\name{ROCnp}
\alias{ROCnp}
\title{Test the ROC curve by Empirical Likelihood}
\usage{
ROCnp(t1, d1, t2, d2, b0, t0)
}
\arguments{
    \item{t1}{a vector of length n. Observed times, may be right censored.}
    \item{d1}{a vector of length n, censoring status.
              d=1 means t is uncensored; d=0 means t is right censored. }
    \item{t2}{a vector of length m. Observed times, may be right censored.}
    \item{d2}{a vector of length m, censoring status.}
    \item{b0}{a scalar between 0 and 1. }
    \item{t0}{a scalar, betwenn 0 and 1. }
}
\description{
    Use empirical likelihood ratio to test the
    hypothesis Ho: (1-b0)th quantile of sample 1 = (1-t0)th quantile 
    of sample 2.
    This is the same as testing Ho: R(t0)= b0, where R(.) is the ROC curve.

The log empirical likelihood been maximized is
\deqn{ \sum_{d1=1} \log \Delta F_1(t1_i) + \sum_{d1=0} \log [1-F_1(t1_i)] 
     + \sum_{d2=1} \log \Delta F_2(t2_j) + \sum_{d2=0} \log [1-F_2(t2_j)] .}
 
This empirical likelihood ratio has a chi square limit under Ho.
}
\details{
Basically, we first test (1-b0)th quantile of sample 1 = c
and also test (1-t0)th quantile of sample 2 = c. 
This way we obtain two log likelihood ratios. 

Then we minimize the sum of the
two log likelihood ratio over c.
 
See the tech report below for details on a similar setting.
}
\value{
    A list with the following components:
    \item{"-2LLR"}{the -2 loglikelihood ratio; have approximate chisq 
                  distribution under \eqn{H_o}.}
    \item{cstar}{the estimated common quantile.}
}
\references{
    Zhou, M. and Liang, H (2008). 
       Empirical Likelihood for Hybrid Two Sample Problem with
       Censored Data. \emph{Univ. Kentucky Tech. Report.}

     Su, H., Zhou, M. and Liang, H. (2011). Semi-parametric
   hybrid empirical likelihood inference for two-sample comparison with censored data. \emph{Lifetime Data Analysis},
\bold{17}, 533-551. 
}
\author{ Mai Zhou. }
\examples{
#### An example of testing the equality of two medians. No censoring.
ROCnp(t1=rexp(100), d1=rep(1,100), t2=rexp(120), d2=rep(1,120), b0=0.5, t0=0.5)
##########################################################################
#### Next, an example of finding 90 percent confidence interval of R(0.5)
####  Note: We are finding confidence interval for R(0.5). So we are testing  
####  R(0.5)= 0.35, 0.36, 0.37, 0.38, etc. try to find values so that 
####  testing R(0.5) = L , U  has p-value of 0.10,  then [L,  U] is the 90 percent
####  confidence interval for R(0.5).
#set.seed(123)
#t1 <- rexp(200)
#t2 <- rexp(200)
#ROCnp( t1=t1, d1=rep(1, 200), t2=t2, d2=rep(1, 200), b0=0.5, t0=0.5)$"-2LLR"
#### since the -2LLR value is less than  2.705543 = qchisq(0.9, df=1),  so the 
#### confidence interval contains 0.5.
#gridpoints <- 35:65/100
#ELvalues <- gridpoints
#for( i in 1:31 ) ELvalues[i] <- ROCnp(t1=t1, d1=rep(1, 200), 
#                 t2=t2, d2=rep(1, 200), b0=gridpoints[i], t0=0.5)$"-2LLR"
#myfun1 <- approxfun(x=gridpoints, y=ELvalues)
#uniroot( f= function(x){myfun1(x)-2.705543}, interval= c(0.35, 0.5) )
#uniroot( f= function(x){myfun1(x)-2.705543}, interval= c(0.5, 0.65) )
#### So, taking the two roots, we see the 90 percent confidence interval for R(0.5) 
#### in this case is [0.4478081,  0.5889425].
}
\keyword{nonparametric}
\keyword{htest}