Empirical likelihood analysis of the Buckley–James estimator

Abstract

The censored linear regression model, also referred to as the accelerated failure time (AFT) model when the logarithm of the survival time is used as the response variable, is widely seen as an alternative to the popular Cox model when the assumption of proportional hazards is questionable. Buckley and James [Linear regression with censored data, Biometrika 66 (1979) 429−436] extended the least squares estimator to the semiparametric censored linear regression model in which the error distribution is completely unspecified. The Buckley–James estimator performs well in many simulation studies and examples. The direct interpretation of the AFT model is also more attractive than the Cox model, as Cox has pointed out, in practical situations. However, the application of the Buckley–James estimation was limited in practice mainly due to its illusive variance. In this paper, we use the empirical likelihood method to derive a new test and confidence interval based on the Buckley–James estimator of the regression coefficient. A standard chi-square distribution is used to calculate the P-value and the confidence interval. The proposed empirical likelihood method does not involve variance estimation. It also shows much better small sample performance than some existing methods in our simulation studies.

1. Introduction

The Cox proportional hazards regression model [2] is very popular and has been routinely used in modeling covariate effects with right censored survival data. However, there are many cases where the proportional hazards model does not apply. As Cox pointed out in an interview [22], “Of course, another issue is the physical or substantive basis for the proportional hazards model. I think that is one of its weaknesses, that accelerated life models are in many ways more appealing because of their quite direct physical interpretation, particularly in an engineering context”. See also [27] and the references therein.

One of the most promising estimate in the semiparametric accelerated failure time (AFT) model is the Buckley–James estimator [1]. They extended the least squares estimation method to linear regression with right censored data. The Buckley–James estimator is calculated using an iterative algorithm. Cheap, fast computers and ever-improving software in recent years have made the calculation of the Buckley–James estimator a routine business. It has also been observed to perform well in many simulation studies and case studies [14,5,6,24]. However, variance estimation of the Buckley–James estimator remains very difficult. The program bj () within the Design library of Harrell (http://biostat.mc.vanderbilt.edu/s/Design), available for both S-plus and R languages [4], uses a variance estimation formula given by the Buckley–James's original paper which does not have a rigorous justification and indeed may not be consistent as pointed out by Lai and Ying [9]. On the other hand, the variance given by Lai and Ying [9] involves the density and the derivative of the density of the unknown distribution. Estimation of such functions can be highly unstable.

Several approaches have been proposed to tackle this problem [13,8,10,28,21,12,7]. Most recently, Jin et al. [7] used a novel resampling method for estimating the variance of the Buckley–James estimator. Qin and Jing [21] and Li and Wang [12] attempted to use the empirical likelihood (EL) method [18] for censored regression analysis. See Owen [18] for some discussion of bootstrap versus EL method. The EL method was first proposed by Thomas and Grunkemeier [25] to obtain better confidence intervals for a survival probability in connection with the Kaplan–Meier estimator. Owen [16,17] and many others developed this into a general methodology. It has many desirable statistical properties [18]. One of the nice features of the EL method particularly appreciated in censored data analysis is that one can construct confidence intervals without estimating the variance of the statistic. It also has better performance than the traditional normal approximation (Wald) method. However, the applications of Qin and Jing [21] and Li and Wang's [12] methods are hampered by the fact that the limiting distribution of their EL ratio is not a standard chi-square but a linear combination of chi-squares with coefficients depending on the unknown underlying distributions. Moreover, their methods are based on a synthetic data method [8] which does not perform well for small samples with heavy censoring [12].

In this paper, we propose a new EL test procedure for the Buckley–James estimator. In contrast to the synthetic data approach of [21,12], we use the true censored EL and show that the corresponding EL ratio has a standard chi-square limiting distribution. Thus the likelihood ratio test and confidence intervals can be obtained without estimating other quantities. The proposed method has demonstrated much more accurate coverage probability than that of [21,12] in our simulation study.

The rest of the paper is organized as follows. In Section 2, we derive an EL associated with the Buckley–James estimator for a semiparametric linear regression model with censored data. The EL is shown to have a standard chi-square limiting distribution. We also discuss how to extend our method to M-estimators for censored linear regression. Section 3 presents simulation results to evaluate the finite sample performance of the proposed method compared to the synthetic databased method of Qin and Jing [21] and Li and Wang [12]. We finally illustrate our method on the Stanford Heart Transplantation data.

2. The regression model and the empirical likelihood

2.1. The model and notations

Consider the linear regression model

$$y_i={\beta }^t{x}_i+{∊}_{i}i=1,\dots ,n,$$

where for subject i, y_i is the logarithm of the survival time and x_i is the associated vector of q covariates. We assume that ε_i's are independent and identically distributed with an unspecified distribution except with zero mean and finite variance.

Suppose that one observes the following right censored observations:

$${\tilde{y}}_i=\min (y_i,c_i),{\delta }_i=I_{[y_i\le c_i]}\text{and}{x}_i,i=1,\dots ,n,$$

where for subject i, c_i is the censoring time, assumed independent of y_i given x_i.

For any candidate estimator b of β, we define e_i(b) = ${\tilde{y}}_i$ – b^t x_i. Let us order the e_i(b)'s:

$$e_{\left(1\right)}\left(b\right)<\cdot \cdot \cdot <e_{\left(n\right)}\left(b\right)$$

and order δ_i and x_i along with e_i(b). Notice this ordering is dependent on b. For simplicity, we assume for the rest of the paper that the e_i(b)'s are already ordered and thus e_(i)(b) = e_i(b). We will also omit the symbol b as appropriate.

2.2. The empirical likelihood and estimation equation

Let F̂_KM(t) denote the Kaplan–Meier estimator based on the right censored sample (e_i(b), δ_i), i = 1, . . . , n. We form an n × n weight matrix M whose (i, j)th element M[i, j] is defined as follows: if δ_i = 0 then

$$m[i,j]=0,j\le i\text{and}m[i,k]=\frac{\Delta {\widehat{F}}_{\mathrm{KM}}\left(e_k\right)}{1-{\widehat{F}}_{\mathrm{KM}}\left(e_i\right)}\text{for}k>i,$$

if δ_i = 1 then m[i, i] = 1, and m[i, j] = 0; j ≠ i. It is easy to see that M is an upper triangle matrix satisfying the property ${\sum }_{j}m[i,j]=1$ for all i.

Let $w_j={\sum }_{i}m[i,j]∕n$. Then w_j, j = 1, 2, . . . , n, is a probability distribution with support on the uncensored e_i's. Because the Kaplan–Meier estimator is a self-consistent estimator, we have w_j = ΔF̂_KM(e_j), j = 1, . . . , n.

The Buckley–James estimating equation (cf. [23]) can be written as

$$0=\frac{1}{n}\sum _{i=1}^n\left({\delta }_i{x}_i{e}_i\left(b\right)+(1-{\delta }_i)x_i\sum _{j:j>i}{e}_j\left(b\right)m[i,j]\right),$$ (1)

where all the n terms in the above summation are nonzero. The Buckley–James estimator is the solution to the above equation and denoted by ${\widehat{\beta }}_{\mathrm{BJ}}$.

We rewrite the Buckley–James estimating equation (1) according to e_i's:

$$0=\frac{1}{n}\sum _i{\delta }_i{e}_i\left(b\right)\left[x_i+\sum _{k<i}m[k,i]x_k\right]$$ (2)

$$=\sum _i{\delta }_i{e}_i\left(b\right)\frac{\sum _{k}m[k,i]x_k}{n{w}_i}\Delta {\widehat{F}}_{\mathrm{KM}}\left(e_i\right).$$ (3)

Notice the number of nonzero terms in the above summation (3) is now the same as the number of nonzero δ's. This suggests that the equation to be used with the censored data EL (defined in (5) below) should be

$$0=\sum _i{\delta }_i{e}_i\left(b\right)\frac{\sum _{k}m[k,i]x_k}{n{w}_i}{p}_i,$$ (4)

where p_i = ΔF(e_i), i = 1, . . . , n, for any distribution F that has support on the uncensored e_i's.

The censored EL based on (e_i(b), δ_i), i = 1, . . . , n, is defined by

$$EL(b,F)=\prod _{i=1}^n{p}_i^{{\delta }_i}{\left(1-\sum _{e_j\le e_i}{p}_j\right)}^{1-{\delta }_i}.$$ (5)

We are to find a distribution F or p_i's that (a) has support only on the uncensored e_i's; (b) satisfies the estimating equation (4); and (c) maximizes the censored EL (5).

Remark 1

The denominator nw_i in (3) seems a bit arbitrary, but it makes the vector, m[k, i]/(nw_i), into a (conditional) probability measure for each i. Therefore, (3) can be interpreted as an double expectation. Also, (1) and (3) are just two versions of double expectations.

Remark 2

When b = β, the true parameter value, then y_i – β^t x_i are independent and identically distributed. Our censored EL is then the same as the censored EL based on independent identically distributed right censored observations used by [25,11,15,20], among others.

Remark 3

When maximizing the censored EL with respect to the p_i's under the constraint (4), the weight matrix M and w_i remain unchanged for a fixed b.

Remark 4

Clearly, if b = ${\widehat{\beta }}_{\mathrm{BJ}}$ then p_i = ΔF̂_KM(t) satisfies the estimating equation (4), and maximizes the EL (5) among all CDF's. Therefore, the confidence regions based on our EL ratio will be “centered” at ${\widehat{\beta }}_{\mathrm{BJ}}$.

Remark 5

The numerical problem of maximizing the censored EL with respect to p_i's with the constraint (4) is the same as the one faced by the censored EL with a mean constraint: $\sum f\left(t_i\right)p_i=\mu$. Here $f\left(t_i\right)=t_i\frac{{\sum }_{k}m[k,i]x_k}{n{w}_i}$ and μ = 0. This computational problem do not have explicit solution, but can be handled by a modified EM algorithm proposed by Zhou [29].

2.3. The asymptotics

It is worth noting that our constraint equation (4) is different from the mean constraint used in the previous works of EL in that the function f(·) in our problem depends on the data, and thus should be denoted by f_n(·):

$$\int f\left(t\right)dF\left(t\right)\mathrm{v}.\mathrm{s}.\int f_n\left(t\right)dF\left(t\right).$$

Consequently, we need a more general EL Wilks theorem for right censored data which is stated in the following theorem.

Theorem 1. Suppose X_i for i = 1, 2, . . . , n are independent random variables with $P(X_i\le t)$ = F₀(t). By right censored data we mean the pairs T_i, δ_i where T_i = min(X_i, C_i); ${\delta }_i=I_{[X_i\le C_i]}$ with C_i independent of X_i. Define the censored EL

$$EL\left(F\right)=\prod _{i=1}^n{p}_i^{{\delta }_i}{\left(1-\sum _{T_j\le T_i}{p}_j\right)}^{1-{\delta }_i}.$$

In addition, suppose that for every n, f_n(t) is a predictable random function with respect to the standard counting process filtration ${\mathcal{F}}_t$ (see for example [3]) and satisfies the regularity conditions (for g_n) in Lemma 1. Then we have

$$-2\log \frac{{\sup }_{F}EL\left(F\right)}{EL\left({\widehat{F}}_{\mathrm{KM}}\right)}\to {\chi }_{\left(1\right)}^2$$

in distribution as n → ∞ where F̂_KM is the Kaplan–Meier estimator based on T_i, δ_i, and the numerator EL is maximized under the constraint

$$\sum {\delta }_i{p}_i{f}_n\left(T_i\right)={\int }_{-\infty }^{\infty }{f}_n\left(t\right)dF\left(t\right)=0.$$

With the help of Lemma 1, this theorem can be proved similarly to [19,20]. We defer the proof of the theorem to appendix.

Lemma 1. Consider censored data as described in Theorem 1, let g_n(t) be predictable functions such that ${\int }_{-\infty }^{\infty }{g}_n\left(t\right)d{F}_0\left(t\right)=0$ and g_n(t) → g(t) in probability as n → ∞ with g(t) satisfy ${\sigma }_{\mathrm{KM}}^2\left(g\right)<\infty$ (defined below), then we have

$$\sqrt{n}{\int }_{-\infty }^{\infty }{g}_n\left(t\right)d{\widehat{F}}_{\mathrm{KM}}\left(t\right)=\sqrt{n}\sum gn\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\to N(0,{\sigma }_{\mathrm{KM}}^2\left(g\right))$$

in distribution as n → ∞. The asymptotic variance is given by

$${\sigma }_{\mathrm{KM}}^2\left(g\right)={\int }_0^{\infty }{\left\{g\left(x\right)[1-F_0\left(x\right)]-{\int }_{-\infty }^{x}g\left(s\right)d{F}_0\left(s\right)\right\}}^2\frac{d{F}_0\left(x\right)}{{[1-F_0\left(x\right)]}^2[1-G_0\left(x\right)]},$$ (6)

where G₀(x) = lim 1/n ${\sum }_{i=1}^{n}P(C_i\le x)$. Furthermore, denote by $M_n\left(t\right)=\sqrt{n}[{\widehat{F}}_{\mathrm{KM}}\left(t\right)-F_0\left(t\right)]∕[1-F_0\left(t\right)]$ the ${\mathcal{F}}_t\text{-}martingale$ associated with the Kaplan–Meier estimator F̂_KM(t) based on the (T_i, δ_i) i = 1, . . . , n, the asymptotic variance, ${\sigma }_{\mathrm{KM}}^2\left(g\right)$, can be consistently estimated by

$${\int }_{-\infty }^{\infty }{\left\{g_n\left(x\right)[1-{\widehat{F}}_{\mathrm{KM}}\left(x\right)]-{\int }_{-\infty }^x{g}_n\left(s\right)d{\widehat{F}}_{\mathrm{KM}}\left(s\right)\right\}}^{2}d\langle M_n\left(x\right)\rangle .$$

Proof. By the assumption $\int g_n\left(t\right)d{F}_0\left(t\right)=0$ we have

$${\int }_{-\infty }^{\infty }{g}_n\left(t\right)d{\widehat{F}}_{\mathrm{KM}}\left(t\right)={\int }_{-\infty }^{\infty }{g}_n\left(t\right)d[{\widehat{F}}_{\mathrm{KM}}\left(t\right)-F_0\left(t\right)].$$

Integration by parts will give

$$={\int }_{-\infty }^{\infty }\left\{g_n\left(t\right)[1-F_0\left(t\right)]-{\int }_t^{\infty }{g}_n\left(s\right)d{F}_0\left(s\right)\right\}d{M}_n\left(t\right).$$

Use the fact that $\int g_n\left(t\right)d{F}_0\left(t\right)=0$ again, we have

$$={\int }_{-\infty }^{\infty }\left\{g_n\left(t\right)[1-F_0\left(t\right)]+{\int }_{-\infty }^t{g}_n\left(s\right)d{F}_0\left(s\right)\right\}d{M}_n\left(t\right).$$

The integrand inside {} in the above is clearly a predictable function and thus the integration is also a martingale. By the CLT for martingales, it converges to a normal distribution with zero mean and a variance that can be consistently estimated by

$${\int }_{-\infty }^{\infty }{\left\{g_n\left(t\right)[1-F_0\left(t\right)]-{\int }_{-\infty }^t{g}_n\left(s\right)d{F}_0\left(s\right)\right\}}^{2}d\langle M_n\left(t\right)\rangle .$$

Replace F₀(·) by its consistent estimate F̂_KM(·) gives the desired result. □

Lemma 2. The weight function $f_n\left(t_i\right)=t_i\left[{\sum }_{k}m[k,i]x_k\right]∕\left(n{w}_i\right)$ is ${\mathcal{F}}_t$ predictable.

Proof. Notice that if the Kaplan–Meier estimator jumps or not at t is not predictable but we are only concerned here with the size of the jump, if there is one. The size of the next jump of the Kaplan–Meier estimator can always be computed from the history and thus is predictable. More specifically, the next jump size of the Kaplan–Meier estimator, at time t, if there is one, is equal to 1/n × 1/(1 – Ĝ(t–)), where Ĝ is the Kaplan–Meier estimator of the censoring distribution.

Similarly we can infer from the history which part of the jump, if there is one, came from t_j; t_j < t. This proportion is precisely m[j, i]/(nw_i). □

Armed with Theorem 1 and Lemma 2, we are ready to prove the following Wilks theorem for the Buckley–James estimator.

Theorem 2. When β = β₀, the residuals $e=\tilde{y}-{\beta }_{0}x=\widetilde{∊}$ are independent and identically distributed before censoring and the estimating equation (4) can be written as

$$E\left(e{E}^{\ast }(x\mid e=t)\right)\equiv 0,$$

where $E^{\ast }$ denotes the average over the m[j, i]/(nw_i), j = 1, 2, . . . , n. As n → ∞, we have

$$-2\log \frac{{\sup }_{F}EL({\beta }_0,F)}{EL({\widehat{\beta }}_{\mathrm{BJ}},{\widehat{F}}_{\mathrm{KM}})}\to {\chi }_{\left(1\right)}^2$$

in distribution, where both EL is defined in (5) and the numerator EL is maximized under constraint (4).

Proof. Since x is independent of ε, $E\left(e{E}^{\ast }(x\mid e=t)\right)=E\left(x\right)E\left(e\right)\equiv 0$. Then the desired result follows from Lemma and Theorem 1. □

2.4. Empirical likelihood for M-estimators

Our EL method for the Buckley–James estimator can be extended to a class of M-estimators along the same lines. For complete data the regression M-estimator is defined as the minimizer of $\sum \rho (y_i-{\beta }^t{x}_i)$ or the solution to the equation

$$\sum x_i\psi (y_i-{\beta }^t{x}_i)=0,$$

where $\psi \left(t\right)=\frac{d\rho \left(t\right)}{dt}$. Usually we assume ψ is monotone.

For right censored data, the Buckley–James estimating equation (for M-estimate) is

$$0=\sum _{i=1}^n\left({\delta }_i{x}_i\psi (y_i-{\beta }^t{x}_i)+(1-{\delta }_i)x_i\sum _{j:e_j>e_i}\psi (y_j-{\beta }^t{x}_j)m[i,j]\right).$$

A rewriting of the estimating equation according to e_i gives

$$0=\sum _i{\delta }_i\psi (y_i-{\beta }^t{x}_i)\left[x_i+\sum _{k:e_k<e_i}m[k,i]x_k\right].$$

In the EL analysis of the censored Buckley–James regression M-estimator, the definition of the censored EL remains unchanged as in (5). The constraint (or estimating equation) to be used with the EL is

$$0=\sum _{i=1}^n\psi (y_i-{\beta }^t{x}_i)\frac{x_i+\sum _{k:e_k<e_i}m[k,i]x_k}{n{w}_i}{\delta }_i{p}_i.$$

Similar results as in the least squares estimator can be obtained for the regression M-estimator. We omit the details here.

3. Simulations

The computations of this section and the next are done with software R [4] with the added package emplik, available at any CRAN site (e.g. http://cran.us.r-project.org).

We first considered the following regression model y_i = 2x_i + ε_i, where x_i is uniform(0.5, 1.5) and ε_i is uniform(−0.5, 0.5). We further take c_i to be 1 + 3.2 exp(1), where exp(1) represents a random variable with standard exponential distribution. The sample size n is 100.

The −2 log EL ratio is computed for each simulation run for the hypothesis H₀: β = 2. The resulting Q–Q plot shows a good fit of the distribution of the EL ratio to the chi-square distribution with 1 degree of freedom (Fig. 1).

Fig. 1.

Q–Q plot of −2 log EL ratio, 5000 simulation run, sample size=100.

We also did a simulation to evaluate the coverage accuracy of the proposed EL method compared with the synthetic data EL methods of [21,12]. We used the regression model y_i = x_i + ε_i where ε_i ∼ N(0, 0.25), x_i ∼ N(1, 0.5) and C_i ∼ N(μ, 16), with μ = −1.8, 1, 3.1 and 6.1, respectively. This produces samples with censoring percentages equal to 75%, 50%, 30% and 10%, approximately.

Table 1 gives the achieved coverage probabilities of confidence intervals based on our proposed empirical likelihood method for the Buckley–James estimator (ELBJ) and the synthetic data empirical likelihood method (ELSD) [21, 12], respectively. Each entry is based on 5000 simulation runs.

Table 1.

Simulated coverage probabilities of empirical likelihood confidence intervals for β: ELSD refers to the empirical likelihood method of [21,12] based on synthetic data; ELBJ is our proposed empirical likelihood method for the Buckley—James estimator

Sample size	Censoring rate	Nominal level=90%		Nominal level=95%
		ELSD	ELBJ	ELSD	ELBJ
50	0.75	0.79	0.84	0.85	0.90
100	0.75	0.83	0.88	0.90	0.93
200	0.75	0.87	0.89	0.92	0.94
50	0.50	0.84	0.88	0.91	0.93
100	0.50	0.88	0.89	0.95	0.94
200	0.50	0.91	0.90	0.95	0.95
50	0.30	0.87	0.89	0.92	0.94
100	0.30	0.89	0.89	0.93	0.95
200	0.30	0.91	0.89	0.95	0.95
50	0.10	0.85	0.89	0.93	0.94
100	0.10	0.91	0.89	0.94	0.94
200	0.10	0.90	0.88	0.94	0.95

We see from Table 1 that both methods perform reasonably well for large samples. However, our proposed empirical likelihood method (ELBJ) has a noticeable better coverage accuracy than the synthetic data empirical likelihood method (ELSD) for smaller samples (say n = 50), especially in the case of heavy censoring (75%). Besides, the confidence level is much easier to set in ELBJ method due to a simpler limiting distribution.

As discussed in the introduction, the function bj( ) inside the Design package of R will compute a variance estimator proposed by Buckley and James [1]. The next simulation show that this variance estimator is not consistent, thus confirm Lai and Ying [9].

We found that the Wald statistic formed by using the given variance estimator and the Buckley–James estimator is smaller than the corresponding chi-square values under null hypothesis. Moreover, this discrepancy do not diminish as sample size increase. We also plot our EL ratio statistic as a comparison.

The regression model used is Y = 0.5 + 1.5X + ε; censoring variable is distributed as 1 + 2.5 exp(1). The covariates X is distributed as Unif[1, 2]. The error ε is distributed as N(0, sd = 0.5). The censoring percentage was about 49%.

We also computed the same simulation except with sample size n = 2000. The Q–Q plot is almost identical to the above (n = 400) (Fig. 2).

Fig. 2.

Q–Q plot of −2 log ELR (EL) and Wald statistics (BJvar). 1000 simulation runs.

4. An example

We illustrate the EL analysis of the Buckley–James estimator with the Stanford Heart Transplant data. Following [14], we use only 152 cases. The specific AFT model we used is log₁₀(T_i) = β₀ + β₁ age + ε_i.

The Buckley–James estimator of the $({\widehat{\beta }}_0,{\widehat{\beta }}_1)$ was marked by an X on the plot. From the plot we see that the contours are fairly symmetrical and elliptically shaped, indicating that the normal approximation is pretty good for the Buckley–James estimator here.

From the plot we see that the estimator ${\widehat{\beta }}_0$ is strongly negatively correlated with ${\widehat{\beta }}_1$. The 95% confidence interval for the β₁ alone is approximately [−0.0357, −0.0028], the 95% confidence interval for β₀ alone is approximately [2.755, 4.255. These are obtained as the left (right, upper or lower) most point of the contour with level 3.84. They are approximate values because we used a coarse grid points to produce the contour plot, and thus interpolation was used in the plot (Fig. 3).

Fig. 3.

Contour plot for the −2 log EL ratio, Stanford Heart Transplant Data, 152 cases.

From the bj() function from the Design library of Harrell, the following results are obtained:

> bj (Surv(log10(time), status)∼ age, data=stanford5, link=“identity”)
Buckley–James Censored Data Regression
bj (formula=Surv(log10 (time), status)∼ age, data=stanford5, link=“identity”)
Obs Events d.f. error d.f. sigma
152 97 1 95 0.6796
Value Std. Error Z Pr(>|Z|)
Intercept 3.52696 0.299123 11.79 4.344e-32
age −0.01990 0.006632 −3.00 2.700e-03

Our confidence intervals are slightly wider than the ones obtained by the Wald confidence interval using the standard error estimator given by the function bj(). We remind readers that the standard error estimator produced by bj() has no theoretical justifications.

5. Concluding remarks

We developed a new empirical likelihood method for the Buckley–James estimator for censored linear regression. The empirical likelihood can be calculated using a modified EM algorithm proposed by Zhou [29]. Our method is different from the previous empirical likelihood methods for censored linear regression in that we use a “true” likelihood for censored data, while previous methods use an “estimated” empirical likelihood based on synthetic data. Consequently, our empirical likelihood has the standard chi-square null limiting distribution while previous methods do not. The empirical likelihood approach is appealing because it does not involve variance estimation which is difficult for the Buckley–James estimator. Our simulation also showed that our empirical likelihood method has much better small sample performance than previous methods. Finally, our results can be extended to a class of regression M-estimators for censored data.

Appendix

We outline the proof of Theorem 1 here. First of all we define a class of functions

$${\mathcal{H}}_g^{F_0}=\left\{h\mid h\text{is left continuous},\int h^{2}d{F}_0<\infty ,g_n\left(t\right)h\left(t\right)\ge 0\right\}.$$

Furthermore, we define a one-parameter family of distribution functions

$${\mathcal{A}}_h^{F_0}=\left\{F_{\lambda }\left(t\right)\mid F_{\lambda }\left(t\right)=\sum _{i:T_{i\le t}}\Delta F_{\lambda }\left(T_i\right)\right\},$$

where

$$\Delta F_{\lambda }\left(T_i\right)=\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\times \frac{1}{1+\lambda h\left(T_i\right)}\times \frac{1}{C\left(\lambda \right)},i=1,2,\dots ,n,$$ (7)

and C(λ) is just a normalizing constant

$$C\left(\lambda \right)=\sum _{i=1}^n\frac{\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)}{1+\lambda h\left(T_i\right)}.$$

The parameter λ is well defined in a neighborhood of zero and for λ = 0, we get back the Kaplan–Meier: F_λ=0 = F̂_KM. Within this family of distributions, there is only one that satisfy the constraint equation

$$\int g_n\left(t\right)d{F}_{\lambda }\left(t\right)=1∕C\left(\lambda \right)\sum _{i=1}^n\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\frac{g_n\left(T_i\right)}{1+\lambda h\left(T_i\right)}=0.$$ (8)

We denote the parameter for this unique distribution as λ₀.

Finally, we define a class of profile empirical likelihood ratio functions as follows:

$${\mathcal{R}}_h^{F_0}=\left\{\phantom{\mid }\frac{L\left(F_{{\lambda }_0}\right)}{L\left({\widehat{F}}_{\mathrm{KM}}\right)}\mid F\in {\mathcal{A}}_h^{F_0}\right\}.$$

Lemma A. Assume all the conditions in Lemma 1. Then, as n → ∞, (1) λ₀ = O_p(n^−1/2), (2) $n{\lambda }_0^2\to {\chi }_{\left(1\right)}^2\times \frac{{\sigma }_{\mathrm{KM}}^2\left(g\right)}{{\left(\int ghd{F}_0\right)}^2}$ in distribution.

Proof. (outline) Expanding (8), we have

$$\begin{array}{cc}\hfill 0& =\sum _{i=1}^n\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\frac{g_n\left(T_i\right)}{1+{\lambda }_{0}h\left(T_i\right)}\hfill \\ \hfill & =\sum _{i=1}^n{g}_n\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)-{\lambda }_0\sum _{i=1}^n{g}_n\left(T_i\right)h\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)+{\lambda }_0^2\sum _{i=1}^n\frac{g_n\left(T_i\right)h^2\left(T_i\right)}{1+{\lambda }_{0}h\left(T_i\right)}\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right),\hfill \end{array}$$

from there we have

$${\lambda }_0=\frac{\sum _{i=1}^n{g}_n\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)}{\sum _{i=1}^n{g}_n\left(T_i\right)h\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)}+o_p\left(n^{-1∕2}\right).$$

By Lemma 1, as n → ∞,

$$\sqrt{n}\left(\sum _{i=1}^n{g}_n\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\right)\to N(0,{\sigma }_{\mathrm{KM}}^2\left(g\right))$$

in distribution, and

$$\sum _{i=1}^n{g}_n\left(T_i\right)h\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\to {\int }_0^{\infty }g\left(x\right)h\left(x\right)d{F}_0\left(x\right)$$

in probability. By Slutsky's theorem $n{\lambda }_0^2\to {\chi }_{\left(1\right)}^2\times a_h$ in distribution as n → ∞, where

$$a_h={\sigma }_{\mathrm{KM}}^2\left(g\right)/{\left({\int }_0^{\infty }g\left(x\right)h\left(x\right)d{F}_0\left(x\right)\right)}^2.◻$$ (9)

Theorem A. If the conditions in Lemma A hold, then, as n → ∞

$$-2\log {\mathcal{R}}_h^{F_0}\to {\chi }_{\left(1\right)}^2\times r_h$$

in distribution, where

$$r_h=\frac{{\sigma }_{\mathrm{KM}}^2\left(g\right)\times c_h}{{\left(\int ghd{F}_0\right)}^2},$$

and

$$\begin{array}{cc}\hfill c_h=& {\int }_0^{\infty }{h}^2\left(x\right)(1-G_0\left(x\right))d{F}_0\left(x\right)+{\int }_0^{\infty }\frac{{\left({\int }_x^{\infty }h\left(s\right)d{F}_0\left(s\right)\right)}^2}{1-F_0\left(x\right)}d{G}_0\left(x\right)\hfill \\ \hfill & -{\left({\int }_0^{\infty }h\left(x\right)d{F}_0\left(x\right)\right)}^2.\hfill \end{array}$$

Furthermore, inf_h r_h = 1.

Proof. Define

$$f\left(\lambda \right)=\log \prod _{i=1}^n{\left(\Delta F_{\lambda }\left(T_i\right)\right)}^{{\delta }_i}{(1-F_{\lambda }\left(T_i\right))}^{1-{\delta }_i},$$ (10)

where $\mid \lambda \mid \le \mid {\lambda }_0\mid$ and $F\in {\mathcal{A}}_h^{F_0}$. From the definition we can see that

$$C\left(0\right)=1\text{and}f\left(0\right)=\log \prod _{i=1}^n{\left(\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\right)}^{\delta }{(1-{\widehat{F}}_{\mathrm{KM}}\left(T_i\right))}^{1-{\delta }_i}=L\left({\widehat{F}}_{\mathrm{KM}}\right).$$

By Lemma A, λ₀ = O_p(n^−1/2) where λ₀ is the root of (8). Hence we can apply Taylor's expansion for f(λ₀):

$$f\left({\lambda }_0\right)=f\left(0\right)+{\lambda }_0{f}^{\prime }\left(0\right)+\frac{{\lambda }_0^2}{2}{f}^{″}\left(0\right)+\frac{{\lambda }_0^3}{3!}{f}^{\prime ″}\left(\xi \right),\mid \xi \mid \le \mid {\lambda }_0\mid .$$

Substituting (7) in (10),

$$\begin{array}{cc}\hfill f\left(\lambda \right)=& \sum _{i=1}^n{\delta }_i\log \Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)-\sum _{i=1}^n{\delta }_i\log (1+\lambda h\left(T_i\right))-n\log \left(\sum _{i=1}^n\frac{\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)}{1+\lambda h\left(T_i\right)}\right)\hfill \\ \hfill & +\sum _{i=1}^n(1-{\delta }_i)\log \left(\sum _{j:T_j>T_i}\frac{\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_j\right)}{1+\lambda h\left(T_j\right)}\right).\hfill \end{array}$$

Some tedious but straight forward calculation show that f′(0) = 0 and the second derivative of f with respect to λ, evaluated at λ = 0 is

$$\begin{array}{cc}\hfill f^{″}\left(0\right)=& n{\left(\sum _{i=1}^{n}h\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\right)}^2-n\sum _{i=1}^n{h}^2\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\hfill \\ \hfill & +\sum _{i=1}^n(1-{\delta }_i)\frac{\sum _{j:T_j>T_i}{h}^2\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)}{1-{\widehat{F}}_{\mathrm{KM}}\left(T_i\right)}\hfill \\ \hfill & -\sum _{i=1}^n(1-{\delta }_i)\frac{{\left(\sum _{j:T_j>T_i}h\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\right)}^2}{{(1-{\widehat{F}}_{\mathrm{KM}}\left(T_i\right))}^2}.\hfill \end{array}$$

Its not hard to show as n → ∞, the following three quantities all converge in probability:

$$\begin{array}{cc}\hfill & \sum _{i=1}^{n}h\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\to {\int }_0^{\infty }h\left(x\right)d{F}_0\left(x\right),\hfill \\ \hfill & \sum _{i=1}^n{h}^2\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\to {\int }_0^{\infty }{h}^2\left(x\right)d{F}_0\left(x\right),\hfill \\ \hfill & \frac{1}{n}\sum _{i=1}^n(1-{\delta }_i)\frac{{\left(\sum _{j:T_j>T_i}h\left(T_i\right)\Delta {\widehat{F}}_{\mathrm{KM}}\left(T_i\right)\right)}^2}{{(1-{\widehat{F}}_{\mathrm{KM}}\left(T_i\right))}^2}\to {\int }_0^{\infty }\frac{{\left({\int }_x^{\infty }h\left(s\right)d{F}_0\left(s\right)\right)}^2}{1-F_0\left(x\right)}d{G}_0\left(x\right).\hfill \end{array}$$

Hence we have

$$-\frac{1}{n}{f}^{″}\left(0\right)\to c_h$$ (11)

in probability. Finally, by similar calculations we can show that the third derivative of f evaluated at ξ is

$$f″\prime \left(\xi \right)=o_p\left(n^{2∕3}\right).$$ (12)

Now observe

$$\begin{array}{cc}\hfill -2\log {\mathcal{R}}_h^{F_0}& =2\left(f\left(0\right)-f\left(0\right)-{\lambda }_0{f}^{\prime }\left(0\right)-\frac{{\lambda }_0^2}{2}{f}^{″}\left(0\right)-\frac{{\lambda }_0^3}{3!}{f}^{″\prime }\left(\xi \right)\right)\hfill \\ \hfill & =-{\lambda }_0^2{f}^{″}\left(0\right)-\frac{{\lambda }_0^3}{3}{f}^{″\prime }\left(\xi \right).\hfill \end{array}$$

By Lemma A, (11), (12), and Slutsky theorem, we obtain

$$-2\log {\mathcal{R}}_h^{F_0}\to {\chi }_{\left(1\right)}^2\times r_h$$

in distribution.

We now prove the infimum of the constant r_h over h is one. First we notice that

$$\frac{\left(\int h^2(1-G)dF+\int \frac{{\left[{\int }_t^{\infty }h\left(s\right)dF\left(s\right)\right]}^2}{1-F\left(t\right)}dG\left(t\right)-{\left[\int hdF\right]}^2\right)}{{\left(\int ghdF\right)}^2}$$

is precisely the information defined by van der Vaart [26], as i_α in his (4.1).

The infimum of i_α over all one-dimensional sub-models is called “efficient Fisher information”. And in this case (right censored observations), the reciprocal of it is given by the last equation in p. 193 of van der Vaart [26] (as the lower bound for the asymptotic variance of estimating ∫ gd F):

$$\inf i_{\alpha }=\frac{1}{{\Vert \beta \Vert }_F^2}={\left(\int \frac{{\left({R_{\tilde{\chi }}}_F\right)}^2}{1-G}dF\right)}^{-1}.$$

Lastly, we notice that ∫ g_nd F̂_KM is an efficient estimate and therefore we can easily check

$$\text{Asy Var}\left(\int gd{\widehat{F}}_{\mathrm{KM}}\right)=\int \frac{{\left(R_{{\tilde{\chi }}_F}\right)}^2}{1-G}dF.$$

Therefore, inf_h r_h = 1. □