Semiparametric efficient estimators in heteroscedastic error models

Abstract

In the mean regression context, this study considers several frequently encountered heteroscedastic error models where the regression mean and variance functions are specified up to certain parameters. An important point we note through a series of analyses is that different assumptions on standardized regression errors yield quite different efficiency bounds for the corresponding estimators. Consequently, all aspects of the assumptions need to be specifically taken into account in constructing their corresponding efficient estimators. This study clarifies the relation between the regression error assumptions and their, respectively, efficiency bounds under the general regression framework with heteroscedastic errors. Our simulation results support our findings; we carry out a real data analysis using the proposed methods where the Cobb–Douglas cost model is the regression mean.

Appendix

1.1 Proof of Proposition 1

First, note that the construction of t ensures the property

$$\begin{aligned} E(\epsilon t|\mathbf{X})=E\left( \epsilon ^3|\mathbf{X}\right) -E\left( \epsilon ^3|\mathbf{X}\right) E\left( \epsilon ^2|\mathbf{X}\right) -E\left( \epsilon |\mathbf{X}\right) =E\left( \epsilon ^3|\mathbf{X}\right) -E\left( \epsilon ^3|\mathbf{X}\right) =0, \end{aligned}$$

which is crucial for the following proof.

Following the semiparametric consideration in (1), we estimate the parameters ${\varvec{\theta }}$, while the nuisance parameter is the distribution $f_{\epsilon \mid \mathbf{X}}(\epsilon ,\mathbf{x})$, which is subject to the constraints $E(\epsilon |\mathbf{X})=0$ and $E(\epsilon ^2|\mathbf{X})=1$. Note that we do not assume any parametric distribution model on $\epsilon$. Thus, (1) is a semiparametric model. We first write out the joint distribution of $(\epsilon ,\mathbf{X})$

$$\begin{aligned} f_{\epsilon ,\mathbf{X}}(\epsilon ,\mathbf{x})=f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{x})f_{\mathbf{X}}(\mathbf{x})= \eta _1(\mathbf{x})\eta _2(\epsilon ,\mathbf{x}), \end{aligned}$$

where $\eta _1(\cdot )$ and $\eta _2(\cdot )$ denote the infinite-dimensional nuisance parameter function. Since $\eta _1(\cdot )$ and $\eta _2(\cdot )$ are probability density functions, we have

$$\begin{aligned} \int \eta _1(\mathbf{x})\hbox {d}\mathbf{x}=1,\quad \text{ and } \int \eta _2(\epsilon ,\mathbf{x})\hbox {d}\epsilon =1\quad \text{ for } \text{ all } \mathbf{x}. \end{aligned}$$

(15)

In the Hilbert space $\mathcal {H}$ formed by all the mean zero finite variance functions, the nuisance tangent space of a semiparametric model is the mean squared closure of all the parametric submodel nuisance tangent spaces. A parametric submodel is defined as a parametric model included by the original semiparametric model and includes the true density $f(\epsilon ,\mathbf{X};{\varvec{\theta }}_0,{\varvec{\gamma }}_0)$. A nuisance tangent space of the parametric model $f(\epsilon ,\mathbf{X};{\varvec{\theta }},{\varvec{\gamma }})$ is a linear space spanned by the nuisance score vector $S_{{\varvec{\gamma }}}=\partial \mathrm{log}f(\epsilon ,\mathbf{X};{\varvec{\theta }}_0,{\varvec{\gamma }})/\partial {\varvec{\gamma }}|_{{\varvec{\gamma }}_0}$. From the above general concept, we first derive the nuisance tangent space $\Lambda$. From Condition (2), we have

$$\begin{aligned} \int \epsilon \eta _2(\epsilon ,\mathbf{x})\hbox {d}\epsilon =0 \text{ for } \text{ all } \mathbf{x},\quad \text{ and } \int \epsilon ^2\eta _2(\epsilon ,\mathbf{x})\hbox {d}\epsilon =1\quad \text{ for } \text{ all } \mathbf{x}. \end{aligned}$$

(16)

From (15) and (16), it follows that

$$\begin{aligned} \Lambda _1= & {} \{\mathbf{a}(\mathbf{x}):E\{\mathbf{a}(\mathbf{X})\}=\mathbf{0}\}, \\ \Lambda _2= & {} \{\mathbf{b}(\mathbf{x},\epsilon ): E\{\mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}=E\{\epsilon \mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}=E\{\epsilon ^2\mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}=\mathbf{0}\} \\= & {} \{\mathbf{b}(\mathbf{x},\epsilon ):E\{\mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}=E\{\epsilon \mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}=E\{t\mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}=\mathbf{0} \}. \end{aligned}$$

In the above, we use

$$\begin{aligned} E\{t\mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}= & {} E\{\epsilon ^2\mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}-E(\epsilon ^3|\mathbf{X})E\{\epsilon \mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}-E\{\mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}\\= & {} E\{\epsilon ^2\mathbf{b}(\mathbf{X},\epsilon )|\mathbf{X}\}=\mathbf{0}. \end{aligned}$$

Combining $\Lambda _1$ and $\Lambda _2$, the nuisance tangent space $\Lambda$ can be written as

$$\begin{aligned} \Lambda= & {} \Lambda _1\oplus \Lambda _2 \\= & {} \{\mathbf{h}_1(\mathbf{x})+\mathbf{h}_2(\mathbf{x},\epsilon ): E\{\mathbf{h}_1(\mathbf{X})\}=E\{\mathbf{h}_2(\mathbf{x},\epsilon )|\mathbf{X}\}=E\{\epsilon \mathbf{h}_2(\mathbf{x},\epsilon )|\mathbf{X}\}\\&=E\{t\mathbf{h}_2(\mathbf{X},\epsilon )|\mathbf{X}\}=\mathbf{0}\}\\= & {} \{\mathbf{h}(\mathbf{x},\epsilon ): E\{\mathbf{h}(\mathbf{x},\epsilon )\}=E\{\epsilon \mathbf{h}(\mathbf{x},\epsilon )|\mathbf{X}\}=E\{t\mathbf{h}(\mathbf{X},\epsilon )|\mathbf{X}\}=\mathbf{0}\}. \end{aligned}$$

In order to find $\Lambda ^\perp$, we let $\mathbf{K}=\{\mathbf{g}(\mathbf{x},\epsilon ):\mathbf{g}(\mathbf{x},\epsilon )=\mathbf{g}_1(\mathbf{x})\epsilon +\mathbf{g}_2(\mathbf{x})t\}$ and have $\Lambda ^\perp =\mathbf{K}$ by showing that $\mathbf{K}\subset \Lambda ^\perp$ and $\Lambda ^\perp \subset \mathbf{K}$.

For arbitrary $\mathbf{h}(\mathbf{x},\epsilon )\in \Lambda$ and $\mathbf{g}(\mathbf{x},\epsilon )=\mathbf{g}_1(\mathbf{x})\epsilon +\mathbf{g}_2(\mathbf{x})t\in \mathbf{K}$,

$$\begin{aligned}&E\{\mathbf{h}(\mathbf{X},\epsilon )^\mathrm{T}\mathbf{g}(\mathbf{X},\epsilon )\}\\&\quad =E\left[ E\{\mathbf{h}(\mathbf{X},\epsilon )^\mathrm{T}\mathbf{g}(\mathbf{X},\epsilon )|\mathbf{X}\}\right] \\&\quad =E\left[ E\{\epsilon \mathbf{h}(\mathbf{X},\epsilon )|\mathbf{X}\}^\mathrm{T}\mathbf{g}_1(\mathbf{X})\right] +E\left[ E\{t\mathbf{h}(\mathbf{X},\epsilon )|\mathbf{X}\}^\mathrm{T}\mathbf{g}_2(\mathbf{X})\right] =0. \end{aligned}$$

Hence, $\mathbf{g}(\mathbf{x},\epsilon )=\mathbf{g}_1(\mathbf{x})\epsilon +\mathbf{g}_2(\mathbf{x})t\in \Lambda ^\perp$. Thus, $\mathbf{K}\subset \Lambda ^\perp$.

Now, we will show $\Lambda ^\perp \subset \mathbf{K}$.

For arbitrary $\mathbf{h}(\mathbf{x},\epsilon )\in \Lambda ^\perp$, we can decompose $\mathbf{h}(\mathbf{x},\epsilon )$ into $\mathbf{h}(\mathbf{x},\epsilon )=\mathbf{r}_1(\mathbf{x},\epsilon )+\mathbf{r}_2(\mathbf{x},\epsilon ),$ where

$$\begin{aligned} \mathbf{r}_1(\mathbf{X},\epsilon )= & {} \mathbf{h}(\mathbf{X},\epsilon )-E\{\epsilon \mathbf{h}(\mathbf{X},\epsilon )|\mathbf{X}\}\epsilon -\frac{E\{t\mathbf{h}(\mathbf{X},\epsilon )|\mathbf{X}\}}{E(t^2|\mathbf{X})}t,\\ \mathbf{r}_2(\mathbf{X},\epsilon )= & {} E\{\epsilon \mathbf{h}(\mathbf{X},\epsilon )|\mathbf{X}\}\epsilon +\frac{E\{t\mathbf{h}(\mathbf{X},\epsilon )|\mathbf{X}\}}{E(t^2|\mathbf{X})}t. \end{aligned}$$

It is obvious that $\mathbf{r}_2(\mathbf{x},\epsilon )\in \mathbf{K}$. Since $\mathbf{r}_2(\mathbf{x},\epsilon )\in \mathbf{K}\subset \Lambda ^\perp$ and $\mathbf{h}(\mathbf{x},\epsilon )\in \Lambda ^\perp$, we have $\mathbf{r}_1(\mathbf{x},\epsilon )=\mathbf{h}(\mathbf{x},\epsilon )-\mathbf{r}_2(\mathbf{x},\epsilon )\in \Lambda ^\perp$. For $\mathbf{r}_1$, we can easily verify that

$$\begin{aligned} E\{\mathbf{r}_1(\mathbf{X},\epsilon )\}=E\{\epsilon \mathbf{r}_1(\mathbf{X},\epsilon )|\mathbf{X}\}=E\{t\mathbf{r}_1(\mathbf{X},\epsilon )|\mathbf{X}\}=\mathbf{0}. \end{aligned}$$

This indicates that $\mathbf{r}_1(\mathbf{x},\epsilon )\in \Lambda$. Since $\mathbf{r}_1(\mathbf{x},\epsilon )$ is also an element of $\Lambda ^\perp$, it follows that $\mathbf{r}_1(\mathbf{x},\epsilon )=\mathbf{0}$, and we obtain $\mathbf{h}(\mathbf{x},\epsilon )=\mathbf{r}_2(\mathbf{x},\epsilon )\in \mathbf{K}$. We next show that $\mathbf{h}(\mathbf{x},\epsilon )\in \mathbf{K}$ for arbitrary $\mathbf{h}(\mathbf{x},\epsilon )\in \Lambda ^\perp$. Thus, $\Lambda ^\perp \subset \mathbf{K}$.

Consequently, we have

$$\begin{aligned} \Lambda ^\perp =\{\mathbf{g}(\mathbf{x},\epsilon ):\mathbf{g}(\mathbf{x},\epsilon )=\mathbf{g}_1(\mathbf{x})\epsilon +\mathbf{g}_2(\mathbf{x})t\}. \end{aligned}$$

$\square$

1.2 Proof of Theorem 1

In model (1), we obtain the score functions of ${\varvec{\theta }}=({\varvec{\alpha }}^\mathrm{T},{\varvec{\beta }}^\mathrm{T})^\mathrm{T}$ as

$$\begin{aligned} {\varvec{S}}_{{\varvec{\alpha }}}= & {} \frac{\partial \mathrm{log}f_{\mathbf{X},Y}(\mathbf{X},y)}{\partial {\varvec{\alpha }}} =-\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}e^{-\sigma (\mathbf{X},{\varvec{\beta }})}\mathbf{m}'_{\varvec{\alpha }}(\mathbf{X},{\varvec{\alpha }}), \\ {\varvec{S}}_{{\varvec{\beta }}}= & {} \frac{\partial \mathrm{log}f_{\mathbf{X},Y}(\mathbf{X},y)}{\partial {\varvec{\beta }}} =-\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})} \epsilon {\varvec{\sigma }}'_{{\varvec{\beta }}}(\mathbf{X},{\varvec{\beta }}) -{\varvec{\sigma }}'_{{\varvec{\beta }}}(\mathbf{X},{\varvec{\beta }}). \end{aligned}$$

By projecting the above score vectors onto $\Lambda ^\perp$, we can find the efficient score vectors of ${\varvec{\alpha }}$ and ${\varvec{\beta }}$.

First, ${\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}$ and ${\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}$ are the form of $\mathbf{g}_1(\mathbf{x})\epsilon +\mathbf{g}_2(\mathbf{x})t$. Thus, ${\varvec{S}}{_\mathrm{eff}}\in \Lambda ^\perp$.

Now, we verify that ${\varvec{S}}_{{\varvec{\theta }}}-{\varvec{S}}{_\mathrm{eff}}\in \Lambda$. ${\varvec{S}}_{{\varvec{\theta }}}-{\varvec{S}}{_\mathrm{eff}}$ is given by

$$\begin{aligned} {\varvec{S}}_{{\varvec{\alpha }}}-{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}= & {} e^{-\sigma (\mathbf{X},{\varvec{\beta }})}\mathbf{m}'_{{\varvec{\alpha }}} \left\{ -\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}-\epsilon +\frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t \right\} , \\ {\varvec{S}}_{{\varvec{\beta }}}-{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}= & {} {\varvec{\sigma }}'_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}) \left\{ -\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}\epsilon -1-\frac{2t}{E(t^2|\mathbf{X})}\right\} . \end{aligned}$$

We verify ${\varvec{S}}_{{\varvec{\theta }}}-{\varvec{S}}{_\mathrm{eff}}\in \Lambda$ by showing the following.

$$\begin{aligned} E({\varvec{S}}_{{\varvec{\theta }}}-{\varvec{S}}{_\mathrm{eff}})=\mathbf{0}, E\{\epsilon ({\varvec{S}}_{{\varvec{\theta }}}-{\varvec{S}}{_\mathrm{eff}})|\mathbf{X}\}=\mathbf{0}, E\{t({\varvec{S}}_{{\varvec{\theta }}}-{\varvec{S}}{_\mathrm{eff}})|\mathbf{X}\}=\mathbf{0}. \end{aligned}$$

We can check the details as follows.

$$\begin{aligned} E({\varvec{S}}_{{\varvec{\alpha }}}-{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}|\mathbf{X})= & {} e^{-\sigma (\mathbf{X},{\varvec{\beta }})}\mathbf{m}'_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }}) E\left\{ -\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}-\epsilon +\frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t |\mathbf{X}\right\} =\mathbf{0}, \\ E({\varvec{S}}_{{\varvec{\beta }}}-{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}|\mathbf{X})= & {} {\varvec{\sigma }}'_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}) E\left\{ -\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}\epsilon -1-\frac{2t}{E(t^2|\mathbf{X})}|\mathbf{X}\right\} =\mathbf{0}. \end{aligned}$$

The above results imply that $E({\varvec{S}}_{{\varvec{\theta }}}-{\varvec{S}}{_\mathrm{eff}})=\mathbf{0}$.

$$\begin{aligned} E\{\epsilon ({\varvec{S}}_{{\varvec{\alpha }}}-{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}})|\mathbf{X}\}= & {} e^{-\sigma (\mathbf{X},{\varvec{\beta }})}\mathbf{m}'_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }}) \\&E\left\{ -\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}\epsilon -\epsilon ^2 +\frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}\epsilon t |\mathbf{X}\right\} =\mathbf{0},\\ E\{t({\varvec{S}}_{{\varvec{\alpha }}}-{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}})|\mathbf{X}\}= & {} e^{-\sigma (\mathbf{X},{\varvec{\beta }})}\mathbf{m}'_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }}) \\&E\left\{ -\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}t-\epsilon t +\frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t^2 |\mathbf{X}\right\} =\mathbf{0}, \\ E\{\epsilon ({\varvec{S}}_{{\varvec{\beta }}}-{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}})|\mathbf{X}\}= & {} {\varvec{\sigma }}'_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}) E\left\{ -\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}\epsilon ^2-\epsilon -\frac{2\epsilon t}{E(t^2|\mathbf{X})}|\mathbf{X}\right\} =\mathbf{0}, \\ E\{t({\varvec{S}}_{{\varvec{\beta }}}-{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}})|\mathbf{X}\}= & {} {\varvec{\sigma }}'_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}) E\left\{ -\frac{\partial f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})/\partial \epsilon }{f_{\epsilon |\mathbf{X}}(\epsilon ,\mathbf{X})}\epsilon t-t-\frac{2t^2}{E(t^2|\mathbf{X})}|\mathbf{X}\right\} =\mathbf{0}. \end{aligned}$$

The above equations verify that ${\varvec{S}}_{{\varvec{\theta }}}-{\varvec{S}}{_\mathrm{eff}}\in \Lambda$.

Now, we find the optimal efficiency matrix. First, we calculate ${\varvec{S}}_\mathrm{eff}{\varvec{S}}^\mathrm{T}_\mathrm{eff}$.

$$\begin{aligned} {\varvec{S}}_\mathrm{eff}{\varvec{S}}^\mathrm{T}_\mathrm{eff}=\left\{ \begin{array}{cc} {\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}^\mathrm{T}&{}\quad {\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}^\mathrm{T}\\ {\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}^\mathrm{T}&{} \quad {\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}^\mathrm{T}\end{array} \right\} , \end{aligned}$$

where

$$\begin{aligned} {\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}^\mathrm{T}= & {} \left\{ \epsilon - \frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t \right\} ^2 e^{-2\sigma (\mathbf{X},{\varvec{\beta }})} \mathbf{m}'_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }}){\mathbf{m}'}^\mathrm{T}_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }}), \\ {\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}^\mathrm{T}= & {} \frac{2te^{-\sigma (\mathbf{X},{\varvec{\beta }})}}{E(t^2|\mathbf{X})}\left\{ \epsilon - \frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t \right\} \mathbf{m}'_{{\varvec{\alpha }}} (\mathbf{X},{\varvec{\alpha }}){{\varvec{\sigma }}'}^\mathrm{T}_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}), \\ {\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}^\mathrm{T}= & {} \frac{4t^2}{\{E(t^2|\mathbf{X})\}^2} {\varvec{\sigma }}'_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}){{\varvec{\sigma }}'}^\mathrm{T}_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}). \end{aligned}$$

Since $E({\varvec{S}}_\mathrm{eff}{\varvec{S}}^\mathrm{T}_\mathrm{eff})=E\{E({\varvec{S}}{_\mathrm{eff}}{\varvec{S}}{_\mathrm{eff}}^\mathrm{T}|\mathbf{X})\}$, we have each block inside $E({\varvec{S}}_\mathrm{eff}{\varvec{S}}^\mathrm{T}_\mathrm{eff})$ as follows.

• 1. $E({\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}^\mathrm{T})$ is equivalent to

  $$\begin{aligned} E\left( E\left[ \left\{ \epsilon - \frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t \right\} ^2|\mathbf{X}\right] e^{-2\sigma (\mathbf{X},{\varvec{\beta }})}\mathbf{m}'_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }}){\mathbf{m}'}^\mathrm{T}_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }})\right) . \end{aligned}$$

  From the above expression, we can rewrite the part $E\left[ \left\{ \epsilon - \frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t \right\} ^2|\mathbf{X}\right]$ as

  $$\begin{aligned} { E\left[ \left\{ \epsilon - \frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t \right\} ^2|\mathbf{X}\right] =1+\frac{\{E(\epsilon ^3|\mathbf{X})\}^2}{E(t^2|\mathbf{X})}}. \end{aligned}$$

  Thus, $E({\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}^\mathrm{T})$ becomes

  $$\begin{aligned} E\left( \left[ 1+\frac{\{E(\epsilon ^3|\mathbf{X})\}^2}{E(t^2|\mathbf{X})}\right] e^{-2\sigma (\mathbf{X},{\varvec{\beta }})}\mathbf{m}'_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }}){\mathbf{m}'}^\mathrm{T}_{{\varvec{\alpha }}}(\mathbf{X},{\varvec{\alpha }})\right) . \end{aligned}$$

• 2. $E({\varvec{S}}_{\mathrm{eff},{\varvec{\alpha }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}^\mathrm{T})$ can be rewritten as

  $$\begin{aligned}&E\left[ E\left\{ \epsilon t- \frac{E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}t^2 |\mathbf{X}\right\} \frac{2}{E(t^2|\mathbf{X})}e^{-\sigma (\mathbf{X},{\varvec{\beta }})} \mathbf{m}'_{{\varvec{\alpha }}} (\mathbf{X},{\varvec{\alpha }}){{\varvec{\sigma }}'}^\mathrm{T}_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}) \right] \\&\quad =E\left\{ -\frac{2E(\epsilon ^3|\mathbf{X})}{E(t^2|\mathbf{X})}e^{-\sigma (\mathbf{X},{\varvec{\beta }})} \mathbf{m}'_{{\varvec{\alpha }}} (\mathbf{X},{\varvec{\alpha }}){{\varvec{\sigma }}'}^\mathrm{T}_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}) \right\} . \end{aligned}$$

• 3. $E({\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}{\varvec{S}}_{\mathrm{eff},{\varvec{\beta }}}^\mathrm{T})$ is equivalent to

  $$\begin{aligned} E\left( \frac{4E(t^2|\mathbf{X})}{\{E(t^2|\mathbf{X})\}^2}{\varvec{\sigma }}'_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}){{\varvec{\sigma }}'}^\mathrm{T}_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }})\right) =E\left\{ \frac{4}{E(t^2|\mathbf{X})}{\varvec{\sigma }}'_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }}){{\varvec{\sigma }}'}^\mathrm{T}_{\varvec{\beta }}(\mathbf{X},{\varvec{\beta }})\right\} . \end{aligned}$$

Theorem 1 is the immediate consequence of the above calculations. $\square$