Conditional quantile residual lifetime models for right censored data

Abstract

Quantile residual lifetime function is a more comprehensive quantitative measure for residual lifetimes than the mean residual lifetime function. It also incorporates the median residual life function, which is less restrictive than the model based on the mean residual lifetime. In this study, we propose a semiparametric estimator of the conditional quantile residual lifetime under different covariate effects at a specified time point by the reinforcement of the auxiliary models. Two kind of test statistics are proposed to compare two quantile residual lifetimes at fixed time points. Asymptotic properties are also established and a revised bootstrap method is proposed to estimate the asymptotic variance of the estimator. Simulation studies are reported to assess the finite sample properties of the proposed estimator and the performance of test statistics in terms of type I error probabilities and powers at fixed time points. We also compare the proposed method with the method of Jung et al. (Biometrics 65:1203–1212, 2009) through simulation studies. The proposed methods are applied to HIV data and some interesting results are presented.

Appendix

In order to prove the results, we need the following assumptions and regularity conditions:

• (A1) The covariate ${\mathbf {Z}}(\cdot )$ has uniformly bounded total variation.

• (A2) $\varvec{\beta }_0\in \mathcal {B}\subset R^p$ and $\mathcal {B}$ is open, convex and bounded.

• (A3) $\tau =\sup \{t:Y(t)>0\}$, $\Lambda _0(t)$ is continuous and $\Lambda _0(\tau )<\infty$ .

• (A4) $\Omega$, the asymptotic covariance matrix of $\sqrt{n}(\hat{\varvec{\beta }}-\varvec{\beta }_0)$, is positive definite.

For simplicity, we denote $\theta _{\alpha }=\theta _{\alpha }(t_0;{\mathbf {z}}_0)$ and $\hat{\theta }_{\alpha }=\hat{\theta }_{\alpha }(t_0;{\mathbf {z}}_0)$.

Proof of Theorem 1

By the strong consistency of $\hat{\varvec{\beta }}$ and $\hat{\Lambda }_0(\cdot )$, we have

$$\begin{aligned} \sup _{t\in [0,\tau ]}|\hat{S}(t|{\mathbf {z}}_0)-S(t|{\mathbf {z}}_0)|\mathop {\rightarrow }\limits ^{P}0, \end{aligned}$$

which implies that

$$\begin{aligned} \sup _{{\theta _\alpha }\in [0,\tau ]}|\hat{U}(\theta _\alpha )-U(\theta _\alpha )|\mathop {\rightarrow }\limits ^{P}0, \end{aligned}$$

where $U(\theta _{\alpha })=S(t_0+\theta _\alpha |{\mathbf {z}}_0)-\alpha S(t_0|{\mathbf {z}}_0)$ and $\hat{U}(\theta _\alpha )=\hat{S}(t_0+\theta _\alpha |{\mathbf {z}}_0)-\alpha \hat{S}(t_0|{\mathbf {z}}_0)$. By the assumption, $U(\theta _\alpha )=0$ has unique solution $\theta _\alpha$, then as $n\rightarrow \infty$, $\hat{U}(\theta _\alpha )=0$ also has unique solution $\hat{\theta }_\alpha$. For any $\epsilon >0$, we have

$$\begin{aligned} \sup _{|\tilde{\theta }_\alpha -\theta _\alpha |> \epsilon }|\hat{U}(\tilde{\theta }_\alpha )-U(\tilde{\theta }_\alpha )|=o_p(1). \end{aligned}$$

It follows that

$$\begin{aligned} \inf _{|\tilde{\theta }_\alpha -\theta _\alpha |> \epsilon }|\hat{U}(\tilde{\theta }_\alpha )|\ge \inf _{|\tilde{\theta }_\alpha -\theta _\alpha |\ge \epsilon }|U(\tilde{\theta }_\alpha )|-o_p(1)>M, \end{aligned}$$

where $M$ is some positive constant. Hence, $|\hat{\theta }_\alpha -\theta _\alpha |\le \epsilon$ and $\hat{\theta }_\alpha$ is consistent. $\square$

To prove Theorem 2, we need the following technical lemmas:

Lemma 1

Let $B_n\in D[a,b]$ and $A_n \in l^{\infty }([a,b])$ be either cadlag or caglad, and assume that $\sup _{t\in (a,b]}|A_n(t)|\mathop {\rightarrow }\limits ^{P}0$, $A_n$ has uniformly bounded total variation, and $B_n$ converges weakly to a tight, mean zero process with sample paths in $D[a,b]$. Then $\int _a^bA_n(s)dB_n(s)\mathop {\rightarrow }\limits ^{P}0.$

Proof of Lemma 1

See Lemma 4.2 of Kosorok (2006). $\square$

Lemma 2

Assume that the above conditions hold and the true conditional $\alpha$th quantile residual lifetime function at $t_0$ given ${\mathbf {z}}_0$ is $\theta _{\alpha }$ and its estimator $\hat{\theta }_{\alpha }$ is consistent. Then as $n\rightarrow \infty$, we have

$$\begin{aligned} \mid \sqrt{n}(\hat{U}(\hat{\theta }_\alpha )-U(\hat{\theta }_\alpha ))-\sqrt{n}(\hat{U}(\theta _\alpha )-U(\theta _\alpha ))\mid \mathop {\longrightarrow }\limits ^{P}0. \end{aligned}$$

(7.1)

Proof of Lemma 2

We just need to prove that

$$\begin{aligned} \sqrt{n}\mid \hat{S}(t_0+\hat{\theta }_\alpha )-S(t_0+\hat{\theta }_\alpha )-\hat{S}(t_0+\theta _\alpha )+S(t_0+\theta _\alpha )\mid \mathop {\longrightarrow }\limits ^{P}0. \end{aligned}$$

In fact, by the arguments of Andersen and Gill (1982), the asymptotic covariance matrix of $\sqrt{n}(\hat{\varvec{\beta }}-\varvec{\beta }_0)$ is $\Omega$, which is

$$\begin{aligned} \Omega =\mathop \int \limits _0^\infty \left\{ \frac{s^{(2)}(\varvec{\beta }_0,t)}{s^{(0)}(\varvec{\beta }_0,t)}-\bar{z}(\varvec{\beta }_0,t)^{\otimes 2}\right\} s^{(0)}(\varvec{\beta }_0,t)d\Lambda _0(t). \end{aligned}$$

Let $W(t)=\sqrt{n}\{\hat{\Lambda }(t)-\Lambda (t)\}$, and it is asymptotically equivalent to

$$\begin{aligned} \tilde{W}(t)=\frac{1}{\sqrt{n}}\sum _{i=1}^n\left[ \,\mathop \int \limits _0^t\frac{\exp (\varvec{\beta }_0^T{\mathbf {z}}_0(u))}{S^{(0)}(\varvec{\beta }_0,u)}dM_i(u)+h^T(t)\Omega ^{-1} \mathop \int \limits _0^\infty \{{\mathbf {Z}}_i(u)-\bar{Z}(\varvec{\beta }_0,u)\}dM_i(u)\right] , \end{aligned}$$

where $M_i(t)=N_i(t)-\int _0^tY_i(u)\exp (\varvec{\beta }_0^T{\mathbf {Z}}_i(u))d\Lambda _0(u)$ is a martingale and

$$\begin{aligned} h(t)&= \mathop \int \limits _0^t\exp (\varvec{\beta }_0^T{\mathbf {z}}_0(u))\{{\mathbf {z}}_0(u)-\bar{z}(\varvec{\beta },u)\}d\Lambda _0(u). \end{aligned}$$

On the other hand, $\hat{S}(t)=\exp \{-\hat{\Lambda }(t)\}$, by the functional delta method, we have that the process $\sqrt{n}(\hat{S}(t)-S(t))$ is asymptotically equivalent to $-S(t)\tilde{W}(t)$. Then by the consistency of $\hat{\theta }_\alpha$ and the continuity of $S(\cdot )$,

$$\begin{aligned}&\sqrt{n}\mid \hat{S}(t_0+\hat{\theta }_\alpha )-S(t_0+\hat{\theta }_\alpha )-\hat{S}(t_0+\theta _\alpha )+S(t_0+\theta _\alpha )\mid \\&\quad =\mid -S(t_0+\hat{\theta }_\alpha )\tilde{W}(t_0+\hat{\theta }_\alpha )+S(t_0+\theta _\alpha )\tilde{W}(t_0+\theta _\alpha )\mid +o_p(1)\\&\quad =S(t_0+\theta _\alpha )\mid \tilde{W}(t_0+\hat{\theta }_\alpha )-\tilde{W}(t_0+\theta _{\alpha })\mid +o_p(1)\\&\quad \le \mid \tilde{W}(t_0+\hat{\theta }_\alpha )-\tilde{W}(t_0+\theta _{\alpha })\mid +o_p(1). \end{aligned}$$

By the expression of $\tilde{W}(t)$, we have

$$\begin{aligned}&\mid \tilde{W}(t_0+\hat{\theta }_\alpha )-\tilde{W}(t_0+\theta _{\alpha })\mid \\&\quad =\Big |\frac{1}{\sqrt{n}}\sum _{i=1}^n\mathop \int \limits _0^{t_0+\hat{\theta }_\alpha }\frac{\exp (\varvec{\beta }_0^T{\mathbf {z}}_0(u))}{S^{(0)}(\varvec{\beta }_0,u)}dM_i(u)- \frac{1}{\sqrt{n}}\sum _{i=1}^n\mathop \int \limits _0^{t_0+\theta _\alpha }\frac{\exp (\varvec{\beta }_0^T{\mathbf {z}}_0(u))}{S^{(0)}(\varvec{\beta }_0,u)}dM_i(u)\\&\qquad +(h(t_0+\hat{\theta }_\alpha )-h(t_0+\theta _\alpha ))^T\Omega ^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^n\mathop \int \limits _0^\infty \{{\mathbf {Z}}_i(u)-\bar{Z}(\varvec{\beta }_0,u)\}dM_i(u)\Big |\\&\quad \le \Big |\frac{1}{\sqrt{n}}\sum _{i=1}^n\mathop \int \limits _0^{\tau }(I\{u\le t_0+\hat{\theta }_\alpha \}-I\{u\le t_0+\theta _\alpha \})\frac{\exp (\varvec{\beta }_0^T{\mathbf {z}}_0(u))}{S^{(0)}(\varvec{\beta }_0,u)}dM_i(u)\Big |\\&\qquad + \Big |\mathop \int \limits _0^{\tau }(I\{u\le t_0+\hat{\theta }_\alpha \}-I\{u\le t_0+\theta _\alpha \})\exp (\varvec{\beta }_0^T{\mathbf {z}}_0(u))\{{\mathbf {z}}_0(u)-\bar{z}(\varvec{\beta },u)\}d\Lambda _0(u)\\&\qquad \Omega ^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^n\mathop \int \limits _0^\infty \{{\mathbf {Z}}_i(u){-}\bar{Z}(\varvec{\beta }_0,u)\}dM_i(u)\Big |\\&\quad =: I_1+I_2. \end{aligned}$$

For $I_1$, we denote

$$\begin{aligned} A_n(t)=(I\{t\le t_0+\hat{\theta }_\alpha \}-I\{t\le t_0+\theta _\alpha \})\frac{\exp (\varvec{\beta }_0^T{\mathbf {z}}_0(t))}{S^{(0)}(\varvec{\beta }_0,t)} \end{aligned}$$

and $B_n(t)=\frac{1}{\sqrt{n}}\sum _{i=1}^nM_i(t)$. Then $B_n$ converges weakly to mean zero Gaussian process because $M_i(t)$ is a martingale with mean zero and by the fact that the covariate ${\mathbf {Z}}(\cdot )$ is uniformly bounded with bounded variation and the consistency of $\hat{\theta }_{\alpha }$, we have $\sup _{t\in (0,\tau ]}|A_n(t)| \mathop {\rightarrow }\limits ^{P}0$. Then using Lemma 1, we have $\mathop \int \limits _0^\tau A_n(u)dB_n(u) \mathop {\longrightarrow }\limits ^{P}0,$ that is $I_1\mathop {\rightarrow }\limits ^{P}0.$ For $I_2$, because

$$\begin{aligned} \Big |\frac{1}{\sqrt{n}}\sum _{i=1}^n\mathop \int \limits _0^\infty \{{\mathbf {Z}}_i(u)-\bar{Z}(\varvec{\beta }_0,u)\}dM_i(u)\Big |=O_p(1) \end{aligned}$$

and $\hat{\theta }_\alpha \mathop {\longrightarrow }\limits ^{P}\theta _\alpha$, so $I_2\mathop {\longrightarrow }\limits ^{P}0$ is trivial. Then we have the conclusion of 7.1. $\square$

Proof of Theorem 2

By simple calculation,

$$\begin{aligned} \sqrt{n}\hat{U}(\hat{\theta }_{\alpha })&= \sqrt{n}\hat{U}(\theta _{\alpha })+\sqrt{n}(U(\hat{\theta }_\alpha )-U(\theta _\alpha ))+o_p(1)\\&= \sqrt{n}\hat{U}(\theta _{\alpha })-f(t_0+\theta _\alpha |{\mathbf {z}}_0)\sqrt{n}(\hat{\theta }_{\alpha }-\theta _{\alpha })+o_p(1), \end{aligned}$$

which means that

$$\begin{aligned} \sqrt{n}(\hat{\theta }_\alpha -\theta _{\alpha })=f(t_0+\theta _{\alpha }|{\mathbf {z}}_0)^{-1}\sqrt{n}\hat{U}(\theta _{\alpha })+o_p(1), \end{aligned}$$

where

$$\begin{aligned} \sqrt{n}\hat{U}(\theta _{\alpha })&= \sqrt{n}[\hat{S}(t_0+\theta _{\alpha }|{\mathbf {z}}_0)-\alpha \hat{S}(t_0|{\mathbf {z}}_0)]\\&= \sqrt{n}[\hat{S}(t_0+\theta _{\alpha }|{\mathbf {z}}_0)-S(t_0+\theta _{\alpha }|{\mathbf {z}}_0)-\alpha (\hat{S}(t_0|{\mathbf {z}}_0)-S(t_0|{\mathbf {z}}_0))] \end{aligned}$$

By the fact $S(t_0+\theta _{\alpha }|{\mathbf {z}}_0)=\alpha S(t_0|{\mathbf {z}}_0)$, the distribution of $\sqrt{n}\hat{U}(\theta _{\alpha })$ can be approximated by

$$\begin{aligned}&-\alpha S(t)\{\tilde{W}(t_0+\theta _{\alpha })-\tilde{W}(t_0)\}=-\alpha S(t) \frac{1}{\sqrt{n}}\sum _{i=1}^n\big [\mathop \int \limits _{t_0}^{t_0+\theta _{\alpha }}\frac{\exp (\varvec{\beta }_0^T{\mathbf {z}}_0(u))}{S^{(0)}(\varvec{\beta }_0,u)}dM_i(u)\\&+(h(t_0+\theta _{\alpha })-h(t_0))^T\Omega ^{-1}\mathop \int \limits _0^\infty \{{\mathbf {Z}}_i(u)-\bar{Z}(\varvec{\beta }_0,u)\}dM_i(u)\big ]. \end{aligned}$$

By the martingale central limit theorem, it is easy to show that the above process converges weakly to a zero-mean Gaussian process with the covariance function

$$\begin{aligned} \Sigma =\mathop \int \limits _{t_0}^{t_0+\theta _{\alpha }}\frac{\exp (2\varvec{\beta }_0^T{\mathbf {z}}_0(u))}{S^{(0)}(\varvec{\beta }_0,u)}d\Lambda _0(u)+(h(t_0+\theta _{\alpha })-h(t_0))^T\Omega ^{-1}(h(t_0+\theta _{\alpha })-h(t_0)). \end{aligned}$$

and thereby

$$\begin{aligned} \sqrt{n}\hat{U}(\theta _{\alpha })\mathop {\rightarrow }\limits ^\mathcal{L} N(0,\alpha ^2S(t_0)^2\Sigma ). \end{aligned}$$

Therefore the asymptotic variance of $\sqrt{n}( \hat{\theta }_{\alpha }(t_0) -\theta _{\alpha }(t_0))$ is

$$\begin{aligned} \sigma ^2=\frac{\alpha ^2S(t_0|{\mathbf {z}}_0)^2}{f(t_0+\theta _{\alpha }|{\mathbf {z}}_0)^2}\Sigma . \end{aligned}$$

(7.2)

Then the conclusion of Theorem 1 can be proved. $\square$