Quantile is an important population characteristic. In some instances the quantile approach is feasible and useful when other approaches are out of the question. For example, to estimate the parameter of a Cauchy distribution, with density $f(x)=1/\pi[1+ (x-\mu)^2], -\infty< x < \infty$, the sample mean $\overline{X}$ is not a consistent estimate of the location parameter $\mu$. However, the sample median $\theta_{1/2}$ is $AN(\mu, \pi^2/4n)$ and thus quite well-behaved.

Let $X_1, X_2, \cdots, X_n$ be a random sample from the unknown distribution $F(x)$ with density $f(x)$. Given $0 < q < 1$, we define the $q$-th quantile by $F^{-1}(q)=\inf\{x: F(x)>q\}$.

In this paper, we investigate how to apply empirical likelihood methods for inference about $\theta_q=F^{-1}(q)$ under right censorship. Assume that the variable $X$ is censored randomly on the right by some censoring variable $C$ and hence cannot be observed completely. One observes only

$\begin{equation} Y=\min\{X, C\}, \delta=I(X\leq C), \end{equation}$

(1.1)

where $I(A)$ is the indicator function of event $A$. Supposed that, $C$ is independent of $X$. The observations are $\{Y_i, \delta_i\}_{i=1}^n$, which is a random sample from the population $(Y, \delta)$.

Empirical likelihood methods were first used by Thomas and Grunkemeier [1] and popularized by Owen [2-3]. It is well-known that Owen's empirical likelihood is based on linear constraints and hence has very general applicability such as in smooth functions of means (see DiCiccio et al. [4]), quantile estimation (see Chen [5]), estimating equation (see Qin and Lawless [6]), empirical likelihood confidence interval (see [7-16]) and so on. For more details, we refer to Owen [17]. However, most of the references on empirical likelihood are concerned with complete data set. In practice, censoring data occurs in opinion polls, market research surveys, mail enquires, social-economic investigations, medical studies and other scientific experiments. Once the censoring values are imputed, the data set can then be analyzed using standard techniques for complete data.

The rest of this paper is arranged as follows. In Section 2, we propose a empirical likelihood method to quantiles. We obtain the empirical log-likelihood ratio statistics of the quantiles and show that it is asymptotically chi-square. The proof is arranged to Section 3.

If $\theta_q$ is $q$-quantile for $F(x)$, we know that $\theta_q$ coincides with the $M$-estimates defined by the equation

$\begin{equation} E[\phi(X-\theta_q)]=\int_{-\infty}^{\infty}\phi(x-\theta_q)dF(x)=0 \end{equation}$

(2.1)

with

$\begin{equation*} \phi(z)=\left\{ \begin{aligned} -1, z\leq 0, \\ q/(1-q), z>0. \end{aligned} \right. \end{equation*}$

Let $U_i(\theta_q)=\frac{\phi(Y_i-\theta_q)\delta_i}{1-G(Y_i)}$, where $G(\cdot)$ is the cumulative distribution function of the censoring variable $C$. Obviously $\{U_i(\theta_q)\}_{i=1}^n$ are independent and identically distributed random variables. Furthermore,

$\begin{equation*} E(U_i(\theta_q))=E\left[\frac{\phi(Y_i-\theta_q)\delta_i}{1-G(Y_i)}\right]=E[\phi(X-\theta_q)]=0. \end{equation*}$

Thus, following the idea of Owen [2] an empirical likelihood-ratio function can similarly be defined for $\theta_q$ as

$R({{\theta }_{q}})=\underset{{{p}_{1}},\cdots ,{{p}_{n}}}{\mathop{\sup }}\,\left\{ \prod\limits_{i=1}^{n}{(n{{p}_{i}})}:\sum\limits_{i=1}^{n}{{{p}_{i}}}=1,\sum\limits_{i=1}^{n}{{{p}_{i}}}{{U}_{i}}({{\theta }_{q}})=0 \right\}.$

(2.2)

However, we cannot use $R(\theta_q)$ directly to make inference on $\theta_q$ since the distribution function $G(\cdot)$ of $\{Z_i\}_{i=1}^n$ is unknown. To do this, it is natural to replace $G(\cdot)$ by the Kaplan-Meier estimator

${{\hat{G}}_{n}}(y)=1-\prod\limits_{i=1}^{n}{{{\left[ \frac{n-i}{n-i+1} \right]}^{I\{{{Y}_{(i)}}\le y,{{\delta }_{i}}=0\}}}},$

(2.3)

where $Y_{(1)}\leq Y_{(2)}\leq\cdots\leq Y_{(n)}$ is the order statistics of $Y_i$'s. Let $\hat{U}_i(\theta_q)=\frac{\phi(Y_i-\theta_q)\delta_i}{1-\hat{G}_n(Y_i)}$, then an estimated empirical log-likelihood ratio function can be defined as

$\begin{equation} \log\hat{R}(\theta_q)=\sup\limits_{p_1, \cdots, p_n}\left\{\mathop \sum \limits_{i = 1}^n \log(np_i): \mathop \sum \limits_{i = 1}^n p_i=1, \mathop \sum \limits_{i = 1}^n p_i\hat{U}_i(\theta_q)=0\right\}. \end{equation}$

(2.4)

By the method of Lagrange multiplier for (2.4), we may prove that the maximization point occurs with

$\begin{equation} p_i=\frac{1}{n}\ \{1+\lambda(\theta_q)\hat{U}_i(\theta_q)\}^{-1}, i=1, \cdots, n, \end{equation}$

(2.5)

where $\lambda(\theta_q)$ is the solution to

$\begin{equation} \frac{1}{n}\mathop \sum \limits_{i = 1}^n \frac{\hat{U}_i(\theta_q)}{1+\lambda(\theta_q)\hat{U}_i(\theta_q)}=0. \end{equation}$

(2.6)

By (2.4) and (2.5), we can obtain

$ \begin{equation} \log\hat{R}(\theta_q)=-\mathop \sum \limits_{i = 1}^n \log\{1+\lambda(\theta_q)\hat{U}_i(\theta_q)\}. \end{equation}$

(2.7)

Theorem 2.1   If $E(U_i^2(\theta_q))<\infty$ and $\theta_q$ is the true $q$-quantile of $F(\cdot)$, we have

$\begin{equation} -2\log \hat{R}(\theta_q)\longrightarrow^L \chi^2_1, \end{equation}$

(2.8)

where $\rightarrow^L$ represents the convergence in distribution, $\chi^2_1$ is standard Chi-square random variable with $1$ degree of freedom.

Remark   On the basis of Theorem 2.1, $-2\log \hat{R}(\theta_q)$ can be used to construct a confidence region for $\theta_q$,

$\begin{equation*} \hat{I}_{\alpha}(\theta_q)=\{\theta_q: -2\log \hat{R}(\theta_q)\leq c_{\alpha}\}, \end{equation*}$

with $P(\chi^2_1\leq c_\alpha)=1-\alpha$. Then by Theorem 2.1, $\hat{I}_{\alpha}(\theta_q)$ gives a confidence interval for $\theta_q$ with asymptotically correct coverage probability $1-\alpha$.

Throughout this section, we use $c>0$ to represent any constant which may take different values for each appearance.

Lemma 3.1   Under the assumptions of Theorem 2.1, if $\theta_q$ is the true $q$-quantile of $F(\cdot)$, we have

$\begin{equation} \frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n \hat{U}_i(\theta_q)\longrightarrow^L N(0, v^2_q), \end{equation}$

(3.1)

where $v^2_q=E(U_i^2(\theta_q))$.

Proof   By the definition of $\hat{U}_i(\theta_q)$, it is easy to show that

$ \frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n \hat{U}_i(\theta_q)= \frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n U_i(\theta_q)+\frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n (\hat{U}_i(\theta_q)-U_i(\theta_q))\\ =\frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n U_i(\theta_q)+\frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n U_i(\theta_q)\frac{\hat{G}_n(Y_i)-G_n(Y_i)}{1-\hat{G}_n(Y_i)}. $

Since $\{U_i(\theta_q)\}_{i=1}^n$ are independent and identically distributed random variables and $E(U_i^2(\theta_q))<\infty$. They imply $\frac{1}{\sqrt{n}}\sum\limits_{i=1}^n|U_i(\theta_q)|=O(1)$. Zhou [18] proved

$\begin{equation} \sup\limits_{Y_i\leq Y_{(n)}}\left|\frac{\hat{G}_n(Y_i)-G_n(Y_i)}{1-\hat{G}_n(Y_i)}\right|=O_p(n^{-1/2}), \end{equation}$

(3.2)

where $Y_{(n)}=\max\limits_{1\leq i\leq n}Y_i$. So we have

$\begin{equation*} \frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n U_i(\theta_q)\frac{\hat{G}_n(Y_i)-G_n(Y_i)}{1-\hat{G}_n(Y_i)}=o_p(1). \end{equation*}$

By the central limit theory of independent and identically distributed random variables,

$\frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n U_i(\theta_q)\longrightarrow^LN(0, v^2_q).$

This completes the proof.

Lemma 3.2   Under the assumptions of Theorem 2.1, if $\theta_q$ is the true $q$-quantile of $F(\cdot)$, we have

$\begin{equation} \frac{1}{n}\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q)\longrightarrow^P v^2_q. \end{equation}$

(3.3)

Proof   By the definition of $\hat{U}_i(\theta_q)$, it is easy to show that

$ \frac{1}{n}\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q)= \frac{1}{n}\mathop \sum \limits_{i = 1}^n U_i^2(\theta_q)+\frac{1}{n}\mathop \sum \limits_{i = 1}^n (\hat{U}_i(\theta_q)-U_i(\theta_q))^2\\ +\frac{2}{n}\mathop \sum \limits_{i = 1}^n U_i(\theta_q)(\hat{U}_i(\theta_q)-U_i(\theta_q))=: I_1+I_2+I_3. $

Next, from the law of large numbers and (3.2), we can get $n^{-1}\sum\limits_{i=1}^nU_i^2(\theta_q)\rightarrow E(U_i^2(\theta_q)), $ a.s., then we have

$ I_2=\frac{1}{n}\mathop \sum \limits_{i = 1}^n (\hat{U}_i(\theta_q)-U_i(\theta_q))^2=\frac{1}{n}\mathop \sum \limits_{i = 1}^n U_i^2(\theta_q)\left|\frac{\hat{G}_n(Y_i)-G_n(Y_i)}{1-\hat{G}_n(Y_i)}\right|^2=o_p(1). $

By using the similar arguments as $I_2=o_p(1)$, we can also obtain $I_3=o_p(1)$. So, the law of large numbers implies that

$\begin{equation*} \frac{1}{n}\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q) =\frac{1}{n}\mathop \sum \limits_{i = 1}^n U_i^2(\theta_q)+o_p(1)\longrightarrow^p v^2_q. \end{equation*}$

This completes the proof.

Lemma 3.3   Under the assumptions of Theorem 2.1, if $\theta_q$ is the true $q$-quantile of $F(\cdot)$, we have

$\begin{equation} \mathop {\max }\limits_{1 \le i \le n}|\hat{U}_i(\theta_q)|=o_p(n^{1/2}), \lambda(\theta_q)=O_p(n^{-1/2}). \end{equation}$

(3.4)

Proof   It is well known that for any sequence of independent and identically distributed random variables $\{\xi\}_{i=1}^n$ with $E(\xi_i^2)<\infty$, we have

$\begin{equation*} \mathop {\max }\limits_{1 \le i \le n}\frac{|\xi_i|}{\sqrt{n}}\longrightarrow 0. \end{equation*}$

This implies that $\max\limits_{1\leq i\leq n}|U_i(\theta_q)|=o_p(n^{1/2})$. From (3.2), we have

$ \mathop {\max }\limits_{1 \le i \le n}|\hat{U}_i(\theta_q)| =\mathop {\max }\limits_{1 \le i \le n}|U_i(\theta_q)|\cdot \left|\frac{\hat{G}_n(Y_i)-G_n(Y_i)}{1-\hat{G}_n(Y_i)}\right|+\mathop {\max }\limits_{1 \le i \le n}|U_i(\theta_q)| =o_p(n^{1/2}). $

Next, we prove $\lambda(\theta_q)=O_p(n^{-1/2})$. Let $\lambda(\theta_q)=\alpha|\lambda(\theta_q)|$, where $\alpha=1$ or $-1$. Note $\bar{\Lambda}=\frac{1}{n}\sum\limits_{i=1}^n\hat{U}_i(\theta_q)$, $\Lambda^{*}=\max\limits_{1\leq i\leq n}|\hat{U}_i(\theta_q)|$, $S=\frac{1}{n}\sum\limits_{i=1}^n\hat{U}_i^2(\theta_q)$. From (2.6), we have

$ 0=\frac{1}{n}\mathop \sum \limits_{i = 1}^n \frac{\hat{U}_i(\theta_q)}{1+\lambda(\theta_q)\hat{U}_i(\theta_q)} =\frac{1}{n}\mathop \sum \limits_{i = 1}^n \frac{\alpha\hat{U}_i(\theta_q)}{1+|\lambda(\theta_q)|\alpha\hat{U}_i(\theta_q)}\\ =\frac{1}{n}\mathop \sum \limits_{i = 1}^n \alpha\hat{U}_i(\theta_q) -|\lambda(\theta_q)|\cdot\frac{1}{n}\mathop \sum \limits_{i = 1}^n \frac{\hat{U}_i^2(\theta_q)} {1+|\lambda(\theta_q)|\alpha\hat{U}_i(\theta_q)}\\ \leq\alpha\bar{\Lambda}-\frac{|\lambda(\theta_q)|}{1+|\lambda(\theta_q)|\Lambda^{*}} \cdot\frac{1}{n}\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q) =\alpha\bar{\Lambda}-\frac{|\lambda(\theta_q)|}{1+|\lambda(\theta_q)|\Lambda^{*}} \cdot S, $

where we have used $0<1+\lambda(\theta_q)\hat{U}_i(\theta_q)\leq 1+|\lambda(\theta_q)|\Lambda^{*}$, which yields from

$p_i=\frac{1}{n}(1+\lambda(\theta_q)\hat{U}_i(\theta_q))^{-1}\geq 0.$

Therefore, $|\lambda(\theta_q)|(S-\alpha\bar{\Lambda}\Lambda^{*})\leq|\alpha\bar{\Lambda}|$. By Lemma 3.1 and Lemma 3.2, we know $\Lambda^{*}=o_p(n^{1/2})$ and $\Lambda=O_p(n^{-1/2})$, we have $ |\lambda(\theta_q)|(S+o_p(1))\leq|\alpha\bar{\Lambda}|. $ Lemma 3.2 implies that $S=\Sigma+o_p(1)$, hence $|\lambda(\theta_q)|=O_p(n^{-1/2})$. This completes the proof.

Proof of Theorem 2.1   Applying a Taylor expansion to equation (2.6) and (2.7), we can obtain

$\begin{eqnarray} -2\log\hat{R}(\theta_q)=2\mathop \sum \limits_{i = 1}^n \log\{1+\lambda(\theta_q)\hat{U}_i(\theta_q)\}\nonumber\\ =2\lambda(\theta_q)\mathop \sum \limits_{i = 1}^n \hat{U}_i(\theta_q)-\lambda^2(\theta_q)\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q)+r_n, \end{eqnarray}$

(3.5)

where

$ |r_n|=C\mathop \sum \limits_{i = 1}^n |\lambda(\theta_q)\hat{U}_i(\theta_q)|^3\leq|\lambda(\theta_q)|^3\mathop {\max }\limits_{1 \le i \le n}|\hat{U}_i(\theta_q)|\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q)\\ =O_p(n^{-3/2})\cdot o_p(n^{1/2})\cdot O_p(n^{-1})=o_p(1). $

From (2.6), we can get

$\begin{eqnarray} 0=\mathop \sum \limits_{i = 1}^n \frac{\hat{U}_i\theta_q)}{1+\lambda(\theta_q)\hat{U}_i(\theta_q)} =\mathop \sum \limits_{i = 1}^n \left\{1-\lambda(\theta_q)\hat{U}_i(\theta_q) +\frac{[\lambda(\theta_q)\hat{U}_i(\theta_q)]^2}{1+\lambda(\theta_q)\hat{U}_i(\theta_q)}\right\} \hat{U}_i(\theta_q)\nonumber\\ =\mathop \sum \limits_{i = 1}^n \hat{U}_i(\theta_q)-\lambda(\theta_q)\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q) +\mathop \sum \limits_{i = 1}^n \frac{\lambda^2(\theta_q)\hat{U}_i^3(\theta_q)}{1+\lambda(\theta_q)\hat{U}_i(\theta_q)}. \end{eqnarray}$

(3.6)

From Lemma 3.2 and Lemma 3.3, by simple calculation, we have

$\mathop \sum \limits_{i = 1}^n \frac{\lambda^2(\theta_q)\hat{U}_i^3(\theta_q)}{1+\lambda(\theta_q)\hat{U}_i(\theta_q)}=o_p(n^{1/2}).$

It follows that

$\lambda(\theta_q)=\frac{\sum\limits_{i=1}^n\hat{U}_i(\theta_q)}{\sum\limits_{i=1}^n\hat{U}_i^2(\theta_q)}+o_p(n^{-1/2}).$

Furthermore, from (3.6), we can get that

$\begin{equation} \lambda(\theta_q)\mathop \sum \limits_{i = 1}^n \hat{U}_i(\theta_q)-\lambda^2(\theta_q)\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q)=o_p(1). \end{equation}$

(3.7)

By (3.5), (3.6) and (3.7), we have

$ -2\log\hat{R}(\theta_q)=\lambda(\theta_q)\mathop \sum \limits_{i = 1}^n \hat{U}_i(\theta_q)+o_p(1)\\ =\left(\frac{1}{\sqrt{n}}\mathop \sum \limits_{i = 1}^n \hat{U}_i(\theta_q)\right)^2\times\left(\frac{1}{n}\mathop \sum \limits_{i = 1}^n \hat{U}_i^2(\theta_q)\right)^{-1}+o_p(1) \rightarrow^L\chi^2_1. $

This completes the proof.