Adjusted empirical likelihood for right censored lifetime data

Abstract

Adjusted empirical likelihood (AEL) is a method to improve the performance of the empirical likelihood (EL) particularly in the construction of the confidence interval based on completely observed data. In this paper, we extend AEL approach to the analysis of right censored data by adopting an influence function method. The main results include that the adjusted log-likelihood ratio is asymptotically Chi-squared distributed. Simulation results indicate that the proposed AEL-based confidence intervals perform better compared with normality-based or EL-based confidence intervals specifically for small sample size within the right-censoring setting. The proposed method is illustrated by analysis of survival time of patients after operation for spinal tumors.

Appendix

Proof of Theorem 2:

To determine $R({\theta }_0;a_n)$, we solve for the Lagrange multipliers $\mu$ and $\lambda$ in

$$\begin{aligned} A=\sum _{i=1}^{n+1} \log ((n+1)p_i) - (n+1)\lambda \sum _{i=1}^{n+1} p_iW_{ni}-\mu \big (1-\sum _{i=1}^{n+1} p_i\big ). \end{aligned}$$

Then $\mu =-n-1$ and $p_{i}=\dfrac{1}{n+1}(1+\lambda W_{ni})^{-1},\ i=1,2,\ldots ,n,$ where $\lambda$ is the solution of

$$\begin{aligned} \sum _{i=1}^{n+1}\frac{W_{ni}}{1+\lambda {W_{ni}}}=0. \end{aligned}$$

(6.1)

As $W_{n{n+1}}$ and $\overline{W}_n=n^{-1}\sum _{i=1}^nW_{ni}$ are on the opposite sides of 0, the solution of (6.1) exists.

The EL ratio of ${\theta }$ can be written as

$$\begin{aligned} R({\theta }_0;a_n)=\prod _{i=1}^{n+1}((n+1)p_{i})=\prod _{i=1}^{n+1}(1+\lambda W_{ni})^{-1}. \end{aligned}$$

(6.2)

Let $\sigma ^2=Var(W_i)$ for $W_i$ defined in (3.3) and $\lambda$ be the solution of (6.1). We prove that $\lambda =O_p(n^{-\frac{1}{2}})$.

Denote $W^{*}={\max }_{1\le {i}\le {n}}|W_{ni}|$. From Lemma 4.3 in He et al. (2012), we have

$$\begin{aligned}&W^{*}=o_p(n^{\frac{1}{2}}), \nonumber \\&\overline{W}_n \equiv n^{-1}\sum _{i=1}^nW_{ni}=O_p(n^{-\frac{1}{2}}), \nonumber \\&n^{-1}\sum _{i=1}^nW_{ni}^2=\sigma ^2+o_p(1). \end{aligned}$$

(6.3)

Let $\rho =|\lambda |, \hat{\lambda }=\lambda /\rho$. Multiplying $n^{-1}\hat{\lambda }$ to both sides of (6.1), we have

$$\begin{aligned} 0&= \frac{\hat{\lambda }}{n}\sum _{i=1}^{n+1}\frac{W_{ni}}{1+\lambda {W_{ni}}} \nonumber \\&= \frac{\hat{\lambda }}{n}\sum _{i=1}^{n+1}W_{ni}-\frac{\rho }{n}\sum _{i=1}^{n+1}\frac{(\hat{\lambda }W_{ni})^2}{1+\rho \hat{\lambda }{W_{ni}}} \nonumber \\&\le \hat{\lambda }\overline{W}_n(1-{a_n}/{n})-\frac{\rho }{n(1+\rho {W^*})}\sum _{i=1}^{n}(\hat{\lambda }W_{ni})^2. \end{aligned}$$

(6.4)

Using $a_n=o_p(n)$ and (6.3), we get

$$\begin{aligned} \frac{\rho }{1+\rho {W^*}}&\le \hat{\lambda }\overline{W}_n(1-a_n/n)\Big (n^{-1}\sum _{i=1}^nW_{ni}^2 \Big )^{-1}\\&= O_p(n^{-1/2})(1-o_p(1))(\sigma ^2+o_p(1))^{-1}\\&= O_p(n^{-1/2}). \end{aligned}$$

It follows that $\rho =|\lambda |=O_p(n^{-1/2})$.

Denoting $\hat{V}_n=n^{-1}\sum _{i=1}^{n}W_{ni}^2$, we get

$$\begin{aligned} 0&= \frac{1}{n}\sum _{i=1}^{n+1}\frac{W_{ni}}{1+\lambda {W_{ni}}}\\&= \frac{1}{n}\sum _{i=1}^{n+1}W_{ni}-\frac{1}{n}\sum _{i=1}^{n+1}\frac{\lambda {W_{ni}}^2}{1+\lambda {W_{ni}}}\\&= \overline{W}_n-\lambda \hat{V}_n+o_p(n^{-\frac{1}{2}}). \end{aligned}$$

As $n\rightarrow {\infty }$, we have $\lambda =\hat{V}_n^{-1}\overline{W}_n+o_p(n^{-\frac{1}{2}})$.

Finally, we use Taylor expansion for $l(\theta _0;a_n)=-2\log R(\theta _0;a_n)$ and have

$$\begin{aligned} -2\log R(\theta _0;a_n)&= 2\sum _{i=1}^{n+1}\log (1+\lambda {W}_{ni})\\&= 2\sum _{i=1}^{n+1}\{\lambda {W}_{ni}-\frac{(\lambda {W}_{ni})^2}{2}\}+o_p(1). \end{aligned}$$

Using $\lambda =\hat{V}_n^{-1}\overline{W}_n+o_p(n^{-\frac{1}{2}})$, we have

$$\begin{aligned} -2\log R(\theta _0;a_n)=n\overline{W}_n^2\hat{V}_n^{-1}+o_p(1). \end{aligned}$$

From Lemma 4.3 of He et al. (2012), we see $n^{\frac{1}{2}}\overline{W}_n =n^{-\frac{1}{2}}\sum _{i=1}^nW_{ni}\rightarrow {N(0, \sigma ^2)}$ in distribution. Hence $n\overline{W}_n^2\hat{V}_n^{-1}\rightarrow {\chi _1^2}$ in distribution.