Modified $C_p$ Criterion in Widely Applicable Models

Abstract

A risk function based on the mean square error of prediction is a widely used measure of the goodness of a candidate model in model selection. A modified $C_p$ criterion referred to as an $MC_p$ criterion is an unbiased estimator of the risk function. The original $MC_p$ criterion was proposed by Fujikoshi and Satoh (1997) in multivariate linear regression models. Thereafter, many authors have proposed $MC_p$ criteria for various candidate models. A purpose of this paper is to propose an $MC_p$ criterion for a wide class of candidate models, including results in previous studies.

Appendix: Mathematical Details

1.1 A.1 The Proof of Lemma 1

Let ${\boldsymbol{\mathcal E}}$ be an $n\times p$ random matrix defined by

$$\begin{aligned} {\boldsymbol{\mathcal E}}=({\boldsymbol{Y}}-{\boldsymbol{\varGamma }}){\boldsymbol{\varOmega }}, \quad {\boldsymbol{\varOmega }}={\boldsymbol{\varSigma }}^{-1/2}. \end{aligned}$$

(A.1)

Notice that

$$\begin{aligned} d({\boldsymbol{Y}}_\textrm{F},\hat{{\boldsymbol{Y}}})=d({\boldsymbol{Y}}_\textrm{F},{\boldsymbol{M}}{\boldsymbol{Y}})+d({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})+2\mathcal L, \end{aligned}$$

(A.2)

where $\hat{{\boldsymbol{Y}}}$ is given by (4), and $\mathcal L=\text{ tr }\{({\boldsymbol{Y}}_\textrm{F}-{\boldsymbol{M}}{\boldsymbol{Y}}){\boldsymbol{\varOmega }}^2({\boldsymbol{M}}{\boldsymbol{Y}}-\hat{{\boldsymbol{Y}}})'\}$. Since ${\boldsymbol{Y}}_\textrm{F}$ is independent of ${\boldsymbol{Y}}$ and distributed according to the same distribution as ${\boldsymbol{Y}}$, and ${\boldsymbol{M}}$ satisfies (2), we have

$$\begin{aligned} E\left[ d({\boldsymbol{Y}}_\textrm{F},{\boldsymbol{M}}{\boldsymbol{Y}})\right] =E\left[ d({\boldsymbol{Y}}_\textrm{F},{\boldsymbol{\varGamma }})\right] +E\left[ \text{ tr }({\boldsymbol{\mathcal E}}'{\boldsymbol{M}}{\boldsymbol{\mathcal E}})\right] =(n+m)p. \end{aligned}$$

(A.3)

Notice that ${\boldsymbol{Y}}_\textrm{F}$, ${\boldsymbol{M}}{\boldsymbol{\mathcal E}}$, and ${\boldsymbol{G}}$ in (3) are mutually independent, and

$$ \mathcal L=\text{ tr }\left\{ \left( \left( {\boldsymbol{Y}}_\textrm{F}-{\boldsymbol{\varGamma }}\right) {\boldsymbol{\varOmega }}-{\boldsymbol{M}}{\boldsymbol{\mathcal E}}\right) \left( {\boldsymbol{M}}{\boldsymbol{\mathcal E}}-{\boldsymbol{H}}{\boldsymbol{\mathcal E}}{\boldsymbol{\varOmega }}^{-1}{\boldsymbol{G}}{\boldsymbol{\varOmega }}+\left( {\boldsymbol{\varGamma }}-{\boldsymbol{H}}{\boldsymbol{\varGamma }}{\boldsymbol{G}}\right) {\boldsymbol{\varOmega }}\right) '\right\} . $$

The above equations and (2) imply

$$\begin{aligned} E[\mathcal L]&=-E\left[ \text{ tr }\left\{ {\boldsymbol{M}}{\boldsymbol{\mathcal E}}\left( {\boldsymbol{M}}{\boldsymbol{\mathcal E}}-{\boldsymbol{H}}{\boldsymbol{\mathcal E}}{\boldsymbol{\varOmega }}^{-1}{\boldsymbol{G}}{\boldsymbol{\varOmega }}\right) '\right\} \right] \nonumber \\&=-E\left[ \text{ tr }({\boldsymbol{\mathcal E}}'{\boldsymbol{M}}{\boldsymbol{\mathcal E}})\right] +E\left[ \mathcal L_{1}\right] =-mp+E\left[ \mathcal L_{1}\right] , \end{aligned}$$

(A.4)

where $\mathcal L_{1}=\text{ tr }({\boldsymbol{\mathcal E}}'{\boldsymbol{H}}{\boldsymbol{\mathcal E}}{\boldsymbol{\varOmega }}{\boldsymbol{G}}'{\boldsymbol{\varOmega }}^{-1})$. Since ${\boldsymbol{\mathcal E}}'{\boldsymbol{H}}{\boldsymbol{\mathcal E}}$ is independent of ${\boldsymbol{G}}$ and $\text{ tr }({\boldsymbol{G}})=q$, the following equation is obtained.

$$\begin{aligned} E\left[ \mathcal L_{1}\right] =\text{ tr }({\boldsymbol{H}})E\left[ \text{ tr }\left( {\boldsymbol{\varOmega }}{\boldsymbol{G}}'{\boldsymbol{\varOmega }}^{-1}\right) \right] =\text{ tr }({\boldsymbol{H}})E\left[ \text{ tr }\left( {\boldsymbol{G}}'\right) \right] =q\text{ tr }({\boldsymbol{H}}). \end{aligned}$$

(A.5)

From (A.2), (A.3), (A.4), and (A.5), Lemma 1 is proved.

1.2 A.2 The Proof of Lemma 2

Notice that $\text{ tr }\{({\boldsymbol{Y}}-{\boldsymbol{M}}{\boldsymbol{Y}}){\boldsymbol{S}}^{-1}({\boldsymbol{M}}{\boldsymbol{Y}}-\hat{{\boldsymbol{Y}}})'\}=0$ because of

$$ \text{ tr }\{({\boldsymbol{Y}}-{\boldsymbol{M}}{\boldsymbol{Y}}){\boldsymbol{S}}^{-1}({\boldsymbol{M}}{\boldsymbol{Y}}-\hat{{\boldsymbol{Y}}})'\}=\text{ tr }\left\{ ({\boldsymbol{I}}_n-{\boldsymbol{M}}){\boldsymbol{Y}}{\boldsymbol{S}}^{-1}({\boldsymbol{Y}}-{\boldsymbol{H}}{\boldsymbol{Y}}{\boldsymbol{G}})'{\boldsymbol{M}}\right\} . $$

It follows from this result and a calculation similar to the one used to find (A.2) that

$$\begin{aligned} \hat{d}({\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})=\hat{d}({\boldsymbol{Y}},{\boldsymbol{M}}{\boldsymbol{Y}})+\hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}}). \end{aligned}$$

(A.6)

It is easy to see from the definition of ${\boldsymbol{S}}$ in (3) that

$$\begin{aligned} \hat{d}({\boldsymbol{Y}},{\boldsymbol{M}}{\boldsymbol{Y}})=(n-m)\text{ tr }({\boldsymbol{S}}{\boldsymbol{S}}^{-1})=(n-m)p. \end{aligned}$$

(A.7)

Substituting (A.7) into (A.6) yields

$$\begin{aligned} \hat{d}({\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})=(n-m)p+\hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}}). \end{aligned}$$

(A.8)

By (7), (A.8), and the definition of the bias, Lemma 2 is proved.

1.3 A.3The Proof of Lemma 3

Let ${\boldsymbol{W}}$ be a $p\times p$ random matrix defined by

$$\begin{aligned} {\boldsymbol{W}}=(n-m){\boldsymbol{\varOmega }}{\boldsymbol{S}}{\boldsymbol{\varOmega }}, \end{aligned}$$

(A.9)

where ${\boldsymbol{\varOmega }}$ is given by (A.1). Then, ${\boldsymbol{W}}\sim \mathcal W_p(n-m,{\boldsymbol{I}}_p)$. Let ${\boldsymbol{Q}}_2$ be a $p\times (p-q)$ matrix satisfying ${\boldsymbol{Q}}_2'{\boldsymbol{Q}}_2={\boldsymbol{I}}_{p-q}$ and ${\boldsymbol{Q}}_1'{\boldsymbol{Q}}_2={\boldsymbol{O}}_{q,p-q}$, where ${\boldsymbol{Q}}_1$ is given by (10), and let ${\boldsymbol{Q}}$ be the $p\times p$ orthogonal matrix defined by ${\boldsymbol{Q}}=({\boldsymbol{Q}}_1,{\boldsymbol{Q}}_2)$. Then, the three matrices ${\boldsymbol{C}}_1$, ${\boldsymbol{C}}_2$, and ${\boldsymbol{C}}_3$ given in (9) can be rewritten as

$$\begin{aligned} {\boldsymbol{C}}_1&={\boldsymbol{Q}}_1\left( {\boldsymbol{Q}}_1'{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_1\right) ^{-1}{\boldsymbol{Q}}_1'{\boldsymbol{W}}^{-1}({\boldsymbol{Q}}_1,{\boldsymbol{Q}}_2){\boldsymbol{Q}}', \nonumber \\ {\boldsymbol{C}}_2&=(n-m){\boldsymbol{Q}}\begin{pmatrix}{\boldsymbol{Q}}_1' \\ {\boldsymbol{Q}}_2' \end{pmatrix}{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_1\left( {\boldsymbol{Q}}_1'{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_1\right) ^{-1}{\boldsymbol{Q}}_1'{\boldsymbol{W}}^{-1}({\boldsymbol{Q}}_1,{\boldsymbol{Q}}_2){\boldsymbol{Q}}', \\ {\boldsymbol{C}}_3&= {\boldsymbol{Q}}\begin{pmatrix}{\boldsymbol{Q}}_1' \\ {\boldsymbol{Q}}_2' \end{pmatrix}{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_1\left( {\boldsymbol{Q}}_1'{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_1\right) ^{-2}{\boldsymbol{Q}}_1'{\boldsymbol{W}}^{-1}({\boldsymbol{Q}}_1,{\boldsymbol{Q}}_2){\boldsymbol{Q}}'. \nonumber \end{aligned}$$

(A.10)

Let ${\boldsymbol{V}}$ be a $p\times p$ random matrix defined by ${\boldsymbol{V}}={\boldsymbol{Q}}'{\boldsymbol{W}}{\boldsymbol{Q}}$, and let ${\boldsymbol{Z}}_1$ and ${\boldsymbol{Z}}_2$ be respectively $(n-m)\times q$ and $(n-m)\times (p-q)$ independent random matrices distributed as $({\boldsymbol{Z}}_1,{\boldsymbol{Z}}_2)\sim \mathcal N_{(n-m)\times p}({\boldsymbol{O}}_{n-m,p},{\boldsymbol{I}}_{(n-m)p})$. By using ${\boldsymbol{V}}\sim \mathcal W_p(n-m,{\boldsymbol{I}}_p)$, the partitioned ${\boldsymbol{V}}$ can be rewritten in ${\boldsymbol{Z}}_1$ and ${\boldsymbol{Z}}_2$ as

$$\begin{aligned} {\boldsymbol{V}}=\begin{pmatrix}{\boldsymbol{V}}_{11} &{} {\boldsymbol{V}}_{12} \\ {\boldsymbol{V}}_{12}' &{} {\boldsymbol{V}}_{22} \end{pmatrix} = \begin{pmatrix}{\boldsymbol{Z}}_1'{\boldsymbol{Z}}_1 &{} {\boldsymbol{Z}}_1'{\boldsymbol{Z}}_2 \\ {\boldsymbol{Z}}_2'{\boldsymbol{Z}}_1 &{} {\boldsymbol{Z}}_2'{\boldsymbol{Z}}_2 \end{pmatrix}. \end{aligned}$$

Hence, it follows from ${\boldsymbol{V}}^{-1}={\boldsymbol{Q}}'{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}$ and the general formula for the inverse of a partitioned matrix, e.g., th. 8.5.11 in [2], that

$$\begin{aligned} {\boldsymbol{V}}^{-1}&=\begin{pmatrix}{\boldsymbol{Q}}_1'{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_1 &{} {\boldsymbol{Q}}_1'{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_2 \\ {\boldsymbol{Q}}_2'{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_1 &{} {\boldsymbol{Q}}_2'{\boldsymbol{W}}^{-1}{\boldsymbol{Q}}_2 \end{pmatrix} \nonumber \\&= \begin{pmatrix}{\boldsymbol{V}}_{11\cdot 2}^{-1} &{} &{} -{\boldsymbol{V}}_{11\cdot 2}^{-1}{\boldsymbol{V}}_{12}{\boldsymbol{V}}_{22}^{-1} \\ -{\boldsymbol{V}}_{22}^{-1}{\boldsymbol{V}}_{12}'{\boldsymbol{V}}_{11\cdot 2}^{-1} &{} ~ &{} {\boldsymbol{V}}_{22}^{-1}+{\boldsymbol{V}}_{22}^{-1}{\boldsymbol{V}}_{12}'{\boldsymbol{V}}_{11\cdot 2}^{-1}{\boldsymbol{V}}_{12}{\boldsymbol{V}}_{22}^{-1} \end{pmatrix}, \end{aligned}$$

(A.11)

where ${\boldsymbol{V}}_{11\cdot 2}={\boldsymbol{V}}_{11}-{\boldsymbol{V}}_{12}{\boldsymbol{V}}_{22}^{-1}{\boldsymbol{V}}_{12}'$. Substituting (A.11) into (A.10) yields

$$\begin{aligned} {\boldsymbol{C}}_1&={\boldsymbol{Q}}_1 \left( {\boldsymbol{I}}_q,-{\boldsymbol{V}}_{12}{\boldsymbol{V}}_{22}^{-1} \right) {\boldsymbol{Q}}', \nonumber \\ {\boldsymbol{C}}_2&=(n-m){\boldsymbol{Q}}\left\{ {\boldsymbol{V}}^{-1}-\begin{pmatrix}{\boldsymbol{O}}_{q,q} &{} {\boldsymbol{O}}_{q,p-q} \\ {\boldsymbol{O}}_{p-q,q} &{} {\boldsymbol{V}}_{22}^{-1}\end{pmatrix}\right\} {\boldsymbol{Q}}', \\ {\boldsymbol{C}}_3&= {\boldsymbol{Q}}\begin{pmatrix}{\boldsymbol{I}}_q &{} &{} -{\boldsymbol{V}}_{12}{\boldsymbol{V}}_{22}^{-1} \\ - {\boldsymbol{V}}_{22}^{-1}{\boldsymbol{V}}_{12}'&{} ~ &{} {\boldsymbol{V}}_{22}^{-1}{\boldsymbol{V}}_{12}'{\boldsymbol{V}}_{12}{\boldsymbol{V}}_{22}^{-1}\end{pmatrix}{\boldsymbol{Q}}'. \nonumber \end{aligned}$$

(A.12)

By using the independence of ${\boldsymbol{Z}}_1$ and ${\boldsymbol{Z}}_2$, and formulas of expectations of the matrix normal distribution and Wishart distribution, we have

$$\begin{aligned}&E\left[ {\boldsymbol{V}}^{-1}\right] =\frac{c_1}{n-m}{\boldsymbol{I}}_p, \quad E\left[ {\boldsymbol{V}}_{22}^{-1}\right] =\frac{c_2}{n-m}{\boldsymbol{I}}_{p-q}, \\&E\left[ {\boldsymbol{V}}_{12}{\boldsymbol{V}}_{22}^{-1}\right] =E\left[ {\boldsymbol{Z}}_{1}'\right] E\left[ {\boldsymbol{Z}}_{2}\left( {\boldsymbol{Z}}_2'{\boldsymbol{Z}}_2\right) ^{-1}\right] ={\boldsymbol{O}}_{q,p-q}, \end{aligned}$$

where $c_1$ and $c_2$ are given by (11). It follows from the result $E[{\boldsymbol{Z}}_1{\boldsymbol{Z}}_1']=q{\boldsymbol{I}}_{n-m}$ and the independence of ${\boldsymbol{Z}}_1$ and ${\boldsymbol{Z}}_2$ that

$$\begin{aligned} E\left[ {\boldsymbol{V}}_{22}^{-1}{\boldsymbol{V}}_{12}'{\boldsymbol{V}}_{12}{\boldsymbol{V}}_{22}^{-1}\right]&= E\left[ \left( {\boldsymbol{Z}}_2'{\boldsymbol{Z}}_2\right) ^{-1}{\boldsymbol{Z}}_2'{\boldsymbol{Z}}_1{\boldsymbol{Z}}_1'{\boldsymbol{Z}}_2\left( {\boldsymbol{Z}}_2'{\boldsymbol{Z}}_2\right) ^{-1}\right] \\&=qE\left[ \left( {\boldsymbol{Z}}_2'{\boldsymbol{Z}}_2\right) ^{-1}{\boldsymbol{Z}}_2'{\boldsymbol{Z}}_2\left( {\boldsymbol{Z}}_2'{\boldsymbol{Z}}_2\right) ^{-1}\right] \\&=qE\left[ {\boldsymbol{V}}_{22}^{-1}\right] =c_3 {\boldsymbol{I}}_{p-q}, \end{aligned}$$

where $c_3$ is given by (11). By using the above expectations, (A.12), and ${\boldsymbol{Q}}_2{\boldsymbol{Q}}_2'={\boldsymbol{I}}_p-{\boldsymbol{Q}}_1{\boldsymbol{Q}}_1'$, we have

$$\begin{aligned} E\left[ {\boldsymbol{C}}_1\right]&={\boldsymbol{Q}}_1 \left( {\boldsymbol{I}}_q,{\boldsymbol{O}}_{q,p-q} \right) '\begin{pmatrix}{\boldsymbol{Q}}_1' \\ {\boldsymbol{Q}}_2'\end{pmatrix}={\boldsymbol{Q}}_1{\boldsymbol{Q}}_1', \nonumber \\ E\left[ {\boldsymbol{C}}_2\right]&={\boldsymbol{Q}}\left\{ c_1{\boldsymbol{I}}_p-\begin{pmatrix}{\boldsymbol{O}}_{q,q} &{} &{} {\boldsymbol{O}}_{q,p-q} \\ {\boldsymbol{O}}_{p-q,q} &{}~ &{} c_2{\boldsymbol{I}}_{p-q}\end{pmatrix}\right\} {\boldsymbol{Q}}' =c_2{\boldsymbol{Q}}_1{\boldsymbol{Q}}_1'+c_1c_3{\boldsymbol{I}}_p, \\ E\left[ {\boldsymbol{C}}_3\right]&= {\boldsymbol{Q}}\begin{pmatrix}{\boldsymbol{I}}_q &{} &{} {\boldsymbol{O}}_{p-q,q} \\ {\boldsymbol{O}}_{q,p-q} &{} ~ &{} c_3{\boldsymbol{I}}_{p-q} \end{pmatrix}{\boldsymbol{Q}}' =\frac{1}{c_1}E\left[ {\boldsymbol{C}}_2\right] . \nonumber \nonumber \end{aligned}$$

Therefore, Lemma 3 is proved.

1.4 A.4 The Proof of Lemma 4

Let ${\boldsymbol{U}}_0$, ${\boldsymbol{U}}_1$ and ${\boldsymbol{U}}_2$ be $p\times p$ symmetric random matrices defined by

$$ {\boldsymbol{U}}_0={\boldsymbol{\varOmega }}{\boldsymbol{Y}}'{\boldsymbol{M}}{\boldsymbol{Y}}{\boldsymbol{\varOmega }}, \quad {\boldsymbol{U}}_1={\boldsymbol{\varOmega }}{\boldsymbol{Y}}'{\boldsymbol{H}}{\boldsymbol{Y}}{\boldsymbol{\varOmega }}, \quad {\boldsymbol{U}}_2={\boldsymbol{\varOmega }}{\boldsymbol{Y}}'{\boldsymbol{H}}^2{\boldsymbol{Y}}{\boldsymbol{\varOmega }}, $$

where ${\boldsymbol{\varOmega }}$ is given by (A.1). It follows from the definitions of ${\boldsymbol{C}}_1$, ${\boldsymbol{C}}_2$, and ${\boldsymbol{C}}_3$ in (9) that

$$\begin{aligned}&{\boldsymbol{\varOmega }}^{-1}{\boldsymbol{G}}{\boldsymbol{\varOmega }}={\boldsymbol{C}}_1', \quad {\boldsymbol{C}}_1'{\boldsymbol{C}}_1={\boldsymbol{C}}_3, \quad {\boldsymbol{\varOmega }}^{-1}{\boldsymbol{S}}^{-1}{\boldsymbol{\varOmega }}^{-1}{\boldsymbol{C}}_1={\boldsymbol{C}}_2, \\&{\boldsymbol{C}}_1'{\boldsymbol{\varOmega }}^{-1}{\boldsymbol{S}}^{-1}{\boldsymbol{\varOmega }}^{-1}{\boldsymbol{C}}_1={\boldsymbol{C}}_2. \end{aligned}$$

By using these results, the definitions of d and $\hat{d}$ in (5), and the assumptions of ${\boldsymbol{M}}$ and ${\boldsymbol{H}}$ in (2), $d({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})$ and $\hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})$ in (8) can be rewritten as

$$\begin{aligned} d({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})&=\text{ tr }({\boldsymbol{U}}_0)-2\text{ tr }({\boldsymbol{U}}_1{\boldsymbol{C}}_1)+\text{ tr }({\boldsymbol{U}}_2{\boldsymbol{C}}_3), \\ \hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})&=(n-m)\text{ tr }({\boldsymbol{U}}_0{\boldsymbol{W}}^{-1})-2\text{ tr }({\boldsymbol{U}}_1{\boldsymbol{C}}_2)+\text{ tr }({\boldsymbol{U}}_2{\boldsymbol{C}}_2), \end{aligned}$$

where ${\boldsymbol{W}}$ is given by (A.9). Notice that ${\boldsymbol{U}}_0$ and ${\boldsymbol{S}}$, ${\boldsymbol{U}}_1$ and ${\boldsymbol{S}}$, and ${\boldsymbol{U}}_2$ and ${\boldsymbol{S}}$ are independent because of ${\boldsymbol{M}}({\boldsymbol{I}}_n-{\boldsymbol{M}})={\boldsymbol{O}}_{n,n}$ and ${\boldsymbol{H}}({\boldsymbol{I}}_n-{\boldsymbol{M}})={\boldsymbol{O}}_{n,n}$, and ${\boldsymbol{C}}_1$, ${\boldsymbol{C}}_2$, and ${\boldsymbol{C}}_3$ are random matrices in which ${\boldsymbol{S}}$ is the only random variable. These imply that

$$\begin{aligned} E\left[ d({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})\right]&=\text{ tr }\left( {\boldsymbol{\varDelta }}_0\right) -2\text{ tr }\left( {\boldsymbol{\varDelta }}_1E\left[ {\boldsymbol{C}}_1\right] \right) +\text{ tr }\left( {\boldsymbol{\varDelta }}_2E\left[ {\boldsymbol{C}}_3\right] \right) , \\ E\left[ \hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})\right]&=(n-m)\text{ tr }\left( {\boldsymbol{\varDelta }}_0E\left[ {\boldsymbol{W}}^{-1}\right] \right) -2\text{ tr }\left( {\boldsymbol{\varDelta }}_1E\left[ {\boldsymbol{C}}_2\right] \right) +\text{ tr }\left( {\boldsymbol{\varDelta }}_2E\left[ {\boldsymbol{C}}_2\right] \right) , \end{aligned}$$

where ${\boldsymbol{\varDelta }}_0=E[{\boldsymbol{U}}_0]$, ${\boldsymbol{\varDelta }}_1=E[{\boldsymbol{U}}_1]$, and ${\boldsymbol{\varDelta }}_2=E[{\boldsymbol{U}}_2]$. Notice that $(n-m)E[{\boldsymbol{W}}^{-1}]=c_1{\boldsymbol{I}}_p$, where $c_1$ is given by (11). By using this result and Lemma 3, we have

$$\begin{aligned} \begin{aligned} E\left[ d({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})\right]&=\text{ tr }\left( {\boldsymbol{\varDelta }}_0\right) -2\text{ tr }\left( {\boldsymbol{\varDelta }}_1{\boldsymbol{Q}}_1{\boldsymbol{Q}}_1'\right) +\frac{1}{c_1}\text{ tr }\left( {\boldsymbol{\varDelta }}_2 E\left[ {\boldsymbol{C}}_2\right] \right) , \\ E\left[ \hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})\right]&=c_1\text{ tr }\left( {\boldsymbol{\varDelta }}_0\right) -2\text{ tr }\left( {\boldsymbol{\varDelta }}_1E\left[ {\boldsymbol{C}}_2\right] \right) +\text{ tr }\left( {\boldsymbol{\varDelta }}_2E\left[ {\boldsymbol{C}}_2\right] \right) , \end{aligned} \end{aligned}$$

(A.13)

where $c_2$ and $c_3$ are given by (11). Let $L=\text{ tr }\{{\boldsymbol{U}}_1((n-m){\boldsymbol{W}}^{-1}-{\boldsymbol{C}}_2)\}$. It is easy to see that $c_2^{-1}=c_1^{-1}+c_2^{-1}c_3$ and $E[L]=c_1\text{ tr }({\boldsymbol{\varDelta }}_1)-\text{ tr }({\boldsymbol{\varDelta }}_1 E[{\boldsymbol{C}}_2])$. Hence, we can derive

$$\begin{aligned} \text{ tr }\left( {\boldsymbol{\varDelta }}_1{\boldsymbol{Q}}_1{\boldsymbol{Q}}_1'\right)&= \frac{1}{c_2}\text{ tr }\left( {\boldsymbol{\varDelta }}_1E\left[ {\boldsymbol{C}}_2\right] \right) -\frac{c_1c_3}{c_2}\text{ tr }\left( {\boldsymbol{\varDelta }}_1\right) \nonumber \\&=\frac{1}{c_1}\text{ tr }\left( {\boldsymbol{\varDelta }}_1E\left[ {\boldsymbol{C}}_2\right] \right) -\frac{c_3}{c_2}E\left[ L\right] . \end{aligned}$$

(A.14)

A simple calculation shows that $c_2^{-1}c_3=q/(n-m)$ and $L=\text{ tr }({\boldsymbol{R}})$, where ${\boldsymbol{R}}$ is given by (12). Using these results, (A.13), and (A.14) yields

$$ E\left[ d({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})\right] =\frac{1}{c_1}E\left[ \hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})\right] +\frac{2q}{n-m}E\left[ \text{ tr }({\boldsymbol{R}})\right] . $$

Consequently, Lemma 4 is proved.

1.5 A.5 The Proof of Theorem 1

From Lemmas 1 and 4, an unbiased estimator of $R_p$ in (6) can be given by

$$\begin{aligned} \hat{R}_p=(n-m)p+2q\text{ tr }({\boldsymbol{H}})+\hat{D}, \end{aligned}$$

(A.15)

where $\hat{D}$ is given by (13). Notice that $\{1-(p+1)/(n-m)\}(n-m)p=(n-m)p-p(p+1)$. Hence, the result and (A.8) imply that

$$\begin{aligned} \hat{D}=\left( 1-\frac{p+1}{n-m}\right) \hat{d}({\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})-(n-m)p+p(p+1)+\frac{2q}{n-m}\text{ tr }({\boldsymbol{R}}). \end{aligned}$$

(A.16)

Substituting (A.16) into (A.15) yields that an unbiased estimator $\hat{R}_p$ coincides with $MC_p$ in (14).

1.6 A.6 The Proof of Equation in (15)

The $\hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})$ when ${\boldsymbol{H}}={\boldsymbol{M}}$ can be rewritten as

$$ \hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})=\text{ tr }\left\{ {\boldsymbol{Y}}'{\boldsymbol{M}}{\boldsymbol{Y}}({\boldsymbol{I}}_p-{\boldsymbol{G}}){\boldsymbol{S}}^{-1}({\boldsymbol{I}}_p-{\boldsymbol{G}})'\right\} . $$

Since ${\boldsymbol{P}}={\boldsymbol{S}}^{1/2}({\boldsymbol{I}}_p-{\boldsymbol{G}}){\boldsymbol{S}}^{-1/2}$ is a symmetric idempotent matrix, we have

$$ ({\boldsymbol{I}}_p-{\boldsymbol{G}}){\boldsymbol{S}}^{-1}({\boldsymbol{I}}_p-{\boldsymbol{G}})' ={\boldsymbol{S}}^{-1/2}{\boldsymbol{P}}^2{\boldsymbol{S}}^{-1/2} =({\boldsymbol{I}}_p-{\boldsymbol{G}}){\boldsymbol{S}}^{-1}. $$

This implies that $\hat{d}({\boldsymbol{M}}{\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})=\text{ tr }\{{\boldsymbol{Y}}'{\boldsymbol{M}}{\boldsymbol{Y}}({\boldsymbol{I}}_p-{\boldsymbol{G}}){\boldsymbol{S}}^{-1}\}=\text{ tr }({\boldsymbol{R}})$. From this result and (A.8), $\hat{d}({\boldsymbol{Y}},\hat{{\boldsymbol{Y}}})=(n-m)p+\text{ tr }({\boldsymbol{R}})$ is derived.