Simple estimation procedures for regression analysis of interval-censored failure time data under the proportional hazards model

Abstract

Interval-censored failure time data occur in many fields including epidemiological and medical studies as well as financial and sociological studies, and many authors have investigated their analysis (Sun, The statistical analysis of interval-censored failure time data, 2006; Zhang, Stat Modeling 9:321–343, 2009). In particular, a number of procedures have been developed for regression analysis of interval-censored data arising from the proportional hazards model (Finkelstein, Biometrics 42:845–854, 1986; Huang, Ann Stat 24:540–568, 1996; Pan, Biometrics 56:199–203, 2000). For most of these procedures, however, one drawback is that they involve estimation of both regression parameters and baseline cumulative hazard function. In this paper, we propose two simple estimation approaches that do not need estimation of the baseline cumulative hazard function. The asymptotic properties of the resulting estimates are given, and an extensive simulation study is conducted and indicates that they work well for practical situations.

Appendix: Proofs of the asymptotic properties of \(\hat{\beta }_I\)

Proof of consistency of $\hat{\beta }_I$ We first reparametrize the local partial likelihood $l_I(\beta )$ via the transformation $\xi =\beta -\beta _0,$ we just need prove that $\hat{\xi }=\hat{\beta }-\beta _0\rightarrow 0$ in probability, the logarithm of the local partial likelihood function is

$$\begin{aligned} \tilde{l}(\xi )\triangleq&l_I(\xi +\beta _0)\\ =&\sum _{i=1}^n\int \limits _0^\infty (\xi +\beta _0)'X_i\text {d}N_i(t)\\&-\sum _{i=1}^n\int \limits _0^\infty \log \left( \sum _{j=1}^nY_j(t)\exp \left[ (\xi +\beta _0)'X_j\right] \right) \text {d}N_i(t), \end{aligned}$$

and define

$$\begin{aligned} M(\xi )=\frac{1}{n}(\tilde{l}(\xi )-\tilde{l}(0))&= \frac{1}{n}\sum _{i=1}^n\int \limits _0^\infty \left( \xi 'X_i-\log \frac{S_{n}^{(0)}(t,\xi +\beta _0)}{S_{n}^{(0)}( t,\beta _0)}\right) \text {d}N_i(t). \end{aligned}$$

Denote $\bar{N}(t)=\sum _{i=1}^n N_i(t)$ and $\bar{X}(t,\xi )=S_{n}^{(1)}(t,\xi +\beta _0)/S_{n}^{(0)}(t,\xi +\beta _0),$ clearly,

$$\begin{aligned} \frac{\partial ^2M(\xi )}{\partial \xi \partial \xi '}&= -\frac{1}{n}\sum _{i=1}^n\int \limits _0^\infty \frac{S_n^{(2)}(t,\xi +\beta _0)S_n^{(0)}(t,\xi +\beta _0)-\left( S_n^{(1)}(t,\xi +\beta _0)\right) ^{\otimes 2}}{\left( S_n^{(0)}(t,\xi +\beta _0)\right) ^2}dN_i(t)\\&= -\frac{1}{n}\sum _{j=1}^n\int \limits _0^\infty \left( X_j-\bar{X}(t,\xi )\right) ^{\otimes 2}Y_j(t)e^{(\xi +\beta _0)'X_j}\frac{d\bar{N}(t)}{S_{n}^{(0)}(t,\xi +\beta _0)}, \end{aligned}$$

which is strictly negative semi-definite in probability. That is, $\tilde{l}(\xi )-\tilde{l}(0)$ is concave in $\xi$ with the unique maximizer $\hat{\xi }.$

Using a similar method used in proving Lemma A.1 in the paper of Fan et al. (2006), we can show that

$$\begin{aligned} \sup _{0\le t\le \tau }\parallel S_{n}^{(k)}(t,\beta _0+\xi )-s_k(t,\beta _0+\xi )\parallel \overset{P}{\longrightarrow }0, \end{aligned}$$

where $s_k(t,\beta _0+\xi )=E\{P_{\beta _0}(U\ge t|X)\exp ((\beta _0+\xi )'X)X^{\otimes k}\}.$ Combining the strong law of large numbers, it follows that $M(\xi )$ converges almost surely to

$$\begin{aligned} \mathcal M (\xi )\triangleq E\left[ \int \limits _0^\infty \xi 'X_1d N_1(t)-\int \limits _0^\infty \log \frac{s_0(t,\xi +\beta _0)}{s_0(t)}dN_1(t)\right] . \end{aligned}$$

A straightforward computation then yields

$$\begin{aligned}&\frac{\partial \mathcal M (\xi )}{\partial \xi }=E\left[ \int \limits _0^\infty \left( X_1-\frac{s_1(t,\xi +\beta _0)}{s_0(t,\xi +\beta _0)}\right) dN_1(t)\right] ,\\&\frac{\partial ^2\mathcal M (\xi )}{\partial \xi \partial \xi '}\\&\ =-E\left[ \int \limits _0^{\infty }\frac{s_2(t,\xi +\beta _0)s_0(t,\xi +\beta _0)-s_1(t,\xi +\beta _0)(s_1(t,\xi +\beta _0))'}{s_0^2(t,\xi +\beta _0)}dN_1(t)\right] . \end{aligned}$$

It can be easily seen that $\left( \partial ^2\mathcal M ({ 0})/\partial \xi \partial \xi '\right)$ is strictly negative definite,and $\left( \partial \mathcal M ( 0)/\partial \xi \right) =0.$ Hence, we obtain that $\mathcal M (\xi )$ has maximum value at $\xi =0.$ By the convexity of $\tilde{l}(\xi )-\tilde{l}(0)$ and the concavity lemma, it follows that $\hat{\xi }\rightarrow 0,$ the maximizer of $\mathcal M (\xi )$ in probability.

This completes the proof.

Proof of the asymptotic normality of $\hat{\beta }_I$ Start from the equality

$$\begin{aligned} \frac{1}{\sqrt{n}}\dot{l}_I({\beta _0})&= \frac{1}{\sqrt{n}}\sum _{i=1}^n\int \limits _0^\infty \left( X_i -\frac{S_{n}^{(1)}(t,\beta _0)}{S_{n}^{(0)}(t,\beta _0)}\right) \text {d}N_i(t)\\&= \frac{1}{\sqrt{n}}\sum _{i=1}^n\int \limits _0^\infty \left( X_i -\frac{s_1(t)}{s_0(t)}\right) \text {d}N_i(t)+o_p(1). \end{aligned}$$

We obtain that

$$\begin{aligned} \qquad \qquad \qquad \qquad n^{-\frac{1}{2}}\ \dot{l}_I(\beta _0)\overset{D}{\longrightarrow }N(0,\varGamma _I(\beta _0)), \end{aligned}$$

(2)

where $\Gamma _I(\beta _0)=E(\int _0^\infty \left( X -s_1(t)/s_0(t)\right) \text {d}N_1(t))^2,$ and which can be consistently estimated by

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\int \limits _0^\infty \left( X_i-\frac{S_n^{(1)}(t,\hat{\beta })}{S_n^{(0)}(t,\hat{\beta })}\right) ^2dN_i(t). \end{aligned}$$

Since $\hat{\beta }$ maximizes the log-likelihood function, and by the Taylor’s expansion, we have

$$\begin{aligned} \dot{l}_I(\beta _0)=\dot{l}_I(\hat{\beta })-\dot{l}_I( \beta _0)=( \ddot{l}_I(\gamma ))'(\hat{\beta }-\beta _0), \end{aligned}$$

where $\gamma$ is between $\hat{\beta }$ and $\beta _0.$ Since $\hat{\beta }\rightarrow \beta _0$ in probability, we have $\ddot{l}_I(\gamma )\rightarrow \ddot{l}_I(\beta _0)$ and

$$\begin{aligned} \qquad \qquad \quad \sqrt{n}(\hat{\beta }-\beta _0)=-\left( \frac{1}{n} \ddot{l}_I(\beta _0)\right) ^{-1}\left( \frac{1}{\sqrt{n}} \dot{l}_I(\beta _0)\right) +o_p(1). \end{aligned}$$

(3)

We now need to prove that $\ddot{l}_I(\beta _0)/n$ converges to an invertible matrix. Note that

$$\begin{aligned} \qquad \frac{1}{n}\ddot{l}_I(\beta _0)&= -\frac{1}{n}\int \limits _0^\infty \sum _{i=1}^n\frac{s_2(t)s_0(t)-s_1(t)(s_1(t))^T}{(s_0(t))^2}\text {d}N_i(t)+o_p(1) \nonumber \\&= -\int \limits _0^\infty \frac{s_2(t)s_0(t)-s_1(t)(s_1(t))^T}{(s_0(t))^2}\text {d}F_n(t)+o_p(1) \nonumber \\&= -\int \limits _0^\infty \frac{s_2(t)s_0(t)-s_1(t)(s_1(t))^T}{(s_0(t))^2}\text {d}F_U(t)+o_p(1) \nonumber \\&\triangleq -R_I(\beta _0)+o_p(1), \end{aligned}$$

(4)

where $F_n(t)=\frac{1}{n}\sum \nolimits _{i=1}^nI(U_i\le t,\delta _i=1),$ and $F_U(t)=P\{U\le t,\delta =1\}.$ It is easy to see that $R_I(\beta _0)^{-1}$ is positive definite. Combining (2), (3) and (4), and by the Slutsky’s theorem we obtain that

$$\begin{aligned} \sqrt{n}(\hat{\beta }-\beta _0)\overset{D}{\longrightarrow }N(\mathbf{0},R_I(\beta _0)^{-1}\varGamma _I(\beta _0)(R_I(\beta _0)^{-1})'). \end{aligned}$$

This completes the proof.