Function-based hypothesis testing in censored two-sample location-scale models

Abstract

Function-based hypothesis testing in two-sample location-scale models has been addressed for uncensored data using the empirical characteristic function. A test of adequacy in censored two-sample location-scale models is lacking, however. A plug-in empirical likelihood approach is used to introduce a test statistic, which, asymptotically, is not distribution free. Hence for practical situations bootstrap is necessary for performing the test. A multiplier bootstrap and a model appropriate resampling procedure are given to approximate critical values from the null asymptotic distribution. Although minimum distance estimators of the location and scale are deployed for the plug-in, any consistent estimators can be used. Numerical studies are carried out that validate the proposed testing method, and real example illustrations are given.

Appendix

1.1 Review of the MDE of $\varvec{\theta }$

See also Bhattacharya and Subramanian (2014), where semiparametric estimation in two-sample location-scale models is investigated. Let $\mathcal{Q}_{\varvec{\theta }}(s) = {\hat{Q}}_2(s)-\varphi _{\varvec{\theta }}({\hat{Q}}_1(s))$, where ${\hat{Q}}_i(s), i=1,2$, are the Kaplan–Meier quantiles. Let G(s) denote a positive measure on (0, 1). To obtain the MDE ${\hat{\varvec{\theta }}}$, one minimizes $\mathcal{S}(\varvec{\theta })$, a Cramér-von Mises type discrepancy (Zhang and Yu 2002), defined as

$$\begin{aligned} \mathcal{S}(\varvec{\theta })= & {} \int \mathcal{Q}^2_{\varvec{\theta }}(s)dG(s):= \langle \mathcal{Q}^2_{\varvec{\theta }},G\rangle . \end{aligned}$$

(A.1)

Define ${\varvec{d}}=(d_1,d_2)^T$ where $d_l=\langle {\hat{Q}}_1^{l-1}{\hat{Q}}_2, G\rangle$, $l=1,2$, and let ${\varvec{A}}=(A_{il})_{2\times 2}$ where $A_{il}=\langle {\hat{Q}}_1^{i-1}{\hat{Q}}_1^{l-1}, G\rangle$. Minimizing $\mathcal{S}(\varvec{\theta })$ given by Eq. (A.1) yields ${\varvec{A}}\varvec{\theta }= {\varvec{d}}$. When ${\hat{Q}}_1$ is not a constant, A is non-singular (Zhang and Yu 2002). Then ${\hat{\varvec{\theta }}}={\varvec{A}}^{-1}{\varvec{d}}$. More specifically, let

$$\begin{aligned} {\hat{\varvec{U}}}\ =\ ({\hat{U}}_1, {\hat{U}}_2, {\hat{U}}_3, {\hat{U}}_4)^T= & {} \left( \langle {\hat{Q}}_1, G\rangle , \langle {\hat{Q}}_1^2, G\rangle , \langle {\hat{Q}}_2, G\rangle , \langle {\hat{Q}}_1{\hat{Q}}_2, G\rangle \right) ^T.\nonumber \\ \end{aligned}$$

(A.2)

Then ${\hat{\varvec{\theta }}}\equiv {\varvec{g}}({\hat{\varvec{U}}})=(g_1({\hat{\varvec{U}}}),g_2({\hat{\varvec{U}}}))^T$, where $g_1$ and $g_2$ are functions of ${\varvec{u}}=(u_1, u_2,u_3,u_4)^T$:

$$\begin{aligned} g_1({\varvec{u}}) = \frac{u_2u_3-u_1u_4}{cu_2-u_1^2}; \quad g_2({\varvec{u}}) = \frac{cu_4-u_1u_3}{cu_2-u_1^2}, \end{aligned}$$

where $c=\langle 1, G\rangle$. Zhang and Yu (2002) derived a large-sample representation for ${\hat{\varvec{\theta }}}$. To develop an equivalent martingale representation for ${\hat{\varvec{\theta }}}$, let ${\varvec{D}}=[D_{il}]_{2\times 4}=\frac{d{\varvec{g}}}{d{\varvec{u}}}\vert _{{\varvec{u}}={\varvec{U}}}$, where

$$\begin{aligned} {\varvec{U}}= & {} (U_1, U_2, U_3, U_4)^T = \left( \langle Q_1, G\rangle , \langle Q_1^2, G\rangle , \langle Q_2, G\rangle , \langle Q_1Q_2, G\rangle \right) ^T; \\ \frac{d{\varvec{g}}}{d{\varvec{u}}}= & {} \begin{pmatrix} \frac{-u_4(cu_2-{u_1}^2)+2u_1(u_2u_3-u_1u_4)}{(cu_2-{u_1}^2)^2} &{}\quad \frac{u_3(cu_2-{u_1}^2)-c(u_2u_3-u_1u_4)}{(cu_2-{u_1}^2)^2} &{}\quad \frac{u_2}{cu_2-{u_1}^2} &{}\quad \frac{-u_1}{cu_2-{u_1}^2} \\ \frac{-u_3(cu_2-{u_1}^2)+2u_1(-u_1u_3+cu_4)}{(cu_2-{u_1}^2)^2} &{}\quad \frac{c(u_1u_3-cu_4)}{(cu_2-{u_1}^2)^2} &{}\quad \frac{-u_1}{cu_2-{u_1}^2} &{}\quad \frac{c}{cu_2-{u_1}^2} \end{pmatrix}. \end{aligned}$$

Then ${\hat{\varvec{\theta }}}-\varvec{\theta }_0 = {\varvec{D}}({\hat{\varvec{U}}}-{\varvec{U}}) + o_{{\mathbb {P}}}(n^{-1/2})$. Let $\tau _{i,l}$ and $\tau _{i,u}$ denote the infimum and supremum of the support set for $G\circ F_i(\cdot )\equiv G(F_i(\cdot ))$. To deal with ${\hat{\varvec{U}}}-{\varvec{U}}$, for $s \in (\tau _{i,l}, \tau _{i,u})$, define

$$\begin{aligned} e_i(s)= & {} \left\langle S_iI_{(s,\tau _{i,u})}/f_i, G\circ F_i\right\rangle ,\quad i=1,2, \\ {\tilde{e}}_{il}(s)= & {} \left\langle (Q_l\circ F_i)S_iI_{(s,\tau _{i,u})}/f_i, G\circ F_i\right\rangle ,\quad i=1,2,\;\; l=1,2. \end{aligned}$$

For the first component of ${\hat{\varvec{U}}}-{\varvec{U}}$, say, apply Eq. (2.4), then apply Theorem II.6.2 of Andersen et al. (1993) for ${\hat{F}}_1-F_1$, and finally interchange the order of integration to obtain

$$\begin{aligned} \langle {\hat{Q}}_1 - Q_1, G\rangle= & {} -\int e_1(s)d\left( {\hat{\Lambda }}_1(s) - \Lambda _1(s)\right) + o_{{\mathbb {P}}}(n^{-1/2}). \end{aligned}$$

The other components of ${\hat{\varvec{U}}}-{\varvec{U}}$ can be treated likewise. Let $\varvec{W}(s)\equiv [W_{kl}(s)]_{4\times 2}$, where $W_{11}(s)=e_1(s), W_{21}(s)=2{\tilde{e}}_{11}(s)$, $W_{32}(s)=e_2(s)$, $W_{41}(s)={\tilde{e}}_{12}(s)$, $W_{42}(s)={\tilde{e}}_{21}(s)$, and $W_{12}(s)=W_{22}(s)=W_{31}(s)=0$. It follows that ${\hat{\varvec{\theta }}}- \varvec{\theta }_0$ has the large-sample representation

$$\begin{aligned} {\hat{\varvec{\theta }}}-\varvec{\theta }_0= & {} -{\varvec{D}}\int \varvec{W}(s)\, d \begin{pmatrix} \hat{\Lambda }_1(s) - \Lambda _1(s) \\ \hat{\Lambda }_2(s) - \Lambda _2(s) \end{pmatrix} + o_{{\mathbb {P}}}(n^{-1/2}) \nonumber \\= & {} -{\varvec{D}}\int \varvec{W}(s)\, \begin{pmatrix} dM_{1\cdot }(s)/(n_1y_1(s)) \\ dM_{2\cdot }(s)/(n_2y_2(s)) \end{pmatrix} + o_{{\mathbb {P}}}(n^{-1/2}) . \end{aligned}$$

(A.3)

Note that the last step above is obtained by applying Eq. (2.1). We will exploit Eq. (A.3) when we derive the large-sample null distribution of the plug-in EL.

Remark A.1

For a purely location model, $\theta _2=1$. Then $\theta _1\equiv \theta$ is the only parameter and its MDE is ${\hat{\varvec{\theta }}}\equiv {\hat{\theta }} = \langle {\hat{Q}}_2 - {\hat{Q}}_1, G\rangle$. We can write $\hat{\theta }- \theta _0={\varvec{D}}^T({\hat{\varvec{U}}}-{\varvec{U}})$, where $D=(-1,1)^T$ and ${\hat{\varvec{U}}}-{\varvec{U}}=(\langle {\hat{Q}}_1 - Q_1, G\rangle , \langle {\hat{Q}}_2 - Q_2, G\rangle )^T$. Then ${\hat{\theta }} - \theta _0$ has the large-sample representation

$$\begin{aligned} {\hat{\theta }}-\theta _0= & {} \frac{1}{n_1}\sum _{j=1}^{n_1} \int \frac{e_1(s)}{y_1(s)}dM_{1j}(s) - \frac{1}{n_2}\sum _{j=1}^{n_2} \int \frac{e_2(s)}{y_2(s)}dM_{2j}(s) + o_{{\mathbb {P}}}(n^{-1/2}),\nonumber \\ \end{aligned}$$

(A.4)

which one can exploit when testing the adequacy of a location model, see Remark 7.

Recall that ${\tilde{\kappa }}_i(t)$ is the number of distinct uncensored lifetimes by time t for each sample. In the notation to handle ties, the two cumulative hazard estimators are given by

$$\begin{aligned} \hat{\Lambda }_i(t)= & {} \sum _{j=1}^{\tilde{\kappa }_i(t)}\left( \frac{d_{ij} }{r_{ij}}\right) ;\quad {\hat{\zeta }}_i(t)\ =\ - \sum _{j=1}^{\tilde{\kappa }_i(t)}\log \left( 1 - \frac{d_{ij}}{r_{ij}}\right) ,\;\; i=1,2. \end{aligned}$$

(A.5)

Let $\Vert h\Vert _{\alpha _1}^{\alpha _2} = \sup _{t\in [\alpha _1,\alpha _2]}|h(t)|$. Lemma 1 gives the asymptotic equivalence between ${\hat{\Lambda }}_i$ and ${\hat{\zeta }}_i$.

Lemma 1

Assume that $S_i(t), i=1,2$, are continuous. Then $\Vert \hat{\Lambda }_i-{\hat{\zeta }}_i\Vert _{\alpha _1}^{\alpha _2} = o_{{\mathbb {P}}}(n^{-1/2})$.

Proof

Following McKeague and Zhao (2005), see their Eq. (A.6), we can show that

$$\begin{aligned} n_i^{1/2}|\hat{\Lambda }_i(t) - {\hat{\zeta }}_i(t)|\le & {} \left\{ \max _{j \le \tilde{\kappa }_i(t)}\left( \frac{n_i^{1/2}d_{ij}}{r_{ij}}\right) \right\} \times \left\{ \sum _{j=1}^{\tilde{\kappa }_i(t)}\left( \frac{d_{ij}}{r_{ij}}\right) \right\} . \end{aligned}$$

(A.6)

By continuity of $S_i$, the first quantity on the right hand side (RHS) of inequality (A.6) is

$$\begin{aligned} \max _{j \le \tilde{\kappa }_i(t)}\left( n_i^{1/2}/r_{ij}\right)\le & {} n_i^{-1/2} \Vert n_i/Y_{i\cdot }\Vert _{0}^{\alpha _2}\ =\ O\left( n_i^{-1/2}\right) O_{{\mathbb {P}}}(1) \ =\ o_{{\mathbb {P}}}(1). \end{aligned}$$

(A.7)

From Eq. (A.5), the second quantity on the RHS of inequality (A.6) is bounded above by $\hat{\Lambda }_i(\alpha _2)=O_{{\mathbb {P}}}(1)$. Combined with inequalities (A.6) and (A.7), this completes the proof. $\square$

Lemma 2

The Lagrange multiplier, ${\hat{\lambda }}_{\varvec{\theta }}(t)$, solving Eq. (2.10), satisfies

$$\begin{aligned} |{\hat{\lambda }}_{\varvec{\theta }}(t)|\le & {} \frac{n_2({\hat{\zeta }}_1(t) - \hat{\Lambda }_2(\varphi _{\varvec{\theta }}(t)))}{\hat{\Lambda }_2(\varphi _{\varvec{\theta }}(t)))},\quad \mathrm{when\ }{\hat{\lambda }}_{\varvec{\theta }}(t) < 0. \end{aligned}$$

(A.8)

$$\begin{aligned} |{\hat{\lambda }}_{\varvec{\theta }}(t)|\le & {} \frac{n_1({\hat{\zeta }}_2(\varphi _{\varvec{\theta }}(t)) - \hat{\Lambda }_1(t))}{\hat{\Lambda }_1(t)},\quad \mathrm{when\ }{\hat{\lambda }}_{\varvec{\theta }}(t) > 0. \end{aligned}$$

(A.9)

Proof

From Eq. (2.10), ${\hat{\lambda }}_{\varvec{\theta }}(t)$ satisfies the equation $I_1(t)-I_2(t)=0$, where

$$\begin{aligned} I_1(t)= \sum _{j=1}^{{\tilde{\kappa }}_1(t)}\log \left( 1- \frac{d_{1j}}{r_{1j}-{\hat{\lambda }}_{\varvec{\theta }}(t)}\right) ; \quad I_2(t)=\sum _{j=1}^{{\tilde{\kappa }}_2(\varphi _{\varvec{\theta }}(t))}\log \left( 1-\frac{d_{2j}}{r_{2j}+{\hat{\lambda }}_{\varvec{\theta }}(t)}\right) . \end{aligned}$$

Consider ${\hat{\lambda }}_{\varvec{\theta }}(t)<0$. Follow Li (1995), leading to his Eq. (2.12) on p. 101, to obtain

$$\begin{aligned} -I_2(t)\ge & {} \sum _{j=1}^{{\tilde{\kappa }}_2(\varphi _{\varvec{\theta }}(t))}\frac{d_{2j}}{r_{2j}}\left( \frac{n_2}{n_2-|{\hat{\lambda }}_{\varvec{\theta }}(t)|}\right) \ =\ \hat{\Lambda }_2(\varphi _{\varvec{\theta }}(t))\left( \frac{1}{1-|{\hat{\lambda }}_{\varvec{\theta }}(t)|/n_2}\right) . \end{aligned}$$

Applying some elementary calculations (cf. Ahmed and Subramanian, 2015), we can show that, for large enough $n_2$, $0<|{\hat{\lambda }}_{\varvec{\theta }}(t)|/n_2<1$ almost surely. Now, as in McKeague and Zhao (2005), use the fact that $1/(1-x)\ge 1+x$ for $0<x<1$, to obtain

$$\begin{aligned} -I_2(t)\ge & {} \hat{\Lambda }_2(\varphi _{\varvec{\theta }}(t)) + \frac{\hat{\Lambda }_2(\varphi _{\varvec{\theta }}(t))|{\hat{\lambda }}_{\varvec{\theta }}(t)|}{n_2}. \end{aligned}$$

(A.10)

Since $-{\hat{\lambda }}_{\varvec{\theta }}(t)>0$, follow Li (1995) leading to his Eq. (2.13) on p. 101, and obtain

$$\begin{aligned} I_1(t)\ge & {} -\sum _{j=1}^{{\tilde{\kappa }}_1(t)}\frac{d_{1j}}{r_{1j}}\left( \frac{n_1}{n_1+|{\hat{\lambda }}_{\varvec{\theta }}(t)|}\right) + \sum _{j=1}^{{\tilde{\kappa }}_1(t)}\left( \log \left( 1-\frac{d_{1j}}{r_{1j}}\right) +\frac{d_{1j}}{r_{1j}}\right) \nonumber \\= & {} -\hat{\Lambda }_1(t)\left( \frac{n_1}{n_1+|{\hat{\lambda }}_{\varvec{\theta }}(t)|}\right) - {\hat{\zeta }}_1(t) + \hat{\Lambda }_1(t) \nonumber \\= & {} \hat{\Lambda }_1(t)\left( \frac{|{\hat{\lambda }}_{\varvec{\theta }}(t)|}{n_1+|{\hat{\lambda }}_{\varvec{\theta }}(t)|}\right) - {\hat{\zeta }}_1(t)\ \ge \ - {\hat{\zeta }}_1(t). \end{aligned}$$

(A.11)

Combine inequalities (A.10) and (A.11) to obtain inequality (A.8) from

$$\begin{aligned} 0= & {} I_1(t)-I_2(t) \ \ge \ - {\hat{\zeta }}_1(\varphi _{\varvec{\theta }}(t)) + \hat{\Lambda }_2(t) + \frac{\hat{\Lambda }_2(t)|{\hat{\lambda }}_{\varvec{\theta }}(t)|}{n_2}. \end{aligned}$$

When ${\hat{\lambda }}_{\varvec{\theta }}(t)>0$, the above approach can be repeated by applying Li’s (1995) technique in reverse order. First obtain the inequality

$$\begin{aligned} I_2(t)\ge & {} -\hat{\Lambda }_2(\varphi _{\varvec{\theta }}(t))\left( \frac{n_2}{n_2+|{\hat{\lambda }}_{\varvec{\theta }}(t)|}\right) - {\hat{\zeta }}_2(\varphi _{\varvec{\theta }}(t)) + \hat{\Lambda }_2(\varphi _{\varvec{\theta }}(t)) \nonumber \\= & {} \hat{\Lambda }_2(\varphi _{\varvec{\theta }}(t))\left( \frac{|{\hat{\lambda }}_{\varvec{\theta }}(t)|}{n_2+|{\hat{\lambda }}_{\varvec{\theta }}(t)|}\right) - {\hat{\zeta }}_2(\varphi _{\varvec{\theta }}(t))\ \ge \ -{\hat{\zeta }}_2(\varphi _{\varvec{\theta }}(t)). \end{aligned}$$

(A.12)

Then obtain the inequality

$$\begin{aligned} -I_1(t)\ge & {} \hat{\Lambda }_1(t)\left( \frac{n_1}{n_1-|{\hat{\lambda }}_{\varvec{\theta }}(t)|}\right) \ \ge \ \hat{\Lambda }_1(t)\left( 1+|{\hat{\lambda }}_{\varvec{\theta }}(t)|/n_1\right) . \end{aligned}$$

(A.13)

Combine inequalities (A.12) and (A.13) to obtain inequality (A.9) from

$$\begin{aligned} 0= & {} I_2(t)-I_1(t) \ge \ -{\hat{\zeta }}_2(\varphi _{\varvec{\theta }}(t)) + \hat{\Lambda }_1(t) + \frac{\hat{\Lambda }_1(t)|{\hat{\lambda }}_{\varvec{\theta }}(t)|}{n_1}. \end{aligned}$$

$\square$

Remark A.3

Define $\mu (\cdot )=-\log (\cdot )$, which is a continuously differentiable function. Then ${\hat{\zeta }}_2(t)=\mu ({\hat{S}}_2(t))$. It follows by Lemma 1 of Ying et al. (1995) that

$$\begin{aligned} \sup _{t \in [\alpha _1,\alpha _2]}\left| {\hat{\zeta }}_2(\varphi _{{\hat{\varvec{\theta }}}}(t))+\log S_2(\varphi _{{\hat{\varvec{\theta }}}}(t))-{\hat{\zeta }}_2(\varphi _{\varvec{\theta }_0}(t)) - \log S_2(\varphi _{\varvec{\theta }_0}(t))\right|= & {} o_{{\mathbb {P}}}(n^{-1/2}).\nonumber \\ \end{aligned}$$

(A.14)

Lemma 3

The estimate ${\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)$, solving Eq. (2.10) with $\varvec{\theta }={\hat{\varvec{\theta }}}$, satisfies $\Vert {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}\Vert _{\alpha _1}^{\alpha _2} = O_{{\mathbb {P}}}(n^{1/2})$.

Proof

The denominators on the RHS of inequalities (A.8) and (A.9) are each $O_{{\mathbb {P}}}(1)$, uniformly for $t\in [\alpha _1,\alpha _2]$. This follows from $\Vert \hat{\Lambda }_i-\Lambda _i\Vert _0^{\tau _i}=o_{{\mathbb {P}}}(1)$, where $\tau _1=\alpha _2$ and $\tau _2=\varphi _{\varvec{\theta }_0}(\alpha _2)$, see Sect. 2.2. Furthermore, the denominators on the RHS of inequalities (A.8) and (A.9) are bounded away from 0 for $t\in [\alpha _1, \alpha _2]$. It remains to show that the numerators on the RHS of inequalities (A.8) and (A.9) are each $O_{{\mathbb {P}}}(n^{1/2})$ uniformly for $t\in [\alpha _1,\alpha _2]$. Consider the numerator on the RHS of inequality (A.8). Note that under the null hypothesis, $-\log (S_2(\varphi _{\varvec{\theta }_0}(t))=\Lambda _2(\varphi _{\varvec{\theta }_0}(t))=\Lambda _1(t)$. Therefore,

$$\begin{aligned} {\hat{\zeta }}_1(t) - \hat{\Lambda }_2(\varphi _{{\hat{\varvec{\theta }}}}(t))= & {} \left[ {\hat{\zeta }}_1(t)-{\hat{\Lambda }}_1(t)\right] +\left[ {\hat{\Lambda }}_1(t)-\Lambda _1(t)\right] \\&-\,\left[ \hat{\Lambda }_2(\varphi _{{\hat{\varvec{\theta }}}}(t))-{\hat{\zeta }}_2(\varphi _{{\hat{\varvec{\theta }}}}(t))\right] \\&-\,\left[ {\hat{\zeta }}_2(\varphi _{{\hat{\varvec{\theta }}}}(t)) + \log S_2(\varphi _{{\hat{\varvec{\theta }}}}(t)) - {\hat{\zeta }}_2(\varphi _{\varvec{\theta }_0}(t)) - \log S_2(\varphi _{\varvec{\theta }_0}(t))\right] \\&+ \left[ \log S_2(\varphi _{{\hat{\varvec{\theta }}}}(t)) - \log S_2(\varphi _{\varvec{\theta }_0}(t))\right] \\&-\,\left[ {\hat{\zeta }}_2(\varphi _{\varvec{\theta }_0}(t)) + \log S_2(\varphi _{\varvec{\theta }_0}(t))\right] . \end{aligned}$$

By Lemma 1, $\Vert {\hat{\zeta }}_1-{\hat{\Lambda }}_1\Vert _{\alpha _1}^{\alpha _2}=o_{{\mathbb {P}}}(n^{-1/2})$. Also, it is standard that $\Vert {\hat{\Lambda }}_1-\Lambda _1\Vert _{\alpha _1}^{\alpha _2}=O_{{\mathbb {P}}}(n^{-1/2})$. By Remark A.2, the third expression within square brackets is $o_{{\mathbb {P}}}(n^{-1/2})$ uniformly for $t \in [\alpha _1, \alpha _2]$. By Remark A.3, the fourth expression within square brackets is $o_{{\mathbb {P}}}(n^{-1/2})$ uniformly for $t \in [\alpha _1, \alpha _2]$. By a Taylor expansion around $\varphi _{\varvec{\theta }_0}(t)$ and the $n^{1/2}$ consistency of the MDE ${\hat{\varvec{\theta }}}$ of $\varvec{\theta }$, the fifth expression within square brackets is seen to be $O_{{\mathbb {P}}}(n^{-1/2})$ uniformly for $t\in [\alpha _1, \alpha _2]$. By a Taylor expansion, the last expression, which equals $-\log {\hat{S}}_2(\varphi _{\varvec{\theta }_0}(t)) + \log S_2(\varphi _{\varvec{\theta }_0}(t))$, is also $O_p(n^{-1/2})$ uniformly for $t\in [\alpha _1, \alpha _2]$. Likewise, it can be shown that the numerator on the RHS of inequality (A.9) is also $O_{{\mathbb {P}}}(n^{-1/2})$ uniformly for $t\in [\alpha _1, \alpha _2]$. $\square$

In the proof of Theorem 1, we will show that $-2R_{{\hat{\varvec{\theta }}}}(t)$ is asymptotically equivalent to the square of $n^{1/2}{\hat{{\mathbb {V}}}}(t)$ scaled by the reciprocal of $\sigma _{\mathrm{c}}^2(t)$ defined by Eq. (2.14),

Lemma 4

Under the null hypothesis, ${\hat{{\mathbb {V}}}}(t)$ is asymptotically linear with influence function $\sum _{i=1}^2 J_{i1}(t)/\sqrt{\rho _i}$, where $J_{11}(t)$ and $J_{21}(t)$ are given by Eqs. (2.15) and (2.16) respectively.

Proof

First apply a Taylor’s expansion for $\log {\hat{S}}_1(t)$ about $S_1(t)$. Then apply the Duhamel equation (Andersen et al. 1993) and Eq. (2.1) to yield an asymptotic representation for the first term of ${\hat{{\mathbb {V}}}}(t)$ as shown below:

$$\begin{aligned} -{\hat{\zeta }}_1(t)= & {} \log S_1(t) + \left( \log {\hat{S}}_1(t) - \log S_1(t)\right) \nonumber \\= & {} \log S_1(t) + \frac{{\hat{S}}_1(t) - S_1(t)}{S_1(t)} + o_{{\mathbb {P}}}(n^{-1/2}) \nonumber \\= & {} -\Lambda _1(t) - \frac{1}{n_1}\sum _{j=1}^{n_1} \int _0^t \frac{1}{y_1(u)}dM_{1j}(u) + o_{{\mathbb {P}}}(n^{-1/2}). \end{aligned}$$

(A.15)

For the second term of ${\hat{{\mathbb {V}}}}(t)$, applying Eq. (A.14) of Remark A.3, it follows that

$$\begin{aligned} {\hat{\zeta }}_2(\varphi _{{\hat{\varvec{\theta }}}}(t))= & {} -\log S_2(\varphi _{{\hat{\varvec{\theta }}}}(t)) - \log {\hat{S}}_2(\varphi _{\varvec{\theta }_0}(t)) + \log (S_2(\varphi _{\varvec{\theta }_0}(t)) + o_{{\mathbb {P}}}(n^{-1/2}) \nonumber \\= & {} \Lambda _2(\varphi _{{\hat{\varvec{\theta }}}}(t)) - \Lambda _2(\varphi _{\varvec{\theta }_0}(t)) - \log {\hat{S}}_2(\varphi _{\varvec{\theta }_0}(t)) + o_{{\mathbb {P}}}(n^{-1/2}). \end{aligned}$$

(A.16)

Applying the delta method to the first and second terms on the RHS of Eq. (A.16) yields

$$\begin{aligned} \Lambda _2(\varphi _{{\hat{\varvec{\theta }}}}(t)) - \Lambda _2(\varphi _{\varvec{\theta }_0}(t))= & {} \lambda _2(\varphi _{\varvec{\theta }_0}(t))\, (\varphi _{{\hat{\varvec{\theta }}}}(t)-\varphi _{\varvec{\theta }_0}(t)) + o_{{\mathbb {P}}}(n^{-1/2}) \nonumber \\= & {} \lambda _2(\varphi _{\varvec{\theta }_0}(t))\,{\varvec{c}}_t^T \left( {\hat{\varvec{\theta }}}-\varvec{\theta }_0\right) + o_{{\mathbb {P}}}(n^{-1/2}) \nonumber \\= & {} -\lambda _2(\varphi _{\varvec{\theta }_0}(t))\,{\varvec{c}}_t^T {\varvec{D}}\int \varvec{W}(u)\,\begin{pmatrix} dM_{1\cdot }(u)/(n_1y_1(u)) \\ dM_{2\cdot }(u)/(n_2y_2(u)) \end{pmatrix} \nonumber \\&+\, o_{{\mathbb {P}}}(n^{-1/2}),\quad \end{aligned}$$

(A.17)

the last step that leads to Eq. (A.17) following from Eq. (A.3), where $\lambda _2$ is the hazard function associated with $F_2$ and ${\varvec{c}}_t=(1,t)^T$. The third term on the RHS of Eq. (A.16) can be treated exactly the way the first term of ${\hat{{\mathbb {V}}}}(t)$ yielded Eq. (A.15):

$$\begin{aligned} - \log {\hat{S}}_2(\varphi _{\varvec{\theta }_0}(t))= & {} -\log S_2(\varphi _{\varvec{\theta }_0}(t)) - \log {\hat{S}}_2(\varphi _{\varvec{\theta }_0}(t)) + \log S_2(\varphi _{\varvec{\theta }_0}(t)) \nonumber \\= & {} \Lambda _2(\varphi _{\varvec{\theta }_0}(t)) -\frac{{\hat{S}}_2(\varphi _{\varvec{\theta }_0}(t)) - S_2(\varphi _{\varvec{\theta }_0}(t))}{S_2(\varphi _{\varvec{\theta }_0}(t))} + o_{{\mathbb {P}}}(n^{-1/2})\nonumber \\= & {} \Lambda _2(\varphi _{\varvec{\theta }_0}(t)) + \frac{1}{n_2}\sum _{j=1}^{n_2} \int _0^{\varphi _{\varvec{\theta }_0}(t)} \frac{1}{y_2(u)}dM_{2j}(u) + o_{{\mathbb {P}}}(n^{-1/2}).\nonumber \\ \end{aligned}$$

(A.18)

Since $\Lambda _1(t)=\Lambda _2(\varphi _{\varvec{\theta }_0}(t))$, from Eqs. (2.13)–(A.18) we obtain

$$\begin{aligned} n^{1/2}{\hat{{\mathbb {V}}}}(t)=\sum _{i=1}^2 \left\{ n_i^{-1/2} \sum _{j=1}^{n_i}\frac{J_{ij}(t)}{\sqrt{\rho _i}}\right\} + o_{{\mathbb {P}}}(1). \end{aligned}$$

$\square$

1.2 Proof of Theorem 1

Let $\breve{\kappa }_1(t)={\tilde{\kappa }}_1(t)$, $\breve{\kappa }_2(t)={\tilde{\kappa }}_2(\varphi _{{\hat{\varvec{\theta }}}}(t))$. From Eq. (2.10), $h_1(-{\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t))-h_2({\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t))=0$, where

$$\begin{aligned} h_i(\lambda )= & {} \sum _{j=1}^{\breve{\kappa }_i(t)}\log \left( 1 - \frac{d_{ij}}{r_{ij} + \lambda }\right) . \end{aligned}$$

(A.19)

Note that $h_1(0) = - {\hat{\zeta }}_1(t)$ and $h_2(0)=- {\hat{\zeta }}_2(\varphi _{{\hat{\varvec{\theta }}}}(t))$ [cf. Eq. (A.5)]. Note also that

$$\begin{aligned} h_i^{'}(\lambda )= & {} \sum _{j=1}^{\breve{\kappa }_i(t)}\frac{d_{ij}}{(r_{ij}+\lambda )(r_{ij}+\lambda - d_{ij})},\quad h_i^{''}(\lambda )= \sum _{j=1}^{\breve{\kappa }_i(t)}\frac{d_{ij}(2(r_{ij} +\lambda )- d_{ij} )}{(r_{ij}+\lambda )^2(r_{ij}+\lambda -d_{ij})^2}. \end{aligned}$$

Clearly, $n_1h_1^{'}(0)={\hat{\sigma }}_1^2(t)$ and $n_2h_2^{'}(0)={\hat{\sigma }}_2^2(\varphi _{{\hat{\varvec{\theta }}}}(t))$, see Eq. (2.2). Let $|{\hat{\xi }}_i|\le |{\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)|, i=1,2$. Taylor’s expansion about 0 yields

$$\begin{aligned} h_1(-{\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t))= & {} - {\hat{\zeta }}_1(t) -\frac{{\hat{\sigma }}_1^2(t){\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)}{n_1} + \frac{1}{2}h^{''}_1({\hat{\xi }}_1)({\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t))^2, \end{aligned}$$

(A.20)

$$\begin{aligned} h_2\left( {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)\right)= & {} - {\hat{\zeta }}_2(\varphi _{{\hat{\varvec{\theta }}}}(t)) + \frac{{\hat{\sigma }}_2^2(\varphi _{{\hat{\varvec{\theta }}}}(t)){\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)}{n_2} + \frac{1}{2}h^{''}_1({\hat{\xi }}_2)\left( {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)\right) ^2. \end{aligned}$$

(A.21)

Applying the Glivenko–Cantelli lemma to $r_{ij}$, we have $\Vert h_i''({\hat{\xi }}_i)\Vert _{\alpha _1}^{\alpha _2}=O_{{\mathbb {P}}}(n_i^{-2})$. Therefore, by Lemma 3, it follows that $\Vert h^{''}_i({\hat{\xi }}_i)({\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t))^2\Vert _{\alpha _1}^{\alpha _2}=O_{{\mathbb {P}}}(n_i^{-1})=O_{{\mathbb {P}}}(n^{-1})$. Recalling ${\hat{{\mathbb {V}}}}(t)$ defined by Eq. (2.13), it follows from Eqs. (A.20), (A.21), and (2.14) that

$$\begin{aligned} 0= & {} -{\hat{\zeta }}_1(t) + {\hat{\zeta }}_2(\varphi _{{\hat{\varvec{\theta }}}}(t)) - {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)\left( \frac{{\hat{\sigma }}_1^2(t)}{n_1} +\frac{{\hat{\sigma }}_2^2(\varphi _{{\hat{\varvec{\theta }}}}(t))}{n_2}\right) + O_{{\mathbb {P}}}\left( \frac{1}{n}\right) \\= & {} -{\hat{{\mathbb {V}}}}(t) - \frac{{\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t) {\hat{\sigma }}^2_{\mathrm{c}}(t)}{n} + O_{{\mathbb {P}}}\left( \frac{1}{n}\right) . \end{aligned}$$

Solving for ${\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)$, we obtain

$$\begin{aligned} {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)= & {} \frac{n}{{\hat{\sigma }}^2_{\mathrm{c}}(t)}\left\{ -{\hat{{\mathbb {V}}}}(t)+O_{{\mathbb {P}}}\left( \frac{1}{n}\right) \right\} . \end{aligned}$$

(A.22)

To complete the proof of Theorem 1, consider Eq. (2.12). Using Taylor expansions of $\log (1+x)$ and $\log (1-x)$ about 0 and applying Eq. (A.22), the leading term of $-2R_{{\hat{\varvec{\theta }}}}(t)$ equals

$$\begin{aligned} \left( {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)\right) ^2\left[ \sum _{i=1}^2\sum _{j=1}^{\breve{\kappa }_i(t)}\frac{d_{ij}}{r_{ij}(r_{ij}-d_{ij})}\right]= & {} \left( {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)\right) ^2\left( \frac{{\hat{\sigma }}_1^2(t)}{n_1} +\frac{{\hat{\sigma }}_2^2(\varphi _{{\hat{\varvec{\theta }}}}(t))}{n_2}\right) \nonumber \\= & {} \frac{n}{\left( {\hat{\sigma }}^2_{\mathrm{c}}(t)\right) ^2}\left\{ {\hat{{\mathbb {V}}}}(t)\right. \nonumber \\&\left. +O_{{\mathbb {P}}}\left( \frac{1}{n}\right) \right\} ^2\left( \frac{{\hat{\sigma }}_1^2(t)}{n_1/n} +\frac{{\hat{\sigma }}_2^2(\varphi _{{\hat{\varvec{\theta }}}}(t))}{n_2/n}\right) \nonumber \\= & {} \frac{1}{\left( {\hat{\sigma }}^2_{\mathrm{c}}(t)\right) ^2}\left\{ n^{1/2}{\hat{{\mathbb {V}}}}(t)+o_{{\mathbb {P}}}(1)\right\} ^2 {\hat{\sigma }}^2_{\mathrm{c}}(t) + o_{{\mathbb {P}}}(1) \nonumber \\= & {} \frac{1}{\sigma ^2_{\mathrm{c}}(t)}\left\{ n^{1/2}{\hat{{\mathbb {V}}}}(t)\right\} ^2 + o_{{\mathbb {P}}}(1), \end{aligned}$$

(A.23)

uniformly over $[\alpha _1,\alpha _2]$. The subsequent terms of $-2R(t)$ are proportional to

$$\begin{aligned} \left( {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)\right) ^l\left[ \sum _{i=1}^2\sum _{j=1}^{\breve{\kappa }_i(t)}\left( \frac{1}{(r_{ij}-d_{ij})^{l-1}} - \frac{1}{r_{ij}^{l-1}}\right) \right] , \quad l=3, 4, \ldots , \end{aligned}$$

each of which is $o_{{\mathbb {P}}}(1)$, uniformly for $t\in [\alpha _1,\alpha _2]$. For example, when $l=3$, we have

$$\begin{aligned} \frac{2}{3}\left( {\hat{\lambda }}_{{\hat{\varvec{\theta }}}}(t)\right) ^3\sum _{j=1}^{\breve{\kappa }_i(t)}\left( \frac{1}{r_{ij}-d_{ij}} - \frac{1}{r_{ij}}\right) \left( \frac{1}{r_{ij} - d_{ij}} + \frac{1}{r_{ij}}\right) ,\quad i=1,2, \end{aligned}$$

which, uniformly for $t\in [\alpha _1,\alpha _2]$, equals

$$\begin{aligned} O_{{\mathbb {P}}}(n^{3/2})O_{{\mathbb {P}}}(n_i^{-1})\sum _{j=1}^{\breve{\kappa }_1(t)}\left( \frac{1}{r_{1j} - d_{1j}} - \frac{1}{r_{1j}}\right)= & {} O_{{\mathbb {P}}}(n^{1/2})O_{{\mathbb {P}}}(n_1^{-1})\ =\ o_{{\mathbb {P}}}(1); \\ O_{{\mathbb {P}}}(n^{3/2})O_{{\mathbb {P}}}(n_i^{-1})\sum _{j=1}^{\breve{\kappa }_2(t)}\left( \frac{1}{r_{2j} - d_{2j}} - \frac{1}{r_{2j}}\right)= & {} O_{{\mathbb {P}}}(n^{1/2})O_{{\mathbb {P}}}(n_2^{-1})\ =\ o_{{\mathbb {P}}}(1). \end{aligned}$$

From Eq. (A.23) and Lemma 4 the proof of Theorem 1 is completed. $\square$