el.cen.kmc1d: Empirical likelihood ratio for 1 mean constraint with right...

In emplik: Empirical Likelihood Ratio for Censored/Truncated Data

Empirical likelihood ratio for 1 mean constraint with right censored data

Description

This program uses a fast recursive formula to compute the maximized (wrt p_i) empirical log likelihood ratio for right censored data with one MEAN constraint:

\sum_{d_i=1} p_i f(x_i) = \int f(t) dF(t) = \mu .

Where p_i = \Delta F(x_i) is a probability, d_i is the censoring indicator, 1(uncensored), 0(right censored). It also returns those p_i.

The empirical log likelihood been maximized is

\sum_{d_i=1} \log \Delta F(x_i) + \sum_{d_i=0} \log [1-F(x_i)] .

Usage

el.cen.kmc1d(x, d, fun, mu, tol = .Machine$double.eps^0.5, step=0.001, ...)

Arguments

x	a vector containing the observed survival times.
d	a vector containing the censoring indicators, 1-uncensored; 0-right censored.
fun	a left continuous (weight) function used to calculate the mean as in H_0. fun(t) must be able to take a vector input t.
mu	a real number used in the constraint, the mean value of f(X).
tol	a small positive number, for the uniroot error tol.
step	a small positive number, for use in the uniroot function (as interval) to find lambda root. Sometimes uniroot will find the wrong root or no root, resulting a negative "-2LLR" or NA. Change the step to a different value often can fix this (but not always). Another sign of wrong root is that the sum of probabilities not sum to one, or has negative probability values.
...	additional arguments, if any, to pass to fun.

Details

This function is similar to the function in package kmc, but much simpler, i.e. all implemented in R and only for one mean. This implementation have two for-loops in R. A faster version would use C to do the for-loop part. But this version seems fast enough and is easier to port to Splus.

We return the log likelihood all the time. Sometimes, (for right censored 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 is the place 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.

The input step is used in uniroot function to find a root of lambda. Sometimes a step value may lead to no root or result in a wrong root. You may try several values for the step to see. If the probabilities returned do not sum to one, then the lambda root is a wrong root. We want the root closest to zero.

In order to get a proper distribution as NPMLE, we automatically change the d for the largest observation to 1 (even if it is right censored). \mu is a given constant. When the given constants \mu is too far away from the NPMLE, there will be no distribution satisfy the constraint. In this case the computation will stop or return something ridiculas, (as negative -2LLR). The -2 Log empirical likelihood ratio may be +infinite.

The constant mu must be inside ( \min f(x_i) , \max f(x_i) ) (with uncensored 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 mu closer to the NPMLE —

\sum_{d_i=1} p_i^0 f(x_i)

p_i^0 taken to be the jumps of the NPMLE of CDF. Or use a different fun.

Value

A list with the following components:

loglik	the maximized empirical log likelihood under the constraint. Note, here the tied observations are not collapsed into one obs. with weight 2 (as in el.cen.EM), so the value may differ from those that do collapse the tied obs. In any case, the -2LLR should not differ (whether collaps or not).
times	locations of CDF that have positive mass.
prob	the jump size of CDF at those locations.
"-2LLR"	If available, it is minus two times the empirical Log Likelihood Ratio. Should be approximately chi-square distributed under Ho. If you got NA or negative value, then something is wrong, most likely the uniroot has found the wrong root. Suggest: use el.cen.EM2() which uses EM algorithm. It is more stable but slower.
Pval	The P-value of the test, using chi-square approximation.
lam	The Lagrange multiplier.

Author(s)

Mai Zhou

References

Zhou, M. and Yang, Y. (2015). A recursive formula for the Kaplan-Meier estimator with mean constraints and its application to empirical likelihood. Computational Statistics Vol. 30, Issue 4 pp. 1097-1109.

Zhou, M. (2005). Empirical likelihood ratio with arbitrary censored/truncated data by EM algorithm. Journal of Computational and Graphical Statistics, 14(3), 643-656.

Examples

x <- c(1, 1.5, 2, 3, 4.2, 5, 6.1, 5.3, 4.5, 0.9, 2.1, 4.3)
d <- c(1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1)
ff <- function(x) {
    x - 3.7
}
el.cen.kmc1d(x=x, d=d, fun=ff, mu=0)
#######################################
## 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.EM(x,d,mu=3.5)
## we should get "-2LLR" = 1.2466....
myfun5 <- function(x, theta, eps) {
u <- (x-theta)*sqrt(5)/eps 
INDE <- (u < sqrt(5)) & (u > -sqrt(5)) 
u[u >= sqrt(5)] <- 0 
u[u <= -sqrt(5)] <- 1 
y <- 0.5 - (u - (u)^3/15)*3/(4*sqrt(5)) 
u[ INDE ] <- y[ INDE ] 
return(u)
}
el.cen.EM(x, d, fun=myfun5, mu=0.5, theta=3.5, eps=0.1)
## example of using wt in the input. Since the x-vector contain
## two 5 (both d=1), and two 2(both d=0), we can also do
xx <- c(1, 1.5, 2, 3, 4, 5, 6, 4, 1, 4.5)
dd <- c(1,   1, 0, 1, 0, 1, 1, 1, 0,   1)
wt <- c(1,   1, 2, 1, 1, 2, 1, 1, 1,   1)
el.cen.EM(x=xx, d=dd, wt=wt, mu=3.5)
## this should be the same as the first example.

emplik documentation built on Sept. 8, 2023, 5:06 p.m.