ROCnp {emplik} R Documentation

## Test the ROC curve by Empirical Likelihood

### 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

 \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.

### Usage

ROCnp(t1, d1, t2, d2, b0, t0)


### Arguments

 t1 a vector of length n. Observed times, may be right censored. d1 a vector of length n, censoring status. d=1 means t is uncensored; d=0 means t is right censored. t2 a vector of length m. Observed times, may be right censored. d2 a vector of length m, censoring status. b0 a scalar between 0 and 1. t0 a scalar, betwenn 0 and 1.

### 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:

 "-2LLR" the -2 loglikelihood ratio; have approximate chisq distribution under H_o. cstar the estimated common quantile.

Mai Zhou.

### References

Zhou, M. and Liang, H (2008). Empirical Likelihood for Hybrid Two Sample Problem with Censored Data. 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. Lifetime Data Analysis, 17, 533-551.

### 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].


[Package emplik version 1.1-1 Index]