\name{el.cen.EM2}
\alias{el.cen.EM2}
\title{Empirical likelihood ratio test for several means 
with right, left or doubly censored data, by EM algorithm}
\usage{
el.cen.EM2(x,d,xc=rep(1,length(x)),fun,mu,maxit=25,error=1e-9,...)
}
\description{
This function is similar to \code{el.cen.EM()}, but for multiple constraints.
This program uses EM algorithm to maximize 
(wrt \eqn{p_i}) empirical
log likelihood function for right, left or doubly censored data with 
the MEAN constraint:
\deqn{ \sum_{d_i=1} p_i f_j(x_i) = \int f_j(t) dF(t) = \mu_j ~, }
for \eqn{j = 1,2,...,k}.
Where \eqn{p_i = \Delta F(x_i)} is a probability,
\eqn{d_i} is the censoring indicator, 1(uncensored), 0(right censored),
2(left censored). 
It also returns those \eqn{p_i}. 
The log likelihood function is defined as
\deqn{ \sum_{d_i=1} \log p_i + \sum_{d_i=2} \log F(x_i) 
     + \sum_{d_i=0} \log [ 1-F(x_i)] ~.}
}
\arguments{
    \item{x}{a vector containing the observed survival times.}
    \item{d}{a vector containing the censoring indicators, 
           1-uncensored; 0-right censored; 2-left censored.}
    \item{xc}{an optional vector of collapsing control values. 
              If xc has different values then two (x,d) will not 
              merge into one obs. even if they are identical.}
    \item{fun}{a continuous (weight) function that returns a matrix. 
         The columns (=k) of the matrix is used to calculate
         the means as in \eqn{H_0}.
         \code{fun(t)} must be able to take a vector input \code{t}.} 
    \item{mu}{a vector of length k. Used in the constraint, 
                    mean value of \eqn{f(X)}.}
    \item{error}{an optional positive real number specifying the tolerance of
       iteration error. This is the bound of the
       \eqn{L_1} norm of the differnence of two successive weights.}
    \item{maxit}{an optional integer, used to control maximum number of
             iterations. }
    \item{...}{additional inputs to pass to \code{fun()}.}
}
\value{
    A list with the following components:
    \item{loglik}{the maximized empirical log likelihood under the constraint.}
    \item{times}{locations of CDF that have positive mass.}
    \item{prob}{the jump size of CDF at those locations.}
    \item{"-2LLR"}{If available, it is Minus two times the 
                   Empirical Log Likelihood Ratio.
                   Should be approx. chi-square distributed under Ho.}
    \item{Pval}{The P-value of the test, using chi-square approximation.}
}
\details{

This implementation is all in R and have several for-loops in it. 
A faster version would use C to do the for-loop part.
(but this version is easier to port to Splus, and seems faster enough). 

We return the log likelihood all the time. Sometimes, (for right censored
and no censor case) we also return the -2 log likelihood ratio.
In other cases, you have to plot a curve with many values of the 
parameter, mu, to
find out where the log likelihood becomes maximum.
And from there you can get -2 log likelihood ratio between
the maximum location and your current parameter in Ho.

In order to get a proper distribution as NPMLE, we automatically
change the \eqn{d} for the largest observation to 1
(even if it is right censored), similar for the left censored, 
smallest observation.
\eqn{\mu} is a given constant vector. 
When the given constants \eqn{\mu} is too far
away from the NPMLE, there will be no distribution
satisfy the constraint.
In this case the computation will stop.
The -2 Log empirical likelihood ratio
should be infinite. 

The constant vector \code{mu} must be inside 
\eqn{( \min f(x_i) , \max f(x_i) ) }
for the computation to continue. 
It is always true that the NPMLE values are feasible. So when the
computation stops, try move the \code{mu} closer
to the NPMLE --- 
\deqn{ \sum_{d_i=1} p_i^0 f(x_i) } 
\eqn{p_i^0} taken to be the jumps of the NPMLE of CDF. 
Or use a different \code{fun}. 

}
\author{ Mai Zhou }
\references{
    Zhou, M. (2002). 
        Computing censored empirical likelihood ratio 
        by EM algorithm. 
    \emph{Tech Report, Univ. of Kentucky, Dept of Statistics}

}
\examples{
## example with tied observations
x <- c(1, 1.5, 2, 3, 4, 5, 6, 5, 4, 1, 2, 4.5)
d <- c(1,   1, 0, 1, 0, 1, 1, 1, 1, 0, 0,   1)
el.cen.EM2(x,d, fun=function(x){x}, mu=3.5)
## we should get, same as in el.cen.EM;  "-2LLR" = 1.2466....
}
\keyword{nonparametric}
\keyword{survival}
\keyword{htest}