An empirical likelihood check with varying coefficient fixed effect model with panel data

Abstract

Semiparametric models are often used to analyze panel data for a good trade-off between parsimony and flexibility. In this paper, we investigate a fixed effect model with a possible varying coefficient component. On the basis of empirical likelihood method, the coefficient functions are estimated as well as their confidence intervals. The estimation procedures are easily implemented. An important problem of the statistical inference with the varying coefficient model is to check the constant coefficient about the regression functions. We further develop checking procedures by constructing empirical likelihood ratio statistics and establishing the Wilks theorems. Finally, some numerical simulations and a real data analysis is presented to assess the finite sample performance.

Appendix A. Proof of the lemmas and theorems.

It is to start with some preliminary results about the naive empirical likelihood ratio. The proof of the theorems can be obtained by extending some of them, which is shown as the following lemmas and proof for them.

Lemma 1

Under some conditions (C1)-(C6), then when $n\rightarrow \infty$, we have

$$\begin{aligned} \text {E}\Big [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]=b(t_0)+o_p(1), \end{aligned}$$

and

$$\begin{aligned} \text {Cov} \Big [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ] = \Sigma ^*(t_0)+o_p(1). \end{aligned}$$

where $\Sigma ^*(t_0)=\nu ^2_0(t_0)\Gamma (t_0).$

Proof

Denote ${\tilde{\zeta }}(t_0) = \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))$ and its l-th component being ${\tilde{\zeta }}_l(t_0)= \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_{i,l}(\theta (t_0)).$ Hence, together with the Eqs. (2.5) and (2.6), it is shown that

$$\begin{aligned} \begin{aligned} {\tilde{\eta }}_{i,l}(\theta (t_0))&=\sum _{j=1}^{m_i}\sum _{k=1}^p \Big \{X_{ij,l}^*X_{ij,k}^{*T}\Big [\theta _k(t_{ij}) -\theta _k(t_0)\Big ]+X_{ijl}^*\epsilon _{ij}\Big \}K_h(t_{ij}-t_0)\\&\triangleq \sum _{j=1}^{m_i} {\tilde{\zeta }}_{il}(t_{ij},t_0)K_h(t_{ij}-t_0). \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} E[{\tilde{\eta }}_{il}(\theta (t_0))] =m_ih\sum _{k=1}^p\int \Big [\theta _k(hu-t_0)-\theta _k(t_0)\Big ]\times \gamma _{lk}(hu-t_0)f(hu-t_0)K(u)du. \end{aligned} \end{aligned}$$

By the conditions (C1) and Taylor expansion, an algebra calculation of the right side of the above formula leads to

$$\begin{aligned} \text {E}\Big [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]=b(t_0)+o_p(1). \end{aligned}$$

For the second item in Lemma 1, the (l, k)-th component of the covariance matrix is

$$\begin{aligned} \text {Cov}\Big [{\tilde{\zeta }}_l(t_0),{\tilde{\zeta }}_k(t_0)\Big ] =E\Big [{\tilde{\zeta }}_l(t_0){\tilde{\zeta }}_k(t_0)\Big ]-E\Big [{\tilde{\zeta }}_l(t_0)\Big ]E\Big [{\tilde{\zeta }}_k(t_0)\Big ]. \end{aligned}$$

So after plugging the notation of ${\tilde{\zeta }}(t_0)$, the first term on the right side is

$$\begin{aligned} \begin{aligned}&E\bigg \{ \bigg [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_{il}(\theta (t_0)) \bigg ] \bigg [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_{ik}(\theta (t_0)) \bigg ]\bigg \}\\&\quad =\frac{1}{Nh} \bigg \{ \sum _{i=1}^n E[{\tilde{\eta }}_{il}(\theta (t_0)){\tilde{\eta }}_{ik}(\theta (t_0))]+ \sum _{i_1 \ne i_2} E [{\tilde{\eta }}_{i_1l}(\theta (t_0)){\tilde{\eta }}_{i_2k}(\theta (t_0))]\bigg \}. \end{aligned} \end{aligned}$$

(A.1)

As to the first term in (A.1),

$$\begin{aligned} \begin{aligned} {\tilde{\eta }}_{i,l}(\theta (t_0)){\tilde{\eta }}_{i,k}(\theta (t_0))&=\sum _{j=1}^{m_i}{\tilde{\zeta }}_{i,l}(t_{ij},t_0){\tilde{\zeta }}_{i,k}(t_{ij},t_0) K_h^2(t_{ij}-t_0)\\&\quad + \sum _{j_1 \ne j_2}{\tilde{\zeta }}_{i,l}(t_{ij_1},t_0){\tilde{\zeta }}_{i,k}(t_{ij_2},t_0)K_h(t_{ij_1}-t_0)K_h(t_{ij_2}-t_0). \end{aligned} \end{aligned}$$

(A.2)

And as $n\rightarrow \infty$, a straightforward calculation leads to that

$$\begin{aligned} \begin{aligned}&E\Big [{\tilde{\zeta }}_{i,l}(t_{ij},t_0){\tilde{\zeta }}_{i,k}(t_{ij},t_0)|t_{ij}=t\Big ] \\&\quad = \sum _{m=1}^p\bigg \{ \Big [ \theta _m(t)-\theta _m(t_0)\Big ]^2 \times E\Big [(X_{ij,l}^*X_{ij,m}^{*})(X_{ij,k}^*X_{ij,m}^{*})|t_{ij}=t\Big ]\bigg \}\\&\qquad +\sigma ^2(t)E\Big [(X_{ij,l}^*X_{ij,r}^{*})|t_{ij}=t\Big ]+\sum _{m_1 \ne m_2}\bigg \{\Big [ \theta _{m_1}(t)-\theta _{m_1}(t_0)\Big ] \Big [ \theta _{m_2}(t)-\theta _{m_2}(t_0)\Big ] \\&\qquad \times E\Big [(X_{ij,l}^*X_{ij,m_1}^{*})(X_{ij,k}^*X_{ij,m_2}^{*})|t_{ij}=t\Big ]\bigg \}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}&E\bigg [ \sum _{j=1}^{m_i}{\tilde{\zeta }}_{il}(t_{ij},t_0){\tilde{\zeta }}_{ik}(t_{ij},t_0)K_h^2(t_{ij}-t_0)\bigg ]\\&\quad =\sum _{j=1}^{m_i}\int E\Big [{\tilde{\zeta }}_{il}(t_{ij},t_0){\tilde{\zeta }}_{ik}(t_{ij},t_0)|t_{ij}=t\Big ]K_h^2(t_{ij}-t)f(t)dt\\&\quad ={m_i}h\sigma ^2(t_0)\gamma _{lr}(t_0)f(t_0)\int K^2(t)dt+o_p(h). \end{aligned} \end{aligned}$$

(A.3)

With a similar derivation of the second term in (A.2), we can obtain

$$\begin{aligned} E\Big [{\tilde{\zeta }}_{i,l}(t_{ij_1},t_0){\tilde{\zeta }}_{i,k}(t_{ij_2},t_0)|t_{ij_1}=t_1,t_{ij_2}=t_2\Big ]\rightarrow \rho _{\epsilon }(t_0) \gamma _{lr}(t_0), \end{aligned}$$

and a further expectation follows that

$$\begin{aligned} \begin{aligned}&E\bigg [ \sum _{j_1 \ne j_2}{\tilde{\zeta }}_{i,l}(t_{ij_1},t_0){\tilde{\zeta }}_{i,k}(t_{ij_2},t_0)K_h(t_{ij_1}-t_0)K_h(t_{ij_2}-t_0)\bigg ]\\&\quad ={m_i}({m_i}-1)h^2\rho _{\epsilon }(t_0)\gamma _{lr}(t_0)f^2(t_0)+o_p({m_i}({m_i}-1)h^2). \end{aligned} \end{aligned}$$

(A.4)

Combining (A.3) and (A.4) with (A.2), when $n\rightarrow \infty$, we have

$$\begin{aligned} \begin{aligned}&(Nh)^{-1}\sum _{i=1}^nE\Big [{\tilde{\eta }}_{i,l}(\theta (t_0)){\tilde{\eta }}_{i,k}(\theta (t_0))\Big ]\\&\quad =\sigma ^2(t_0)\gamma _{lk}(t_0) f(t_0)\int K^2(t)dt+ h^*\gamma _{lk}(t_0)\rho _{\epsilon }(t_0)f^2(t_0)+o_p(h^*). \end{aligned} \end{aligned}$$

(A.5)

where $h^*=hN^{-1}(\sum _{i=1}^nm_i^2-N)$. With $h=N^{-1/5}h_0$ and $\lim _{n\rightarrow \infty }N^{-6/5}\sum _{i=1}^n m_i^2=\lambda$, we have

$$\begin{aligned} hN^{-1}\bigg (\sum _{i=1}^nm_i^2-N\bigg )\rightarrow \lambda h_0,~~(n\rightarrow \infty ). \end{aligned}$$

Following a similar context of (A.13) in Wu et al. (1998), it follows that

$$\begin{aligned}&\bigg |(Nh)^{-1}\sum _{{i_1}\ne i_2} E\Big [{\tilde{\eta }}_{i,l}(\theta (t_0)){\tilde{\eta }}_{i,k}(\theta (t_0))\Big ]-E\Big [ (Nh)^{-1/2} \sum _{i=1}^n{\tilde{\eta }}_{i,l}(\theta (t_0))\Big ] \nonumber \\&\qquad \times E\Big [ (Nh)^{-1/2} \sum _{i=1}^n{\tilde{\eta }}_{i,k}(\theta (t_0))\Big ]\bigg |\rightarrow 0. \end{aligned}$$

(A.6)

Sum up (A.5), (A.6), the second result in Lemma 1 is concluded. $\square$

Lemma 2

Under conditions ($\text {C1}$)–(C6) hold, then

$$\begin{aligned} \max _{1\le i\le n} \Vert {\tilde{\eta }}_i(\theta (t_0))\Vert =o_p((Nh)^{-\frac{1}{2}}). \end{aligned}$$

Proof

The proof is similar to that one of Lemma A.1 in Xue and Zhu (2007). $\square$

Lemma 3

Suppose that conditions (C1)–(C6) hold, and for a given $t_0$,

$$\begin{aligned} {\sqrt{Nh}}\Big ({\tilde{\theta }}_{\text {NAEL}}(t_0)-\theta (t_0)\Big )-B(t_0)\xrightarrow {{\mathcal {L}}} N(0,\Sigma (t_0)), \end{aligned}$$

(A.7)

where $B(t_0)=f(t_0)^{-1}\Gamma ^{-1}(t_0)b(t_0)$, $\Sigma (t_0)=f(t_0)^{-2}\nu ^2(t_0)\Gamma ^{-1}(t_0)$, $\Gamma (t_0)$ is defined in condition (C6), $b(t_0)=(b_0(t_0), \ldots ,b_p(t_0))$, $b_l(t_0)$ and $\nu ^2(t_0)$ are defined by

$$\begin{aligned} b_l(t_0)= & {} h_0^{5/2}\sum _{k=1}^p[\theta _k'(t_0)\gamma _{lk}'(t_0)f(t_0)+\theta _k'(t_0)\gamma _{lk}(t_0)f'(t_0) \\&+(1/2)\theta _k''(t_0)\gamma _{lk}(t_0)f(t_0)]\int u^2K(u)du, \end{aligned}$$

and

$$\begin{aligned} \nu ^2(t_0)=\sigma ^2(t_0)f(t_0)\int K^2(t)dt+\lambda h_0 \rho _{\epsilon }(t_0)f^2(t_0), \end{aligned}$$

where $h_0$ and $\lambda$ are defined in conditions (C1) and (C2).

Proof

With a similar proof of (A.1) in Wu et al. (1998) and a straightforward calculation, we can derive that

$$\begin{aligned} {\tilde{\theta }}_{\text {NAEL}}(t_0)-\theta (t_0) = (f(t_0))^{-1}\Gamma ^{-1}(t_0)\bigg [\frac{1}{\sqrt{Nh}}\sum _{i=1}^n {\tilde{\eta }}_i(\theta (t_0))\bigg ]+o_p(1). \end{aligned}$$

(A.8)

By Lemma 1, we can derive that

$$\begin{aligned} \frac{1}{\sqrt{Nh}}\sum _{i=1}^n {\tilde{\eta }}_i(\theta (t_0))\xrightarrow {{\mathcal {L}}}N(b(t_0),\Sigma ^*(t_0)). \end{aligned}$$

(A.9)

Invoking (A.8) and (A.9), we can show the result in Lemma 3 . $\square$

Lemma 4

Suppose that conditions (C2)–(C6) hold and $Nh\rightarrow \infty$, $h=o(N^{-1/5})$, if $\theta (t_0)$ is the true parameter, then

$$\begin{aligned} {\tilde{\ell }}(\theta (t_0)){\mathop {\longrightarrow }\limits ^{{\mathcal {L}}}}\chi _p^2, \end{aligned}$$

(A.10)

where ${\mathop {\longrightarrow }\limits ^{{\mathcal {L}}}}$ stands for convergence in distribution and $\chi _p^2$ means the chi-squared distribution with freedom p.

Proof

It is easy to see that the proof of Lemma 4 requires the following three asymptotic results as

$$\begin{aligned}&\frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\sim N(0,\Sigma ^*(t_0)),\end{aligned}$$

(A.11)

$$\begin{aligned}&\frac{1}{Nh}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0)){\tilde{\eta }}_i^T(\theta (t_0))\rightarrow \Sigma ^*(t_0), \end{aligned}$$

(A.12)

$$\begin{aligned}&{\tilde{\ell }} (\theta (t)) \approx \Big [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]^T {\tilde{D}}(\theta (t_0))^{-1}\Big [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]. \end{aligned}$$

(A.13)

Firstly, by Lemma 1, when $Nh\rightarrow \infty$, $Nh^5\rightarrow 0$, we can obtain (A.11).

Secondly, according to the proof of (A.6) in Lemma 1, we have (A.12).

Following the Taylor expansion of $\frac{1}{Nh}\sum _{i=1}^n\frac{{\tilde{\eta }}_i(\theta (t_0))}{1+\lambda ^T {\tilde{\eta }}_i(\theta (t_0))}=0$, then

$$\begin{aligned} \begin{aligned} 0&=\sum _{i=1}^n\frac{{\tilde{\eta }}_i(\theta (t_0))}{1+\lambda ^T {\tilde{\eta }}_i(\theta (t_0))}\\&=\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))-\sum _{i=1}^n \lambda ^T {\tilde{\eta }}_i(\theta (t_0)){\tilde{\eta }}_i(\theta (t_0))+ \sum _{i=1}^n\frac{{\tilde{\eta }}_i(\theta (t_0))[\lambda ^T {\tilde{\eta }}_i(\theta (t_0))]^2}{1+\lambda ^T {\tilde{\eta }}_i(\theta (t_0))}. \end{aligned} \end{aligned}$$

(A.14)

And by the proof of (2.16) in Owen (1998), the last term, on the right side of the foregoing equation, has a norm bounded by

$$\begin{aligned} \Vert \lambda \Vert ^2\Vert {\tilde{\eta }}_i(\theta (t_0))\Vert ^3 |1+\lambda {\tilde{\eta }}_i(\theta (t_0))|^{-1}= o_p((Nh)^{\frac{1}{2}})O_p((Nh))^{-1}O_p(1)=o_p((Nh)^{-\frac{1}{2}}). \end{aligned}$$

Then,

$$\begin{aligned} \lambda =\Big [\sum _{i=1}^n{\tilde{\eta }}_i^T(\theta (t_0)){\tilde{\eta }}_i(\theta (t_0))\Big ]^{-1} \sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))+o_p((Nh)^{\frac{1}{2}}). \end{aligned}$$

(A.15)

Again, applying the Taylor expansion to (2.8), we obtain that

$$\begin{aligned} \begin{aligned} {\tilde{\ell }} (\theta (t_0))&=2\sum _{i=1}^n\log (1+\lambda ^T{\tilde{\eta }}_i(\theta (t_0)))\\&=2\sum _{i=1}^n\lambda ^T{\tilde{\eta }}_i(\theta (t_0))-\sum _{i=1}^n[\lambda ^T {\tilde{\eta }}_i(\theta (t_0))]^2+o_p(1)\\&=2\lambda Nh\Big [\frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]^T{\tilde{D}}(\theta (t_0))^{-1} \Big [\frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]\\ {}&~~- Nh\Big [\frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]^T{\tilde{D}}(\theta (t_0))^{-1} {\tilde{D}}(\theta (t_0)){\tilde{D}}(\theta (t_0))^{-1} \\&\qquad \Big [\frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]+o_p(1). \end{aligned} \end{aligned}$$

(A.16)

Following the proof of (A.12) in Wu et al. (1998), we obtain that ${\tilde{D}}(\theta (t_0))\xrightarrow {{\mathcal {P}}}\Sigma ^*(t_0)$. And taking together (A.8), (A.9) and (A.10), it can be proven that

$$\begin{aligned} {\tilde{\ell }} (\theta (t_0)) \approx \Big [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]^T {\tilde{D}}(\theta (t_0))^{-1}\Big [ \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\tilde{\eta }}_i(\theta (t_0))\Big ]\xrightarrow {{\mathcal {L}}}\chi _p^2. \end{aligned}$$

(A.17)

$\square$

To prove Theorem 4, we show the following Lemma 5, which can be verified similarly to the proof of Lemma 7.2 in Zhu et al. (2007).

Lemma 5

Suppose that conditions (C1)-(C6) hold, then under $H_{1n}$,

• (a) if $n^\gamma C_n\rightarrow$a (a is a constant), $0\le \gamma <\frac{1}{2}$, $\sqrt{n}{\hat{\psi }}_1\rightarrow \infty$, as $n \rightarrow \infty$;

• (b) if $n^{\frac{1}{2}} C_n\rightarrow 1$, $\sqrt{n}{\tilde{\psi }}_1\xrightarrow {{\mathcal {L}}}N(\mu ,\sigma ^2)$, where $\mu = E\left\{ W(X, \theta )X^{(2)T}\theta ^{(2)}(t)\right\}$.

$\square$

Proof of Theorem 1

Firstly, by a proof similar to that of Lemma 1, we can derive that

$$\begin{aligned} \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\hat{\eta _i}}(\theta (t_0))\sim N(0,\Sigma ^*(t_0)). \end{aligned}$$

(A.18)

Specially, according to the estimator ${\hat{\theta }}_{\text {RAEL}}(t_0)$, we have

$$\begin{aligned} {\hat{\theta }}_{\text {RAEL}}(t_0)-\theta (t_0)={\hat{V}}^{-1}(t_0)\frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\hat{\eta _i}}(\theta (t_0)). \end{aligned}$$

Furthermore, together with the proof context of Lemma 3, it is shown that

$$\begin{aligned} {\hat{\theta }}_{\text {RAEL}}(t_0)-\theta (t_0) = (f(t_0))^{-1}\Gamma ^{-1}(t_0)\bigg [\frac{1}{\sqrt{Nh}}\sum _{i=1}^n {\hat{\eta }}_i(\theta _i(t_0))\bigg ]+o_p(1). \end{aligned}$$

(A.19)

Hence, this thereupon completes the proof of Theorem 1. $\square$

Proof of Theorem 2

To prove that under some regularity conditions,

$$\begin{aligned} {\hat{\ell }}(\theta (t_0))\sim \chi _p^2. \end{aligned}$$

Now, we have

$$\begin{aligned} \begin{aligned}&\frac{1}{\sqrt{nh}}\sum _{i=1}^n{\hat{\eta }}_i(\theta (t_0)) \\&\quad = \frac{1}{\sqrt{nh}}\sum _{i=1}^n\sum _{j=1}^mX_{ij}^*\Big \{Y_{ij}^*-X_{ij}^{*T}\theta (t_0)-X_{ij}^{*T}\big [{\tilde{\theta }}_{\text {NAEL}}(t_{ij})- {\tilde{\theta }}_{\text {NAEL}}(t_0)\big ]\Big \}K_h(t_{ij}-t_0)\\&\quad =\frac{1}{\sqrt{nh}}\sum _{i=1}^n\sum _{j=1}^mX_{ij}^*\epsilon _{ij}+\frac{1}{\sqrt{nh}}\sum _{i=1}^n\sum _{j=1}^m X_{ij}^*X_{ij}^{*T}\big [\theta (t_{ij})-\theta (t_0)\big ]K_h(t_{ij}-t_0)\\&\qquad -\frac{1}{\sqrt{nh}}\sum _{i=1}^n\sum _{j=1}^mX_{ij}^{*}\Big \{ X_{ij}^{*T}\big [{\tilde{\theta }}_{\text {NAEL}}(t_{ij})- {\tilde{\theta }}_{\text {NAEL}}(t_0)\big ]\Big \}K_h(t_{ij}-t_0)\\&\quad =I_1+I_2+I_3. \end{aligned} \end{aligned}$$

(A.20)

By the Taylor expansion and $\theta _l'(t_0)-{\hat{\theta _l}}'(t_0)\xrightarrow {P}0$, we derive $I_2+I_3\xrightarrow {P}0.$ So, together with the fact that

$$\begin{aligned} \frac{1}{\sqrt{Nh}}\sum _{i=1}^n{\hat{\eta _i}}(\theta (t_0))\sim N(0,\Sigma ^*(t_0)). \end{aligned}$$

(A.21)

and

$$\begin{aligned} \frac{1}{Nh}\sum _{i=1}^n{\hat{\eta _i}}(\theta (t_0)){\hat{\eta _i^T}}(\theta (t_0))\rightarrow \Sigma ^*(t_0), \end{aligned}$$

(A.22)

Theorem 2 can be derived along with the argument in the proof in Lemma 4. $\square$

Proof of Theorem 3

From the fixed effect-corrected technique in Sect. 2, we have $E\big [Y_{ij}^*-X_{ij}^{^*T}\theta |X_{ij}^*\big ] = 0$ and further assume a auxiliary vector as

$$\begin{aligned} \psi _i = \sum _{j=1}^{m_i} w(X_{ij}^*,\theta )(Y_{ij}-X_{ij}^{^*T}\theta ). \end{aligned}$$

Denote an estimator for $\theta$ be ${\tilde{\theta }}_n$, ${\tilde{w}}_{ij} \triangleq w(X_{ij}^*,{\tilde{\theta }}_n)$ and $w_{ij} \triangleq w(X_{ij}^*,\theta )$, then

$$\begin{aligned} {\tilde{\psi }}_i= \sum _{j=1}^{m_i} {\tilde{w}}_{ij}(Y_{ij}-X_{ij}^{^*T}{\tilde{\theta }}_n). \end{aligned}$$

(A.23)

Consider

$$\begin{aligned} \begin{aligned} \frac{1}{\sqrt{N}} \sum _{i=1}^{n}{\tilde{\psi }}_i&= \frac{1}{\sqrt{N}} \sum _{i=1}^{n}\sum _{j=1}^{m_i} {\tilde{w}}_{ij}(Y_{ij}-X_{ij}^{^*T}{\tilde{\theta }}_n)\\&= \frac{1}{\sqrt{N}} \sum _{i=1}^{n}\sum _{j=1}^{m_i} \big [ \varepsilon _{ij} + X_{ij}^{^*T}(\theta _0-{\tilde{\theta }}_n)\big ] \big [{\tilde{w}}_{ij} + {\tilde{w}}_{ij}'(\theta _0-{\tilde{\theta }}_n)\big ]\\&= \frac{1}{\sqrt{N}} \sum _{i=1}^{n}\sum _{j=1}^{m_i}\big [ {\tilde{w}}_{ij}\varepsilon _{ij}-{\tilde{w}}_{ij}X_{ij}^{^*T}(\theta _0-{\tilde{\theta }}_n)\big ]+o_P(1). \end{aligned} \end{aligned}$$

By using a similar proof of the Lemma 7.1 in Zhu et al. (2007), we have

$$\begin{aligned} \frac{1}{\sqrt{N}} \sum _{i=1}^{n} {\tilde{\psi }}_i\rightarrow N(0, \sigma ^2), \end{aligned}$$

(A.24)

where $\sigma ^2 =\lim _{n\rightarrow \infty }n^{-1}\sum _{i=1}^{n}\sum _{j=1}^{m_i} E \big [ {\tilde{w}}_{ij}\varepsilon _{ij}-{\tilde{w}}_{ij}X_{ij}^{^*T}L(X_{ij}^*,\theta _0)\big ]^2$ and $L(X_{ij}^*,\theta _0)$ is the asymptotical expression of $(\theta _n-{\tilde{\theta }}_0)$ proposed in Zhu et al. (2007). Then, by the law of Large Numbers, we can obtain that

$$\begin{aligned} \frac{1}{ N} \sum _{i=1}^{n} {\tilde{\psi }}_i{\tilde{\psi }}_i^T\rightarrow \sigma _1^2, \end{aligned}$$

(A.25)

where $\sigma _1^2 = \lim _{N\rightarrow \infty }N^{-1}\sum _{i=1}^{n}\sum _{j=1}^{m_i} E ( {\tilde{w}}_{ij}\varepsilon _{ij})^2$.

Together with the standard procedure in the empirical likelihood literature, one can find that

$$\begin{aligned} \lambda = O_P(N^{-\frac{1}{2}}). \end{aligned}$$

(A.26)

Then combined with with (A.25), we derive that

$$\begin{aligned} \begin{aligned} {\tilde{R}}_n&= \sum _{i=1}^n \lambda ^T {\tilde{\psi }}_i {\tilde{\psi }}_i^T\lambda +o_p(1)\\&=(n^{-1/2}\sum _{i=1}^{n} {\tilde{\psi }}_i)^T(n^{-1}\sum _{i=1}^n{\tilde{\psi }}_i{\tilde{\psi }}_i^T)^{-1} (n^{-1/2}\sum _{i=1}^{n} {\tilde{\psi }}_i)+o_p(1)\\&=\frac{\sigma ^2}{\sigma _1^{2}}(\sigma ^{-1}n^{-1/2} \sum _{i=1}^{n} {\tilde{\psi }}_i)^2+o_p(1). \end{aligned} \end{aligned}$$

(A.27)

Therefore, by (A.24) and (A.27), we have

$$\begin{aligned} {\tilde{R}}_n \rightarrow \frac{\sigma ^2}{\sigma _1^{2}}\chi ^2_{1}, \end{aligned}$$

(A.28)

where $\chi ^2_{1}$ is a random variable of chi-squared distribution with degree freedom of 1 . $\square$

Proof of Theorem 4

According to the definition of ${\hat{w}}_{ij}$ in Sect. 3, we have

$$\begin{aligned} {\hat{w}}_{ij} = 1 - \sum _{u=1}^n\sum _{v=1}^{m_i}\Big [K_h(t_{uv}-t_{ij})X_{uv}^{*T}\Big ] \Big [\sum _{u=1}^n\sum _{v=1}^{m_i}K_h(t_{uv}-t_{ij})X_{uv}^*X_{uv}^{*T}\Big ]^{-1}X_{ij}^*. \end{aligned}$$

Then,

$$\begin{aligned} \begin{aligned} \frac{1}{\sqrt{N}}\sum _{i=1}^n{\hat{\psi _i}}&= \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{j=1}^{m_i}{\hat{w}}_{ij}(y_{ij}-X_{ij}^{*T}\theta _n)\\&= \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{j=1}^{m_i}{\hat{w}}_{ij}\varepsilon _{ij}-{\hat{w}}_{ij}X_{ij}^{*T}(\theta _n-\theta _0)+o_p(1). \end{aligned} \end{aligned}$$

(A.29)

By the feature of ${\hat{w}}_{ij}$ being orthogonal to $X_{ij}^*$, the second term in (A.29) tends to 0 as $n\rightarrow \infty$. And by the CLT, we can easily conclude that

$$\begin{aligned} \frac{1}{\sqrt{N}}\sum _{i=1}^n{\hat{\psi _i}} \rightarrow N(0,\sigma _1^2). \end{aligned}$$

(A.30)

Furthermore, invoking the proof in Theorem 3, one can show that

$$\begin{aligned} \frac{1}{N}\sum _{i=1}^n{\hat{\psi _i{\hat{\psi }}_i^T}} \rightarrow \sigma _1^2~~\text {and}~~\lambda = O_p(n^{-1/2}). \end{aligned}$$

Hence, together with the decomposition

$$\begin{aligned} {\hat{R}}_n =(\sigma _1^{-1}\frac{1}{\sqrt{N}}\sum _{i=1}^{n} {\hat{\psi _i}})^2+o_p(1), \end{aligned}$$

we can prove the Theorem 4. $\square$

Proof of Theorem 5

By Lemma 5, similar to the proof of Theorem 4, we can obtain Theorem 5. $\square$