Jackknife empirical likelihood inference for the accelerated failure time model

Abstract

Accelerated failure time (AFT) model is a useful semi-parametric model under right censoring, which is an alternative to the commonly used proportional hazards model. Making statistical inference for the AFT model has attracted considerable attention. However, it is difficult to compute the estimators of regression parameters due to the lack of smoothness for rank-based estimating equations. Brown and Wang (Stat Med 26(4):828–836, 2007) used an induced smoothing approach, which smooths the estimating functions to obtain point and variance estimators. In this paper, a more computationally efficient method called jackknife empirical likelihood (JEL) is proposed to make inference for the accelerated failure time model without computing the limiting variance. Results from extensive simulation suggest that the JEL method outperforms the traditional normal approximation method in most cases. Subsequently, two real data sets are analyzed for illustration of the proposed method.

Appendix A: Proofs of Theorems

To derive the asymptotic properties of $l(\beta _0)$ and $l^*(\beta _{10})$, we assume some regularity conditions hold.

• (C.1) X is bounded, that is, $P(\left\| X \right\| \le M) = 1$ for some $0<M<\infty$.

• (C.2) The conditional distribution ${F_{{e_1}(\beta )\left| {{X_1}} \right. }}(t)$ of ${e_1}(\beta ) = {Y_1} - {\beta ^T}{X_1}$ given $X_1$ is twice continuously differentiable in t for all X.

• (C.3) For any X, the conditional density function ${{F'}_{{e_1}(\beta )\left| {{X_1}} \right. }}(t) = {f_{{e_1}(\beta )\left| {{X_1}} \right. }}(t) > 0$ for t in a neighborhood of 0.

Firstly, we re-express the smoothed rank estimating function ${{ S}_n}(\beta )$ in (2.2) as a U-statistic with a symmetric kernel function.

$$\begin{aligned} {{S}_n}(\beta )&= \sum \limits _{i = 1}^n {\sum \limits _{j = 1}^n {{\Delta _i}({X_i} - {X_j})\Phi \left[ {\frac{{{e_j}(\beta ) - {e_i}(\beta )}}{{{r_{ij}}}}} \right] } } \\&= \sum \limits _{1 \le i< j \le n} {{\Delta _i}({X_i} - {X_j})\Phi \left[ {\frac{{{e_j}(\beta ) - {e_i}(\beta )}}{{{r_{ij}}}}} \right] }\\&\quad +\, \sum \limits _{1 \le j< i \le n} {{\Delta _i}({X_i} - {X_j})\Phi \left[ {\frac{{{e_j}(\beta ) - {e_i}(\beta )}}{{{r_{ij}}}}} \right] } \\&= \sum \limits _{1 \le i< j \le n} {{\Delta _i}({X_i} - {X_j})\Phi \left[ {\frac{{{e_j}(\beta ) - {e_i}(\beta )}}{{{r_{ij}}}}} \right] }\\&\quad +\, \sum \limits _{1 \le i < j \le n} {{\Delta _j}({X_j} - {X_i})\Phi \left[ {\frac{{{e_i}(\beta ) - {e_j}(\beta )}}{{{r_{ji}}}}} \right] }, \end{aligned}$$

because of $r_{ij}^2 = {({X_i} - {X_j})^T}({X_i} - {X_j})/n$, one has that ${r_{ij}} = {r_{ji}}$,

$$\begin{aligned} {{ S}_n}(\beta )&= \sum \limits _{1 \le i< j \le n} {({X_i} - {X_j})\left\{ {{\Delta _i}\Phi \left[ {\frac{{{e_j}(\beta ) - {e_i}(\beta )}}{{{r_{ij}}}}} \right] - {\Delta _j}\Phi \left[ {\frac{{{e_i}(\beta ) - {e_j}(\beta )}}{{{r_{ij}}}}} \right] } \right\} } \\&= \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) \left[ {{{\left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) }^{ - 1}}\sum \limits _{1 \le i < j \le n} {K({Z_i},{Z_j};\beta )} } \right] \\&\equiv \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) S_n^*(\beta ), \end{aligned}$$

where $S_n^*(\beta )$ is a U-statistic of degree 2

$$\begin{aligned} S_n^*(\beta ) = {\left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) ^{ - 1}}\sum \limits _{1 \le i < j \le n} {K({Z_i},{Z_j};\beta )} \equiv {U_n}(\beta ), \end{aligned}$$

with the kernel function

$$\begin{aligned} K({Z_i},{Z_j};\beta ) = ({X_i} - {X_j})\left\{ {{\Delta _i}\Phi \left[ {\frac{{{e_j}(\beta ) - {e_i}(\beta )}}{{{r_{ij}}}}} \right] - {\Delta _j}\Phi \left[ {\frac{{{e_i}(\beta ) - {e_j}(\beta )}}{{{r_{ij}}}}} \right] } \right\} . \end{aligned}$$

Similarly, we can also derive ${{\tilde{S}}_n}(\beta )$ in (2.1) as a U-statistic with a symmetric kernel function, that is,

$$\begin{aligned} {{\tilde{S}}_n}(\beta )&= \sum \limits _{i = 1}^n {\sum \limits _{j = 1}^n {{\Delta _i}({X_i} - {X_j})I[{e_j}(\beta ) \ge {e_i}(\beta )]} ,} \\&= \sum \limits _{1 \le i< j \le n} {{\Delta _i}({X_i} - {X_j})I[{e_j}(\beta ) \ge {e_i}(\beta )]} + \sum \limits _{1 \le j< i \le n} {{\Delta _i}({X_i} - {X_j})I[{e_j}(\beta ) \ge {e_i}(\beta )]} \\&= \sum \limits _{1 \le i< j \le n} {{\Delta _i}({X_i} - {X_j})I[{e_j}(\beta ) \ge {e_i}(\beta )]} + \sum \limits _{1 \le i< j \le n} {{\Delta _j}({X_j} - {X_i})I[{e_i}(\beta ) \ge {e_j}(\beta )]} \\&= \sum \limits _{1 \le i< j \le n} {({X_i} - {X_j})\left\{ {{\Delta _i}I[{e_j}(\beta ) \ge {e_i}(\beta )] - {\Delta _j}I[{e_i}(\beta ) \ge {e_j}(\beta )]} \right\} } \\&= \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) \left[ {{{\left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) }^{ - 1}}\sum \limits _{1 \le i < j \le n} {H({Z_i},{Z_j};\beta )} } \right] \\&\equiv \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) {{W}_n}(\beta ), \end{aligned}$$

with the kernel function

$$\begin{aligned} H({Z_i},{Z_j};\beta ) = ({X_i} - {X_j})\left\{ {{\Delta _i}I[{e_j}(\beta ) \ge {e_i}(\beta )] - {\Delta _j}I[{e_i}(\beta ) \ge {e_j}(\beta )]} \right\} . \end{aligned}$$

Fygenson and Ritov (1994) pointed out that when evaluated that ${W_n}(\beta _0 )$ is asymptotically normal and has expectation zero. Furthermore, by (A.7) in Appendix of Johnson and Strawderman (2009), we have the asymptotically equivalence of ${{ U}_n}(\beta _0 )$ to ${W_n}(\beta _0 )$, $\sqrt{n} \left\| {{{ U}_n}(\beta _0 ) - {W_n}(\beta _0 )} \right\| \xrightarrow {p} 0$, that is,

$$\begin{aligned} {U_n}(\beta _0 ) = {W_n}(\beta _0 ) + {o_p}({n^{ - 1/2}}). \end{aligned}$$

(A.1)

Then, $E{U_n}(\beta _0 ) = E{W_n}(\beta _0 ) + E\left[ {{o_p}({n^{ - 1/2}})} \right]$. Hence, we can have $E{U_n}({\beta _0}) \rightarrow 0$ as $n \rightarrow \infty$.

Before proving Theorem 2.1, we display similar notations like Li et al. (2016). Define

$$\begin{aligned} \left\{ \begin{array}{l} {{{\hat{V}}}_i}(\beta ) = n{W_n}(\beta ) - (n - 1)W_{n - 1}^{( - i)}(\beta ),\,i = 1,\ldots ,n,\\ {W_n}(\beta ) = \frac{1}{n}\sum \limits _{i = 1}^n {{{{\hat{V}}}_i}(\beta )},\\ G(\beta ) = \frac{1}{n}\sum \limits _{i = 1}^n {{{{\hat{V}}}_i}(\beta ){\hat{V}}_i^T(\beta )}, \\ G^*(\beta ) = \frac{1}{n}\sum \limits _{i = 1}^n {{{{\hat{Q}}}_i}(\beta ){\hat{Q}}_i^T(\beta )}, \\ \phi (z,\beta ) = {({\phi _1}(z,\beta ),\ldots ,{\phi _{_p}}(z,\beta ))^T} = EH(z,{Z_1};\beta ),\\ \psi (x,y,\beta ) = H(x,y;\beta ) - \phi (x,\beta ) - \phi (y,\beta ), \\ g(z,\beta ) = {({g_1}(z,\beta ),\ldots ,{g_{_p}}(z,\beta ))^T} = 2\phi (z,\beta ),\\ \sigma _l^2(\beta ) = Var({\phi _l}({Z_1},\beta )),l = 1,\ldots ,p,\\ \sigma _{st}^2(\beta ) = Cov({\phi _s}({Z_1},\beta ),{\phi _t}({Z_1},\beta )),s,t = 1,\ldots ,p,\\ \Sigma _{p \times p}^{(\beta )}:\mathrm{the\ asymptotic\ variance - covariance\ matrix\ of} \sqrt{n} {W_n}(\beta ), \\ \mathrm{with}\,\mathrm{elements}\,4{\sigma _{st}}(\beta ),\,s,\,t = 1,\ldots ,p. \end{array} \right. \end{aligned}$$

Under conditions (C.1)–(C.3), following Jing et al. (2009) and Li et al. (2016), we will prove Lemmas A.1 to  A.5.

Lemma A.1

Under conditions (C.1)–(C.3), as $n \rightarrow \infty$, one has

$$\begin{aligned} \sqrt{n} {U_n}(\beta _0 ) \xrightarrow {d} N(0,\Sigma _{p \times p}^{(\beta _0 )}). \end{aligned}$$

Proof

From Li et al. (2016), we can conclude that $\sqrt{n} {{W}_n}(\beta _0 )$ tends to have a normal distribution with mean 0 and covariance $\Sigma _{p \times p}^{(\beta _0 )}$. Then, as $n \rightarrow \infty$, by (A.1), we can derive that

$$\begin{aligned} \begin{aligned} E[\sqrt{n} {U_n}({\beta _0})]&= E[\sqrt{n} ({W_n}({\beta _0}) + {o_p}({n^{ - 1/2}}))]\\&= E[\sqrt{n} {W_n}({\beta _0}) + {o_p}(1)] = E[\sqrt{n} {W_n}({\beta _0})] + {o_p}(1) \rightarrow 0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \mathrm{cov}\left( \sqrt{n} {U_n}(\beta _0 )\right)&= \mathrm{cov}\left( \sqrt{n} \left( {W_n}(\beta _0 ) + {o_p}\left( {n^{ - 1/2}}\right) \right) \right) = \mathrm{cov}\left( \sqrt{n} {W_n}(\beta _0 ) + {o_p}(1)\right) \\&= \mathrm{cov}\left( \sqrt{n} {W_n}(\beta _0 )\right) + 2\mathrm{cov}\left( \sqrt{n} {W_n}(\beta _0 ),{o_p}(1)\right) + {\mathrm{cov}} ({o_p}(1))\\&= \mathrm{cov}\left( \sqrt{n} {W_n}(\beta _0 )\right) + 2\left[ E\left( \sqrt{n} {W_n}(\beta _0 ) \times {o_p}(1)\right) - E\left( \sqrt{n} {W_n}(\beta _0 )\right) \right. \\&\quad \left. \times E({o_p}(1)) \right] + {o_p}(1) = \Sigma _{p \times p}^{(\beta _0 )} + {o_p}(1) \rightarrow \Sigma _{p \times p}^{(\beta _0 )}. \end{aligned}$$

Thus, Lemma A.1 holds. $\square$

Lemma A.2

Under conditions (C.1)–(C.3), with probability tending to one as $n \rightarrow \infty$, the zero vector is contained in the interior of the convex hull of $\left\{ {{{{\hat{Q}}}_1}(\beta _0 ),\ldots ,{{\hat{Q}}_n}(\beta _0 )} \right\}$.

Proof

Combining the Hoeffding decomposition, the proof of Lemma A.2 in Owen (1990) and Li et al. (2016), we complete the proof of Lemma A.2. $\square$

Lemma A.3

Under conditions (C.1)–(C.3), one has ${G^*}(\beta _0 ) = \Sigma _{p \times p}^{({\beta _0})} + o(1)$, a.s.

Proof

Combining Lemma A.1 in Li et al. (2016) and strong law of large numbers for U-statistics, we get ${W_n}(\beta _0)=o(1) \ a.s$. For $l=1,\ldots ,r$, let $\sigma _{H,l}^2(\beta _0) = Var({H_l}({Z_1},{Z_2};{\beta _0}))$. Since $E[H_l^2({Z_1},{Z_2};\beta _0 )] < \infty$, $\sigma _{H,l}^2(\beta _0)< \infty$. As a result,

$$\begin{aligned} \begin{aligned} G(\beta _0 )&= \frac{1}{n}\sum \limits _{i = 1}^n {{{\hat{V}}_i}(\beta _0 ){\hat{V}}_i^T(\beta _0 )} \\&= \frac{1}{n}\sum \limits _{i = 1}^n {\left[ {{{{\hat{V}}}_i}(\beta _0 ) - {W_n}(\beta _0 ) + {W_n}(\beta _0 )} \right] {{\left[ {{{\hat{V}}_i}(\beta _0 ) - {W_n}(\beta _0 ) + {W_n}(\beta _0 )} \right] }^T}} \\&= \frac{1}{n}\sum \limits _{i = 1}^n {\left[ {{{{\hat{V}}}_i}(\beta _0 ) - {W_n}(\beta _0 )} \right] {{\left[ {{{{\hat{V}}}_i}(\beta _0 ) - {W_n}(\beta _0 )} \right] }^T}} + {W_n}(\beta _0 )W_n^T(\beta _0 )\\&= \frac{1}{n}\sum \limits _{i = 1}^n \left[ {n{W_n}(\beta _0 ) - (n - 1)W_{n - 1}^{( - i)}(\beta _0 ) - {W_n}(\beta _0 )} \right] \left[ n{W_n}(\beta _0 ) \right. \\&\quad \left. - (n - 1)W_{n - 1}^{( - i)}(\beta _0 ) - {W_n}(\beta _0 ) \right] ^T + {W_n}(\beta _0 )W_n^T(\beta _0 )\\&= \frac{{{{(n - 1)}^2}}}{n}\sum \limits _{i = 1}^n {\left[ {{W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )} \right] {{\left[ {{W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )} \right] }^T}} + o(1)\;a.s. \end{aligned} \end{aligned}$$

(A.2)

From Lemma A.3 in Li et al. (2016), we have that

$$\begin{aligned} G(\beta _0 ) = \Sigma _{p \times p}^{({\beta _0})} + o(1)\;a.s. \end{aligned}$$

Also, since

$$\begin{aligned} \begin{aligned} {W_n}(\beta _0 )&= \frac{1}{n}\sum \limits _{i = 1}^n {{{\hat{V}}_n}(\beta _0 )} = \frac{1}{n}\sum \limits _{i = 1}^n {\left[ {n{W_n}(\beta _0 ) - (n - 1)W_{n - 1}^{( - i)}(\beta _0 )} \right] } \\&= n{W_n}(\beta _0 ) - \frac{{n - 1}}{n}\sum \limits _{i = 1}^n {W_{n - 1}^{( - i)}(\beta _0 )}, \end{aligned} \end{aligned}$$

(A.3)

it leads to

$$\begin{aligned} \sum \limits _{i = 1}^n {W_{n - 1}^{( - i)}(\beta _0 )} = n{W_n}(\beta _0 ). \end{aligned}$$

(A.4)

Furthermore,

$$\begin{aligned} \begin{aligned} {G^*}(\beta _0 )&= \frac{1}{n}\sum \limits _{i = 1}^n {{{\hat{Q}}_i}(\beta _0 )} {\hat{Q}}_i^T(\beta _0 )\\&= \frac{1}{n}\sum \limits _{i = 1}^n {\left[ {{{{\hat{Q}}}_i}(\beta _0 ) - {U_n}(\beta _0 ) + {U_n}(\beta _0 )} \right] {{\left[ {{{\hat{Q}}_i}(\beta _0 ) - {U_n}(\beta _0 ) + {U_n}(\beta _0 )} \right] }^T}} \\&= \frac{1}{n}\sum \limits _{i = 1}^n {\left[ {{{{\hat{Q}}}_i}(\beta _0 ) - {U_n}(\beta _0 )} \right] {{\left[ {{{{\hat{Q}}}_i}(\beta _0 ) - {U_n}(\beta _0 )} \right] }^T}} + {U_n}(\beta _0 )U_n^T(\beta _0 ). \end{aligned} \end{aligned}$$

(A.5)

Note that the first term $\sum \nolimits _{i = 1}^n {{{\hat{Q}}_i}(\beta _0 )} {\hat{Q}}_i^T(\beta _0 )/n$ in Eq. (A.5) can be proved as $\Sigma _{p \times p}^{({\beta _0})} + o(1)\;a.s.$

$$\begin{aligned}&\frac{1}{n}\sum \limits _{i = 1}^n {\left[ {{{\hat{Q}}_i}(\beta _0 ) - {U_n}(\beta _0 )} \right] {{\left[ {{{\hat{Q}}_i}(\beta _0 ) - {U_n}(\beta _0 )} \right] }^T}} \\&\quad = \frac{1}{n}\sum \limits _{i = 1}^n {\left[ {n{U_n}(\beta _0 ) - (n - 1)U_{n - 1}^{( - i)}(\beta _0 ) - {U_n}(\beta _0 )} \right] {{\left[ {n{U_n}(\beta _0 ) - (n - 1)U_{n - 1}^{( - i)}(\beta _0 ) - {U_n}(\beta _0 )} \right] }^T}} \\&\quad = \frac{{{{(n - 1)}^2}}}{n}\sum \limits _{i = 1}^n {\left[ {{U_n}(\beta _0 ) - U_{n - 1}^{( - i)}(\beta _0 )} \right] {{\left[ {{U_n}(\beta _0 ) - U_{n - 1}^{( - i)}(\beta _0 )} \right] }^T}} \\&\quad = \frac{{{{(n - 1)}^2}}}{n}\sum \limits _{i = 1}^n {\left[ {({W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )) + o({n^{ - 1/2}}))} \right] {{\left[ {({W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )) + o({n^{ - 1/2}}))} \right] }^T}} \\&\quad = \frac{{{{(n - 1)}^2}}}{n}\sum \limits _{i = 1}^n {\left[ {{W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )} \right] {{\left[ {{W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )} \right] }^T}} +o(1)\\&\qquad +\, \frac{{2{{(n - 1)}^2}}}{n}\sum \limits _{i = 1}^n {o({n^{ - 1/2}})\left[ {{W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )} \right] } \\&\quad = \frac{{{{(n - 1)}^2}}}{n}\sum \limits _{i = 1}^n {\left[ {{W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )} \right] {{\left[ {{W_n}(\beta _0 ) - W_{n - 1}^{( - i)}(\beta _0 )} \right] }^T}} + o(1)\\&\qquad +\, \frac{{2{{(n - 1)}^2}}}{n}o({n^{ - 1/2}})\left( {n{W_n}(\beta _0 ) - \sum \limits _{i = 1}^n {W_{n - 1}^{( - i)}(\beta _0 )} } \right) \\&\quad = G(\beta _0 ) + o(1)\\&\quad = \Sigma _{p \times p}^{({\beta _0})} + o(1)\;a.s. \end{aligned}$$

Also, based on the strong law of large number for U-statistics, one has ${U_n}(\beta _0 ) = o(1)$ a.s. Therefore, $G^* (\beta _0)= \Sigma _{p \times p}^{({\beta _0})} + o(1)\;a.s$. $\square$

Lemma A.4

Let ${A_n} = {\max _{1 \le i \ne j \le n}}\left\| {K({Z_1},{Z_2};\beta _0 )} \right\|$. Under the condition (C.1), we have ${A_n} = o({n^{1/2}})$ a.s.

Proof

By Borel–Cantelli lemma and Li et al. (2016), ${A_n} = o({n^{1/2}})$ a.s. $\square$

Lemma A.5

Let ${B_n} = {\max _{1 \le i \le n}}\left\| {{{{\hat{Q}}}_i}(\beta _0 )} \right\|$. Under conditions (C.1)–(C.3), ${B_n} = o({n^{1/2}})$ and ${n^{ - 1}}{\sum \nolimits _{i = 1}^n {\left\| {{{{\hat{Q}}}_i}(\beta _0 )} \right\| } ^3} = o({n^{1/2}})$.

Proof

We can check that

$$\begin{aligned} \begin{aligned} {U_n}(\beta _0 )&= \frac{1}{{n(n - 1)}}\sum \limits _{l = 1}^n {\sum \limits _{j = 1,j \ne l}^n {K({Z_l},{Z_j};\beta _0 )} } \\&= \frac{2}{{n(n - 1)}}\sum \limits _{j = 1,j \ne i}^n {K({Z_i},{Z_j};\beta _0 )} + \frac{{n - 2}}{n}U_{n - 1}^{( - i)}(\beta _0 ). \end{aligned} \end{aligned}$$

Then, for any $1 \le i \le n$,

$$\begin{aligned} \begin{aligned} \left\| {{{{\hat{Q}}}_i}(\beta _0 )} \right\|&= \left\| {\frac{2}{{n - 1}}\sum \limits _{j = 1,j \ne i}^n {K({Z_i},{Z_j};\beta _0 )} - U_{n - 1}^{( - i)}(\beta _0 )} \right\| \\&\le 3{\max _{1 \le i \ne j \le n}}\left| {K({Z_i},{Z_j};\beta _0 )} \right| \\&= 3{A_n}. \end{aligned} \end{aligned}$$

(A.6)

Combining (A.6) and the result of Lemma A.4, that is, ${A_n} = o({n^{1/2}})$ a.s., we prove Lemma A.5. $\square$

Proof of Theorem 2.1

We let $\lambda =\rho \theta$, where $\rho \ge 0$ and $\left\| \theta \right\| = 1$. Let $e_j$ be the unit vector on the jth coordinate direction. According to (2.5), we obtain that like Owen (2001) and Lu and Liang (2006),

$$\begin{aligned} \begin{aligned} 0&\ge \left| {{\theta ^ T}f(\rho \theta )} \right| \\&\ge \, \frac{{\rho {\theta ^T}{G^*}(\beta _0)\theta }}{{1 + \rho {B_n}}} - \frac{1}{n}\left| {\sum \limits _{j = 1}^p {e_j^T} \sum \limits _{i = 1}^n {{{{\hat{Q}}}_i}({\beta _0})} } \right| . \end{aligned} \end{aligned}$$

We also have ${G^*}(\beta _0 ) = \Sigma _{p \times p}^{({\beta _0})} + o(1)$ a.s. from Lemma A.3. Thus, one has

$$\begin{aligned} \left\| \lambda \right\| = \rho = {O_p}({n^{ - 1/2}}). \end{aligned}$$

(A.7)

Denote ${\eta _i} = {\lambda ^T}{{{\hat{Q}}}_i}({\beta _0})$. From Lemma A.5 and (A.7), we can obtain

$$\begin{aligned} \lambda = {({G^*}(\beta _0))^{ - 1}}{U_n}({\beta _0}) + \gamma , \end{aligned}$$

where $\left\| \gamma \right\| = {o_p}({n^{ - 1/2}})$. By Taylor expansion, we obtain that

$$\begin{aligned} \begin{aligned} {l}({\beta _0})&= 2 \sum \limits _{i = 1}^n {{\eta _i} - \sum \limits _{i = 1}^n {\eta _i^2} } +{o_p}(1) \\&= nU_n^T({\beta _0}){({G^*}(\beta _0))^{ - 1}}{U_n}({\beta _0}) - n{\gamma ^T}{G^*}(\beta _0)\gamma +{o_p}(1). \end{aligned} \end{aligned}$$

(A.8)

In (A.8), the first term is $nU_n^T(\beta _0 ){(G^*(\beta _0))^{ - 1}}{U_n}(\beta _0 )\mathop \rightarrow \limits ^d \chi _p^2$. Moreover, the second term is $n{\gamma ^T}{G^*}(\beta _0)\gamma = n{o_p}({n^{ - 1/2}}){O_p}(1){o_p}({n^{ - 1/2}}) = {o_p}(1)$. Therefore, $- 2\log R({\beta _0})\mathop \rightarrow \limits ^d \chi _p^2$. $\square$

Proof of Theorem 2.2

We follow the similar arguments in Yu et al. (2011) and Yang and Zhao (2012). Corresponding to $\beta _0=(\beta _{10}^T, \beta _{20}^T)^T$, we denote $Z=(Z_{1}^T, Z_{2}^T)^T$. Recall that $\sqrt{n} (\hat{\beta }- {\beta _0})$ was asymptotically normally distributed with mean zero and variance-covariance matrix $D_n{({\beta _0})^{ - 1}}B_n({\beta _0}){(D_n{({\beta _0})^{ - 1}})^T}$.

Define

$$\begin{aligned} {\bar{D}}({\beta _0}) = \mathop {\lim }\limits _{n \rightarrow \infty } E{\left[ {{{\partial {S_n}} / {\partial {\beta _2}}}} \right] _{{\beta _0}}}. \end{aligned}$$

Since $D_n(\beta _0)$ is positive definite, ${\bar{D}}(\beta _0)$ is of rank $p-q$. Denote

$$\begin{aligned} {{\hat{\beta }}_2} = \arg \mathop {\inf }\limits _{{\beta _2}} l\left[ {{{\left( \beta _{10}^T,\beta _{2}^T\right) }^T}} \right] . \end{aligned}$$

Similar to Qin and Lawless (1994), we can show that

$$\begin{aligned} \sqrt{n} \left( {{\hat{\beta }}_2} - {\beta _{20}}\right)= & {} - {\left( {\bar{D}}{\left( {\beta _0}\right) ^T}{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) ^{ - 1}}{\bar{D}}\left( {\beta _0}\right) \right) ^{ - 1}}\\&\times {\bar{D}}{\left( {\beta _0}\right) ^T}{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) ^{ - 1}}\frac{1}{{\sqrt{n} }}\sum \limits _{i = 1}^n {{{\hat{Q}}_i}\left( {\beta _0}\right) } + {o_p}(1), \end{aligned}$$

and

$$\begin{aligned}&\sqrt{n} {\lambda _2} = \left( {I - {{\left( \Sigma _{p \times p}^{({\beta _0})}\right) }^{ - 1}}{\bar{D}}({\beta _0}){{\left( {\bar{D}}{{\left( {\beta _0}\right) }^T}{{\left( \Sigma _{p \times p}^{({\beta _0})}\right) }^{ - 1}}\bar{D}({\beta _0})\right) }^{ - 1}}{\bar{D}}{{({\beta _0})}^T}} \right) \\&\quad \left( \Sigma _{p \times p}^{({\beta _0})}\right) ^{ - 1}\frac{1}{{\sqrt{n} }}\sum \limits _{i = 1}^n {{{{\hat{Q}}}_i}({\beta _0})} + {o_p}(1), \end{aligned}$$

where $\lambda _2$ is the corresponding Lagrange multiplier. Recall that

$$\begin{aligned} {U_n}(\beta ) = \frac{1}{n}\sum \limits _{i = 1}^n {{{{\hat{Q}}}_i}} (\beta ). \end{aligned}$$

Hence, by Taylor’s expansion, one has that

$$\begin{aligned} {l^*}({\beta _{10}})&= {\left( {\frac{1}{{\sqrt{n} }}\sum \limits _{i = 1}^n {{{{\hat{Q}}}_i}({\beta _0})} } \right) ^T}\left( {{\left( \Sigma _{p \times p}^{({\beta _0})}\right) }^{ - 1}}\right. \\&\quad \left. -\, {{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) }^{ - 1}}{\bar{D}}\left( {\beta _0}\right) {{\left( {\bar{D}}{{\left( {\beta _0}\right) }^T}{{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) }^{ - 1}}\bar{D}\left( {\beta _0}\right) \right) }^{ - 1}}{\bar{D}}{{\left( {\beta _0}\right) }^T}{{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) }^{ - 1}} \right) \\&\quad \times \left( {\frac{1}{{\sqrt{n} }}\sum \limits _{i = 1}^n {{{\hat{Q}}_i}\left( {\beta _0}\right) } } \right) + {o_p}\left( 1\right) \\&= {\left( {{{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) }^{ - 1/2}}\frac{1}{{\sqrt{n} }}\sum \limits _{i = 1}^n {{{{\hat{Q}}}_i}\left( {\beta _0}\right) } } \right) ^T}\Psi \left( {{{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) }^{ - 1/2}}\frac{1}{{\sqrt{n} }}\sum \limits _{i = 1}^n {{{\hat{Q}}_i}\left( {\beta _0}\right) } } \right) + {o_p}\left( 1\right) \\&= {\left( {{{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) }^{ - 1/2}}\sqrt{n} {U_n}\left( {\beta _0}\right) } \right) ^T}\Psi \left( {{{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) }^{ - 1/2}}\sqrt{n} {U_n}\left( {\beta _0}\right) } \right) + {o_p}\left( 1\right) , \end{aligned}$$

where

$$\begin{aligned} \Psi = I - {\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) ^{ - 1/2}}{\bar{D}}\left( {\beta _0}\right) {\left( {\bar{D}}{\left( {\beta _0}\right) ^T}{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) ^{ - 1}}{\bar{D}}\left( {\beta _0}\right) \right) ^{ - 1}}{\bar{D}}{\left( {\beta _0}\right) ^T}{\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) ^{ - 1/2}}. \end{aligned}$$

$\Psi$ is a symmetric matrix with trace q. By Lemma A.1,

$$\begin{aligned} {\left( \Sigma _{p \times p}^{\left( {\beta _0}\right) }\right) ^{ - 1/2}}\sqrt{n} {U_n}\left( {\beta _0}\right) \xrightarrow {d} N\left( 0,{I_{p \times p}}\right) . \end{aligned}$$

Then, we have that

$$\begin{aligned} - 2\log R^*\left( \beta _{10} \right) \mathop \rightarrow \limits ^d \chi _q^2. \end{aligned}$$

The proof of Theorem 2.2 is completed. $\square$