_{(1)}$ are independent, we get

$$\begin{aligned}&{\text {Cov}}\left( {{\varvec{\psi }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}\right) =\text {E}\left\{ \left( {{\varvec{\psi }}}_{(1)}\otimes {\varvec{I}}_{p}\right) \text {E}\left[ {{\varvec{\delta }}}_{(1)}{{\varvec{\delta }}}^{\prime }_{(1)}\right] \left( {{\varvec{\psi }}}^{\prime }_{(1)}\otimes {\varvec{I}}_{p}\right) \right\} \\&\quad =\sigma ^{2}_{\delta }\text {E}\left\{ \left( {{\varvec{\psi }}}_{(1)}{{\varvec{\psi }}}^{\prime }_{(1)} \otimes {\varvec{I}}_{p}\right) \right\} =\sigma ^{2}_{\delta }\sigma ^{2}_{\psi }{\varvec{I}}_{p^{2}}, \end{aligned}$$

and then, Part (6) follows from Lemma A.1. Parts (7) and (8) are established in the similar way. $\square$

Lemma A.3

Let Y be a $p\times q$ random matrix and ${\varvec{Y}}\sim \mathcal {MN}_{p\times q}({\varvec{0}}_{\scriptscriptstyle p\times q},{\varvec{\Lambda }})$, with ${\varvec{\Lambda }}$ a $pq\times pq$ matrix. For $j=1,2, \dots , m$, let $\kappa _{j}$ and $\alpha _{j}$ be $q\times q-$ nonrandom matrices, let $\iota _{j}$ and $\beta _{j}$ be $p\times p$-nonrandom matrices, and let $\varrho _{j}$ be $p\times q$-nonrandom matrices. Then

$\left( \begin{array}{ccc} \iota _{1}{\varvec{Y}}\kappa _{1}+\beta _1{\varvec{Y}}\alpha _1+\varrho _1\\ \iota _{2}{\varvec{Y}}\kappa _{2}+\beta _2{\varvec{Y}}\alpha _2+\varrho _2\\ \vdots \\ \iota _{m}{\varvec{Y}}\kappa _{m}+\beta _m {\varvec{Y}}\alpha _m+\varrho _m \end{array} \right)$ $\sim \mathcal {MN}_{mp\times q}$ $\left( \left( \begin{array}{c} \varrho _1 \\ \varrho _2 \\ \vdots \\ \varrho _{m} \end{array} \right) , \left( \begin{array}{cccc} {\varvec{A}}_{11} &{} {\varvec{A}}_{12}&{}\cdots &{} {\varvec{A}}_{1m} \\ {\varvec{A}}_{21} &{} {\varvec{A}}_{22}&{} \cdots &{} {\varvec{A}}_{2m}\\ \vdots &{} \cdots &{} \cdots &{} \vdots \\ {\varvec{A}}_{m1} &{} {\varvec{A}}_{m2}&{} \cdots &{} {\varvec{A}}_{mm} \end{array} \right) \right) ,$ where $\quad { } {\varvec{A}}_{ji}=({\varvec{A}}_{ij})^{\prime }, i,j=1,2,\dots ,m,$ and

$$\begin{aligned} {\varvec{A}}_{ij}&=(\kappa ^{\prime }_{i}\otimes \iota _{i})\Lambda (\kappa _{j}\otimes \iota ^{\prime }_{j}) +(\kappa ^{\prime }_{i}\otimes \iota _{i})\Lambda (\alpha _j\otimes \beta ^{\prime }_j) +(\alpha ^{\prime }_i\otimes \beta _i){\varvec{\Lambda }}(\kappa _{j}\otimes \iota ^{\prime }_{j}) \nonumber \\&\quad +(\alpha ^{\prime }_i\otimes \beta ^{}_i)\Lambda (\alpha _j\otimes \beta ^{\prime }_j). \end{aligned}$$

Proof

We have vec$\left( \left( \begin{array}{ccc} \iota _{1}{\varvec{Y}}\kappa _{1}+\beta _1{\varvec{Y}}\alpha _1+\varrho _1\\ \iota _{2}{\varvec{Y}}\kappa _{2}+\beta _2{\varvec{Y}}\alpha _2+\varrho _2\\ \vdots \\ \iota _{m}{\varvec{Y}}\kappa _{m}+\beta _m {\varvec{Y}}\alpha _m+\varrho _m \end{array} \right) \right) {=} \left( \begin{array}{c} \kappa ^{\prime }_{1}\otimes \iota _{1}+\alpha ^{\prime }_1\otimes \beta _1\\ \kappa ^{\prime }_{2}\otimes \iota _{2}+\alpha ^{\prime }_2\otimes \beta _2\\ \vdots \\ \kappa ^{\prime }_{m}\otimes \iota _{m}+\alpha ^{\prime }_m\otimes \beta _m \end{array} \right) {\text {vec}}({\varvec{Y}}) +\left( \begin{array}{c} {\text {vec}}(\varrho _1)\\ {\text {vec}}(\varrho _2)\\ \vdots \\ {\text {vec}}(\varrho _m) \end{array} \right)$ then the rest of the proof follows from the properties of normal random vectors along with some algebraic computations, this completes the proof. $\square$

Note that this result is more general than Corollary A.2 in Chen and Nkurunziza (2016). By using this lemma, we establish the following lemma, which is more general than Proposition A.10 and Corollary A.2 in Chen and Nkurunziza (2016). The established lemma is particularly useful in deriving the joint asymptotic normality between $\hat{{\varvec{B}}}_1$, $\hat{{\varvec{B}}}_2$,$\hat{{\varvec{B}}}_3$ and $\hat{{\varvec{B}}}_4$.

Lemma A.4

For $j=1,2,\dots ,m$, let $\{\kappa _{jn}\}_{n=1}^{\infty }$, $\{\iota _{jn}\}_{n=1}^{\infty }$ $\{\alpha _{jn}\}_{n=1}^{\infty }$,$\{\beta _{jn}\}_{n=1}^{\infty }$, $\{\varrho _{jn}\}_{n=1}^{\infty }$, be sequences of random matrices such that $\kappa _{jn}\xrightarrow [n\rightarrow \infty ]{P}\kappa _{j}$, $\iota _{jn}\xrightarrow [n\rightarrow \infty ]{P}\iota _{j}$, $\alpha _{jn}\xrightarrow [n\rightarrow \infty ]{P}\alpha _j$, $\beta _{jn}\xrightarrow [n\rightarrow \infty ]{P}\beta _j$, $\varrho _{jn}\xrightarrow [n\rightarrow \infty ]{P}\varrho _j$, where, for $j=1,2,\dots , m$, $\kappa _j$, $\alpha _j$, $\iota _j$ and $\beta _j$,$\varrho _j$, are non-random matrices as defined in Lemma A.3. If a sequence of $p\times q$ random matrices $\{{\varvec{Y}}_n\}_{n=1}^{\infty }$ is such that ${\varvec{Y}}_n\xrightarrow [n\rightarrow \infty ]{d}{\varvec{Y}}\sim \mathcal {MN}_{p\times q}({\varvec{0}}_{\scriptscriptstyle p\times q},\Lambda )$, where ${\varvec{\Lambda }}$ is a $pq\times pq$ matrix. We have

$\left( \begin{array}{ccc} \iota _{1n} {\varvec{Y}}_n\kappa _{1n}+\beta _{1n}{\varvec{Y}}_n\alpha _{1n}+\varrho _{1n}\\ \iota _{2n}{\varvec{Y}}_n\kappa _{2n}+\beta _{2n}{\varvec{Y}}_n\alpha _{2n}+\varrho _{2n}\\ \vdots \\ \iota _{mn}{\varvec{Y}}_n\kappa _{mn}+\beta _{mn}{\varvec{Y}}_n\alpha _{mn}+\varrho _{mn} \end{array} \right) \xrightarrow [n\rightarrow \infty ]{d} {\varvec{U}}\sim \mathcal {MN}_{mp\times q} \left( {\varvec{\varrho }},\,{\varvec{A}} \right)$

with ${\varvec{\varrho }}= \left( \begin{array}{c} \varrho _1 \\ \varrho _2 \\ \vdots \\ \varrho _m \end{array} \right) ,$  ${\varvec{A}}= \left( \begin{array}{c c c c} {\varvec{A}}_{11} &{} {\varvec{A}}_{12}&{} \cdots &{} {\varvec{A}}_{1m}\\ {\varvec{A}}_{21} &{} {\varvec{A}}_{22}&{} \cdots &{} {\varvec{A}}_{2m}\\ \vdots &{} \cdots &{} \cdots &{} \vdots \\ {\varvec{A}}_{m1} &{} {\varvec{A}}_{m2}&{} \cdots &{} {\varvec{A}}_{mm} \end{array} \right)$,

where ${\varvec{A}}_{ij}$, $i=1,2, \dots , m; j=1,2, \dots , m$ are as defined in Lemma A.3.

Proof

We have

$$\begin{aligned}&\left( \begin{array}{c} {\text {vec}}(\iota _{1n} {\varvec{Y}}_n\kappa _{1n}+\beta _{1n}{\varvec{Y}}_n\alpha _{1n} +\varrho _{1n})\\ {\text {vec}}(\iota _{2n} {\varvec{Y}}_n\kappa _{2n}+\beta _{2n}{\varvec{Y}}_n\alpha _{2n} +\varrho _{2n})\\ \vdots \\ {\text {vec}}(\iota _{mn} {\varvec{Y}}_n\kappa _{mn}+\beta _{mn}{\varvec{Y}}_n\alpha _{mn}+\varrho _{mn}) \end{array} \right) = \left( \begin{array}{c} \kappa ^{\prime }_{1n}\otimes \iota _{1n}+\alpha ^{\prime }_{1n}\otimes \beta _{1n}\\ \kappa ^{\prime }_{2n}\otimes \iota _{2n}+\alpha ^{\prime }_{2n}\otimes \beta _{2n}\\ \vdots \\ \kappa ^{\prime }_{mn}\otimes \iota _{mn}+\alpha ^{\prime }_{mn}\otimes \beta _{mn} \end{array} \right) {\text {vec}}({\varvec{Y}}_n)\\&+ \left( \begin{array}{c} {\text {vec}}(\varrho _{1n})\\ {\text {vec}}(\varrho _{2n})\\ \vdots \\ {\text {vec}}(\varrho _{mn}) \end{array} \right) , \end{aligned}$$

where ${\text {vec}}({\varvec{Y}}_n)\xrightarrow [n\rightarrow \infty ]{d}{\text {vec}}({\varvec{Y}})\sim {\mathcal {N}}_{pq}(0,{\varvec{\Lambda }})$, $\left( \begin{array}{c} {\text {vec}}(\varrho _{1n})\\ {\text {vec}}(\varrho _{2n})\\ \vdots \\ {\text {vec}}(\varrho _{mn}) \end{array} \right) \xrightarrow [n\rightarrow \infty ]{P}\left( \begin{array}{c} {\text {vec}}(\varrho _{1})\\ {\text {vec}}(\varrho _2)\\ \vdots \\ {\text {vec}}(\varrho _{m}) \end{array} \right) ,$

and $\left( \begin{array}{c} \kappa ^{\prime }_{1n}\otimes \iota _{1n}+\alpha ^{\prime }_{1n}\otimes \beta _{1n}\\ \kappa ^{\prime }_{2n}\otimes \iota _{2n}+\alpha ^{\prime }_{2n}\otimes \beta _{2n}\\ \vdots \\ \kappa ^{\prime }_{mn}\otimes \iota _{mn}+\alpha ^{\prime }_{mn}\otimes \beta _{mn} \end{array} \right) \xrightarrow [n\rightarrow \infty ]{P} \left( \begin{array}{c} \kappa ^{\prime }_{1}\otimes \iota _{1}+\alpha ^{\prime }_1\otimes \beta _1\\ \kappa ^{\prime }_{2}\otimes \iota _{2}+\alpha ^{\prime }_2\otimes \beta _2\\ \vdots \\ \kappa ^{\prime }_{m}\otimes \iota _{m}+\alpha ^{\prime }_m\otimes \beta _m \end{array} \right) .$

Then, by using Slutsky’s theorem, we have

$$\begin{aligned} {vec} \left( \begin{array}{ccc} \iota _{1n} {\varvec{Y}}_n\kappa _{1n}+\beta _{1n}{\varvec{Y}}_n\alpha _{1n}+\varrho _{1n}\\ \iota _{2n}{\varvec{Y}}_n\kappa _{2n}+\beta _{2n}{\varvec{Y}}_n\alpha _{2n}+\varrho _{2n}\\ \vdots \\ \iota _{mn}{\varvec{Y}}_n\kappa _{mn}+\beta _{mn}{\varvec{Y}}_n\alpha _{mn}+\varrho _{mn} \end{array} \right) \xrightarrow [n\rightarrow \infty ]{d}{\text {vec}} \left( \begin{array}{ccc} \iota _{1} {\varvec{Y}}\kappa _{1}+\beta _{1}{\varvec{Y}}\alpha _{1}+\varrho _{1}\\ \iota _{2}{\varvec{Y}}\kappa _{2}+\beta _{2}{\varvec{Y}}\alpha _{2}+\varrho _{2}\\ \vdots \\ \iota _{m}{\varvec{Y}}\kappa _{m}+\beta _{m}{\varvec{Y}}\alpha _{m}+\varrho _{m} \end{array} \right) \end{aligned}$$

and then

$$\begin{aligned} \left( \begin{array}{ccc} \iota _{1n} {\varvec{Y}}_n\kappa _{1n}+\beta _{1n}{\varvec{Y}}_n\alpha _{1n}+\varrho _{1n}\\ \iota _{2n}{\varvec{Y}}_n\kappa _{2n}+\beta _{2n}{\varvec{Y}}_n\alpha _{2n}+\varrho _{2n}\\ \vdots \\ \iota _{mn}{\varvec{Y}}_n\kappa _{mn}+\beta _{mn}{\varvec{Y}}_n\alpha _{mn}+\varrho _{mn} \end{array} \right) \xrightarrow [n\rightarrow \infty ]{d} \left( \begin{array}{ccc} \iota _{1} {\varvec{Y}}\kappa _{1}+\beta _{1}{\varvec{Y}}\alpha _{1}+\varrho _{1}\\ \iota _{2}{\varvec{Y}}\kappa _{2}+\beta _{2}{\varvec{Y}}\alpha _{2}+\varrho _{2}\\ \vdots \\ \iota _{m}{\varvec{Y}}\kappa _{m}+\beta _{m}{\varvec{Y}}\alpha _{m}+\varrho _{m} \end{array} \right) \equiv {\varvec{U}}. \end{aligned}$$

Then, the proof follows directly from Lemma A.3. $\square$

From this lemma, we establish the following corollary.

Corollary A.1

Suppose that the conditions Lemma A.4 hold. We have $({\varvec{Y}}^{\prime }_n,({\varvec{Y}}_n+\beta _{2n}{\varvec{Y}}_n\alpha _{2n}+\varrho _{2n})^{\prime })^{\prime }\xrightarrow [n\rightarrow \infty ]{d}({\varvec{Y}}^{\prime },({\varvec{Y}}+\beta _{2} {\varvec{Y}}\alpha _{2}+\varrho _{2})^{\prime })^{\prime }$, with

$$\begin{aligned} \left( \begin{array}{ccc} {\varvec{Y}}\\ {\varvec{Y}}+\beta _{2} {\varvec{Y}}\alpha _{2}+\varrho _{2} \end{array} \right) \sim \mathcal {MN}_{2p\times q} \left( \left( \begin{array}{c} {\varvec{0}}_{\scriptscriptstyle p\times q} \\ \varrho _{2} \end{array} \right) , \left( \begin{array}{c c} {\varvec{V}}_{11} &{} {\varvec{V}}_{12} \\ {\varvec{V}}_{21} &{} {\varvec{V}}_{22} \end{array} \right) \right) \end{aligned}$$

where ${\varvec{V}}_{11}={\varvec{\Lambda }}$;    ${\varvec{V}}_{12}={\varvec{\Lambda }}+{\varvec{\Lambda }} (\alpha ^{\prime }_{2}\otimes \beta _{2})$;    ${\varvec{V}}_{21}=({\varvec{V}}_{12})^{\prime }$;

${\varvec{V}}_{22}=({\varvec{I}}_{pq}+\alpha ^{\prime }_{2}\otimes \beta _{2}){\varvec{\Lambda }}({\varvec{I}}_{pq}+\alpha ^{\prime }_{2}\otimes \beta _{2}).$

The proof follows directly from Lemma A.4 by taking $m=2$, $\kappa _{jn}={\varvec{I}}_{q}$, $\iota _{jn}={\varvec{I}}_{p}$, $\alpha _{1n}={\varvec{0}}$, $\beta _{1n}={\varvec{0}}$ and $\varrho _{1n}={\varvec{0}}$.

Proof of Theorem 3.2

We have

$$\begin{aligned}&(\hat{{\varvec{B}}}(\hat{{\varvec{\Sigma }}})-{\varvec{B}})=(\hat{{\varvec{B}}}_1-{\varvec{B}}) +\hat{{\varvec{\Sigma }}}^{-1}{{\varvec{R}}^{\prime }_1}[{\varvec{R}}_{1}\hat{{\varvec{\Sigma }}}^{-1} {{\varvec{R}}^{\prime }_1}]^{-1} ({\varvec{\theta }}-{\varvec{R}}_{1}\hat{{\varvec{B}}}_1{\varvec{R}}_{2})({{\varvec{R}}^{\prime }_{2}}{\varvec{R}}_{2})^{-1} {{\varvec{R}}^{\prime }_{2}}\\&\quad =(\hat{{\varvec{B}}}_1-{\varvec{B}})-{\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}}){\varvec{R}}_{1} \left( \hat{{\varvec{B}}}_1-{\varvec{B}}\right) {\varvec{R}}_{2}{\varvec{P}}_{n}+{\varvec{G}}_{2n} ({\hat{{\varvec{\Sigma }}}})({\varvec{\theta }}-{\varvec{R}}_{1}B_1{\varvec{R}}_{2}){\varvec{P}}_{n}. \end{aligned}$$

with ${\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}})=\hat{{\varvec{\Sigma }}}^{-1}{{\varvec{R}}^{\prime }_1}[{\varvec{R}}_{1} \hat{{\varvec{\Sigma }}}^{-1}{{\varvec{R}}^{\prime }_1}]^{-1}$ and ${\varvec{P}}_{n}=({{\varvec{R}}^{\prime }_{2}}{\varvec{R}}_{2})^{-1}{{\varvec{R}}^{\prime }_{2}}$. Then, since

${\varvec{R}}_{1}{\varvec{B}}{\varvec{R}}_2={\varvec{\theta }}+{\varvec{\theta }}_{0}\big /\sqrt{n}$, this last relation gives

$$\begin{aligned} n^{\frac{1}{2}}(\hat{{\varvec{B}}}(\hat{{\varvec{\Sigma }}})-{\varvec{B}})=n^{\frac{1}{2}} (\hat{{\varvec{B}}}_1-{\varvec{B}})- {\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}}){\varvec{R}}_{1}\left( n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1 -{\varvec{B}})\right) {\varvec{R}}_{2}{\varvec{P}}_{n}-{\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}})\theta _{0} {\varvec{P}}_{n}. \end{aligned}$$

Hence,

$$\begin{aligned} \left( \begin{array}{c} \sqrt{n}(\hat{{\varvec{B}}}_1-{\varvec{B}})\\ \sqrt{n}(\hat{{\varvec{B}}}(\hat{{\varvec{\Sigma }}})-{\varvec{B}}) \end{array} \right)&= \left( \begin{array}{c} n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})\\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})-{\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}}) {\varvec{R}}_{1}\left( n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})\right) {\varvec{R}}_{2}{\varvec{P}}_{n} \end{array} \right) \\&\quad +\left( \begin{array}{c} {\varvec{0}}_{\scriptscriptstyle p\times q} \\ -{\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}}){\varvec{\theta }}_{0}{\varvec{P}}_{n} \\ \end{array} \right) . \end{aligned}$$

Note that

$$\begin{aligned} {\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}})\xrightarrow [n\rightarrow \infty ]{P} {\varvec{G}}_2({\varvec{Q}}_{0}), \quad { } {\varvec{P}}_{n}\xrightarrow [n\rightarrow \infty ]{P}P \end{aligned}$$

(A.1)

with

$$\begin{aligned} {\varvec{G}}_2({\varvec{Q}}_{0})={\varvec{Q}}_{0}^{-1}{\varvec{R}}^{\prime }_1({\varvec{R}}_{1}{\varvec{Q}}_{0}^{-1} {\varvec{R}}^{\prime }_1)^{-1}, \quad { } \text{ and } \quad { } {\varvec{P}}=({{\varvec{R}}^{\prime }_{2}}{\varvec{R}}_{2})^{-1}{{\varvec{R}}^{\prime }_{2}}. \end{aligned}$$

(A.2)

Further, let ${\varvec{\beta }}_{2}={\varvec{G}}_2({\varvec{Q}}_{0}){\varvec{R}}_{1}$, let ${\varvec{\alpha }}_{2}={\varvec{R}}_{2}P$ and let ${\varvec{\varrho }}={\varvec{G}}_{2}({\varvec{Q}}_{0}){\varvec{\theta }}_{0}{\varvec{P}}$. By using Corollary A.1, we have

this completes the proof. $\square$

Proof of Theorem 3.1

The proof is similar to that of Theorem 3.2 by taking ${\varvec{\mu }}\left( {\varvec{Q}}_{0}\right) ={\varvec{0}}$. For the convenience of the reader, we also write the main steps here. We have

$$\begin{aligned} n^{\frac{1}{2}}(\hat{{\varvec{B}}}(\hat{{\varvec{\Sigma }}})-{\varvec{B}})=n^{\frac{1}{2}} (\hat{{\varvec{B}}}_1-{\varvec{B}})- {\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}}){\varvec{R}}_{1}\left( n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1 -{\varvec{B}})\right) {\varvec{R}}_{2}{\varvec{P}}_{n}. \end{aligned}$$

Hence,

$$\begin{aligned} \left( \begin{array}{c} \sqrt{n}(\hat{{\varvec{B}}}_1-{\varvec{B}})\\ \sqrt{n}(\hat{{\varvec{B}}}(\hat{{\varvec{\Sigma }}})-{\varvec{B}}) \end{array} \right) = \left( \begin{array}{c} n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})\\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})-{\varvec{G}}_{2n}({\hat{{\varvec{\Sigma }}}}) {\varvec{R}}_{1}\left( n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})\right) {\varvec{R}}_{2}{\varvec{P}}_{n} \end{array} \right) . \end{aligned}$$

Hence, by combining Corollary A.1 with relations (A.1) and (A.2), we get

$$\begin{aligned} \sqrt{n}\left( (\hat{{\varvec{B}}}_1-{\varvec{B}})^{\prime },\,(\hat{{\varvec{B}}}(\hat{{\varvec{\Sigma }}})-{\varvec{B}})^{\prime }\right) ^{\prime } \xrightarrow [n\rightarrow \infty ]{d} \left( {\varvec{Y}}^{\prime }, {\varvec{Y}}^{\prime }+{\varvec{\beta }}^{\prime }_{2} {\varvec{Y}}^{\prime }{\varvec{\alpha }}^{\prime }_{2}+{\varvec{\varrho }}^{\prime } \right) ^{\prime } \end{aligned}$$

where ${\varvec{\beta }}_{2}={\varvec{G}}_2({\varvec{Q}}_{0}){\varvec{R}}_{1}$, let ${\varvec{\alpha }}_{2}={\varvec{R}}_{2}P$ and let ${\varvec{\varrho }}={\varvec{0}}$. Then, we have

$$\begin{aligned} \left( \begin{array}{ccc} {\varvec{Y}}\\ {\varvec{Y}}+{\varvec{\beta }}_{2} {\varvec{Y}}{\varvec{\alpha }}_{2}+{\varvec{\varrho }} \end{array} \right) \sim \mathcal {MN}_{2p\times q} \left( {\varvec{0}}_{\scriptscriptstyle 2p\times q}, \left( \begin{array}{c c} {\varvec{\Sigma }}_{11} &{} {\varvec{\Sigma }}_{12}\left( {\varvec{Q}}_{0}\right) \\ {\varvec{\Sigma }}_{21}\left( {\varvec{Q}}_{0}\right) &{} {\varvec{\Sigma }}_{22}\left( {\varvec{Q}}_{0}\right) \end{array} \right) \right) , \end{aligned}$$

as desired result. $\square$

Proof of Theorem 3.3

From (2.7), (2.8) and (2.9), we have

$$\begin{aligned} n^{\frac{1}{2}}(\hat{{\varvec{B}}}_2-{\varvec{B}})=n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) -n^{\frac{1}{2}}{\varvec{G}}_{2n}{\varvec{R}}_{1}(\hat{{\varvec{B}}}_1-{\varvec{B}}){\varvec{R}}_{2}{\varvec{P}}_{n} +n^{\frac{1}{2}}{\varvec{G}}_{2n}(\theta -{\varvec{R}}_{1}{\varvec{B}}{\varvec{R}}_2){\varvec{P}}_{n}\\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_3-{\varvec{B}})=n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) +n^{\frac{1}{2}}{\varvec{G}}_{3n}(\theta -{\varvec{R}}_{1}\hat{{\varvec{B}}}_1{\varvec{R}}_{2}){\varvec{P}}_{n} \\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_4-{\varvec{B}})=n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) -n^{\frac{1}{2}}{\varvec{G}}_{4n}{\varvec{R}}_{1}(\hat{{\varvec{B}}}_1-{\varvec{B}}){\varvec{R}}_{2}{\varvec{P}}_{n} +n^{\frac{1}{2}}{\varvec{G}}_{4n}(\theta -{\varvec{R}}_{1}{\varvec{B}}{\varvec{R}}_2){\varvec{P}}_{n}. \end{aligned}$$

with ${\varvec{G}}_{2n}{=}({\varvec{S}} {\varvec{K}}_{X})^{-1}{{\varvec{R}}^{\prime }_1}[{\varvec{R}}_{1}({\varvec{S}} {\varvec{K}}_{X})^{-1} {{\varvec{R}}^{\prime }_1}]^{-1}$,   ${\varvec{S}}{=}{\varvec{X}}^{\prime }{\varvec{X}}$, ${\varvec{G}}_{3n}{=}{\varvec{S}}^{-1} {{\varvec{R}}^{\prime }_1}[{\varvec{R}}_{1}{\varvec{S}}^{-1}{{\varvec{R}}^{\prime }_1}]^{-1}$,

${\varvec{G}}_{4n}={{\varvec{R}}^{\prime }_1}[{\varvec{R}}_{1}{{\varvec{R}}^{\prime }_1}]^{-1}$,   ${\varvec{P}}_{n}=({\varvec{R}}_{2}^{\prime }{\varvec{R}}_{2})^{-1}{\varvec{R}}_{2}^{\prime }$. Then, since ${\varvec{R}}_{1}{\varvec{B}}{\varvec{R}}_2={\varvec{\theta }}+{\varvec{\theta }}_{0}/\sqrt{n}$, we have

$$\begin{aligned} n^{\frac{1}{2}}(\hat{{\varvec{B}}}_2-{\varvec{B}})=n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) -n^{\frac{1}{2}}{\varvec{G}}_{2n}{\varvec{R}}_{1}(\hat{{\varvec{B}}}_1-{\varvec{B}}){\varvec{R}}_{2}{\varvec{P}}_{n} +{\varvec{G}}_{2n}\theta _{0}{\varvec{P}}_{n}\\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_3-{\varvec{B}})=n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) -n^{\frac{1}{2}}{\varvec{G}}_{3n}{\varvec{R}}_{1}(\hat{{\varvec{B}}}_1-{\varvec{B}}){\varvec{R}}_{2}{\varvec{P}}_{n} +{\varvec{G}}_{3n}\theta _{0}{\varvec{P}}_{n}. \\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_4-{\varvec{B}})=n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) -n^{\frac{1}{2}}{\varvec{G}}_{4n}{\varvec{R}}_{1}(\hat{{\varvec{B}}}_1-{\varvec{B}}){\varvec{R}}_{2}{\varvec{P}}_{n} +{\varvec{G}}_{4n}\theta _{0}{\varvec{P}}_{n}. \end{aligned}$$

Therefore,

$$\begin{aligned} \left( \begin{array}{c} n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) \\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_2-{\varvec{B}}) \\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_3-{\varvec{B}}) \\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_4-{\varvec{B}}) \\ \end{array} \right) =\left( \begin{array}{c} n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) \\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})-{\varvec{G}}_{2n}{\varvec{R}}_{1}\left( n^{\frac{1}{2}} (\hat{{\varvec{B}}}_1-{\varvec{B}})\right) {\varvec{R}}_{2}{\varvec{P}}_{n} +{\varvec{G}}_{2n}\theta _{0}{\varvec{P}}_{n} \\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})-{\varvec{G}}_{3n}{\varvec{R}}_{1}\left( n^{\frac{1}{2}} (\hat{{\varvec{B}}}_1-{\varvec{B}})\right) {\varvec{R}}_{2}{\varvec{P}}_{n} +{\varvec{G}}_{3n}\theta _{0}{\varvec{P}}_{n} \\ n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})-{\varvec{G}}_{4n}{\varvec{R}}_{1}\left( n^{\frac{1}{2}} (\hat{{\varvec{B}}}_1-{\varvec{B}})\right) {\varvec{R}}_{2}{\varvec{P}}_{n} +{\varvec{G}}_{4n}\theta _{0}{\varvec{P}}_{n} \\ \end{array} \right) , \end{aligned}$$

with $n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}}) \xrightarrow [n\rightarrow \infty ]{d}\eta _1\sim \mathcal {MN}_{p \times q}\left( {\varvec{O}}_{\scriptscriptstyle p\times q}, {\varvec{\Sigma }}_{11}\right)$,

${\varvec{G}}_{2n} \xrightarrow [n\rightarrow \infty ]{P}{\varvec{G}}_{2}=({\varvec{\Sigma }} {\varvec{K}})^{-1}{\varvec{R}}^{\prime }_1({\varvec{R}}_{1}({\varvec{\Sigma }} {\varvec{K}})^{-1}{\varvec{R}}^{\prime }_1)^{-1}$, ${\varvec{G}}_{3n} \xrightarrow [n\rightarrow \infty ]{P}{\varvec{G}}_{3}={\varvec{\Sigma }}^{-1} {\varvec{R}}^{\prime }_1({\varvec{R}}_{1} {\varvec{\Sigma }}^{-1}{\varvec{R}}^{\prime }_1)^{-1}$, ${\varvec{G}}_{4n} \xrightarrow [n\rightarrow \infty ]{P}{\varvec{G}}_{4}={\varvec{R}}^{\prime }_1({\varvec{R}}_{1} {\varvec{R}}^{\prime }_1)^{-1}$, ${\varvec{P}}_{n} \xrightarrow [n\rightarrow \infty ]{P}{\varvec{P}}=({\varvec{R}}_{2}^{\prime }{\varvec{R}}_{2})^{-1} {\varvec{R}}_{2}^{\prime }$. Therefore, by using Lemma A.4, we get the statement of the proposition. $\square$

Proof of Theorem 3.4

The first statement follows from Theorem 3.2. Further, we have,

$$\begin{aligned}&\text {ADR}\left( \tilde{{\varvec{B}}}(\hat{{\varvec{\Sigma }}}),{\varvec{B}},{\varvec{W}}\right) =\text {trace}\left[ ({\varvec{W}}\otimes {\varvec{I}}_{q})\text {E}\left[ \text {vec} ({\varvec{\eta }}^{*})\left( \text {vec}({\varvec{\eta }}^{*})\right) ^{\prime }\right] \right] \\&\quad =\text {trace}(({\varvec{W}}\otimes {\varvec{I}}_{q}) ({\varvec{\Sigma }}_{22}({\varvec{Q}}_{0})))+\text {trace}\left( {\varvec{\mu }}^{\prime }({\varvec{Q}}_{0}) {\varvec{W}}{\varvec{\mu }}({\varvec{Q}}_{0})\right) . \end{aligned}$$

This gives

$$\begin{aligned}&\text {ADR}(\tilde{{\varvec{B}}}(\hat{{\varvec{\Sigma }}}),{\varvec{B}},{\varvec{W}})=\text {ADR} (\hat{{\varvec{B}}}_1,{\varvec{B}};{\varvec{W}})-\text {trace}(({\varvec{W}}\otimes {\varvec{I}}_{q})({\varvec{A}}_1 {\varvec{\Lambda }}({\varvec{J}}^{\prime }_1({\varvec{Q}}_{0})\otimes {\varvec{J}}))\\&\quad -\text {trace}(({\varvec{W}}\otimes {\varvec{I}}_{q})(({\varvec{J}}_1({\varvec{Q}}_{0})\otimes {\varvec{J}}){\varvec{\Lambda }} {\varvec{A}}^{\prime }_{1})+\text {trace}(({\varvec{W}}\otimes {\varvec{I}}_{q}) (({\varvec{J}}_1({\varvec{Q}}_{0})\\&\quad \otimes {\varvec{J}}){\varvec{\Lambda }}({\varvec{J}}^{\prime }_1({\varvec{Q}}_{0})\otimes {\varvec{J}})))\\&\quad +\text {trace}(({\varvec{C}}_4{\varvec{C}}^{\prime }_4)\otimes ({\varvec{C}}^{\prime }_3({\varvec{Q}}_{0}){\varvec{W}} {\varvec{C}}_3({\varvec{Q}}_{0})){\text {vec}}({\varvec{\theta }}_0)({\text {vec}} ({\varvec{\theta }}_0))^{\prime }). \end{aligned}$$

Further, one can verify that ${\text {vec}}({\varvec{J}}_1({\varvec{Q}}_{0})\otimes {\varvec{J}})={\text {vec}}(({\varvec{\Sigma }} {\varvec{K}})^{-1}\otimes {\varvec{I}}_q)$. Then, the rest of the proof follows from some algebraic computations. $\square$

Proof of Theorem 3.5

From Theorem 3.4, we have

$$\begin{aligned} \text {ADR}(\tilde{{\varvec{B}}}(\hat{{\varvec{\Sigma }}}),{\varvec{B}},{\varvec{W}}) =\text {ADR}(\hat{{\varvec{B}}}_1,{\varvec{B}},{\varvec{W}})-f_1({\varvec{Q}}_{0}) +({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\varvec{F}}_1({\varvec{Q}}_{0}){\text {vec}} ({\varvec{\theta }}_0).\nonumber \\ \end{aligned}$$

(A.3)

Note that $f_1({\varvec{Q}}_{0})\ge 0$ and obviously, if $f_1({\varvec{Q}}_{0})=0$, ADR$(\tilde{{\varvec{B}}}(\hat{{\varvec{\Sigma }}}), {\varvec{B}}; {\varvec{W}})>$ADR$(\hat{{\varvec{B}}}_1, {\varvec{B}},{\varvec{W}})$ provided that $({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\varvec{F}}_1({\varvec{Q}}_{0}){\text {vec}} ({\varvec{\theta }}_0)>0$. Thus, we only consider the case where $f_1({\varvec{Q}}_{0})>0$. From (A.3), ADR$(\tilde{{\varvec{B}}}(\hat{{\varvec{\Sigma }}}), {\varvec{B}}; W)>$ADR$(\hat{{\varvec{B}}}_1, {\varvec{B}},{\varvec{W}})$ if and only if

$-f_1({\varvec{Q}}_{0})+({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\varvec{F}}_1({\varvec{Q}}_{0}) {\text {vec}}({\varvec{\theta }}_0)>0$. If $f_1({\varvec{Q}}_{0})<({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\varvec{F}}_1({\varvec{Q}}_{0}) {\text {vec}}({\varvec{\theta }}_0),$ then

$$\begin{aligned}&\frac{f_1({\varvec{Q}}_{0})}{({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\varvec{F}}_1({\varvec{Q}}_{0}) {\text {vec}}({\varvec{\theta }}_0)}<1, \, \text{ and } \, ({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\text {vec}}({\varvec{\theta }}_0)\displaystyle \frac{f_1({\varvec{Q}}_{0})}{({\text {vec}}({\varvec{\theta }}_0))^{\prime } {\varvec{F}}_1({\varvec{Q}}_{0}){\text {vec}}({\varvec{\theta }}_0)}\\&\quad <||{\varvec{\theta }}_0||^2. \end{aligned}$$

Further, since $f_1({\varvec{Q}}_{0})>0$, we have

$$\begin{aligned} \displaystyle \frac{({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\varvec{F}}_1({\varvec{Q}}_{0}) {\text {vec}}({\varvec{\theta }}_0)}{({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\text {vec}}({\varvec{\theta }}_0)f_1}>\frac{1}{||{\varvec{\theta }}_0||^2}. \end{aligned}$$

(A.4)

Further, by using Courant Theorem, we have

$$\begin{aligned} \displaystyle \text {ch}_{min}({\varvec{F}}_1({\varvec{Q}}_{0}))<\displaystyle \frac{({\text {vec}} ({\varvec{\theta }}_0))^{\prime }{\varvec{F}}_1({\varvec{Q}}_{0}){\text {vec}}({\varvec{\theta }}_0)}{({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\text {vec}}({\varvec{\theta }}_0)} <\displaystyle \text {ch}_{max}({\varvec{F}}_1({\varvec{Q}}_{0})). \end{aligned}$$

(A.5)

Therefore, for the inequality in (A.4) to hold, it suffices to have

$$\begin{aligned} \frac{1}{||{\varvec{\theta }}_0||^2}<\frac{\text {ch}_{min}({\varvec{F}}_1({\varvec{Q}}_{0}))}{f_1({\varvec{Q}}_{0})}. \end{aligned}$$

That is if $||{\varvec{\theta }}_0||^2>\displaystyle \frac{f_1({\varvec{Q}}_{0})}{\text {ch}_{min}({\varvec{F}}_1 ({\varvec{Q}}_{0}))}$, then $\text {ADR}(\tilde{{\varvec{B}}}(\hat{{\varvec{\Sigma }}}),{\varvec{B}},{\varvec{W}})>\text {ADR}(\hat{{\varvec{B}}}_1, {\varvec{B}};{\varvec{W}}).$ Further, if $f_1({\varvec{Q}}_{0})>({\text {vec}}({\varvec{\theta }}_0))^{\prime }{\varvec{F}}_1({\varvec{Q}}_{0}) {\text {vec}}({\varvec{\theta }}_0)$, by using (A.5), we establish the condition that if $||{\varvec{\theta }}_0||^2<\displaystyle \frac{f_1({\varvec{Q}}_{0})}{\text {ch}_{max}({\varvec{F}}_1 ({\varvec{Q}}_{0}))}$, then $\text {ADR}(\tilde{{\varvec{B}}}(\hat{{\varvec{\Sigma }}}),{\varvec{B}},{\varvec{W}})<\text {ADR} (\hat{{\varvec{B}}}_1,{\varvec{B}};{\varvec{W}}),$ as stated. $\square$

Proposition A.1

Under the Assumption 1, we have

$$\begin{aligned} \begin{aligned}&(1)\quad \displaystyle {\lim _{n\rightarrow \infty }}n^{-1}{\varvec{M}}^{\prime }{\varvec{M}} = {\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}, \quad n^{-1}{\varvec{\Psi }}^{\prime }{\varvec{\Psi }} \xrightarrow [n\rightarrow \infty ]{P} \sigma _\psi ^2{\varvec{I}}_{p}, \quad n^{-1}{\varvec{\Delta }}^{\prime }{\varvec{\Delta }} \xrightarrow [n\rightarrow \infty ]{P} \sigma _\delta ^2 {\varvec{I}}_{p};\\&(2)\quad n^{-1}\left[ {\varvec{M}}^{\prime }{\varvec{E}} \vdots {\varvec{\Psi }}^{\prime }{\varvec{E}} \vdots {\varvec{\Delta }}^{\prime }{\varvec{E}} \right] \xrightarrow [n\rightarrow \infty ]{P}{\varvec{0}}, \quad { } and \quad { } \quad n^{-1}\left[ {\varvec{M}}^{\prime }{\varvec{\Psi }}\vdots {\varvec{M}}^{\prime }{\varvec{\Delta }} \vdots {\varvec{\Psi }}^{\prime }{\varvec{\Delta }} \right] \xrightarrow [n\rightarrow \infty ]{P}{\varvec{0}}. \end{aligned} \end{aligned}$$

Proof

We have ${\varvec{M}}^{\prime }{\varvec{M}}=\displaystyle {\sum \nolimits _{i=1}^{n}}{{\varvec{m}}}_{(i)}{{\varvec{m}}}^{\prime }_{(i)}$. Then, from the condition $({\mathcal {A}}_{5})$ and Cesàro’s mean theorem, we get $\displaystyle {\lim \limits _{n\rightarrow \infty }}n^{-1}{\varvec{M}}^{\prime }{\varvec{M}}={\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}$. Further, from Lemma A.2, we have

$$\begin{aligned}&\left[ n^{-1/2}{\varvec{\Psi }}^{\prime }{\varvec{\Psi }}-n^{1/2} \sigma _\psi ^2{\varvec{I}}_{p}\vdots n^{-1/2}{\varvec{\Delta }}^{\prime }{\varvec{\Delta }}-n^{1/2} \sigma _\delta ^2{\varvec{I}}_{p}\right] =O_{p}(1),\\&\quad { } n^{-1/2}\left[ {\varvec{M}}^{\prime }{\varvec{E}} \vdots {\varvec{\Psi }}^{\prime }{\varvec{E}} \vdots {\varvec{\Delta }}^{\prime }{\varvec{E}} \right] =O_{p}(1), \end{aligned}$$

and $n^{-1/2}\left[ {\varvec{M}}^{\prime }{\varvec{\Psi }}\vdots {\varvec{M}}^{\prime }{\varvec{\Delta }} \vdots {\varvec{\Psi }}^{\prime }{\varvec{\Delta }} \right] =O_{p}(1)$. Then, together with Slutsky’s theorem, we get the stated results. $\square$

From Proposition A.1, we establish the following corollary which plays an important role in deriving the the joint asymptotic normality of the UE and the RE. In the sequel, let

$K_{\scriptscriptstyle 0,n}=\left( n^{-1}{\varvec{M}}^{\prime }{\varvec{M}}+\sigma ^{2}_{\psi }{\varvec{I}}_{p}+\sigma ^{2}_{\delta }{\varvec{I}}_{p}\right) ^{-1} \left( n^{-1}{\varvec{M}}^{\prime }{\varvec{M}}+\sigma ^{2}_{\psi }{\varvec{I}}_{p}\right)$.

Corollary A.2

Suppose that Assumption 1 holds. Then, $\hat{{\varvec{\Sigma }}}_X \xrightarrow [n\rightarrow \infty ]{P}{\varvec{\Sigma }}$, $K_{\scriptscriptstyle 0,n}\xrightarrow [n\rightarrow \infty ]{P} {\varvec{K}},$

$\hat{{\varvec{\Sigma }}}_D \xrightarrow [n\rightarrow \infty ]{P} {\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}$, ${\varvec{K}}_{X}\xrightarrow [n\rightarrow \infty ]{P} {\varvec{K}}$, $n^{-1}({\varvec{X}}^{\prime }{\varvec{Z}})\xrightarrow [n\rightarrow \infty ]{P} \left( {\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}\right) {\varvec{B}}$.

Proof

Since ${\varvec{X}}={\varvec{M}}+{\varvec{\Psi }}+{\varvec{\Delta }}$, then $\hat{{\varvec{\Sigma }}}_X=n^{-1}({\varvec{M}}+{\varvec{\Psi }}+{\varvec{\Delta }})^{\prime }({\varvec{M}}+{\varvec{\Psi }} +{\varvec{\Delta }})$ and then

$$\begin{aligned}\begin{aligned} \hat{{\varvec{\Sigma }}}_X&=n^{-1}({\varvec{M}}^{\prime }{\varvec{M}}+{\varvec{M}}^{\prime }{\varvec{\Psi }}+{\varvec{M}}^{\prime }{\varvec{\Psi }}+{\varvec{\Psi }}^{\prime }{\varvec{M}} +{\varvec{\Psi }}^{\prime }{\varvec{\Psi }}+{\varvec{\Psi }}^{\prime }{\varvec{\Delta }} +{\varvec{\Delta }}^{\prime }{\varvec{M}}+{\varvec{\Delta }}^{\prime }{\varvec{\Psi }}+{\varvec{\Delta }}^{\prime }{\varvec{\Delta }}). \end{aligned} \end{aligned}$$

Then, by using Proposition A.1, we have $n^{-1}\left( {\varvec{M}}^{\prime }{\varvec{\Psi }}+{\varvec{M}}^{\prime }{\varvec{\Psi }}+{\varvec{\Psi }}^{\prime }{\varvec{M}}+{\varvec{\Delta }}^{\prime }{\varvec{M}} +{\varvec{\Psi }}^{\prime }{\varvec{\Delta }}\right) \xrightarrow [n\rightarrow \infty ]{P}0$,

$K_{\scriptscriptstyle 0,n}\xrightarrow [n\rightarrow \infty ]{P} {\varvec{K}}$, and $n^{-1}({\varvec{M}}^{\prime }{\varvec{M}}+{\varvec{\Psi }}^{\prime }{\varvec{\Psi }}+ {\varvec{\Delta }}^{\prime }{\varvec{\Delta }})\xrightarrow [n\rightarrow \infty ]{P} {\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}+\sigma _\delta ^2{\varvec{I}}_{p}$. Hence,

${\hat{\Sigma }}_X\xrightarrow [n\rightarrow \infty ]{P}{\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}+\sigma _\delta ^2{\varvec{I}}_{p}={\varvec{\Sigma }}$. Since ${\hat{\Sigma }}_D={\hat{\Sigma }}_X-{\hat{\sigma }}_\delta ^2{\varvec{I}}_{p}$, where ${\hat{\sigma }}_\delta ^2$ is a consistent estimator for $\sigma _\delta ^2$, then ${\hat{\Sigma }}_D\xrightarrow [n\rightarrow \infty ]{P}{\varvec{\Sigma }} -\sigma _\delta ^2{\varvec{I}}_{p}={\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}$. Thus,

${\varvec{K}}_{X}={\hat{\Sigma }}_X^{-1}{\hat{\Sigma }}_D\xrightarrow [n\rightarrow \infty ]{P} {\varvec{\Sigma }}^{-1}\left( {\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}\right) ={\varvec{K}}$. Further,

$$\begin{aligned} n^{-1}({\varvec{X}}^{\prime }{\varvec{Y}})= & {} n^{-1}[{\varvec{X}}^{\prime }{\varvec{X}}-({\varvec{M}}+{\varvec{\Psi }}+{\varvec{\Delta }})^{\prime }{\varvec{\Delta }}]{\varvec{\beta }}+({\varvec{M}} +{\varvec{\Psi }}+{\varvec{\Delta }})^{\prime }{\varvec{E}} \xrightarrow [n\rightarrow \infty ]{P}({\varvec{\Sigma }} -\sigma _\delta ^2{\varvec{I}}_{p}){\varvec{\beta }}\\= & {} \left( {\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}\right) {\varvec{\beta }}. \end{aligned}$$

$\square$

By using Corollary A.2, we prove the following proposition which proves that the UE is a consistent estimator.

Proposition A.2

If Assumption 1 hold, then $\hat{{\varvec{B}}}_1 \xrightarrow [n\rightarrow \infty ]{P}{\varvec{B}}$.

Proof

We have $\hat{{\varvec{B}}}_1 ={\varvec{K}}_{X}^{-1}\hat{{\varvec{\Sigma }}}_{X}^{-1} \left( \frac{{\varvec{X}}^{\prime }{\varvec{Z}}}{n}\right)$. Further, from Corollary A.2,

$$\begin{aligned}&{\varvec{K}}_{X} \xrightarrow [n\rightarrow \infty ]{P}{\varvec{K}}={\varvec{\Sigma }}^{-1} \left( {\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}\right) , \hat{{\varvec{\Sigma }}}_{X}\xrightarrow [n\rightarrow \infty ]{P}{\varvec{\Sigma }} \text{ and } \\&\quad \frac{{\varvec{X}}^{\prime }{\varvec{Z}}}{n}\xrightarrow [n\rightarrow \infty ]{P} \left( {\varvec{\sigma }}_{\scriptscriptstyle M}{\varvec{\sigma }}^{\prime }_{\scriptscriptstyle M}+\sigma _\psi ^2{\varvec{I}}_{p}\right) {\varvec{B}}. \end{aligned}$$

This proves that $\hat{{\varvec{B}}}_1 \xrightarrow [n\rightarrow \infty ]{P}{\varvec{B}}$, as desired result. $\square$

By combining Corollary A.2, Lemma A.1 and Lemma A.2, we prove Theorem 2.1.

Proof of Theorem 2.1

We have

$$\begin{aligned} \hat{{\varvec{B}}}_1=n^{-\frac{1}{2}}\hat{{\varvec{\Sigma }}}_D^{-1}{\varvec{T}}_{n}+n^{-\frac{1}{2}} \hat{{\varvec{\Sigma }}}_D^{-1}{\varvec{H}}_{n}{\varvec{K}}{\varvec{B}} +\hat{{\varvec{\Sigma }}}_D^{-1}{\hat{{\varvec{\Sigma }}}_D}{\varvec{B}}+O_p(n^{-1}). \end{aligned}$$

Hence $n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})=\hat{{\varvec{\Sigma }}}_D^{-1}[{\varvec{T}}_{n} +{\varvec{H}}_{n}{\varvec{K}}{\varvec{B}}]+{O_p}(n^{-\frac{1}{2}})$ and then,

$$\begin{aligned} n^{\frac{1}{2}}{(\hat{{\varvec{B}}}_1-{\varvec{B}})}^{\prime }=[{\varvec{T}}^{\prime }_{n}+{\varvec{B}}^{\prime }{\varvec{K}} {\varvec{H}}_{n}]\hat{{\varvec{\Sigma }}}_D^{-1}+{O_p}(n^{-\frac{1}{2}}). \end{aligned}$$

Apply the Vec() operator on both sides, we obtain

$$\begin{aligned} {\text {vec}}[n^{\frac{1}{2}}{(\hat{{\varvec{B}}}_1-{\varvec{B}})}^{\prime }] =(\hat{{\varvec{\Sigma }}}_D^{-1}\otimes I_q)[{\text {vec}}({\varvec{T}}^{\prime }_{n})+{\text {vec}}({\varvec{B}}^{\prime }{\varvec{K}} {\varvec{H}}_{n})]+O_p(n^{-\frac{1}{2}}).\qquad \end{aligned}$$

(A.6)

Further, we have $\mathrm {Vec}({\varvec{T}}^{\prime }_{n}) =n^{-\frac{1}{2}}\sum \limits _{i=1}^n {\varvec{U}}_{1i}{\varvec{w}}_{1i}$ where

${\varvec{U}}_{1i}= [{{\varvec{m}}}_{(i)}\otimes {{\varvec{I}}}_q,{{\varvec{I}}}_{pq},{{\varvec{I}}}_{pq},-{{\varvec{m}}}_{(i)}\otimes {\varvec{B}}^{\prime }, -{{\varvec{I}}}_p\otimes {\varvec{B}}^{\prime },-{{\varvec{I}}}_p\otimes {\varvec{B}}^{\prime }]$

are $qp\times (2qp+2p^2+p+q)$ non-random matrices and

$$\begin{aligned} {\varvec{w}}_{1i}={\text {vec}}\left[ ({\varvec{\epsilon }}_{(i)},{\varvec{\epsilon }}_{(i)} {{\varvec{\psi }}^{\prime }}_{(i)},{\varvec{\epsilon }}_{(i)}{{\varvec{\delta }}^{\prime }}_{(i)}, {{\varvec{\delta }}}_{(i)},{\varvec{\Delta }}_{(i)}{{\varvec{\psi }}^{\prime }}_{(i)},{{\varvec{\delta }}}_{(i)} {{\varvec{\delta }}^{\prime }}_{(i)}-\sigma _\delta ^2{{\varvec{I}}}_p)\right] \end{aligned}$$

are $(2qp+2p^2+p+q)$ iid column of random vectors. Similarly, one can verify that

${\text {vec}}\left( {\varvec{B}}^{\prime }{{{\varvec{K}}}}{{\varvec{H}}}_{n}\right) =n^{-\frac{1}{2}}\displaystyle \sum \limits _{i=1}^n {\varvec{U}}_{2i}{\varvec{w}}_{2i}$ where

${\varvec{U}}_{2i}= (I_p\otimes {\varvec{B}}^{\prime }{\varvec{K}}) \left[ ({{\varvec{I}}}_p\otimes {{\varvec{m}}}_{(i)}),({{\varvec{m}}}_{(i)}\otimes {{\varvec{I}}}_p), {{\varvec{I}}}_{p^2},{{\varvec{I}}}_{p^2},{{\varvec{I}}}_{p^2}\right]$

are $pq\times (3p^2+2p)$ non-random matrices and

$$\begin{aligned}\begin{aligned} {\varvec{w}}_{2i}&={\text {vec}}\big [({{\varvec{\psi }}^{\prime }}_{(i)}+{{\varvec{\delta }}^{\prime }}_{(i)}, {{\varvec{\psi }}}_{(i)}+{{\varvec{\delta }}}_{(i)}, {{\varvec{\psi }}}_{(i)}{{\varvec{\delta }}^{\prime }}_{(i)}+{{\varvec{\delta }}}_{(i)}{{\varvec{\psi }}^{\prime }}_{(i)}, {{\varvec{\psi }}}_{(i)}{{\varvec{\psi }}^{\prime }}_{(i)}-\sigma _{{\psi }}^2{{\varvec{I}}}_p,{{\varvec{\delta }}}_{(i)}{{\varvec{\delta }}^{\prime }}_{(i)}-\sigma _\delta ^2{{\varvec{I}}}_p)\big ] \end{aligned} \end{aligned}$$

are $(3p^2+2p)$-column random vectors. Therefore, we have

$\left[ {\text {vec}}({\varvec{T}}^{\prime }_{n})+{\text {vec}}\left( {\varvec{B}}^{\prime }{{{\varvec{K}}}}{{\varvec{H}}}_{n}\right) \right] =n^{-1/2}\sum \limits _{i=1}^n[{\varvec{U}}_{1i},\ {\varvec{U}}_{2i}][{\varvec{w}}^{\prime }_{1i},\ {\varvec{w}}^{\prime }_{2i}]^{\prime } =n^{-1/2}\sum \limits _{i=1}^n {\varvec{U}}_{i}{\varvec{w}}_{i}$.

We also have $\mathrm {E}\left( {\text {vec}}({\varvec{T}}^{\prime }_{n})+{\text {vec}} \left( {\varvec{B}}^{\prime }{{{\varvec{K}}}}{{\varvec{H}}}_{n}\right) \right) =0$. Further, let

${{\varvec{\Lambda }}} =\lim \limits _{n\rightarrow \infty }n^{-1}\mathrm {E}\left[ \left( \sum \limits _{i=1}^n {\varvec{U}}_{i}{\varvec{w}}_{i}\right) \left( \sum \limits _{i=1}^n {\varvec{U}}_{i}{\varvec{w}}_{i}\right) ^{\prime }\right]$.

Then, by Lemma A.1, we get

$\left[ {\text {vec}}({\varvec{T}}^{\prime }_{n})+{\text {vec}}\left( {\varvec{B}}^{\prime }{{{\varvec{K}}}}{{\varvec{H}}}_{n}\right) \right] \xrightarrow [n\rightarrow \infty ]{d}{\varvec{\varsigma }}_{0} \sim {\mathcal {N}}_{pq}({{\varvec{0}}},{{\varvec{\Lambda }}})$.

Therefore, together with (A.6), Corollary A.2 and Slutsky’s theorem,

$$\begin{aligned} {\text {vec}}[n^{\frac{1}{2}}(\hat{{\varvec{B}}}_1-{\varvec{B}})^{\prime }] \xrightarrow [n\rightarrow \infty ]{d}{\text {vec}}({\varvec{\eta }}^{\prime }_1)\sim {\mathcal {N}}_{qp}({\varvec{0}},{\varvec{A}}_1{\varvec{\Lambda }}{\varvec{A}}_1^{\prime }), \end{aligned}$$

this completes the proof. $\square$

Proof of Proposition 2.1

From 2.3, adding and subtracting ${\varvec{X}}\hat{{\varvec{B}}}_{1}$, we get

$$\begin{aligned} {\mathcal {G}}_{2}=\text {trace}({\varvec{Z}}^{\prime }{\varvec{Z}})+\text {trace} (\hat{{\varvec{B}}}^{\prime }_{1}{\varvec{X}}{\varvec{X}}\hat{{\varvec{B}}}_{1}) +\text {trace}((\hat{{\varvec{B}}}_{1}-{\varvec{B}})^{\prime }{\varvec{X}}^{\prime }{\varvec{X}}(\hat{{\varvec{B}}}_{1} -{\varvec{B}}))\nonumber \\ -2 \text { trace}({\varvec{Z}}^{\prime }{\varvec{X}}\hat{{\varvec{B}}}_{1})+2\text {trace}(({\varvec{Z}} -{\varvec{X}}{\varvec{B}})^{\prime }{\varvec{X}}(\hat{{\varvec{B}}}_{1}-{\varvec{B}})) -\text {trace}({\varvec{B}}^{\prime }n{\varvec{\Sigma }}_{X}{\varvec{B}}).\quad {} \end{aligned}$$

(A.7)

Further

$$\begin{aligned} \text {trace}[{\varvec{B}}^{\prime }(n{\varvec{\Sigma }}_{X}){\varvec{B}}-{\varvec{B}}^{\prime }(n\hat{{\varvec{\Sigma }}}_{X}) {\varvec{K}}_{X}{\varvec{B}}]=\text { trace}((\hat{{\varvec{B}}}_{1}-{\varvec{B}})^{\prime }n{\varvec{\Sigma }}_{X} ({\varvec{I}}_{p}-{\varvec{K}}_{X})(\hat{{\varvec{B}}}_{1}-{\varvec{B}}))\nonumber \\ -2\text {trace}(\hat{{\varvec{B}}}^{\prime }_{1}n{\varvec{\Sigma }}_{X}({\varvec{I}}_{p}-{\varvec{K}}_{X}) (\hat{{\varvec{B}}}_{1}-{\varvec{B}})) +\text {trace}(\hat{{\varvec{B}}}^{\prime }_{1}n{\varvec{\Sigma }}_{X}({\varvec{I}}_{p}-{\varvec{K}}_{X}) \hat{{\varvec{B}}}_{1}).\quad { } \nonumber \\ \end{aligned}$$

(A.8)

By combining the relations (A.7) and (A.8), we get

$$\begin{aligned} {\mathcal {G}}_{2}&=\text {trace}({\varvec{Z}}^{\prime }{\varvec{Z}})+\text {trace}(\hat{{\varvec{B}}}^{\prime }_{1}n {\varvec{\Sigma }}_{X}{\varvec{K}}_{X}\hat{{\varvec{B}}}_{1})-2 \text {trace}({\varvec{Z}}^{\prime } {\varvec{X}}\hat{{\varvec{B}}}_{1})\\&\quad +\text {trace}((\hat{{\varvec{B}}}_{1}-{\varvec{B}})^{\prime }n{\varvec{\Sigma }}_{X}{\varvec{K}}_{X} (\hat{{\varvec{B}}}_{1}-{\varvec{B}})) +2\text {trace}(({\varvec{Z}}^{\prime }{\varvec{X}}-\hat{{\varvec{B}}}_{1}n{\varvec{\Sigma }}_{X}{\varvec{K}}_{X}) (\hat{{\varvec{B}}}_{1}-{\varvec{B}})). \end{aligned}$$

The stated result follows from the fact that ${\varvec{Z}}^{\prime }{\varvec{X}}-\hat{{\varvec{B}}}_{1}n{\varvec{\Sigma }}_{X}{\varvec{K}}_{X}={\varvec{0}}$. $\square$