Composite change point estimation for bent line quantile regression

Abstract

The bent line quantile regression describes the situation where the conditional quantile function of the response is piecewise linear but still continuous in covariates. In some applications, the change points at which the quantile functions are bent tend to be the same across quantile levels or for quantile levels lying in a certain region. To capture such commonality, we propose a composite estimation procedure to estimate model parameters and the common change point by combining information across quantiles. We establish the asymptotic properties of the proposed estimator, and demonstrate the efficiency gain of the composite change point estimator over that obtained at a single quantile level through numerical studies. In addition, three different inference procedures are proposed and compared for hypothesis testing and the construction of confidence intervals. The finite sample performance of the proposed procedures is assessed through a simulation study and the analysis of a real data.

Appendix

Lemma 1

Suppose Assumptions A1–A3 hold, then $\hat{\varvec{\theta }}$ is a consistent estimator of $\varvec{\theta }_{0}$.

Proof

At a fixed point u, we need to minimize the following objective function

$$\begin{aligned} n^{-1}\sum _{k=1}^{K}\sum _{i=1}^{n}\rho _{\tau _{k}}\{Y_{i} - Q_{Y}(\tau _{k}; \varvec{\theta }|\mathbf {W}_{i})\}, \end{aligned}$$

which is equivalent to minimize

$$\begin{aligned} n^{-1}\sum _{i=1}^{n}\rho _{\tau _{k}}\{Y_{i} - Q_{Y}(\tau _{k}; \varvec{\eta }, u|\mathbf {W}_{i})\}, \end{aligned}$$

for any $1\le k\le K$. The rest of the proof follows the similar arguments as that of Lemma 1 in Li et al. (2011) and thus is omitted. $\square$

Lemma 2

Suppose Assumptions A1–A3 hold, we have

$$\begin{aligned}&\sup _{\Vert \varvec{\theta }- \varvec{\theta }_{0}\Vert \le C_{2}n^{-1/2}} \Bigg \Vert n^{-1/2}\sum _{k=1}^{K}\sum _{i=1}^{n}[\psi _{\tau _{k}}\{Y_{i} - Q_{Y}(\tau _{k};\varvec{\theta }|\mathbf {W}_{i})\}h_{k}(\mathbf {W}_{i};\varvec{\theta }) -\psi _{\tau _{k}}\{Y_{i} \nonumber \\&\quad - Q_Y(\tau _{k}; \varvec{\theta }_{0}|\mathbf {W}_{i})\} h_{k}(\mathbf {W}_{i};\varvec{\theta }_{0})]-n^{-1/2}E\Bigg [\sum _{k=1}^{K}\sum _{i=1}^{n}\psi _{\tau _{k}}\{Y_{i}-Q_{Y}(\tau _{k};\varvec{\theta }|\mathbf {W}_{i})\} \nonumber \\&\quad \quad \times h_{k}(\mathbf {W}_{i};\varvec{\theta })\Bigg ] \Bigg \Vert =o_{p}(1), \end{aligned}$$

(8)

where $C_{2}$ is some positive constant.

Proof

Let

$$\begin{aligned} u_{i}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})&= \sum _{k=1}^{K}\psi _{\tau _{k}}\{Y_{i}-Q_{Y}(\tau _{k};\varvec{\theta }|\mathbf {W}_{i}) \}h_{k}(\mathbf {W}_{i};\varvec{\theta })\\&\quad \ - \sum _{k=1}^{K}\psi _{\tau _{k}}\{Y_{i}-Q_{Y}(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i})\} h_{k}(\mathbf {W}_{i};\varvec{\theta }_{0}), \end{aligned}$$

where $\mathbf {V}_{i}$ includes all the random variables $Y_{i}$ and $\mathbf {W}_{i}$. Therefore, it can be rewritten as the following

$$\begin{aligned}&u_{i}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})= u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})+u_{i2}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})+u_{i3}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})+u_{i4}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0}), \end{aligned}$$

where $u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})=u_{i}(\mathbf {V}_{i};$ $\varvec{\theta },\varvec{\theta }_{0})I\{X_{i}\le \min (u, u_{0})\}$, $u_{i2}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})=u_{i}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})I(u_{0}$ $< X_{i} \le u)$, $u_{i3}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})=u_{i}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})I(u < X_{i} \le u_{_{0}})$ and $u_{i4}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})=u_{i}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})$ $I\{X_{i}> \max (u, u_{0})\}$.

To obtain (8), it is sufficient to show

$$\begin{aligned}&\sup _{\Vert \varvec{\theta }- \varvec{\theta }_{0}\Vert \le C_{2}n^{-1/2}} \left\| n^{-1/2}\sum _{i=1}^{n}\{B_{i} - E(B_{i})\} \right\| =o_{p}(1), \end{aligned}$$

where $B_{i}$ represents $u_{ij}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})$ for $j=1,2,3,4$. These results follow from Lemma 4.6 in He and Shao (1996), we only show the proof for $B_{i}=u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})$ for instance. To verify this, we need to check the conditions (B1), (B3) and (B5$^{'}$) of He and Shao (1996).

For (B1), the measurability is easy to show.

For (B3), take $r=1$, for any $\Vert \varvec{\theta }- \varvec{\theta }_{0} \Vert \le C_{2}n^{-1/2}$, we have

$$\begin{aligned}&\Vert u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})\Vert \nonumber \\&\quad = \Bigg \Vert \Bigg [\sum _{k=1}^{K}\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i}) \}h_{k}(\mathbf {W}_{i};\varvec{\theta }) \nonumber \\&\qquad - \sum _{k=1}^{K}\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i}) \}h_{k}(\mathbf {W}_{i};\varvec{\theta }_{0}) \Bigg ] \ I\{X_{i}\le \min (u, u_{0})\} \Bigg \Vert \nonumber \\&\quad \le \Bigg \Vert \sum _{k=1}^{K}\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i}) \}\{h_{k}(\mathbf {W}_{i};\varvec{\theta })-h_{k}(\mathbf {W}_{i};\varvec{\theta }_{0}) \}I\{X_{i}\le \min (u, u_{0})\} \Bigg \Vert \nonumber \\&\qquad + \Bigg \Vert \sum _{k=1}^{K}[\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i}) \}-\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i}) \} ]h_{k}(\mathbf {W}_{i};\varvec{\theta }_{0}) \nonumber \\&\quad \quad \times I\{X_{i}\le \min (u, u_{0})\} \Bigg \Vert \nonumber \\&\quad = \Vert I_{1i}\Vert +\Vert I_{2i}\Vert . \end{aligned}$$

(9)

For $I_{1i}$, it is easy to obtain

$$\begin{aligned}&E(\Vert I_{1i}\Vert ^{2}|\mathbf {W}_{i})= n^{-1} O_{p}(1). \end{aligned}$$

(10)

For $I_{2i}$, we have

$$\begin{aligned} \Vert I_{2i}\Vert= & {} \Bigg \Vert \sum _{i=1}^{K}[I\{Y_{i} \le Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i})\} - I\{Y_{i}\le Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i}) \}] h_{k}(\mathbf {W}_{i};\varvec{\theta }_{0}) \\&I\{X_{i}\le \min (u, u_{0})\} \Bigg \Vert \\\le & {} L_{1}\Vert \mathbf {U}_{i}\Vert \sum _{k=1}^{K}I\{ Q_{1}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0})\le Y_{i}\le Q_{2}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0}) \}I\{X_{i}\le \min (u, u_{0})\}, \end{aligned}$$

where $L_{1}$ is some constant, $\mathbf {U}_{i}=(1, X_{i}, \mathbf {Z}^{T}_{i})^{T}$, $Q_{1}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0})= \min \{Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i}),$ $Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i}) \}$ and $Q_{2}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0})= \max \{ Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i}),$ $Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i}) \}$. Thus

$$\begin{aligned}&E(\Vert I_{2i}\Vert ^{2}|\mathbf {W}_{i})\\&\quad \le L^{2}_{1}\Vert \mathbf {U}_{i}\Vert ^{2}I \{X_{i}\le \min (u, u_{0})\}E\bigg [\sum _{k=1}^{K}\sum _{k'=1}^{K}I\{ Q_{1}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0})\le Y_{i}\le Q_{2}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0})\}\\&\quad \quad \times I\{ Q_{1}(\tau _{k'};\varvec{\theta },\varvec{\theta }_{0})\le Y_{i}\le Q_{2}(\tau _{k'};\varvec{\theta },\varvec{\theta }_{0}) \} |\mathbf {W}_i\bigg ]. \end{aligned}$$

Without loss of generality, we assume $Q_{1}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0})< Q_{2}(\tau _{k'};\varvec{\theta },\varvec{\theta }_{0})$ and $Q_{1}(\tau _{k'};\varvec{\theta },\varvec{\theta }_{0})< Q_{2}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0})$. Denote $Q_{1}(\tau _{k},\tau _{k'})= \min \{Q_{1}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0}), Q_{1}(\tau _{k'};\varvec{\theta },\varvec{\theta }_{0}) \}$ and $Q_{2}(\tau _{k},\tau _{k'})= \max \{Q_{2}(\tau _{k};\varvec{\theta },\varvec{\theta }_{0}), Q_{2}(\tau _{k'};\varvec{\theta },\varvec{\theta }_{0}) \}$. Hence

(11)

where the first inequality follows from the mean value theorem with $\xi _{k,k'}$ between $Q_{1}(\tau _{k};\tau _{k'})$ and $Q_{2}(\tau _{k};\tau _{k'})$, the second inequality follows from

$$\begin{aligned}&| \{ Q_{2}(\tau _{k};\tau _{k'}) - Q_{1}(\tau _{k};\tau _{k'})\} I \{X_{i}\le \min (u, u_{0})\} |\\&\quad \le |\{Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i}) - Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i})\} I \{X_{i}\le \min (u, u_{0})\}|\\&\quad \quad + |\{Q_Y(\tau _{k'};\varvec{\theta }|\mathbf {W}_{i}) - Q_Y(\tau _{k'};\varvec{\theta }_{0}|\mathbf {W}_{i})\} I \{X_{i}\le \min (u, u_{0})\}|\\&\quad \le |(\alpha _{\tau _{k}} - \alpha _{\tau _{k},0}) + (\varvec{\beta }_{1,\tau _{k}} - \varvec{\beta }_{1,\tau _{k},0})X_{i} - (\varvec{\beta }_{1,\tau _{k}}u - \varvec{\beta }_{1,\tau _{k},0}u_{0}) + \mathbf {Z}^{T}_{i}(\varvec{\gamma }_{\tau _{k}} - \varvec{\gamma }_{\tau _{k},0})|\\&\quad \quad + |(\alpha _{\tau _{k'}} - \alpha _{\tau _{k'},0}) + (\varvec{\beta }_{1,\tau _{k'}} - \varvec{\beta }_{1,\tau _{k'},0})X_{i} - (\varvec{\beta }_{1,\tau _{k'}}u - \varvec{\beta }_{1,\tau _{k'},0}u_{0}) + \mathbf {Z}^{T}_{i}(\varvec{\gamma }_{\tau _{k'}}\\&\quad \quad \quad - \varvec{\gamma }_{\tau _{k'},0})|\\&\quad \le L_{3} n^{-1/2}\Vert \mathbf {U}_{i}\Vert , \end{aligned}$$

where $L_{2}$ and $L_{3}$ are some positive constants satisfying $L_{2}= L^{2}_{1}L^{2}_{3}$. By Assumptions A3 and A4, combining (9), (10) and (11), for large n we have

$$\begin{aligned}&E\{\Vert u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})\Vert ^{2} |\mathbf {W}_{i})\}\le L n^{-1/2}\Vert \mathbf {U}_{i}\Vert ^{3} \sum _{k=1}^{K}\sum _{k'=1}^{K} f_{i}(\xi _{k,k'}). \end{aligned}$$

It is obvious to obtain (B3) by taking $a_{i}= \sqrt{L\Vert \mathbf {U}_{i}\Vert ^{3} \sum _{k=1}^{K}\sum _{k'=1}^{K}f_{i}(\xi _{k,k'})}$.

For (B5$^{'}$), let $A_{n}=L\sum _{i=1}^{n}\Vert \mathbf {U}_{i}\Vert ^{3}\sum _{k=1}^{K}\sum _{k'=1}^{K}f_{i}(\xi _{k,k'})$. From Assumptions A2 and A3, we have $E(A_{n})= O(n)$. For any positive constant $C_{3}>0$, taking the decreasing sequence of positive number $d_{n}$ satisfying $n^{-1/2}(\log n)^4=o(d_{n})$ and $d_{n}=o(1)$, we can show

$$\begin{aligned}&P \Bigg (\Vert \max _{1\le i \le n}u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})\Vert \ge C_{3}A^{1/2}_{n}d^{1/2}_{n} (\log n)^{-2} \Bigg )\\&\quad \le \sum _{i=1}^{n}P(\Vert u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})\Vert \ge C_{3}A^{1/2}_{n}d^{1/2}_{n} (\log n)^{-2} )\\&\quad \le \sum _{i=1}^{n} \frac{E\Vert u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})\Vert ^{2} }{C_{3}^{2}A_{n}d_{n} (\log n)^{-4}}\\&\quad \le \frac{ n^{-1/2}(\log n)^4 }{ C_{3}^{2}d_{n} }\\&\quad = o(1). \end{aligned}$$

Hence,

$$\begin{aligned}&\max _{1\le i \le n}\Vert u_{i1}(\mathbf {V}_{i};\varvec{\theta },\varvec{\theta }_{0})\Vert =O_{p}(A^{1/2}_{n}d^{1/2}_{n} (\log n)^{-2}), \end{aligned}$$

thus (B5’) is satisfied. This completes the proof of Lemma 2. $\square$

Proof of Theorem 1

By Lemmas 1 and 2, we obtain

$$\begin{aligned}&n^{-1/2}\sum _{k=1}^{K}\sum _{i=1}^{n}[\psi _{\tau _{k}}\{Y_{i} - Q_Y(\tau _{k};\hat{\varvec{\theta }}|\mathbf {W}_{i})\}\varvec{h}_{k}(\mathbf {W}_{i};\hat{\varvec{\theta }})-\psi _{\tau _{k}}\{Y_{i} - Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i})\} \nonumber \\&\quad \times h_{k}(\mathbf {W}_{i};\varvec{\theta }_{0})]-n^{-1/2} \Bigg [E\sum _{k=1}^{K}\sum _{i=1}^{n}\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k}; \varvec{\theta }|\mathbf {W}_{i})\}\varvec{h}_{k}(\mathbf {W}_{i}; \varvec{\theta }) \Bigg ]\Bigg |_{\varvec{\theta }= \hat{\varvec{\theta }}} \nonumber \\&\quad =o_{p}(1). \end{aligned}$$

(12)

Applying the Taylor expansion, we get

$$\begin{aligned}&\Bigg [E\sum _{k=1}^{K}\sum _{i=1}^{n}\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k}; \varvec{\theta }|\mathbf {W}_{i})\}\varvec{h}_{k}(\mathbf {W}_{i}; \varvec{\theta }) \Bigg ]\Bigg |_{\varvec{\theta }= \hat{\varvec{\theta }}} =n\varvec{D}_{n}(\hat{\varvec{\theta }}-\varvec{\theta }_{0}) \nonumber \\&\quad +\, O_{p}(n ( \hat{\varvec{\theta }} - \varvec{\theta }_{0})^2), \end{aligned}$$

(13)

where

$$\begin{aligned} \varvec{D}_{n}= & {} n^{-1}\sum _{k=1}^{K}\sum _{i=1}^{n}\frac{\partial E\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i})\}\varvec{h}_{k}(\mathbf {W}_{i};\varvec{\theta })}{\partial \varvec{\theta }}\Bigg |_{\varvec{\theta }=\varvec{\theta }_{0}}\\= & {} n^{-1}\sum _{k=1}^{K}\sum _{i=1}^{n} \frac{\partial ([\tau _{k} - F_{i}\{Q_Y(\tau _{k};\varvec{\theta }|\mathbf {W}_{i})\}]\varvec{h}_{k}(\mathbf {W}_{i};\varvec{\theta }))}{\partial \varvec{\theta }}\Bigg |_{\varvec{\theta }=\varvec{\theta }_{0}} \\= & {} n^{-1}\sum _{k=1}^{K}\sum _{i=1}^{n} \Bigg ([-f_{i}\{Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i})\} \varvec{h}_{k}(\mathbf {W}_{i};\varvec{\theta }_{0})\varvec{h}^{T}_{k}(\mathbf {W}_{i};\varvec{\theta }_{0})] \\&+\,[\tau _{k} - F_{i}\{Q(\tau _{k})\}]\frac{\partial \varvec{h}_{k}(\mathbf {W}_{i};\varvec{\theta })}{\partial \varvec{\theta }}\Bigg |_{\varvec{\theta }=\varvec{\theta }_{0}} \Bigg ) \\= & {} n^{-1}\sum _{k=1}^{K}\sum _{i=1}^{n} \left[ -f_{i}\{Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i})\} \varvec{h}_{k}(\mathbf {W}_{i};\varvec{\theta }_{0})\varvec{h}^{T}_{k}(\mathbf {W}_{i};\varvec{\theta }_{0})\right] . \end{aligned}$$

In addition, by the subgradient condition of quantile regression (pages 34–38 in Koenker 2005), we have

$$\begin{aligned}&n^{-1/2}\sum _{k=1}^{K}\sum _{i=1}^{n}\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\hat{\varvec{\theta }}|\mathbf {W}_{i})\}\varvec{h}_{k}(\mathbf {W}_{i};\hat{\varvec{\theta }})=o_{p}(1). \end{aligned}$$

(14)

Combining (12), (13) and (14), we have

$$\begin{aligned}&-n^{-1/2}\sum _{k=1}^{K}\sum _{i=1}^{n}[\psi _{\tau _{k}}\{Y_{i} - Q_Y(\tau _{k}; \varvec{\theta }_{0}|\mathbf {W}_{i})\}\varvec{h}_{k}(\mathbf {W}_{i}; \varvec{\theta }_{0}) ] = n^{1/2}\mathbf {D}_{n}( \hat{\varvec{\theta }}-\varvec{\theta }_{0} )\\&\qquad +\, O_{p}(n^{1/2}( \hat{\varvec{\theta }}-\varvec{\theta }_{0} )^{2}) + o_{p}(1). \end{aligned}$$

This results in $\hat{\varvec{\theta }}- \varvec{\theta }_{0}= O_{p}(n^{-1/2})$. Therefore,

$$\begin{aligned} n^{1/2}(\hat{\varvec{\theta }}-\varvec{\theta }_{0})=&-\varvec{D}^{-1}_{n} n^{-1/2}\sum _{k=1}^{K}\sum _{i=1}^{n}\varvec{\psi }_{\tau _{k}}\{Y_{i} - Q_Y(\tau _{k};\varvec{\theta }_{0}|\mathbf {W}_{i})\}h_{k}(\mathbf {W}_{i};\varvec{\theta }_{0}) +o_{p}(1). \end{aligned}$$

Following central limit theorem, by Assumption A4 and some simple calculation, we can obtain that $n^{1/2}(\hat{\varvec{\theta }} - \varvec{\theta }_{0})$ is asymptotically normal with mean zero and variance $\varvec{D}^{-1}\varvec{C}\varvec{D}^{-1}$. This completes the proof of Theorem 1. $\square$

Proof of Theorem 2

Denote

$$\begin{aligned}&T^{*}_{n}=\varvec{S}_{n}^{*T}\varvec{(}V^{*}_{n})^{-1} \varvec{S}_{n}^{*}, \end{aligned}$$

where $\varvec{S}^{*}_{n}= \{S^{*}_{n,1},\ldots , S^{*}_{n,K}\}^{T}$, $S^{*}_{n,k}=n^{-1/2}\sum _{i=1}^{n}p^{*}_{i}(\tau _{k};\varvec{\eta }_{k,0},u_{0}|\mathbf {W}_{i})\psi _{\tau _{k}}(e_{i,\tau _{k}})$, $e_{i,\tau _{k}}=Y_{i}-Q_Y(\tau _{k};\varvec{\eta }_{k,0}, u_{0}|\mathbf {W}_{i})$, $\varvec{V}^{*}_{n}$ is a $K\times K$ matrix with the $(k,k^{'})$th element

$$\begin{aligned}&\mathrm{Cov}(S^{*}_{n,k}, S^{*}_{n,k^{'}})\\&\quad = \mathrm{Cov} \left\{ n^{-1/2}\sum _{i=1}^{n}p^{*}_{i}(\tau _{k}; \varvec{\eta }_{k,0}, u_{0})\psi _{\tau _{k}}(e_{i,\tau _{k}}),\ n^{-1/2}\sum _{i=1}^{n}p^{*}_{i}(\tau _{k^{'}}; \varvec{\eta }_{k^{'},0}, u_{0})\psi _{\tau _{k^{'}}}(e_{i,\tau _{k^{'}}}) \right\} \\&\quad = n^{-1}\sum _{i=1}^{n}\mathrm{Cov} \{ p^{*}_{i}(\tau _{k}; \varvec{\eta }_{k,0}, u_{0})\psi _{\tau _{k}}(e_{i,\tau _{k}}),\ p^{*}_{i}(\tau _{k^{'}}; \varvec{\eta }_{k^{'},0}, u_{0})\psi _{\tau _{k^{'}}}(e_{i,\tau _{k^{'}}}) \} \\&\quad = n^{-1}\sum _{i=1}^{n} \{\min (\tau _{k}, \tau _{k^{'}})-\tau _{k}\tau _{k^{'}}\} p^{*}_{i}(\tau _{k}; \varvec{\eta }_{k,0}, u_{0})p^{*}_{i}(\tau _{k^{'}}; \varvec{\eta }_{k^{'},0}, u_{0}). \end{aligned}$$

Following central limit theorem, we have $\varvec{S}^{*}_{n}\ \xrightarrow {d}\ N(0,\ \varvec{V}^{*}_{n}).$ Therefore, we can obtain

$$\begin{aligned}&\varvec{T}_{n}^{*} \xrightarrow {d}\ \chi ^{2}_{K}. \end{aligned}$$

To obtain the desired result, we need to show

$$\begin{aligned}&\varvec{V}_{n}= \varvec{V}^{*}_{n}+o_{p}(1), \end{aligned}$$

(15)

and

$$\begin{aligned}&\varvec{S}_{n}= \varvec{S}^{*}_{n}+o_{p}(1), \end{aligned}$$

(16)

It is easy to show that (15) holds since by Theorem 1,

$$\begin{aligned}&\Vert \hat{\varvec{\eta }}_{k}(u_{0}) - \varvec{\eta }_{k,0}\Vert =O_{p}(n^{-1/2}). \end{aligned}$$

(17)

For (16), it is sufficient to show that $S_{n,k}=S^{*}_{n,k}+o_{p}(1)$ for any $1\le k \le K$. Denote $S_{n,k}(\varvec{\eta }_{k})=n^{-1/2}\sum _{i=1}^{n}p^{*}_{i}(\tau _{k}; \varvec{\eta }_{k},u_{0})\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\eta }_{k}, u_{0}|\mathbf {W}_{i})\}$, because we have

$$\begin{aligned} S_{n,k}(\varvec{\eta }_{k})-S^{*}_{n,k}= & {} n^{-1/2}\sum _{i=1}^{n}p^{*}_{i}(\tau _{k};\varvec{\eta }_{k},u_{0})\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\eta }_{k}, u_{0})\}\\&-n^{-1/2}\sum _{i=1}^{n}p^{*}_{i}(\tau _{k};\varvec{\eta }_{k,0}, u_{0})\psi _{\tau _{k}}\{Y_{i}-Q_Y(\tau _{k};\varvec{\eta }_{k,0}, u_{0})\}. \end{aligned}$$

Because $E(S^{*}_{n,k})=0$, following He and Shao (2000), we have

$$\begin{aligned}&\sup _{\Vert \varvec{\eta }_{k} -\varvec{\eta }_{k,0}\Vert \le C_{4}n^{-1/2} } \Vert S_{n,k}(\varvec{\eta }_{k})-S^{*}_{n,k} - E\{S_{n,k}(\varvec{\eta }_{k})\} \Vert =o_{p}(1), \end{aligned}$$

(18)

where $C_{4}$ is some positive constant. For any $\varvec{\eta }_k$ such that $\Vert \varvec{\eta }_k-\varvec{\eta }_{k,0}\Vert \le C_{4}n^{-1/2}$ , by the Taylor expansion, we get

$$\begin{aligned}&\ E\{S_{n,k}(\varvec{\eta }_{k}) \} = n^{-1/2}\sum _{i=1}^{n}E (p^{*}_{i}(\tau _{k};\varvec{\eta }_{k}, u_{0} )[ \tau _{k} - F_{i}\{Q_{Y}(\tau _{k};\varvec{\eta }_{k},u_{0})|\mathbf {W}_{i}\} ] ) \nonumber \\&\quad = n^{-1/2}\sum _{i=1}^{n}E ( p^{*}_{i}(\tau _{k};\varvec{\eta }_{k}, u_{0}) \ [ - f_{i}\{Q_{Y} (\tau _{k};\varvec{\eta }_{k,0},u_{0})|\mathbf {W}_{i} \} \mathbf {m}^{T}(\mathbf {W}_{i}, u_{0})(\varvec{\eta }_{k}-\varvec{\eta }_{k,0}) \nonumber \\&\quad \quad - f^{'}_{i}\{Q_{Y}(\tau _{k};\varvec{\eta }_{k,0},u_{0})\}\{\mathbf {m}^{T}(\mathbf {W}_{i}, u_{0})(\varvec{\eta }_{k} - \varvec{\eta }_{k,0})\}^{2} + o_{p}(\Vert \varvec{\eta }_{k} - \varvec{\eta }_{k,0}\Vert ^{2})] ) \nonumber \\&\quad = -n^{-1/2}\sum _{i=1}^{n}E (\varvec{p}^{*}_{i}(\tau _{k}; \varvec{\eta }_{k},u_{0} ) f^{'}_{i}\{Q_{Y}(\tau _{k};\varvec{\eta }_{k,0},u_{0})|\mathbf {W}_{i} \}\nonumber \\&\qquad [\{\mathbf {m}^{T}(\mathbf {W}_{i}, u_{0}) ( \varvec{\eta }_{k}-\varvec{\eta }_{k,0})\}^{2}+o(1)] ) \nonumber \\&\quad =o(1), \end{aligned}$$

(19)

where the third equality follows from the orthogonalization between $\varvec{p}^{*}_{i}(\tau _{k}; \varvec{\eta }_{k}, u_{0})$ and $\mathbf {m}(\mathbf {W}_{i},$ $u_{0})$, the fourth equality is based on $\Vert n^{-1}\sum _{i=1}^{n}E \varvec{p}^{*}_{i}(\tau _{k}; \varvec{\eta }_{k}, u_{0}) \Vert \le \Vert n^{-1}\sum _{i=1}^{n} E \varvec{p}_{i}(\tau _{k};\varvec{\eta }_{k},$ $u_{0}) \Vert \le o(1)$. Let $\varvec{\eta }_{k} = \hat{\varvec{\eta }}_{k}(u_{0})$, note that $S_{n,k}= S_{n,k}\{ \hat{\varvec{\eta }}_{k}(u_{0}) \}$. By combining (16) with (19), we obtain (16). This completes the proof of Theorem 2. $\square$