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emplik (version 1.3-2)

el.test.wt2: Weighted Empirical Likelihood ratio for mean(s), uncensored data

Description

This program is similar to el.test( ) except it takes weights.

The mean constraints are: $$ \sum_{i=1}^n p_i x_i = \mu . $$ Where \(p_i = \Delta F(x_i)\) is a probability. Plus the probability constraint: \( \sum p_i =1\).

The weighted log empirical likelihood been maximized is $$ \sum_{i=1}^n w_i \log p_i. $$

Usage

el.test.wt2(x, wt, mu, maxit = 25, gradtol = 1e-07, Hessian = FALSE, 
    svdtol = 1e-09, itertrace = FALSE)

Value

A list with the following components:

lambda

the Lagrange multiplier. Solution.

wt

the vector of weights.

grad

The gradian at the final solution.

nits

number of iterations performed.

prob

The probabilities that maximized the weighted empirical likelihood under mean constraint.

Arguments

x

a matrix (of size nxp) or vector containing the observations.

wt

a vector of length n, containing the weights. If weights are all 1, this is very simila to el.test. wt have to be positive.

mu

a vector of length p, used in the constraint. weighted mean value of \(f(X)\).

maxit

an integer, the maximum number of iteration.

gradtol

a positive real number, the tolerance for a solution

Hessian

logical. if the Hessian needs to be computed?

svdtol

tolerance in perform SVD of the Hessian matrix.

itertrace

TRUE/FALSE, if the intermediate steps needs to be printed.

Author

Mai Zhou

Details

This function used to be an internal function. It becomes external because others may find it useful.

It is similar to the function el.test( ) with the following differences:

(1) The output lambda in el.test.wts, when divided by n (the sample size or sum of all the weights) should be equal to the output lambda in el.test.

(2) The Newton step of iteration in el.test.wts is different from those in el.test. (even when all the weights are one).

References

Owen, A. (1990). Empirical likelihood ratio confidence regions. Ann. Statist. 18, 90-120.

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

Zhou, M. (2002). Computing censored empirical likelihood ratio by EM algorithm. Tech Report, Univ. of Kentucky, Dept of Statistics

Examples

Run this code
## 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)

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