# ROCnp2: Test the ROC curve by Empirical Likelihood In emplik: Empirical Likelihood Ratio for Censored/Truncated Data

## 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

∑_{d1=1} \log Δ F_1(t1_i) + ∑_{d1=0} \log [1-F_1(t1_i)] + ∑_{d2=1} \log Δ F_2(t2_j) + ∑_{d2=0} \log [1-F_2(t2_j)] .

This empirical likelihood ratio has a chi square limit under Ho.

## Usage

 `1` ```ROCnp2(t1, d1, t2, d2, b0, t0) ```

## Arguments

 `t1` a vector of length n. Observed times, sample 1. 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, sample 2. 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

First, we 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 ratios over c.

This version use an exhaust search for the minimum (over c). Since the objective (log lik) are piecewise constants, the optimum( ) function in R do not work well. 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

Su, Haiyan; Zhou, Mai and Liang, Hua (2011). Semi-parametric Hybrid Empirical Likelihood Inference for Two Sample Comparison with Censored Data. Lifetime Data Analysis, 17, 533-551.

## Examples

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37``` ```#### An example of testing the equality of two medians. #### No censoring. # ROCnp2(t1=rexp(100), d1=rep(1,100), t2=rexp(120), # d2=rep(1,120), b0=0.5, t0=0.5) ############################################################### #### This example do not work on the Solaris Sparc machine. #### But works fine on other platforms. ########### #### 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] <- ROCnp2(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 [0.4457862, 0.5907723]. ############################################### ```

emplik documentation built on Aug. 18, 2018, 1:05 a.m.