emplikHs.test2: Two sample empirical likelihood ratio test for hazards with right censored, left truncated data. Many constraints.

Description

Use empirical likelihood ratio and Wilks theorem to test the null hypothesis that $$ \int{f_1(t) dH_1(t)} - \int{f_2(t) dH_2(t)} = \theta $$ where $H_*(t)$ is the (unknown) cumulative hazard functions for sample 1 and sample 2; $f_*(t)$ can be any predictable functions of $t$. $\theta$ is a vector of parameters (dim=q). The given value of $\theta$ in these computation are the value to be tested. The data can be right censored and left truncated.

The two samples are assumed independent of each other.

When the given constants $\theta$ is too far away from the NPMLE, there will be no hazard function satisfy this constraint and the -2 Log empirical likelihood ratio will be infinite. In this case the computation will stop.

Usage

emplikHs.test2(x1, d1, y1= -Inf, x2, d2, y2 = -Inf,
          theta, fun1, fun2, maxit=25,tola = 1e-7,itertrace =FALSE)

Value

A list with the following components:

"-2LLR": The -2Log empirical Likelihood ratio.
lambda: the final value of the Lagrange multiplier.
"-2LLR(sample1)": The -2Log empirical likelihood ratio for sample one only. Useful in one sample problems.
"Llog(sample1)": The numerator only of the above "-2LLR(sample1)", without -2.

Arguments

x1: a vector of length n1, the observed survival times, sample 1.
d1: a vector, the censoring indicators, 1-uncensor; 0-censor.
y1: optional vector, the left truncation times.
x2: a vector of length n2, the observed survival times, sample 2.
d2: a vector, the censoring indicators, 1-uncensor; 0-censor.
y2: optional vector, the left truncation times.
fun1: a predictable function used to calculate the weighted discrete hazard to form the null hypothesis $H_0$. fun1(x) must be able to take a vector input (length n1) x, and output a matrix of n1 x q. When q=1, the output can also be a vector.
fun2: Ditto. but for length n2
tola: an optional positive real number, the tolerance of iteration error in solve the non-linear equation needed in constrained maximization.
theta: a given vector of length q. for Ho constraint.
maxit: integer, maximum number of Newton-Raphson type iterations.
itertrace: Logocal, if the results of each iteration needs to be printed.

Author

Mai Zhou

Details

The log likelihood been maximized is the Poisson likelihood: $$ \sum D_{1i} \log w_i - \sum R_{1i} w_i + \sum D_{2j} \log v_j - \sum R_{2j} v_j $$ where $w_i = \Delta H_1(t_i)$ is the jump of the cumulative hazard function at $t_i$ (for first sample), $D_{1i}$ is the number of failures observed at $t_i$, $R_{1i}$ is the number of subjects at risk at time $t_i$. Dido for sample two.

For (proper) discrete distributions, the jump size of the cumulative hazard at the last jump is always 1. So, in the likelihood ratio, it cancels. But the last jump of size 1 still matter when computing the constraint.

The constants theta must be inside the so called feasible region for the computation to continue. This is similar to the requirement that in testing the value of the mean, the value must be inside the convex hull of the observations. It is always true that the NPMLE values are feasible. So when the computation stops, try move the theta closer to the NPMLE, which we print out first thing in this function, even when other later computations do not go. When the computation stops, the -2LLR should have value infinite.

This function uses the llog etc. function, so sometimes it may produce different result from the one sample result. which use the regular log function. The advantage is that we avoid the possible log(0) situation.

You can also use this function for one sample problems. You need to artificially supply data for sample two of minimal size (like size 2q+2), and specify a fun2() that ALWAYS return 0's (zero vector, with length=n2 vector length, or zero matrix, with dim n2 x q as the input). Then, look for -2LLR(sample1) in the output.

References

Zhou and Fang (2001). ``Empirical likelihood ratio for 2 sample problems for censored data''. Tech Report, Univ. of Kentucky, Dept of Statistics

Examples

Run this code

if(require("boot", quietly = TRUE)) {
####library(boot)
data(channing)
ymale <- channing[1:97,2]
dmale <- channing[1:97,5]
xmale <- channing[1:97,3]
yfemale <- channing[98:462,2]
dfemale <- channing[98:462,5]
xfemale <- channing[98:462,3]
fun1 <- function(x) { as.numeric(x <= 960) }
########################################################
fun2 <- function(x){ cbind(as.numeric(x <= 960), as.numeric(x <= 860))}
############ fun2 has matrix output ###############
emplikHs.test2(x1=xfemale, d1=dfemale, y1=yfemale, 
 x2=xmale, d2=dmale, y2=ymale, theta=c(0,0), fun1=fun2, fun2=fun2)
}
#############################################
###################### Second example:
if(require("KMsurv", quietly = TRUE)) {
####library(KMsurv)
data(kidney)
### these functions counts the risk set size, so delta=1 always ###
temp1 <- Wdataclean3(z=kidney$time[kidney[,3]==1], d=rep(1,43) )
temp2 <- DnR(x=temp1$value, d=temp1$dd, w=temp1$weight)
TIME <- temp2$times
RISK <- temp2$n.risk
fR1 <- approxfun(x=TIME, y=RISK, method="constant", yright=0, rule=2, f=1)
temp1 <- Wdataclean3(z=kidney$time[kidney[,3]==2], d=rep(1,76) )
temp2 <- DnR(x=temp1$value, d=temp1$dd, w=temp1$weight)
TIME <- temp2$times
RISK <- temp2$n.risk
fR2 <- approxfun(x=TIME, y=RISK, method="constant", yright=0, rule=2, f=1)

### the weight function for two sample Gehan-Wilcoxon type test ###
fun <- function(t){ fR1(t)*fR2(t)/((76*43)*sqrt(119/(76*43)) )}
### Here comes the test: ###
emplikHs.test2(x1=kidney[kidney[,3]==1,1],d1=kidney[kidney[,3]==1,2],
   x2=kidney[kidney[,3]==2,1],d2=kidney[kidney[,3]==2,2],
   theta=0, fun1= fun, fun2=fun)
### The results should include this ###
#$"-2LLR"
#[1] 0.002473070
#
#$lambda
#[1] -0.1713749
#######################################
######### the weight function for log-rank test #####
funlogrank <- function(t){sqrt(119/(76*43))*fR1(t)*fR2(t)/(fR1(t)+fR2(t))}
##### Now the log-rank test ###
emplikHs.test2(x1=kidney[kidney[,3]==1,1],d1=kidney[kidney[,3]==1,2],
  x2=kidney[kidney[,3]==2,1],d2=kidney[kidney[,3]==2,2],
  theta=0, fun1=funlogrank, fun2=funlogrank)
##### The result of log rank test should include this ###
#
#$"-2LLR"
#[1] 2.655808
#
#$lambda
#[1] 3.568833
#######################################################
###### the weight function for both type test ####
funBOTH <- function(t) {
       cbind(sqrt(119/(76*43))*fR1(t)*fR2(t)/(fR1(t)+fR2(t)), 
                 fR1(t)*fR2(t)/((76*43)*sqrt(119/(76*43)))) }
#### The test that combine both tests ###
emplikHs.test2(x1=kidney[kidney$type==1,1],d1=kidney[kidney$type==1,2],
    x2=kidney[kidney$type==2,1],d2=kidney[kidney$type==2,2],
    theta=c(0,0), fun1=funBOTH, fun2=funBOTH)
#### the result should include this ###
#
#$"-2LLR"
#[1] 13.25476
#
#$lambda
#[1]  14.80228 -21.86733
##########################################
}

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