Semiparametric simultaneous confidence bands for the difference of survival functions

Abstract

In the analysis of censored survival data, simultaneous confidence bands are useful devices to help determine the efficacy of a treatment over a control. Semiparametric confidence bands are developed for the difference of two survival curves using empirical likelihood and compared with the nonparametric counterpart. Simulation studies are presented to show that the proposed semiparametric approach is superior, with the new confidence bands giving empirical coverage closer to the nominal level. Further comparisons reveal that the semiparametric confidence bands are tighter and, hence, more informative. For censoring rates between 10 and 40 %, the semiparametric confidence bands provide a relative reduction in enclosed area amounting to between 2 and 10 % over their nonparametric bands, with increased reduction attained for higher censoring rates. The methods are illustrated using an University of Massachusetts AIDS data set.

Appendix

Proof of Eq. (2.13)

Then, suppressing the dependence on t, we can write $R = \tilde{R}_1 + \tilde{R}_2 + \tilde{R}_3 +\tilde{R}_4$, where

$$\begin{aligned} \tilde{R}_1= & {} \sum _{j=1}^{\tilde{\kappa }_1(t)}d_{1j}\hat{m}_{1j}\log \left( \frac{d_{1j}\hat{m}_{1j}}{r_{1j}+\hat{\lambda }\hat{S}_1(T_{1j})}\right) - \sum _{j=1}^{\tilde{\kappa }_1(t)}d_{1j}\hat{m}_{1j}\log \left( \frac{d_{1j}\hat{m}_{1j}}{r_{1j}}\right) , \\ \tilde{R}_2= & {} \sum _{j=1}^{\tilde{\kappa }_1(t)}(r_{1j} - d_{1j}\hat{m}_{1j})\log \left( 1 -\frac{d_{1j}\hat{m}_{1j}}{r_{1j}+\hat{\lambda }\hat{S}_1(T_{1j})}\right) \\&- \sum _{j=1}^{\tilde{\kappa }_1(t)}(r_{1j} - d_{1j}\hat{m}_{1j})\log \left( 1 -\frac{d_{1j}\hat{m}_{1j}}{r_{1j}}\right) ,\\ \tilde{R}_3= & {} \sum _{j=1}^{\tilde{\kappa }_2(t)}d_{2j}\hat{m}_{2j}\log \left( \frac{d_{2j}\hat{m}_{2j}}{r_{2j} - \hat{\lambda }\hat{S}_2(T_{2j})}\right) - \sum _{j=1}^{\tilde{\kappa }_2(t)}d_{2j}\hat{m}_{2j}\log \left( \frac{d_{2j}\hat{m}_{2j}}{r_{2j}}\right) ,\\ \tilde{R}_4= & {} \sum _{j=1}^{\tilde{\kappa }_2(t)}(r_{2j} - d_{2j}\hat{m}_{2j})\log \left( 1 -\frac{d_{2j}\hat{m}_{2j}}{r_{2j}-\hat{\lambda }\hat{S}_2(T_{2j})}\right) \\&- \sum _{j=1}^{\tilde{\kappa }_2(t)}(r_{2j} - d_{2j}\hat{m}_{2j})\log \left( 1 -\frac{d_{2j}\hat{m}_{2j}}{r_{2j}}\right) . \end{aligned}$$

However, writing $\tilde{R}_1 = R_1 - \breve{R}_1$, where

$$\begin{aligned} \breve{R}_1= & {} \sum _{j=1}^{\tilde{\kappa }_1(t)}(r_{1j} - d_{1j}\hat{m}_{1j})\log \left( \frac{d_{1j}\hat{m}_{1j}}{r_{1j}+\hat{\lambda }\hat{S}_1(T_{1j})}\right) - \sum _{j=1}^{\tilde{\kappa }_1(t)}(r_{1j} - d_{1j}\hat{m}_{1j})\log \left( \frac{d_{1j}\hat{m}_{1j}}{r_{1j}}\right) , \end{aligned}$$

we see that

$$\begin{aligned} R_1\equiv & {} \sum _{j=1}^{\tilde{\kappa }_1(t)}r_{1j}\log \left( \frac{d_{1j} \hat{m}_{1j}}{r_{1j}+\hat{\lambda }\hat{S}_1(T_{1j})}\right) - \sum _{j=1}^{\tilde{\kappa }_1(t)}r_{1j}\log \left( \frac{d_{1j} \hat{m}_{1j}}{r_{1j}}\right) \\= & {} -\sum _{j=1}^{\tilde{\kappa }_1(t)}r_{1j}\log \left( 1+\frac{\hat{\lambda }\hat{S}_1(T_{1j})}{r_{1j}}\right) . \end{aligned}$$

Next we write $\tilde{R}_1 + \tilde{R}_2 \equiv R_1 + R_2$, where $R_2=\tilde{R}_2 - \breve{R}_1$. Then, after some algebra, we obtain

$$\begin{aligned} R_2= & {} \sum _{j=1}^{\tilde{\kappa }_1(t)}(r_{1j}-d_{1j}\hat{m}_{1j})\log \left( 1 + \frac{\hat{\lambda }\hat{S}_1({T_{1j}})}{r_{1j}-d_{1j}\hat{m}_{1j}}\right) . \end{aligned}$$

Applying exactly the same technique, we have that $\tilde{R}_3+\tilde{R}_4\equiv R_3+R_4$, where

$$\begin{aligned} R_3= & {} -\sum _{j=1}^{\tilde{\kappa }_2(t)}r_{2j}\log \left( 1-\frac{\hat{\lambda }\hat{S}_2(T_{2j})}{r_{2j}}\right) ,\\ R_4= & {} \sum _{j=1}^{\hat{\kappa }_2(t)}(r_{2j}-d_{2j}\hat{m}_{2j})\log \left( 1 - \frac{\hat{\lambda }\hat{S}_2(T_{2j})}{r_{2j}-d_{2j}\hat{m}_{2j}}\right) . \end{aligned}$$

Therefore, we have shown that $R(t)=R_1+R_2+R_3+R_4$. It now follows that

We prove a number of lemmas. Recall that $\tilde{\kappa }_i(t)=\sum _{j=1}^{n_i}I(Z_{ij}\le t),i=1,2$. Define

$$\begin{aligned} \hat{\zeta }_i(t)= & {} \sum _{j=1}^{\tilde{\kappa }_i(t)}\left( \frac{d_{ij} \hat{m}_{ij}}{r_{ij}}\right) \hat{S}_i(T_{ij}) \equiv \int _0^t \hat{S}_i(s)d\hat{\Lambda }_i(s),\qquad i=1,2. \end{aligned}$$

(5.1)

$\square$

Lemma 1

The process $n_i^{1/2}(\hat{\zeta }_i-F_i)$ is asymptotically equivalent to $S_i\cdot n_i^{1/2}(\hat{\Lambda }_i-\Lambda _i)$.

Proof

Add and subtract $\int _0^t S_i(s) d\hat{\Lambda }_i(s)$ and apply the Duhamel equation to get

$$\begin{aligned} n_i^{1/2}\left( \hat{\zeta }_i(t) - F_i(t) \right)= & {} n_i^{1/2}\left( \int _0^t\hat{S}_i(s)d\hat{\Lambda }_i(s)- \int _0^t S_i(s)d\Lambda _i(s) \right) \\= & {} - n_i^{1/2}\int _0^t S_i(s) \left( \int _0^s \frac{\hat{S}_i(u-)}{S_i(u)}d\left( \hat{\Lambda }_i(u) - \Lambda _i(u)\right) \right) d\Lambda _i(s)\\&+\, n_i^{1/2} \int _0^t S_i(s)d \left( \hat{\Lambda }_i(s) - \Lambda _i(s) \right) \ :=\ I_1(t)+I_2(t). \end{aligned}$$

Interchanging the order of integration, we get, uniformly for $t\in [0,t_i]$ such that $y_i(t_i)>0$,

$$\begin{aligned} I_1(t)= & {} - n_i^{1/2} \int _0^t \left( \int _u^t S_i(s) d\Lambda _i(s) \right) \frac{\hat{S}_i(u-)}{S_i(u)} d \left( \hat{\Lambda }_i(u) - \Lambda _i(u) \right) . \end{aligned}$$

Applying Lenglart’s inequality, it follows that

$$\begin{aligned} I_1(t)= & {} - n_i^{1/2} \int _0^t \left( S_i(u) - S_i(t) \right) d \left( \hat{\Lambda }_i(u) - \Lambda _i(u) \right) + o_p(1)\\= & {} -I_2(t) + S_i(t)n_i^{1/2}\left( \hat{\Lambda }_i(t) - \Lambda _i(t)\right) + o_p(1). \end{aligned}$$

$\square$

Noting that $\int _0^t S_i^2(s) d\Lambda _i(s)=(1-S_i^2(t))/2$, next define

$$\begin{aligned} \hat{\zeta }_i^{(2)}(t)= & {} \sum _{j=1}^{\tilde{\kappa }_i(t)}\left( \frac{d_{ij} \hat{m}_{ij}}{r_{ij}}\right) \hat{S}_i^2(T_{ij})\ \equiv \ \int _0^t \hat{S}_i^2(s)d\hat{\Lambda }_i(s),\qquad i=1,2. \end{aligned}$$

(5.2)

Lemma 2

The process $n_i^{1/2}\left( \hat{\zeta }_i^{(2)}-\frac{1-S_i^2}{2}\right)$ is asymptotically equivalent to $S_i^2n_i^{1/2} (\hat{\Lambda }_i-\Lambda _i)$.

Proof

Add and subtract $\int _0^t S_i^2(s) d\hat{\Lambda }_i(s)$ and apply the Duhamel equation to get

$$\begin{aligned}&n_i^{1/2}\left( \hat{\zeta }_i^{(2)}(t)-\int _0^t S_i^2(s)d\Lambda _i(s)\right) \\&\quad = - n_i^{1/2}\int _0^t 2S_i^2(s) \left( \int _0^s \frac{\hat{S}_i(u-)}{S_i(u)}d\left( \hat{\Lambda }_i(u) - \Lambda _i(u)\right) \right) d\Lambda _i(s) \\&\qquad +\, n_i^{1/2} \int _0^t S_i^2(s)d \left( \hat{\Lambda }_i(s) - \Lambda _i(s) \right) \ :=\ I_3(t)+I_4(t). \end{aligned}$$

Interchanging the order of integration, we get, uniformly for $t\in [0,t_i]$ such that $y_i(t_i)>0$,

$$\begin{aligned} I_3(t)= & {} - n_i^{1/2} \int _0^t \left( \int _u^t 2S_i^2(s) d\Lambda _i(s) \right) d \left( \hat{\Lambda }_i(u) - \Lambda _i(u) \right) + o_p(1). \end{aligned}$$

Note that the integrand above equals $S_i^2(u) - S_i^2(t)$. It follows that

$$\begin{aligned} I_3(t)= & {} S_i^2(t)n_i^{1/2}\left( \hat{\Lambda }_i(t) - \Lambda _i(t)\right) - I_4(t) + o_p(1). \end{aligned}$$

$\square$

Recall that $\Vert h\Vert _{\tau _1}^{\tau _2} = \sup _{t\in [\tau _1,\tau _2]}|h(t)|$. For our next result, define

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

(5.3)

Lemma 3

For $i=1,2$, we have $\Vert \hat{\zeta }_i-\tilde{\zeta }_i\Vert _{\tau _1}^{\tau _2} = o_p(n^{-1/2})$.

Proof

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

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

(5.4)

Under continuity of $H_i$, the first quantity on the right side of Eq. (5.4) is bounded above by

$$\begin{aligned} \max _{j \le \tilde{\kappa }_i(t)}\left( n_i^{1/2}/r_{ij}\right)\le & {} n_i^{-1/2} \Vert 1/\bar{Y}_i\Vert _{0}^{\tau _2}\ =\ O\left( n_i^{-1/2}\right) O_p(1) \ =\ o_p(1). \end{aligned}$$

(5.5)

From Eq. (5.1), the second quantity on the right side of Eq. (5.4) is bounded above by $\hat{\zeta }_i(\tau _2)\ {\buildrel \mathrm{a.s.} \over \longrightarrow }\ F_i(\tau _2)$, see also Theorem 2.4 of Dikta (1998) and Lemma 1. Thus, $\hat{\zeta }_i(\tau _2)=O_p(1)$ which, combined with Eqs. (5.4) and (5.5), completes the proof. $\square$

Lemma 4

The Lagrange multiplier, $\hat{\lambda }(t)\equiv \hat{\lambda }$, solving Eq. (2.12), satisfies

$$\begin{aligned} |\hat{\lambda }|\le & {} \frac{n_2(\alpha (t) + \tilde{\zeta }_1(t) -\hat{\zeta }_2(t))}{\sum _{j=1}^{\tilde{\kappa }_2(t)}d_{2j}\hat{m}_{2j} \hat{S}_2^2(T_{2j})/r_{2j}}, \mathrm{when }\hat{\lambda }> 0; \end{aligned}$$

(5.6)

$$\begin{aligned} |\hat{\lambda }|\le & {} \frac{n_1(-\alpha (t) + \tilde{\zeta }_2(t) - \hat{\zeta }_1(t))}{\sum _{j=1}^{\tilde{\kappa }_1(t)} d_{1j}\hat{m}_{1j}\hat{S}_1^2(T_{1j})/r_{1j}}, \mathrm{when\ }\hat{\lambda }<0. \end{aligned}$$

(5.7)

Proof

For our semiparametric setting, we adapt the approach of McKeague and Zhao (2005), who follow Li (1995). From Eq. (2.12), write $\alpha (t)=I_1(t)+I_2(t)$, where

$$\begin{aligned} I_i(t)= & {} (-1)^{i+1}\sum _{j=1}^{\tilde{\kappa }_i(t)}\log \left( 1 - \frac{d_{ij}\hat{m}_{ij}}{r_{ij}+(-1)^{i+1}\hat{\lambda }\hat{S}_i(T_{ij})} \right) \hat{S}_i(T_{ij}),\qquad i=1,2. \end{aligned}$$

First assume that $\hat{\lambda }> 0$. Since $\mathrm{log}(1-x) + x$ is decreasing for $x \in (0,1)$, we have that

$$\begin{aligned} \log \left( 1 - \frac{d_{1j} \hat{m}_{1j}}{r_{1j} + \hat{\lambda }\hat{S}_1(T_{1j})}\right) + \frac{d_{1j} \hat{m}_{1j}}{r_{1j}+\hat{\lambda }\hat{S}_1(T_{1j})}\ge & {} \log \left( 1 - \frac{d_{1j} \hat{m}_{1j}}{r_{1j}}\right) + \frac{d_{1j} \hat{m}_{1j} }{r_{1j}}. \end{aligned}$$

It follows that

$$\begin{aligned} I_1(t)\ge & {} \sum _{j=1}^{\tilde{\kappa }_1(t)}\left[ -\frac{d_{1j} \hat{m}_{1j}}{r_{1j}+\hat{\lambda }\hat{S}_1(T_{1j})}+ \log \left( 1 - \frac{d_{1j} \hat{m}_{1j}}{r_{1j}}\right) + \frac{d_{1j} \hat{m}_{1j} }{r_{1j}} \right] \hat{S}_1(T_{1j}).\nonumber \\ \end{aligned}$$

(5.8)

Since, for $x>0$, the inequality $n_1/(n_1+x)\le n_2/(n_2+x)$ holds whenever $n_1\le n_2$, we have the following lower bound for the first term on the right hand side of Eq. (5.8):

$$\begin{aligned} -\sum _{j=1}^{\tilde{\kappa }_1(t)} \frac{d_{1j} \hat{m}_{1j}}{r_{1j}+\hat{\lambda }\hat{S}_1(T_{1j})}\hat{S}_1(T_{1j})= & {} -\sum _{j=1}^{\tilde{\kappa }_1(t)}\left( \frac{d_{1j} \hat{m}_{1j}}{r_{1j}}\right) \left( \frac{r_{1j}}{r_{1j}+\hat{\lambda }\hat{S}_1(T_{1j})}\right) \hat{S}_1(T_{1j}) \nonumber \\\ge & {} -\sum _{j=1}^{\hat{\kappa }_1(t)}\left( \frac{d_{1j}\hat{m}_{1j}}{r_{1j}}\right) \left( \frac{n_1}{n_1+|\hat{\lambda }|\hat{S}_1(T_{1j})}\right) \hat{S}_1(T_{1j}).\nonumber \\ \end{aligned}$$

(5.9)

Note that the second term on the right hand side of Eq. (5.8) equals $-\tilde{\zeta }_1(t)$, see Eq. (5.3). Using the fact that $-\mathrm{log} (1-x) \ge x$ when $0 \le x < 1$, we obtain

$$\begin{aligned} I_2(t)\ge & {} \sum _{j=1}^{\tilde{\kappa }_2(t)}\left( \frac{d_{2j} \hat{m}_{2j}}{r_{2j}-\hat{\lambda }\hat{S}_2(T_{2j})}\right) \hat{S}_2(T_{2j})\\= & {} \sum _{j=1}^{\tilde{\kappa }_2(t)}\left( \frac{d_{2j} \hat{m}_{2j}}{r_{2j}}\right) \left( \frac{r_{2j}}{r_{2j}-|\hat{\lambda }|\hat{S}_2(T_{2j})}\right) \hat{S}_2(T_{2j}). \end{aligned}$$

Since $n_1/(n_1-x)\ge n_2/(n_2-x)$, when $x>0$ and $n_1\le n_2$, we obtain the lower bound

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

(5.10)

From Eqs. (5.8)–(5.10), we obtain

$$\begin{aligned} \alpha (t)\ge & {} - \sum _{j=1}^{\tilde{\kappa }_1(t)}\left( \frac{d_{1j}\hat{m}_{1j}}{r_{1j}}\right) \left( \frac{n_1}{n_1+|\hat{\lambda }|\hat{S}_1(T_{1j})}\right) \hat{S}_1(T_{1j}) -\tilde{\zeta }_1(t)\nonumber \\&+ \sum _{j=1}^{\tilde{\kappa }_1(t)} \left( \frac{d_{1j}\hat{m}_{1j}}{r_{1j}} \right) \hat{S}_1(T_{1j}) \nonumber \\&+ \sum _{j=1}^{\tilde{\kappa }_2(t)}\left( \frac{d_{2j}\hat{m}_{2j}}{r_{2j}}\right) \left( \frac{n_2}{n_2-|\hat{\lambda }|\hat{S}_2(T_{2j})} \right) \hat{S}_2(T_{2j}). \end{aligned}$$

(5.11)

From the right hand side of Eq. (5.11), combining the first and third terms gives a nonnegative number. Since $1/(1 - x) \ge 1 + x$ when $x < 1$, the fourth term is bounded below by

$$\begin{aligned}&\sum _{j=1}^{\tilde{\kappa }_2(t)}\left( \frac{d_{2j}\hat{m}_{2j}}{r_{2j}}\right) \hat{S}_2(T_{2j}) \left( 1 + |\hat{\lambda }|\hat{S}_2(T_{2j})/n_2\right) \nonumber \\&\quad = \hat{\zeta }_2(t) + \frac{1}{n_2}\sum _{j=1}^{\tilde{\kappa }_2(t)} \left( \frac{d_{2j}\hat{m}_{2j}}{r_{2j}}\right) \hat{S}_2^2(T_{2j})|\hat{\lambda }|, \end{aligned}$$

provided that $|\hat{\lambda }|\hat{S}_2(T_{2j})/n_2<1$ almost surely. To show this, assume no ties and note from Eq. (2.14) that $0<\hat{\lambda }<\min _{j:T_{2j}\le t}\{(r_{2j}-d_{2j}\hat{m}_{2j})/\hat{S}_2(T_{2j})\}$. Also, $\bar{Y}_2(t)/\hat{S}_2(t){\buildrel \mathrm{a.s.} \over \longrightarrow }1-G_2(t)<1$ uniformly over $[\tau _1,\tau _2]$. For large enough $n_2$, $\epsilon$ sufficiently small, and some $T_{2l}\ne 0$, we have

$$\begin{aligned} \frac{\hat{\lambda }}{n_2}< & {} \min _{j:T_{2j}\le t}\frac{r_{2j}/n_2}{\hat{S}_2(T_{2j})}\le \frac{\bar{Y}_2(T_{2l})}{\hat{S}_2(T_{2l})} < 1-G_2(T_{2l})+\epsilon \ <\ 1 \end{aligned}$$

almost surely. We therefore obtain a lower bound for $\alpha (t)$ given by

$$\begin{aligned} \alpha (t)\ge & {} - \tilde{\zeta }_1(t) + \hat{\zeta }_2(t) + \frac{1}{n_2}\sum _{j=1}^{\tilde{\kappa }_2(t)} \left( \frac{d_{2j}\hat{m}_{2j}}{r_{2j}}\right) \hat{S}_2^2(T_{2j})|\hat{\lambda }|, \end{aligned}$$

from which Eq. (5.6) follows. Proof of Eq. (5.7) can be shown by analogous techniques. $\square$

Lemma 5

The Lagrange multiplier, $\hat{\lambda }(t)\equiv \hat{\lambda }$, solving Eq. (2.12), satisfies $\Vert \hat{\lambda }\Vert _{\tau _1}^{\tau _2} = O_p(n^{1/2})$.

Proof

That the denominators of Eqs. (5.6) and (5.7) are each $O_p(1)$, uniformly for $t\in [\tau _1,\tau _2]$, follows from $\Vert \hat{\Lambda }_i-\Lambda _i\Vert _0^{\tau _2}=o(1)$ almost surely (cf. Theorem 2.4 of Dikta 1998) and Lemma 2. It remains to show that the numerators of Eqs. (5.6) and (5.7) are each $O_p(n^{-1/2})$, uniformly for $t\in [\tau _1,\tau _2]$. By applying Lemma 3, it suffices to show that $\alpha (t) + \hat{\zeta }_1(t) - \hat{\zeta }_2(t) = O_p(n^{-1/2})$. Let $n_i/n\rightarrow p_i$ as $n\rightarrow \infty$. We then have by Lemma 1 and results from Sect. 2.1 that

$$\begin{aligned} n^{1/2}\left( \alpha (t) + \hat{\zeta }_1(t) - \hat{\zeta }_2(t)\right)&= \sum _{i=1}^2 S_i(t)\frac{n_i^{1/2}(\hat{\Lambda }_i(t)-\Lambda _i(t))}{(n_i/n)^{1/2}} + o_p(1) \nonumber \\&{\buildrel {\mathcal {D}} \over \longrightarrow }\sum _{i=1}^2\frac{S_i(t)U_i(t)}{\sqrt{p}_i}\ \equiv \ W(t), \end{aligned}$$

(5.12)

where W is the zero-mean Gaussian process with covariance function given by Eq. (2.17). $\square$

1.1 Proof of Theorem 1

From Eq. (2.12), we can write $\alpha (t)=f_1(\hat{\lambda })-f_2(-\hat{\lambda })$, where

$$\begin{aligned} f_i(\lambda )= & {} \sum _{j=1}^{\tilde{\kappa }_i(t)}\log \left( 1 - \frac{ d_{ij} \hat{m}_{ij}}{r_{ij} + \lambda \hat{S}_i(T_{ij})}\right) \hat{S}_i(T_{ij}). \end{aligned}$$

(5.13)

Note that $f_i(0) = - \tilde{\zeta }_i(t)$ [cf. Eq. (5.3)]. Before applying a Taylor’s expansion, we note that

$$\begin{aligned} f_i^{'}(\lambda )= & {} \sum _{j=1}^{\tilde{\kappa }_i(t)}\frac{d_{ij} \hat{m}_{ij}}{(r_{ij}+\lambda \hat{S}_i(T_{ij}))(r_{ij}+\lambda \hat{S}_i(T_{ij})- d_{ij}\hat{m}_{ij})}\hat{S}^2_i(T_{ij}),\\ f_i^{''}(\lambda )= & {} \sum _{j=1}^{\tilde{\kappa }_i(t)}\frac{d_{ij}\hat{m}_{ij}(2(r_{ij} +\lambda \hat{S}_i(T_{ij}))- d_{ij} \hat{m}_{ij})}{(r_{ij}+\lambda \hat{S}_i(T_{ij}))^2(r_{ij}+\lambda \hat{S}_i(T_{ij})-d_{ij} \hat{m}_{ij})^2}\hat{S}^3_i(T_{ij}). \end{aligned}$$

Recall that $\gamma _i(t)$ is defined by Eq. (2.15). Write $n_if_i^{'}(0)=\tilde{\gamma }_i(t)$, so that

$$\begin{aligned} \tilde{\gamma }_i(t) = n_i\sum _{j:T_{ij} \le t} \frac{d_{ij}\hat{m}_{ij}}{r_{ij}(r_{ij} - d_{ij}\hat{m}_{ij})}\hat{S}^2_i(T_{ij}). \end{aligned}$$

For $i=1,2$, let $|\hat{\xi }_i|\le |\hat{\lambda }|$. Taylor’s expansion about 0 yields

$$\begin{aligned} f_i\left( (-1)^{i-1}\hat{\lambda }\right)= & {} - \tilde{\zeta }_i(t) + (-1)^{i-1}\frac{\tilde{\gamma }_i(t)\hat{\lambda }}{n_i} + \frac{1}{2}f^{''}_i(\hat{\xi }_i)\hat{\lambda }^2. \end{aligned}$$

(5.14)

Applying the Glivenko–Cantelli lemma to $r_{ij}$ and Lemma 5, we have $\Vert f_i''(\hat{\xi }_i)\Vert _{\tau _1}^{\tau _2}=O_p(n_i^{-2})$. Therefore, by Lemma 5, it follows that $\Vert f^{''}_i(\hat{\xi }_i)\hat{\lambda }^2\Vert _{\tau _1}^{\tau _2}=O_p(n_i^{-1})=O_p(n^{-1})$. Furthermore, $\tilde{\gamma }_i(t)$ is uniformly consistent for $\gamma _i(t)$ over $[0,\tau _2]$. It follows from Eq. (5.14) and Eq. (2.16) that

$$\begin{aligned} \alpha (t)= & {} -\tilde{\zeta }_1(t) + \tilde{\zeta }_2(t) + \hat{\lambda }\left( \frac{\tilde{\gamma }_1(t)}{n_1} +\frac{\tilde{\gamma }_2(t)}{n_2}\right) + O_p\left( \frac{1}{n}\right) \\= & {} -\tilde{\zeta }_1(t) + \tilde{\zeta }_2(t) + \hat{\lambda }n^{-1}\sigma ^2_d(t) + O_p(n^{-1}). \end{aligned}$$

Solving for $\hat{\lambda }$, we obtain

$$\begin{aligned} \hat{\lambda }= & {} n\sigma ^{-2}_\mathrm{d}(t)\left( \alpha (t) +\tilde{\zeta }_1(t)-\tilde{\zeta }_2(t)+O_p(n^{-1})\right) . \end{aligned}$$

(5.15)

To complete the Proof of Theorem 1, consider Eq. (2.13). Using Taylor expansions of $\log (1+x)$ and $\log (1-x)$ about 0, the leading term of $-2R(t)$ is a product of $\hat{\lambda }^2$ and a random scaling factor expressed as a double sum, and equals

$$\begin{aligned} \hat{\lambda }^2\sum _{i=1}^2\sum _{j=1}^{\tilde{\kappa }_i(t)}\frac{d_{ij}\hat{m}_{ij} \hat{S}_i^2(T_{ij})}{r_{ij}(r_{ij}-d_{ij}\hat{m}_{ij})}. \end{aligned}$$

Each single sum can be expressed as an integral as in the Proof of Lemma 1. Applying Eq. (2.6.10) of Andersen et al. (1993), it follows that the double sum equals $\sigma _\mathrm{d}^2(t)/n +o_p(1)$, uniformly over $[0, \tau _2]$. Therefore, applying Eq. (5.15), the leading term of $-2R(t)$ equals

$$\begin{aligned} n \hat{\sigma }^{-2}_\mathrm{d}(t)\left( \alpha (t)+\tilde{\zeta }_1(t)-\tilde{\zeta }_2(t)+O_p(n^{-1})\right) ^2 + o_p(1)= & {} \left( \frac{W(t)}{\sigma _\mathrm{d}(t)}\right) ^2 + o_p(1).\nonumber \\ \end{aligned}$$

(5.16)

The other terms of $-2R(t)$ are proportional to

$$\begin{aligned} \hat{\lambda }^l\left[ \sum _{i=1}^2\sum _{j=1}^{\tilde{\kappa }_1(t)}\hat{S}^l_i(T_{ij}) \left( \frac{1}{(r_{ij}-d_{ij}\hat{m}_{ij})^{l-1}} - \frac{1}{r_{ij}^{l-1}}\right) \right] , \qquad l=3, 4, \ldots , \end{aligned}$$

each of which is $o_p(1)$, uniformly for $t\in [0,\tau _2]$. For example, when $l=3$, we have

$$\begin{aligned} \frac{2}{3}\hat{\lambda }^3\sum _{j=1}^{\tilde{\kappa }_i(t)}\hat{S}^3_i(T_{1j}) \left( \frac{1}{r_{ij}-d_{ij}\hat{m}_{ij}} - \frac{1}{r_{ij}}\right) \left( \frac{1}{r_{ij} - d_{ij}\hat{m}_{ij}} + \frac{1}{r_{ij}}\right) ,\qquad i=1,2, \end{aligned}$$

which, uniformly for $t\in [0,\tau _2]$, equals

$$\begin{aligned} \frac{2}{3}O_p(n^{3/2})O_p(n_i^{-1})\sum _{j=1}^{\tilde{\kappa }_i(t)} \hat{S}^2_i(T_{ij})\left( \frac{1}{r_{ij} - d_{ij}\hat{m}_{ij}} - \frac{1}{r_{ij}}\right) = O_p(n^{1/2})\tilde{\gamma }_i(t)/n_i = o_p(1). \end{aligned}$$

Now apply Lemma 3 and Eq. (5.12) to complete the Proof of Theorem 1. $\square$

1.2 Large sample justification of the multiplier bootstrap

Write $\mathbb {\hat{H}}^*_i(t) = L^*_{n_i, 1}(t) + L^*_{n_i, 2}(t)$, where $L^*_{n_i, 1}(t)$ and $L^*_{n_i, 2}(t)$ are defined by Eqs. (2.20) and (2.21). To show that ${\mathbb {W}}^*$ defined by Eq. (2.19) has the limit distribution as that of $W/\sigma _\mathrm{d}$, it suffices to show that ${\hat{\mathbb {H}}}^*_i(\cdot )$ has the same weak limit as $\hat{\mathbb {H}}_i(t)=L_{n_i,1}(t)+L_{n_i,2}(t)+o_p(1)$. Let ${\mathbb {P}}_{n_i}$, ${\mathbb {E}}_{n_i}$, $\mathrm{Cov}_{n_i}$, and $\mathrm{Var}_{n_i}$ be the probability measure, expectation, covariance, and variance with respect to the bootstrap, that is, conditioned on the sample $\{(Z_{ij}, \delta _{ij}), j = 1, \ldots , n_i\}$. To show that $\hat{\mathbb {H}}_i^*(t)$ has the limiting covariance structure given by Eq. (2.1), note that

$$\begin{aligned} \mathrm{Cov}_{n_i}(\hat{\mathbb {H}}_i^*(s), \hat{\mathbb {H}}_i^*(t))= & {} {\mathbb {E}}_{n_i}(L^*_{n_i,1}(s) L^*_{n_i,1}(t)) + {\mathbb {E}}_{n_i}(L^*_{n_i,2}(s) L^*_{n_i,2}(t)) \nonumber \\&+\,{\mathbb {E}}_{n_i}(L^*_{n_i,1}(s) L^*_{n_i,2}(t)) + {\mathbb {E}}_{n_i}(L^*_{n_i,1}(t) L^*_{n_i,2}(s)).\nonumber \\ \end{aligned}$$

(5.17)

Strong consistency of $\hat{\varvec{\theta }}_i$, assumption $A_6$ of Dikta (1998) and the arguments in the Proof of Theorem 2.4 of Dikta (1998) imply that $\Vert m_i(\cdot , \varvec{\hat{\theta }}_i)-m_i(\cdot , \varvec{\theta }_i)\Vert _{0}^{\tau _2}=o(1)$ almost surely. Likewise, $\hat{\alpha }(\cdot ,\cdot )$ is strongly uniformly consistent over $[0,\tau _2]\times [0,\tau _2]$. Finally, $\Vert \bar{Y}_i-y_i\Vert _0^{\tau _2}=o(1)$ almost surely. The first quantity on the right hand side of Eq. (5.17) can be computed to yield

$$\begin{aligned} {\mathbb {E}}_{n_i}(L^*_{n_i,1}(s), L^*_{n_i,2}(t))= & {} \frac{1}{n_i} \sum _{j=1}^{n_i}\frac{\hat{m}_{ij}^2}{\bar{Y}_i^2(Z_{ij})}I(Z_{ij} < s \wedge t)\ =\ \frac{1}{n_i} \sum _{j=1}^{n_i}\frac{m_{ij}^2}{y_i^2(Z_{ij})}\\&\qquad I(Z_{ij} < s \wedge t) + o(1). \end{aligned}$$

By the strong law of large numbers, for almost all sample sequences $\{Z_{ij},\delta _{ij}, 1\le j\le n_i\}$, ${\mathbb {E}}_{n_i}(L^*_{n_i,1}(s), L^*_{n_i,2}(t))$ converges to the first term on the right hand side of Eq. (2.1). The second quantity on the right hand side of Eq. (5.17) can be computed to yield

$$\begin{aligned} {\mathbb {E}}_{n_i}(L^*_{n_i,2}(s), L^*_{n_i,2}(t))= & {} \frac{1}{n_i} \sum _{j=1}^{n_i} \left( \frac{\delta _{ij} - \hat{m}_{ij}}{\hat{m}_{ij}\hat{\bar{m}}_{ij}}\right) ^2 \int _0^{s} \int _0^{t} \frac{\hat{\alpha }(u, Z_{ij})}{\bar{Y}_i(u)} \frac{\hat{\alpha }(v, Z_{ij})}{\bar{Y}_i(v)}\\&\qquad d\hat{H}_i(u)d\hat{H}_i(v)\\= & {} \frac{1}{n_i} \sum _{j=1}^{n_i} \left( \frac{\delta _{ij} - m_{ij}}{m_{ij}\bar{m}_{ij}}\right) ^2 \int _0^{s} \int _0^{t} \frac{\alpha (u, Z_{ij})}{y_i(u)} \frac{\alpha (v, Z_{ij})}{y_i(v)}\\&\qquad dH_i(u)dH_i(v) +o(1). \end{aligned}$$

By the strong law of large numbers, for almost all sample sequences $\{Z_{ij},\delta _{ij}, 1\le j\le n_i\}$, ${\mathbb {E}}_{n_i}(L^*_{n_i,2}(s), L^*_{n_i,2}(t))$ converges to the second term on the right hand side of Eq. (2.1). One of the cross-product moment terms on the right hand side of Eq. (5.17) is given by

$$\begin{aligned} E_{n_i}(L^*_{n_i,1}(t_1), L^*_{n_i,2}(t_2)) = \frac{1}{n_i} \sum _{j=1}^{n_i} \frac{(\delta _{ij} - \hat{m}_{ij}) \hat{m}_{ij}}{\hat{m}_{ij}\hat{\bar{m}}_{ij}}\frac{I(Z_{ij} < t_1)}{\bar{Y}_i(Z_{ij})} \int _0^{t_2} \frac{\hat{\alpha }(u, Z_{ij})}{\bar{Y}_i(u)}d\hat{H}_i(u),\nonumber \\ \end{aligned}$$

(5.18)

with the other term given in an analogous way. The aforementioned arguments, followed by applying iterated conditional expectation with conditioning by $Z_{i1}$, implies that the two cross-moment terms in Eq. (5.17) are each zero.

To show that $\hat{\mathbb {H}}^*_i(\cdot )$ converges weakly to a zero-mean Gaussian process we verify Lindeberg’s condition and tightness. We can write $\mathbb {\hat{H}}^*_i(t) = L^*_{n_i,1}(t) + L^*_{n_i,2}(t) = \sum _{j=1}^{n_i}B_{ij}(t)G_{ij}$, where

$$\begin{aligned} B_{ij}(t)= n_i^{-1/2} \left[ \frac{\hat{m}_{ij}}{\bar{Y}_i(Z_{ij})}I(Z_{ij} \le t) + \frac{\delta _{ij} - \hat{m}_{ij}}{\hat{m}_{ij}\hat{\bar{m}}_{ij}} \int _0^t \frac{\hat{\alpha }(u, Z_{ij})}{\bar{Y}_i(u)}d\hat{H}_i(u) \right] . \end{aligned}$$

Let $s^2_i = \sum _{j=1}^{n_i} \mathrm{Var}_{n_i}[B_{ij}(t)G_{ij}] = \sum _{j=1}^{n_i} B_{ij}^2(t)$. As in Mondal and Subramanian (2015), it can be shown that for almost all sample sequences $\{Z_{ij},\delta _{ij}, 1\le j\le n_i\}$, for any $\eta _i > 0$,

$$\begin{aligned} \frac{1}{s^2_i}\sum _{j=1}^{n_i}{\mathbb {E}}_{n_1} [B_{ij}^2(t)G_{ij}^2 I(|B_{ij}(t)G_{ij}| > s_i\eta _i)] \rightarrow 0 \text{ as } n \rightarrow \infty . \end{aligned}$$

Let $K \ge 3$. To verify tightness, we follow Mondal and Subramanian (2015) to show that

$$\begin{aligned}&\underset{n\rightarrow \infty }{\overline{\lim }}[\hat{\mathbb {H}}_i^*(t) - \hat{\mathbb {H}}_i^*(s) ]^4 \le K\underset{n\rightarrow \infty }{\overline{\lim }}\left[ \sum _{j=1}^{n_i}(B_{ij}(t) - B_{ij}(s))^2\right] ^2, \quad s < t.\quad \end{aligned}$$

(5.19)

The sum on the right hand side of inequality (5.19) is given by

$$\begin{aligned} \sum _{j=1}^{n_i} (B_{ij}(t) - B_{ij}(s))^2= & {} \frac{1}{n_i} \sum _{j=1}^{n_i}\left[ \frac{\hat{m}_{ij}}{\bar{Y}_i(Z_{ij})}I(s<Z_{ij} \le t)\right. \\&\left. + \frac{\delta _{ij} - \hat{m}_{ij}}{\hat{m}_{ij}\hat{\bar{m}}_{ij}} \int _{s}^{t} \frac{\hat{\alpha }(u, Z_{ij})}{\bar{Y}_i(u)}d\hat{H}_i(u) \right] ^2, \end{aligned}$$

which can be shown equal to

$$\begin{aligned} \int _s^t\frac{m_i^{2}(u)}{y_i^2(u)}dH_i(u)+\int _s^t\int _s^t \frac{\alpha _i(u,v)}{y_i(u)y_i(v)}dH_i(u)dH_i(v) + o_p(1). \end{aligned}$$

Therefore, the left hand side of inequality (5.19) is finite and tightness is verified.