Empirical likelihood for semivarying coefficient model with measurement error in the nonparametric part

Abstract

A semivarying coefficient model with measurement error in the nonparametric part was proposed by Feng and Xue (Ann Inst Stat Math 66:121–140, 2014), but its inferences have not been systematically studied. This paper applies empirical likelihood method to construct confidence regions/intervals for the regression parameter and coefficient function. When some auxiliary information about the parametric part is available, the empirical log-likelihood ratio statistic for the regression parameter is introduced based on the corrected local linear estimator of the coefficient function. Furthermore, corrected empirical log-likelihood ratio statistic for coefficient function is also investigated with the use of auxiliary information. The limiting distributions of the resulting statistics both for the regression parameter and coefficient function are shown to have standard Chi-squared distribution. Simulation experiments and a real data set are presented to evaluate the finite sample performance of our proposed method.

Appendix A: Assumptions and proofs

For the convenience and simplicity, let $Z=(X_1,\ldots ,X_n)^\mathrm{T}$, $\varepsilon =(\varepsilon _1,\ldots ,\varepsilon _n)^\mathrm{T}$, $\eta =(\eta _1,\ldots ,\eta _n)^\mathrm{T}$, $\widetilde{Z}=(I-S)Z$, $\widetilde{M}=(I-S)M$, $\widetilde{\varepsilon }=(I-S)\varepsilon$, $\widetilde{\eta }=(I-S)\eta$, $a_n=\{\frac{\log (1/h_1)}{nh_1}\}^{1/2}+h_1^2$ and $C$ denote positive constant whose value may vary at each occurrence. Before proving the main theorems, we begin this section with making the following assumptions.

• (C1) The kernel $K(\cdot )$ is a symmetric probability function with bounded support.

• (C2) The variable $U$ has a bounded support $\fancyscript{U}$ and its density function $p(u)>0$ is Lipschitz continuous and bounded away from zero on $\fancyscript{U}$.

• (C3) The matrixes $\Gamma (u)$ is non-singular. $\Gamma ^{-1}(u)$, $\Phi (u)$ and $E(X_1X_1^\mathrm{T}|U_1=u)$ are all Lipschitz continuous. $E\Vert Z_1\Vert ^{2s}<\infty$, $E\Vert X_1\Vert ^{2s}<\infty$, $E|\varepsilon _1|^{2s}<\infty$ and $E\Vert \eta _1\Vert ^{2s}<\infty$ for some $s>2$, where $\Vert \cdot \Vert$ is the $L_2$ norm.

• (C4) $\{\alpha _j(\cdot ),j=1,\ldots ,q\}$ have continuous second derivative on $\fancyscript{U}$.

• (C3) There exist a $\delta <2-s^{-1}$ such that $\lim _{n\rightarrow \infty }n^{2\delta -1}h_1=\infty .$

• (C4) The bandwidth $h_1$ satisfies that $nh_1^2(\log n)^{-2}\rightarrow \infty$, and $nh_1^8\rightarrow 0$.

• (C5) The bandwidth $h_2$ satisfies that $nh_2\rightarrow \infty$ and $h_2\rightarrow 0.$

Lemma 5.1

Assume that conditions (C1)–(C5) are satisfied. Then we have

$$\begin{aligned}&(D_u^W)^\mathrm{T} w_u D_u^W-\Omega =nf(u)\Gamma (u)\otimes \left( \begin{array}{l@{\quad }l} 1&{} \mu _1\\ \mu _1 &{} \mu _2 \end{array}\right) \{1+O_p(a_n)\},\\&(D_u^W)^\mathrm{T} w_u D_u^X =nf(u)\Phi (u)\otimes (1,\mu _1)\{1+O_p(a_n)\},\\&(D_u^W)^\mathrm{T} w_u D_u^Z =nf(u)\Gamma (u)\otimes (1,\mu _1)\{1+O_p(a_n)\},\\&n^{-1}\sum _{i=1}^n(\widetilde{X}_i\widetilde{X}_i^\mathrm{T}-X^\mathrm{T}Q_i^\mathrm{T}\Sigma _{\eta } Q_iX)\longrightarrow \Delta _1, \quad \hbox {a.s.} \end{aligned}$$

Lemma 5.1 can be proved as Lemmas 2 and 3 in Feng and Xue (2014).

Lemma 5.2

Let $D_1,\ldots ,D_n$ be independent and identical distributed random variables. If $E|D_1|^s<\infty$ for $s>1,$ then $\max _{1\le i\le n}|D_i|=o(n^{1/s})$ a.s.

Lemma 5.2 can be proved as Lemma 3 in Owen (1990).

Lemma 5.3

Under the conditions of Theorem 2.1, if $\beta$ is the true value of the parameter$,$ then we have

$$\begin{aligned} \mathrm{(i)}\quad&\frac{1}{\sqrt{n}}\sum _{i=1}^n\psi _i(\beta ) \mathop {\longrightarrow }\limits ^{\mathcal {D}}N(0,V_{AI}),\\ \mathrm{(ii)}\quad&\frac{1}{n}\sum _{i=1}^n \psi _i(\beta ) \psi _i^\mathrm{T}(\beta )\mathop {\longrightarrow }\limits ^{\mathcal {P}}V_{AI},\\ \mathrm{(iii)}\quad&\max _{1\le i\le n}\Vert \psi _i(\beta )\Vert =o_p(n^{1/2}), \quad \lambda _1=O_p(n^{-1/2}), \end{aligned}$$

where $V_{AI}=\bigg (\begin{array}{ll} V_{A}&{}0\\ 0&{}V_1 \end{array}\bigg ),$ $V_1 =E\big \{\big [E(X_1X_1^\mathrm{T}|U_1)-\Phi ^\mathrm{T}(U_1)\Gamma ^{-1}(U_1)\Phi (U_1)\big ](\varepsilon _1-\eta _1^\mathrm{T}\alpha (U_1))^2\big \} +\sigma ^2E\big \{\Phi ^\mathrm{T}(U_1)\Gamma ^{-1}(U_1)\Sigma _{\eta }\Gamma ^{-1}(U_1)\Phi (U_1)\big \} +E\big \{\Phi ^\mathrm{T}(U_1)\Gamma ^{-1}(U_1)(\eta _1\eta _1^\mathrm{T}-\Sigma _{\eta })\big \}^{\otimes 2}$ and $V_A=E\{A(X_1)A^\mathrm{T}(X_1)\}$.

Proof

From the definition of $\varphi _i(\beta )$ and Lemma 5.1, by simply calculation and the proof in Lemma 4 in Feng and Xue (2014), we have

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n\varphi _i(\beta )&= \frac{1}{\sqrt{n}}\sum _{i=1}^n\big \{\widetilde{X}_i(\widetilde{Z}_i^\mathrm{T}\alpha (U_i)+\widetilde{\varepsilon }_i)-X^\mathrm{T}Q_i^\mathrm{T}\Sigma _{\eta } Q_i(M+\varepsilon )\big \} \nonumber \\&= \frac{1}{\sqrt{n}}\sum _{i=1}^n\big \{[X_i-\Phi ^\mathrm{T}(U_i)\Gamma ^{-1}(U_i)Z_i][\varepsilon _i-\eta _i^\mathrm{T}\alpha (U_i)] \nonumber \\&-\Phi ^\mathrm{T}(U_i)\Gamma ^{-1}(U_i)\eta _i\varepsilon _i+\Phi ^\mathrm{T}(U_i)\Gamma ^{-1}(U_i)(\eta _i\eta _i^\mathrm{T}-\Sigma _{\eta })\alpha (U_i)\big \}+o_p(1) \nonumber \\&:= \frac{1}{\sqrt{n}}\sum _{i=1}^n G_{i}+o_p(1). \end{aligned}$$

(5.1)

It is easy to see that $G_{i}$ is independent and identical distributed with mean zero and $\text{ Var }(G_{i})=V_1$. Thus, by the Slutsky theorem and the central limit theorem, we obtain

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n\varphi _i(\beta )\mathop {\longrightarrow }\limits ^{\mathcal {D}}N(0,V_{1}). \end{aligned}$$

(5.2)

Also, we find $\frac{1}{\sqrt{n}}\sum _{i=1}^nA(X_i)\mathop {\longrightarrow }\limits ^{\mathcal {D}}N(0,V_{A})$ and $\mathrm{Cov}\big (\frac{1}{\sqrt{n}}\sum _{i=1}^nA(X_i), \frac{1}{\sqrt{n}}\sum _{i=1}^n\varphi _i(\beta )\big ) \rightarrow 0$ by (5.1), which together with (5.2) and the central limit theorem yields Lemma 5.3(i).

As to Lemma 5.3(ii), observe that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n \psi _i(\beta )\psi _i^\mathrm{T}(\beta )= \frac{1}{n}\sum _{i=1}^n\left( \begin{array}{ll} A(X_i)A^\mathrm{T}(X_i)&{} A(X_i)\varphi _i^\mathrm{T}(\beta )\\ \varphi _i(\beta )A^\mathrm{T}(X_i)&{}\varphi _i(\beta )\varphi _i^\mathrm{T}(\beta ) \end{array}\right) . \end{aligned}$$

The law of large numbers implies $\frac{1}{n}\sum _{i=1}^nA(X_i)A^\mathrm{T}(X_i)\mathop {\longrightarrow }\limits ^{\mathcal {P}}V_A.$ On the other hand, by Lemma 5.1 and the proof in (5.1), it follows that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n \varphi _i(\beta )\varphi _i^\mathrm{T}(\beta )&= \frac{1}{n} \sum _{i=1}^n \big \{\big [X_i-\Phi ^\mathrm{T}(U_i)\Gamma ^{-1}(U_i)Z_i][\varepsilon _i-\eta _i^\mathrm{T}\alpha (U_i)]\\&-\Phi ^\mathrm{T}(U_i)\Gamma ^{-1}(U_i)\eta _i\varepsilon _i+\Phi ^\mathrm{T}(U_i)\Gamma ^{-1}(U_i) (\eta _i\eta _i^\mathrm{T}-\Sigma _{\eta })\alpha (U_i)\big \}^{\otimes 2}\\&+\,o_p(1)\\&= V_1+o_p(1). \end{aligned}$$

Similar to the proof of (5.1), we can derive

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^nA(X_i)\varphi _i^\mathrm{T}(\beta )=\frac{1}{n}\sum _{i=1}^nA(X_i)G^\mathrm{T}_i+o_p(1)\mathop {\longrightarrow }\limits ^{\mathcal {P}}0, \end{aligned}$$

follows from $E\Vert n^{-1}\sum _{i=1}^nA(X_i)G^\mathrm{T}_i\Vert ^2=n^{-1}E\Vert A(X_i)G_i^\mathrm{T}\Vert ^2\rightarrow 0$ and Markov inequality. Thus, Lemma 5.3(ii) holds.

As to Lemma 5.3(iii), note that $A(X_i)$ is i.i.d. and $E\{A(X_i)A^\mathrm{T}(X_i)\}$ is positive definite, then by Lemma 5.2, we have $\max _{1\le i\le n}\Vert A(X_i)\Vert =o(n^{1/2})$ $a.s.$. From (5.1) and standard calculations, when $n$ is large enough, we can derive that

$$\begin{aligned}&\max _{1\le i\le n}\Vert \varphi _i(\beta )\Vert \le C\Big (\max _{1\le i\le n} \Vert X_i\Vert +\max _{1\le i\le n}\Vert \Phi ^\mathrm{T}(U_1)\Gamma ^{-1}(U_1)(Z_i+\eta _i)\Vert \Big )\\&\qquad \times \Big (\max _{1\le i\le n}|\varepsilon _i|+\max _{1\le i\le n}\Vert X_i+\eta _i\Vert + \max _{1\le i\le n}|Z_i^\mathrm{T}\alpha (U_i)|+\max _{1\le i\le n}|\eta _i^\mathrm{T}\alpha (U_i)|\Big )\\&\quad =o_p(n^{1/2}). \end{aligned}$$

Thus, $\max _{1\le i\le n}\Vert \psi _i(\beta )\Vert =o_p(n^{1/2})$. Furthermore, one can obtain that $\lambda _1=O_p(n^{-1/2})$ by Lemma 5.3(ii) and using the arguments similar to Owen (1990). The proof of Lemma 5.1 is thus completed. $\square$

Set $\lambda _2=(\lambda _{21}^\mathrm{T},\lambda _{22}^\mathrm{T})^\mathrm{T}$, where $\lambda _{21}$ and $\lambda _{22}$ are $r$-dimensional and $q$-dimensional column vectors.

Lemma 5.4

Under the conditions of Theorem 2.3, for a given $u,$ if $\alpha (u)$ is the true value of the parameter$,$ then

$$\begin{aligned} \mathrm{(i)}\quad&\sum _{i=1}^n\widehat{\xi }_i(\alpha (u))\mathop {\longrightarrow }\limits ^{\mathcal {D}}N(0,\Sigma _{AI}(u)),\\ \mathrm{(ii)}\quad&\frac{1}{n}\sum _{i=1}^n \widehat{\xi }_i(\alpha (u)) \widehat{\xi }^\mathrm{T}_i(\alpha (u))\mathop {\longrightarrow }\limits ^{\mathcal {P}}\Sigma _{AI}(u),\\ \mathrm{(iii)}\quad&\max _{1\le i\le n}\Vert \widehat{\xi }_i(\alpha (u))\Vert =o_p(1),\quad \lambda _{21}=O_p(n^{-1/2}),\quad \lambda _{22}=O_p((nh_2)^{-1/2}), \end{aligned}$$

where $\widehat{\xi }_i(\alpha (u))=\bigg (\begin{array}{l} \frac{1}{\sqrt{n}}A(X_i)\\ \frac{1}{\sqrt{nh_2}}\widehat{\zeta }_i(\alpha (u)) \end{array}\bigg )$, $\Sigma _{AI}(u)=\bigg (\begin{array}{ll} V_{A}&{}0\\ 0&{}\Sigma _1(u) \end{array}\bigg )$ and $\Sigma _1(u)=\Big \{(\Gamma (u)+\Sigma _{\eta })E(\varepsilon _1^2|U_1=u)+\alpha ^\mathrm{T}(u)\Sigma _{\eta }\alpha (u)\Gamma (u)\Big \}f(u)\int K^2(x)\mathrm{d}x.$

Proof

Observe that

$$\begin{aligned} \frac{1}{\sqrt{nh_2}}\sum _{i=1}^n\hat{\zeta }_i(\alpha (u))&= \frac{1}{\sqrt{nh_2}}\sum _{i=1}^n \big [\varepsilon _i(Z_i+\eta _i)+\eta _i^\mathrm{T}\alpha (U_i)Z_i\big ]K\Big (\frac{U_i-u}{h_2}\Big ) \nonumber \\&+\frac{1}{\sqrt{nh_2}}\sum _{i=1}^nX_i^\mathrm{T}(\beta -\hat{\beta })W_iK\Big (\frac{U_i-u}{h_2}\Big )\nonumber \\&+ \frac{1}{\sqrt{nh_2}}\sum _{i=1}^nW_i^\mathrm{T}(\tilde{\alpha }(u)-\alpha (u))W_iK\Big (\frac{U_i-u}{h_2}\Big ) \nonumber \\&+\frac{1}{\sqrt{nh_2}}\sum _{i=1}^nZ_i^\mathrm{T}(\alpha (U_i)-\tilde{\alpha }(U_i))W_iK\Big (\frac{U_i-u}{h_2}\Big )\nonumber \\&-\frac{1}{\sqrt{nh_2}}\sum _{i=1}^n(\eta _i\eta _i^\mathrm{T}-\Sigma _{\eta })\tilde{\alpha }(U_i)K\Big (\frac{U_i-u}{h_2}\Big ) \nonumber \\&-\frac{1}{\sqrt{nh_2}}\sum _{i=1}^n\Sigma _{\eta }(\tilde{\alpha }(U_i)-\alpha (U_i))K\Big (\frac{U_i-u}{h_2}\Big )\nonumber \\&+\frac{1}{\sqrt{nh_2}}\sum _{i=1}^n\eta _i^\mathrm{T}(\alpha (U_i)-\tilde{\alpha }(U_i))Z_iK\Big (\frac{U_i-u}{h_2}\Big ) \nonumber \\&:= \sum _{i=1}^7B_{in}. \end{aligned}$$

(5.3)

Note that $\big \{\big [\varepsilon _i(Z_i+\eta _i)+\eta _i^\mathrm{T}\alpha (U_i)Z_i\big ]K\big (\frac{U_i-u}{h_2}\big ),1\le i\le n\big \}$ is independent and identical distributed with mean zero and

$$\begin{aligned} \text{ Var }(B_{1n})&= h_2^{-1}E\bigg \{\Big [E(\varepsilon _1^2|U_1)\big [E(Z_1Z_1^\mathrm{T}|U_1)+\Sigma _{\eta }\big ]\\&+E\big [\alpha ^\mathrm{T}(U_1)\Sigma _{\eta }\alpha (U_1)|U_1\big ]E(Z_1Z_1^\mathrm{T}|U_1) \Big ]K^2\Big (\frac{U_i-u}{h_2}\Big )\bigg \}\\&= \Big \{(\Gamma (u)+\Sigma _{\eta })E(\varepsilon _1^2|U_1=u)+\alpha ^\mathrm{T}(u)\Sigma _{\eta }\alpha (u)\Gamma (u)\Big \}f(u)\int K^2(x)\mathrm{d}x\\&= \Sigma _1(u). \end{aligned}$$

Then we have $B_{1n}\mathop {\longrightarrow }\limits ^{\mathcal {D}}N(0,\Sigma _1(u))$.

Theorems 2.2 and 2.4, Lemma 5.2 and condition (C3) imply that $\hat{\beta }-\beta =O_p(n^{-1/2})$, $\tilde{\alpha }(u)-\alpha (u)=O_p(n^{-1/2})$ and $\eta _i\eta _i^\mathrm{T}-\Sigma _{\eta }=o(n^{1/s})$ $a.s.$ Then by some simple calculations we obtain $B_{in}=o_p(1)$ for $i=2,\ldots ,7.$ Invoking the Slutsky theorem and (5.3), we get

$$\begin{aligned} \frac{1}{\sqrt{nh_2}}\sum _{i=1}^n\hat{\zeta }_i(\alpha (u))\mathop {\longrightarrow }\limits ^{\mathcal {D}}N(0,\Sigma _1(u)). \end{aligned}$$

(5.4)

Note that $\frac{1}{\sqrt{n}}\sum _{i=1}^nA(X_i)\mathop {\longrightarrow }\limits ^{\mathcal {D}}N(0,V_{A})$ and $\mathrm{Cov}\big (\frac{1}{\sqrt{n}}\sum _{i=1}^nA(X_i), \frac{1}{\sqrt{nh_2}}\sum _{i=1}^n\hat{\zeta }_i(\alpha (u))\big ) \rightarrow 0$, which together with (5.4) and the central limit theorem leads to Lemma 5.4(i).

Analogously to the proof of Lemma 5.3(ii), we can verify Lemma 5.4(ii) easily. As to Lemma 5.4(iii), we find

$$\begin{aligned} \max _{1\le i\le n}\Vert \widehat{\zeta }_i(\alpha (u))\Vert&\le \max _{1\le i\le n}\Big \Vert \big [\varepsilon _i(Z_i+\eta _i)+\eta _i^\mathrm{T}\alpha (U_i)Z_i\big ]K\Big (\frac{U_i-u}{h_2}\Big )\Big \Vert \\&+\max _{1\le i\le n}\Big \Vert X_i^\mathrm{T}(\beta -\hat{\beta }_\mathrm{ME})W_iK\Big (\frac{U_i-u}{h_2}\Big )\Big \Vert \\&+ \max _{1\le i\le n}\Big \Vert W_i^\mathrm{T}(\tilde{\alpha }(u)-\alpha (u))W_iK\Big (\frac{U_i-u}{h_2}\Big )\Big \Vert \\&+\max _{1\le i\le n}\Big \Vert Z_i^\mathrm{T}(\alpha (U_i)-\tilde{\alpha }(U_i))W_iK\Big (\frac{U_i-u}{h_2}\Big )\Big \Vert \\&+\max _{1\le i\le n}\Big \Vert (\eta _i\eta _i^\mathrm{T}-\Sigma _{\eta })\tilde{\alpha }(U_i)K\Big (\frac{U_i-u}{h_2}\Big )\Big \Vert \\&+\max _{1\le i\le n}\Big \Vert \Sigma _{\eta }(\tilde{\alpha }(U_i)-\alpha (U_i))K\Big (\frac{U_i-u}{h_2}\Big )\Vert \\&+\max _{1\le i\le n}\Big \Vert \eta _i^\mathrm{T}(\alpha (U_i)-\tilde{\alpha }(U_i))Z_iK\Big (\frac{U_i-u}{h_2}\Big )\Big \Vert \\&:= \sum _{i=1}^7J_{in}. \end{aligned}$$

From Markov inequality and conditions (C3) and (C5), one can obtain that

$$\begin{aligned} P(J_{1n}\ge \sqrt{nh_2})&\le (nh_2)^{-s}\sum _{i=1}^n E\Big [\Big (\varepsilon _i(Z_i+\eta _i)+\eta _i^\mathrm{T}\alpha (U_i)Z_i\Big )K\Big (\frac{U_i-u}{h_2}\Big )\Big ]^{2s}\\&\le C(nh_2)^{1-s}\rightarrow 0, \end{aligned}$$

which implies that $D_1=o_p(\sqrt{nh_2})$. Similarly, by $\hat{\beta }-\beta =O_p(n^{-1/2})$, $\tilde{\alpha }(u)-\alpha (u)=O_p(n^{-1/2})$ and $\eta _i\eta _i^\mathrm{T}-\Sigma _{\eta }=o(n^{1/s})$ $a.s.$, it can be shown that $J_{in}=o_p(\sqrt{nh_2})$ $i=2,\ldots ,7$. Therefore we obtain that $\max _{1\le i\le n}\Vert \widehat{\zeta }_i(\alpha (u))\Vert =o_p((nh_2)^{1/2})$, which together with $\max _{1\le i\le n}\Vert A(X_i)\Vert =o(n^{1/2})$ $a.s.$, gives $\max _{1\le i\le n}\Vert \widehat{\xi }_i(\alpha (u))\Vert =o_p(1)$.

Applying Lemma 5.4(ii) and the proof in Owen (1990) one can derive that $\lambda _{21}=O_p(n^{-1/2})$ and $\lambda _{22}=O_p((nh_2)^{-1/2})$, which completes the proof of Lemma 5.4. $\square$

Proof of Theorem 2.1

Applying the Taylor expansion to (2.8) and invoking Lemma 5.3, we obtain that

$$\begin{aligned} {\mathcal {L}}_{1n,AI}(\beta _0)=2\sum _{i=1}^n\{\lambda _1^\mathrm{T}\psi _i(\beta _0)-[\lambda _1^\mathrm{T}\psi _i(\beta _0)]^2/2\}+o_p(1). \end{aligned}$$

From (2.9), we have

$$\begin{aligned} 0&= \frac{1}{n}\sum _{i=1}^n\frac{\psi _i(\beta _0)}{1+\lambda _1^\mathrm{T}\psi _i(\beta _0)}\\&= \frac{1}{n}\sum _{i=1}^n\psi _i(\beta _0)-\frac{1}{n}\sum _{i=1}^n\psi _i(\beta _0)\psi ^\mathrm{T}_i(\beta _0)\lambda _1 +\frac{1}{n}\sum _{i=1}^n\frac{\psi _i(\beta _0)[\lambda _1^\mathrm{T} \psi _i(\beta _0)]^2}{1+\lambda _1^\mathrm{T}\psi _i(\beta _0)}. \end{aligned}$$

Using Lemma 5.3, we find

$$\begin{aligned} \left\| \frac{1}{n}\sum _{i=1}^n\frac{\psi _i(\beta _0)[\lambda _1^\mathrm{T}\psi _i(\beta _0)]^2}{1+\lambda _1^\mathrm{T}\psi _i(\beta _0)}\right\|&\le \frac{1}{n}\sum _{i=1}^n\frac{\Vert \psi _i(\beta _0)\Vert ^3\Vert \lambda _1\Vert ^2}{|1+\lambda _1^\mathrm{T} \psi _i(\beta _0)|}\\&\le \Vert \lambda _1\Vert ^2\max _{1\le i\le n}\Vert \psi _i(\beta _0)\Vert \frac{1}{n}\sum _{i=1}^n\Vert \psi _i(\beta _0)\Vert ^2\\&= O_p(n^{-1})o_p(n^{1/2})O_p(1)=o_p(n^{-1/2}). \end{aligned}$$

Then $\sum _{i=1}^n[\lambda _1^\mathrm{T} \psi _i(\beta _0)]^2=\sum _{i=1}^n\lambda _1^\mathrm{T} \psi _i(\beta _0)+o_p(1),$ and

$$\begin{aligned} \lambda _1=\left[ \sum _{i=1}^n\psi _i(\beta _0)\psi _i^\mathrm{T}(\beta _0)\right] ^{-1}\sum _{i=1}^n\psi _i(\beta _0) +o_p(n^{-1/2}). \end{aligned}$$

Thus,

$$\begin{aligned} {\mathcal {L}}_{1n,AI}(\beta _0)&= \left( \frac{1}{\sqrt{n}}\sum _{i=1}^n\psi _i(\beta _0)\right) ^\mathrm{T} \left( \frac{1}{n}\sum _{i=1}^n\psi _i(\beta _0)\psi ^\mathrm{T}_i(\beta _0)\right) ^{-1} \left( \frac{1}{\sqrt{n}}\sum _{i=1}^n\psi _i(\beta _0)\right) \\&+\,o_p(1). \end{aligned}$$

This together with Lemma 5.3 completes the proof. $\square$

Proof of Theorem 2.2

Note that $\hat{\beta }_\mathrm{ME}$ and $\hat{\lambda }_1=\lambda (\hat{\beta }_\mathrm{ME})$ satisfy $H_{1n}(\hat{\beta }_\mathrm{ME},\hat{\lambda }_1)=0$ and $H_{2n}(\hat{\beta }_\mathrm{ME},\hat{\lambda }_1)=0$, where

$$\begin{aligned} H_{1n}(\beta ,\lambda _1)&= \frac{1}{n}\sum _{i=1}^n\frac{\varphi _i(\beta )}{1+\lambda _1^\mathrm{T}\varphi _i(\beta )} \quad \text{ and } \\ H_{2n}(\beta ,\lambda _1)&= \frac{1}{n}\sum _{i=1}^n\frac{1}{1+\lambda _1^\mathrm{T}\varphi _i(\beta )}\left( \frac{\partial \varphi _i(\beta )}{\partial \beta ^\mathrm{T}}\right) ^\mathrm{T}\lambda _1. \end{aligned}$$

Then by expanding $H_{1n}(\hat{\beta }_\mathrm{ME},\hat{\lambda }_1)=0$ and $H_{2n}(\hat{\beta }_\mathrm{ME},\hat{\lambda }_1)=0$ at $(\beta _0,0)$, we derive that

$$\begin{aligned} 0&= H_{1n}(\hat{\beta }_\mathrm{ME},\hat{\lambda }_1)=H_{1n}(\beta _0,0)+\frac{\partial H_{1n}(\beta _0,0)}{\partial \beta ^\mathrm{T}}(\hat{\beta }_\mathrm{ME}-\beta _0)\\&+\,\frac{\partial H_{1n}(\beta _0,0)}{\partial \lambda ^\mathrm{T}_1}(\hat{\lambda }-0)+o_p(\delta _n),\\ 0&= H_{2n}(\hat{\beta }_\mathrm{ME},\hat{\lambda }_1)=H_{2n}(\beta _0,0)+\frac{\partial H_{2n}(\beta _0,0)}{\partial \beta ^\mathrm{T}}(\hat{\beta }_\mathrm{ME}-\beta _0)\\&+\,\frac{\partial H_{2n}(\beta _0,0)}{\partial \lambda ^\mathrm{T}_1}(\hat{\lambda }_1-0)+o_p(\delta _n), \end{aligned}$$

where $\delta _n=\Vert \hat{\beta }_\mathrm{ME}-\beta _0\Vert +\Vert \hat{\lambda }_1\Vert$. Then, we find

$$\begin{aligned} \left( \begin{array}{l} \hat{\lambda }_1\\ \hat{\beta }_\mathrm{ME}-\beta _0 \end{array}\right)&= \left( \begin{array}{ll} \frac{\partial H_{1n}(\beta ,\lambda _1)}{\partial \lambda ^\mathrm{T}_1} &{} \frac{\partial H_{1n}(\beta ,\lambda _1)}{\partial \beta ^\mathrm{T}}\\ \frac{\partial H_{2n}(\beta ,\lambda _1)}{\partial \lambda ^\mathrm{T}_1} &{} \frac{\partial H_{2n}(\beta ,\lambda _1)}{\partial \beta ^\mathrm{T}} \end{array}\right) _{(\beta _0,0)}^{-1} \left( \begin{array}{l} -H_{1n}(\beta _0,0)\!+\!o_p(\delta _n)\\ -H_{2n}(\beta _0,0)\!+\!o_p(\delta _n) \end{array}\right) \\&= \left( \begin{array}{ll} -\frac{1}{n}\sum _{i=1}^n \varphi _i(\beta _0)\varphi ^\mathrm{T}_i(\beta _0) &{} \frac{1}{n}\sum _{i=1}^n\frac{\partial \varphi _i(\beta _0)}{\partial \beta ^\mathrm{T}}\\ \frac{1}{n}\sum _{i=1}^n\frac{\partial \varphi _i(\beta _0)}{\partial \beta ^\mathrm{T}} &{} 0 \end{array}\right) ^{-1} \left( \begin{array}{l} -H_{1n}(\beta _0,0)+o_p(\delta _n)\\ -H_{2n}(\beta _0,0)+o_p(\delta _n) \end{array}\right) \\&= \left( \begin{array}{ll} -\frac{1}{n}\sum _{i=1}^n \varphi _i(\beta _0)\varphi ^\mathrm{T}_i(\beta _0) &{} -\widehat{\Delta }_1\\ -\widehat{\Delta }_1 &{} 0 \end{array}\right) ^{-1} \left( \begin{array}{l} -H_{1n}(\beta _0,0)+o_p(\delta _n)\\ -H_{2n}(\beta _0,0)+o_p(\delta _n) \end{array}\right) \end{aligned}$$

where $\widehat{\Delta }_1=\frac{1}{n}\sum _{i=1}^n(\widetilde{X}_i\widetilde{X}_i^\mathrm{T}-X^\mathrm{T} Q_i^\mathrm{T} \Sigma _{\eta } Q_i X)$. Lemma 5.3 and $H_{1n}(\beta _0,0)=n^{-1}\sum _{i=1}^n\varphi _i(\beta _0)=O_p(n^{-1/2})$ imply $\delta _n=O_p(n^{-1/2})$. Therefore,

$$\begin{aligned} \sqrt{n}(\hat{\beta }_\mathrm{ME}-\beta _0)=\widehat{\Delta }_1^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^n\varphi _i(\beta _0)+o_p(1). \end{aligned}$$

This together with (5.3), Lemmas 5.1 and Slutsky theorem yields the result of Theorem 2.2. $\square$

Theorem 2.3 can be proved by using the same argument used in Theorems 2.1. Theorem 2.4 can be verified by the proof of Theorem 3 in Feng and Xue (2014) and the asymptotic normalities of $\tilde{\beta }$ and $\hat{\beta }_\mathrm{ME}$. We omit the details here.