Empirical likelihood method for non-ignorable missing data problems

Abstract

Missing response problem is ubiquitous in survey sampling, medical, social science and epidemiology studies. It is well known that non-ignorable missing is the most difficult missing data problem where the missing of a response depends on its own value. In statistical literature, unlike the ignorable missing data problem, not many papers on non-ignorable missing data are available except for the full parametric model based approach. In this paper we study a semiparametric model for non-ignorable missing data in which the missing probability is known up to some parameters, but the underlying distributions are not specified. By employing Owen (1988)’s empirical likelihood method we can obtain the constrained maximum empirical likelihood estimators of the parameters in the missing probability and the mean response which are shown to be asymptotically normal. Moreover the likelihood ratio statistic can be used to test whether the missing of the responses is non-ignorable or completely at random. The theoretical results are confirmed by a simulation study. As an illustration, the analysis of a real AIDS trial data shows that the missing of CD4 counts around two years are non-ignorable and the sample mean based on observed data only is biased.

Appendix

1.1 Derivation of loglikelihood $\ell (\beta )$ by profiling out $(\eta ,\theta )$

Differentiating $\ell (\varvec{\lambda },\varvec{\nu },\xi )$ w.r.t. $(\eta ,\theta )$ and using (4) and (5) we have

$$\begin{aligned} \frac{\partial \ell (\varvec{\lambda },\varvec{\nu },\xi )}{\partial \eta }= & {} -\frac{n_0}{1-\eta }-\sum _{i=1}^{n_1}\frac{ \varvec{\lambda }^{\mathrm { T} }\frac{\partial }{\partial \eta }\psi (x_i,y_i)}{1+\varvec{\lambda }^{\mathrm { T} }\psi (x_i,y_i)},\\ \frac{\partial \ell (\varvec{\lambda },\varvec{\nu },\xi )}{\partial \theta }= & {} -\sum _{i=1}^{n_1}\frac{\frac{\partial }{\partial \theta }\psi ^{\mathrm { T} }(x_i,y_i)\varvec{\lambda }}{1+\varvec{\lambda }^{\mathrm { T} }\psi (x_i,y_i)}+\sum _{j=n_1+1}^n \frac{ \varvec{\nu }}{1+\varvec{\nu }^{\mathrm { T} }\{\phi (x_j)-\theta \}}. \end{aligned}$$

Equating the above derivatives to zeros and noting $\frac{\partial }{\partial \eta } \psi (x,y;\xi )= (-1, \theta ^{\mathrm { T} }, 0)^{\mathrm { T} }$ and $\frac{\partial }{\partial \theta } \psi ^{\mathrm { T} }(x,y;\xi )= \left( \varvec{0}, -(1-\eta )I, \varvec{0}\right)$ we have

$$\begin{aligned} 0= & {} -\frac{n_0}{1-\eta }+(\lambda _0-\theta ^{\mathrm { T} }\lambda ^{(1)}) \sum _{i=1}^{n_1}\frac{1}{1+\varvec{\lambda }^{\mathrm { T} }\psi (x_i,y_i)}=(\lambda _0-\theta ^{\mathrm { T} }\lambda ^{(1)}){n_1}-\frac{n_0}{1-\eta },\\ 0= & {} \sum _{i=1}^{n_1}\frac{(1-\eta )\varvec{\lambda }^{(1)}}{1+\varvec{\lambda }^{\mathrm { T} }\psi (x_i,y_i)}+\sum _{j=n_1+1}^n \frac{ \varvec{\nu }}{1+\varvec{\nu }^{\mathrm { T} }\{\phi (x_j)-\theta \}} =n_1(1-\eta )\varvec{\lambda }^{(1)} +n_0\varvec{\nu }. \end{aligned}$$

Therefore we have proved (6) which imply

$$\begin{aligned} 1+\varvec{\lambda }^{\mathrm { T} }\psi (x_i,y_i)= & {} \lambda _0[\pi (x, y;\beta )-\eta ]+[1-\pi (x, y;\beta )]\varvec{\lambda }^{(1)}\phi (x)-(1-\eta )\theta ^{\mathrm { T} }\varvec{\lambda }^{(1)}\\&+\varvec{\lambda }_2[\mu (x)-\bar{\mu }]\\= & {} \lambda _0[\pi (x, y;\beta )-1]+[1-\pi (x, y;\beta )]\varvec{\lambda }^{(1)}\phi (x)+\frac{n_0}{n_1}\\&+\varvec{\lambda }_2[\mu (x)-\bar{\mu }]\\= & {} \frac{n_0}{n_1}-\lambda _0[1-\pi (x, y;\beta )]+[1-\pi (x, y;\beta )]\varvec{\lambda }^{(1)}\phi (x)\\&+\varvec{\lambda }_2[\mu (x)-\bar{\mu }]\\= & {} \frac{n_0}{n_1}+\varvec{\lambda }^{\mathrm { T} }\gamma (x, y;\beta ),\\ 1+\varvec{\nu }^{\mathrm { T} }\{\phi (x_j)-\theta \}= & {} 1-\frac{n_1(1-\eta )}{n_0}\{\phi (x_j)-\theta \}^{\mathrm { T} }\varvec{\lambda }^{(1)}\\= & {} 1-\frac{n_1(1-\eta )}{n_0} \phi ^{\mathrm { T} }(x_j)\varvec{\lambda }^{(1)}+\frac{n_1(1-\eta )}{n_0}\theta ^{\mathrm { T} }\varvec{\lambda }^{(1)}\\= & {} 1-\frac{n_1(1-\eta )}{n_0} \phi ^{\mathrm { T} }(x_j)\varvec{\lambda }^{(1)}+\frac{n_1(1-\eta )}{n_0} \lambda _0-1\\= & {} \frac{n_1(1-\eta )}{n_0} \lambda _0-\frac{n_1(1-\eta )}{n_0} \phi ^{\mathrm { T} }(x_j)\varvec{\lambda }^{(1)}\\= & {} \frac{n_1(1-\eta )}{n_0}\varvec{\lambda }_1^{\mathrm { T} }\rho (x_j;\beta ). \end{aligned}$$

Plugging these into $\ell (\varvec{\lambda },\varvec{\nu },\xi )$ we obtain $\ell (\varvec{\lambda },\varvec{\nu },\xi )=\tilde{\ell }(\varvec{\omega })-n_0\log (n_1/n_0)$. So we arrive at (7) by ignoring the parameter-free term. The expressions for $S({\varvec{\omega }})$ and $\hat{p}_i$ follow from the above results.

In rest of this section we present proofs of the Theorems in Sect. 2. We assume that $\eta =\eta _0$ and ${\varvec{\omega }}=(\varvec{\lambda }^{\mathrm { T} },\beta ^{\mathrm { T} })^{\mathrm { T} }={\varvec{\omega }}_0=(\varvec{\lambda }^{\mathrm { T} }_0,\beta ^{\mathrm { T} }_0)^{\mathrm { T} }$. We denote the conditional nonresponse probability by $\bar{\pi }(x, y)=1-\pi (x, y)$. To simplify notations we will suppress the dependence of some quantities on the parameter $\beta$ and even the random variables X and/or Y.

1.2 Proof of Theorem 1

It is easy to prove that when $\bar{\mu }=\mu _0$, $\eta =\eta _0$ and ${\varvec{\omega }}={\varvec{\omega }}_0$, the estimating functions are $S({\varvec{\omega }})=\sum _{i=1}^n z_i +\mathcal{{O}}_P(1)$, where $z_i=g(x_i, y_i, d_i)$, $g=(g_{1}^{\mathrm { T} }, g_{2}^{\mathrm { T} }, g_{3}^{\mathrm { T} })^{\mathrm { T} }$ and

$$\begin{aligned} g_{1}(x,y,d)&= \eta \left( d\left[ \frac{\rho (x)}{\pi (x,y)}+\text{ E }\left\{ \frac{\bar{\pi }(X,Y)\rho (X)}{\eta \pi (X,Y)}\right\} \right] -\rho (x)- \text{ E }\left\{ \frac{\bar{\pi }(X,Y)\rho (X)}{\pi (X,Y)}\right\} \right) ,\\ g_2(x,y,d)&= -\eta \left( d\left[ \frac{\mu (x)-\mu _0}{\pi (x,y)}+\text{ E }\left\{ \frac{\mu (X)-\mu _0}{\eta \pi (X,Y)}\right\} \right] - \text{ E }\left\{ \frac{\mu (X)-\mu _0}{\pi (X,Y)}\right\} \right) , \\ g_3(x,y,d)&= -\frac{1}{\eta }\text{ E }\left\{ \frac{\dot{\pi }_\beta (X,\,Y)}{\pi (X, \,Y)}\right\} (d-\eta ). \end{aligned}$$

It can be shown that in distribution

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n z_i\rightarrow N(0, U), \end{aligned}$$

(9)

where U is the variance-covariance matrix of $Z=g(X,Y,D)$. It is easy to see that when $\bar{\mu }=\mu _0$, $\eta =\eta _0$ and ${\varvec{\omega }}={\varvec{\omega }}_0$, the entries of $U=(u_{ij})_{3\times 3}$ are

$$\begin{aligned} u_{11}= & {} {\eta ^2}\text{ E }\left( \frac{\bar{\pi }}{\pi }\rho \rho ^{\mathrm { T} }\right) +\eta (1-\eta )\text{ E }\left( \frac{\bar{\pi }}{\pi }\rho \right) \text{ E }\left( \frac{\bar{\pi }}{\pi }\rho ^{\mathrm { T} }\right) \nonumber \\&+\eta \left\{ \text{ E }\left( \bar{\pi }\rho \right) \text{ E }\left( \frac{\bar{\pi }}{\pi }\rho ^{\mathrm { T} }\right) +\text{ E }\left( \frac{\bar{\pi }}{\pi }\rho \right) \text{ E }\left( \bar{\pi }\rho ^{\mathrm { T} }\right) \right\} , \end{aligned}$$

(10)

$$\begin{aligned} u_{21}= & {} u_{12}^{\mathrm { T} }= -\eta ^2 \left[ \text{ E }\left\{ \frac{\bar{\pi }}{\pi }(\mu -\mu _0)\rho ^{\mathrm { T} }\right\} +\frac{1-\eta }{\eta } \text{ E }\left( \frac{\mu -\mu _0}{\pi }\right) \right. \nonumber \\&\left. \text{ E }\left( \frac{1-\eta +\pi }{\pi }\bar{\pi }\rho ^{\mathrm { T} }\right) \right] , \end{aligned}$$

(11)

$$\begin{aligned} u_{31}= & {} u_{13}^{\mathrm { T} }=-\text{ E }\left( \frac{\dot{\pi }_\beta }{\pi }\right) \text{ E }\left( \frac{1-\eta +\pi }{\pi }\bar{\pi }\rho ^{\mathrm { T} }\right) , \end{aligned}$$

(12)

$$\begin{aligned} u_{22}= & {} \eta (1-\eta )\text{ E }\left( \frac{\mu -\mu _0}{\pi }\right) \text{ E }\left( \frac{\mu -\mu _0}{\pi }\right) ^{\mathrm { T} }\nonumber \\&+\eta ^2\text{ E }\left\{ \frac{(\mu -\mu _0)(\mu -\mu _0)^{\mathrm { T} }}{\pi }\right\} , \end{aligned}$$

(13)

$$\begin{aligned} u_{32}= & {} u_{23}^{\mathrm { T} }=(1-\eta )\text{ E }\left( \frac{\dot{\pi }_\beta }{\pi }\right) \text{ E }\left( \frac{\mu -\mu _0}{\pi }\right) ^{\mathrm { T} }, \end{aligned}$$

(14)

$$\begin{aligned} u_{33}= & {} \frac{1-\eta }{\eta }\text{ E }\left( \frac{\dot{\pi }_\beta }{\pi }\right) \text{ E }\left( \frac{\dot{\pi }_\beta ^{\mathrm { T} }}{\pi }\right) . \end{aligned}$$

(15)

The entries of the Hessian matrix of $\tilde{\ell }({\varvec{\omega }})$, $H({\varvec{\omega }})=\{H_{ij}({\varvec{\omega }})\}_{3\times 3}$, are

$$\begin{aligned} H_{11}({\varvec{\omega }})= & {} \sum _{i=1}^{n_1}\frac{\bar{\pi }^2(x_i,y_i;\beta )\rho (x_i)\rho ^{\mathrm { T} }(x_i)}{\{n/n_1+\varvec{\lambda }^{\mathrm { T} }\gamma (x_i,y_i)\}^2} +\sum _{j=n_1+1}^n \frac{\rho (x_j)\rho ^{\mathrm { T} }(x_j)}{\{\varvec{\lambda }_{1}^{\mathrm { T} }\rho (x_j;\beta )\}^2}, \\ H_{12}({\varvec{\omega }})= & {} -\sum _{i=1}^{n_1}\frac{\bar{\pi }(x_i,y_i;\beta )\rho (x_i)\{\mu (x_i)-\bar{\mu }\}^{\mathrm { T} }}{\{n/n_1+\varvec{\lambda }^{\mathrm { T} }\gamma (x_i,y_i)\}^2},\\ H_{13}({\varvec{\omega }})= & {} \sum _{i=1}^{n_1}\frac{\left[ \{n/n_1+\varvec{\lambda }^{\mathrm { T} }\gamma (x_i,y_i)\}I+\bar{\pi }(x_i,y_i)\rho (x_i)\varvec{\lambda }_{1}^{\mathrm { T} }\right] }{\{n/n_1+\varvec{\lambda }^{\mathrm { T} }\gamma (x_i,y_i)\}^2}\\&\quad \quad \cdot \Big . \left[ \bar{\pi }(x_i,y_i)\dot{\rho }_{\beta ^{\mathrm { T} }}(x_i)-\rho (x_i)\dot{\pi }_\beta ^{\mathrm { T} }(x_i,y_i)\right] \\&-\sum _{j=n_1+1}^n \frac{\left\{ \varvec{\lambda }_1^{\mathrm { T} }\rho (x_j)I-\rho (x_j)\varvec{\lambda }_{1}^{\mathrm { T} }\right\} \dot{\rho }_{\beta ^{\mathrm { T} }}(x_j)}{\{\varvec{\lambda }_{1}^{\mathrm { T} }\rho (x_j;\beta )\}^2},\\ H_{22}({\varvec{\omega }})= & {} \sum _{i=1}^{n_1}\frac{\{\mu (x_i)-\bar{\mu }\}\{\mu (x_i)-\bar{\mu }\}^{\mathrm { T} }}{\{n/n_1+\varvec{\lambda }^{\mathrm { T} }\gamma (x_i,y_i)\}^2},\\ H_{23}({\varvec{\omega }})= & {} -\sum _{i=1}^{n_1}\frac{\{\mu (x_i)-\bar{\mu }\}\varvec{\lambda }_{1}^{\mathrm { T} }\left[ \bar{\pi }(x_i,y_i)\dot{\rho }_{\beta ^{\mathrm { T} }}(x_i)-\rho (x_i)\dot{\pi }_\beta ^{\mathrm { T} }(x_i,y_i)\right] }{\{n/n_1+\varvec{\lambda }^{\mathrm { T} }\gamma (x_i,y_i)\}^2},\\ H_{33}({\varvec{\omega }})= & {} \sum _{i=1}^{n_1}\frac{\pi (x_i,y_i;\beta )\ddot{\pi }_{\beta \beta }(x_i,y_i;\beta )-\dot{\pi }_{\beta }(x_i,y_i;\beta )\dot{\pi }_{\beta }^{\mathrm { T} }(x_i,y_i;\beta )}{\pi ^2(x_i,y_i,\beta )}\\&-\sum _{i=1}^{n_1}\frac{\left\{ n/n_1+\varvec{\lambda }^{\mathrm { T} }\gamma (x_i,y_i)\right\} \frac{\partial }{\partial \beta ^{\mathrm { T} }}\dot{\gamma }^{\mathrm { T} }_\beta (x_i,y_i) \varvec{\lambda } -\dot{\gamma }^{\mathrm { T} }_\beta (x_i,y_i) \varvec{\lambda }\left\{ \dot{\gamma }^{\mathrm { T} }_\beta (x_i,y_i) \varvec{\lambda }\right\} ^{\mathrm { T} }}{\left\{ n/n_1+\varvec{\lambda }^{\mathrm { T} }\gamma (x_i,y_i)\right\} ^2}\\&+\sum _{j=n_1+1}^n\frac{ \varvec{\lambda }_1^{\mathrm { T} }\rho (x_j;\beta )\frac{\partial }{\partial \beta ^{\mathrm { T} }}\dot{\rho }^{\mathrm { T} }_\beta (x_j)\varvec{\lambda }_1 -\dot{\rho }^{\mathrm { T} }_\beta (x_j)\varvec{\lambda }_1\left\{ \dot{\rho }^{\mathrm { T} }_\beta (x_j)\varvec{\lambda }_1\right\} ^{\mathrm { T} }}{\left\{ \varvec{\lambda }_1^{\mathrm { T} }\rho (x_j;\beta )\right\} ^2}. \end{aligned}$$

It can also be shown that when ${\varvec{\omega }}={\varvec{\omega }}_0$, $\eta =\eta _0$, and either $\bar{\mu }=\mu _0$ or $\bar{\mu }=n^{-1}\sum _{i=1}^n\mu (x_i)$, we have

$$\begin{aligned} V_n\equiv \frac{1}{n}H(\varvec{\omega })=\frac{1}{n}\frac{\partial S({\varvec{\omega }})}{\partial {\varvec{\omega }}} \rightarrow V=(v_{ij})_{3\times 3}, \;\text{ a.s.. } \end{aligned}$$

(16)

Let $\Sigma _\mu =\text{ var }\{\mu (X)\}$. We have

$$\begin{aligned} V= \left( \begin{array}{ll} V_{11} &{} -\eta \text{ E }\left( \frac{1}{\pi }\Phi \dot{\pi }_\beta ^{\mathrm { T} }\right) \\ -\eta \text{ E }\left( \frac{1}{\pi }\dot{\pi }_\beta \Phi ^{\mathrm { T} }\right) &{} \mathbf 0 _{(p+1)\times (p+1)} \\ \end{array} \right) , \end{aligned}$$

(17)

where

$$\begin{aligned} V_{11}\equiv \left( \begin{array}{ll} v_{11} &{} v_{12} \\ v_{21} &{} v_{22} \\ \end{array} \right) =\eta ^2\text{ E }\left( \frac{\bar{\pi }}{\pi }\Phi \Phi ^{\mathrm { T} }\right) +\eta ^2\left( \begin{array}{ll} 0 &{} 0\\ 0 &{} \Sigma _\mu \\ \end{array} \right) . \end{aligned}$$

Clearly, under the conditions of the theorem, matrix V is positive definite. It is also true that

$$\begin{aligned} \sqrt{n}(\hat{{\varvec{\omega }}}-{\varvec{\omega }})=-V_n^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^n z_i+\mathcal{O}_P(n^{-1/2}). \end{aligned}$$

(18)

Therefore, Part (i) of the theorem follows from (9) and (16).

If $\bar{\mu }=\frac{1}{n}\sum _{i=1}^n\mu (x_i)$, $\eta =\eta _0$ and ${\varvec{\omega }}={\varvec{\omega }}_0$, then

$$\begin{aligned}S_3({\varvec{\omega }})=\sum _{i=1}^{n} \left( z_i+z_i^* \right) +\mathcal{{O}}_P(1),\end{aligned}$$

where $z_i^*=[0,\eta \{\mu (x_i)-\mu _0\}^{\mathrm { T} },0]^{\mathrm { T} }$. So the asymptotic variance of $S({\varvec{\omega }})$ in this case is $U+W$, where $W=(w_{ij})_{3\times 3}$ is a symmetric matrix whose entries are

$$\begin{aligned} w_{ij}= & {} 0, \;\; \text{ if } i\ne 2 \hbox { or } j\ne 2, \end{aligned}$$

(19)

$$\begin{aligned} w_{21}= & {} w_{12}^{\mathrm { T} }=\eta ^2\text{ E }\left( \frac{\pi \mu }{\eta }-\mu _0\right) \text{ E }\left( \frac{\bar{\pi }}{\pi }\rho ^{\mathrm { T} }\right) , \end{aligned}$$

(20)

$$\begin{aligned} w_{22}= & {} -\eta ^2\left\{ \text{ E }\left( \frac{\pi \mu }{\eta }-\mu _0\right) \text{ E }\left( \frac{\mu -\mu _0}{\pi }\right) ^{\mathrm { T} }\right. \nonumber \\&\left. +\text{ E }\left( \frac{\mu -\mu _0}{\pi }\right) \text{ E }\left( \frac{\pi \mu }{\eta }-\mu _0\right) ^{\mathrm { T} }\right\} -\eta ^2\text{ var }(\mu ) \end{aligned}$$

(21)

$$\begin{aligned} w_{32}= & {} w_{23}^{\mathrm { T} }= -\eta \text{ E }\left( \frac{\dot{\pi }_\beta }{\pi }\right) \text{ E }\left( \frac{\pi \mu }{\eta }-\mu _0\right) . \end{aligned}$$

(22)

So Part (ii) is also proved.

1.3 Proof of Theorem 2

Expanding $\hat{\mu }_Y$ as a function of $\hat{\varvec{\omega }}$ at ${\varvec{\omega }}={\varvec{\omega }}_0$, we have

$$\begin{aligned} \sqrt{n}(\hat{\mu }_Y-\mu _Y)= & {} \frac{1}{\sqrt{n}}\sum _{i=1}^{n} \left[ \left\{ \frac{d_iy_i}{\pi (x_i,y_i;\beta )}-\mu _Y\right\} +\frac{1}{\eta }\text{ E }\left\{ \frac{Y}{\pi (X,Y;\beta )}\right\} (d_i-\eta )\right] \\&- \frac{1}{\sqrt{n}}K^{\mathrm { T} }(\beta ,\eta ) V^{-1}S({\varvec{\omega }})+\mathcal{O}_P(n^{-1/2}), \end{aligned}$$

where

$$\begin{aligned} K(\beta ,\eta )= \text{ E }\left\{ \frac{Y}{\pi (X,Y;\beta )} \left( \begin{array}{c} \eta \gamma (X,Y;\beta )\\ \dot{\pi }_\beta (X,Y;\beta ) \end{array} \right) \right\} . \end{aligned}$$

Thus if $\bar{\mu }=\mu _0$, $\sqrt{n}(\hat{\mu }_Y-\mu _Y)\rightarrow N(0,\sigma ^2),$ where

$$\begin{aligned} \sigma ^2= & {} \frac{1-\eta }{\eta }\left\{ \text{ E }\left( \frac{Y}{\pi }\right) -\mu _Y\right\} ^2 +\left\{ \text{ E }\left( \frac{Y^2}{\pi }\right) -\frac{\mu _Y^2}{\eta }\right\} \nonumber \\&+K^{\mathrm { T} }(\beta ,\eta )\Sigma K(\beta ,\eta ) \nonumber \\&-2K^{\mathrm { T} }(\beta ,\eta )V^{-1}\left\{ \text{ E }(YZ)+\frac{1}{\eta }\text{ E }\left( \frac{Y}{\pi }\right) \text{ E }\left( {\pi }Z\right) \right\} . \end{aligned}$$

(23)

If $\bar{\mu }=\frac{1}{n}\sum _{i=1}^n\mu (x_i)$, $\sqrt{n}(\hat{\mu }_Y-\mu _Y)\rightarrow N(0,\sigma ^2_*),$ where

$$\begin{aligned} \sigma _*^2= & {} \sigma ^2+K^{\mathrm { T} }(\beta ,\eta )V^{-1}WV^{-1} K(\beta ,\eta ) \nonumber \\&-2K^{\mathrm { T} }(\beta ,\eta )V^{-1}\left[ \text{ E }\{YZ^*\}+\frac{1}{\eta }\text{ E }\left( \frac{Y}{\pi }\right) \text{ E }\left\{ {\pi }Z^*\right\} \right] , \end{aligned}$$

(24)

and $Z^*=[0,\eta \{\mu (X)-\mu _0\},0]^{\mathrm { T} }$. The proof of Theorem 2 is complete.

1.4 Proof of Theorem 3

Without loss of generality, we assume $p=\text{ dim }(\beta _2)=1$ and $r(y)=y$. Under the null hypothesis $H_0$: $\beta _2=0$, $\pi (x,y;\beta ) =\eta _0.$ Replacing $\psi (x,y,\xi )$ by $\psi _1(x,y,\xi )$ and using the above model for $\pi =P(D=1)$, we have, by Taylor expansion, when $\eta =\eta _0$

$$\begin{aligned} \tilde{\ell }(\hat{\varvec{\omega }}^*)=\tilde{\ell }({\varvec{\omega }}^*)-\frac{1}{2n}\tilde{S}^{\mathrm { T} }({\varvec{\omega }}^*)\tilde{V}^{-1}\tilde{S}({\varvec{\omega }}^*)+o_P(1), \end{aligned}$$

(25)

where ${\varvec{\omega }}^*=(\lambda ^{\mathrm { T} }, \beta _1)^{\mathrm { T} }$ takes on $({\mathbf \lambda }^{\mathrm { T} }_0, \beta _{10})^{\mathrm { T} }$, $\tilde{S}({\varvec{\omega }}^*)=\{\tilde{S}_1^{\mathrm { T} }({\varvec{\omega }}^*),\tilde{S}_2({\varvec{\omega }}^*)\}^{\mathrm { T} }$,

$$\begin{aligned} \tilde{S}_1({\varvec{\omega }}^*)= & {} \sum _{i=1}^{n}\left( (d_i-\eta )\left[ \rho (x_i)+\frac{1-\eta }{\eta }\text{ E }\left\{ \rho (X)\right\} \right] \right) +\mathcal{{O}}_P(1),\\ \tilde{S}_2({\varvec{\omega }}^*)= & {} \frac{1-\eta }{\eta } \sum _{i=1}^{n}(d_i-\eta )+\mathcal{{O}}_P(1), \end{aligned}$$

and

$$\begin{aligned}\tilde{V}={\eta (1-\eta )} \left( \begin{array}{ll} A &{} B \\ B^{\mathrm { T} }&{} 0\\ \end{array} \right) ,\quad A=\text{ E }(\rho \rho ^{\mathrm { T} }) ,\quad B=\text{ E }(\rho ). \end{aligned}$$

While using model (1) we have

$$\begin{aligned} \tilde{\ell }(\hat{\varvec{\omega }})=\tilde{\ell }({\varvec{\omega }})-\frac{1}{2n}S^{\mathrm { T} }({\varvec{\omega }})V^{-1}S({\varvec{\omega }})+o_P(1), \end{aligned}$$

(26)

where ${\varvec{\omega }}=(\lambda ^{\mathrm { T} },\beta ^{\mathrm { T} })^{\mathrm { T} }$ takes on $(\lambda ^{\mathrm { T} }_0,\beta ^{\mathrm { T} }_0)^{\mathrm { T} }$ with $\beta _0=(\beta _{10},\beta _{20}=0)^{\mathrm { T} }$,

$$\begin{aligned}S({\varvec{\omega }})= \left( \begin{array}{l} \tilde{S}({\varvec{\omega }}^*) \\ n\mu _Y\frac{1-\eta }{\eta } (\tilde{\eta }-\eta )\\ \end{array} \right) ,\;\;V= \left( \begin{array}{ll} \tilde{V} &{} V_0 \\ V_0^{\mathrm { T} }&{} 0 \\ \end{array} \right) ,\\ V_0=\eta (1-\eta )(C^{\mathrm { T} },0)^{\mathrm { T} },\;\;C^{\mathrm { T} }= \left\{ \mu _Y, -\text{ E }(y\phi ^{\mathrm { T} }) \right\} . \end{aligned}$$

Inverting the partitioned matrix, we obtain

$$\begin{aligned} V^{-1} =\left( \begin{array}{ll} \tilde{V}^{-1} &{}\quad \mathbf 0 _{(d+1)\times 1} \\ \mathbf 0 ^{\mathrm { T} }_{(d+1)\times 1} &{}\quad 0 \\ \end{array} \right) -\frac{1}{\varsigma ^2} \left( \begin{array}{ll} \tilde{V}^{-1}V_0V_0^{\mathrm { T} }\tilde{V}^{-1} &{}\quad -{\tilde{V}^{-1}V_0} \\ -{V_0^{\mathrm { T} }\tilde{V}^{-1}} &{}\quad 1 \\ \end{array} \right) , \end{aligned}$$

where $\varsigma ^2= V_0^{\mathrm { T} }\tilde{V}^{-1}V_0 = \eta (1-\eta ) \left( C^{\mathrm { T} }A^{-1}C-\mu _Y^2\right) .$ It follows from (18) that

$$\begin{aligned} \tilde{S}({\varvec{\omega }})=-n\tilde{V} ({\varvec{\omega }}^*-{\varvec{\omega }})+\mathcal{O}_P(1),\;\; \tilde{\eta }-\eta =-\eta ^2B^{\mathrm { T} }\tilde{\lambda }+\mathcal{O}_P(n^{-1}), \end{aligned}$$

where $\tilde{\eta }=n_1/n$. By (26) and (26) we can get

$$\begin{aligned} 2\{\tilde{\ell }(\hat{\varvec{\omega }})-\tilde{\ell }(\hat{{\varvec{\omega }}}^*)\}= & {} \frac{n}{\varsigma ^2}\left\{ \eta (1-\eta )D^{\mathrm { T} }\tilde{\lambda }\right\} ^2, \end{aligned}$$

where $D=C-\mu _YB$. By Theorem 2, $\sqrt{n}D^{\mathrm { T} }\tilde{\lambda }\rightarrow N(0, \sigma _0^2)$ in distribution, where

$$\begin{aligned} \sigma _0^2= & {} (D^{\mathrm { T} },0)\tilde{V}^{-1}\tilde{U}\tilde{V}^{-1}(D^{\mathrm { T} },0)^{\mathrm { T} },\\ \tilde{U}= & {} \frac{1-\eta }{\eta } \left( \begin{array}{ll} \tilde{A} &{} (1-\eta ) B \\ (1-\eta ) B^{\mathrm { T} }&{} (1-\eta )^2 \\ \end{array} \right) , \;\tilde{A}=A-(1-\eta ^2) \left( \begin{array}{ll} 0 &{} 0\\ 0 &{} \Sigma _\phi \\ \end{array} \right) , \end{aligned}$$

and $\Sigma _\phi =\text{ var }_F(\phi )$. Simple matrix algebra shows that

$$\begin{aligned} \sigma _0^2= & {} \frac{1}{\eta ^3(1-\eta )} D^{\mathrm { T} }A^{-1}\tilde{A}A^{-1}D =\frac{\varsigma ^2}{\{\eta (1-\eta )\}^2}. \end{aligned}$$

Consequently, $R=2\{\tilde{\ell }(\hat{\varvec{\omega }})-\tilde{\ell }(\hat{{\varvec{\omega }}}^*)\}$ converges in distribution to $\chi ^2(p)$ distribution.