Empirical likelihood ratio for mean with right, left or doubly censored data, by EM algorithm
Description
This program uses EM algorithm to compute the maximized (wrt p_i) empirical log likelihood function for right, left or doubly censored data with the MEAN constraint:
∑_{d_i=1} p_i f(x_i) = \int f(t) dF(t) = μ .
Where p_i = Δ F(x_i) is a probability, d_i is the censoring indicator, 1(uncensored), 0(right censored), 2(left censored). It also returns those p_i.
The empirical log likelihood been maximized is
∑_{d_i=1} \log Δ F(x_i) + ∑_{d_i=0} \log [1F(x_i)] + ∑_{d_i=2} \log F(x_i) .
Usage
1 
Arguments
x 
a vector containing the observed survival times. 
d 
a vector containing the censoring indicators, 1uncensored; 0right censored; 2left censored. 
wt 
a weight vector (case weight). positive. same length as d 
fun 
a left continuous (weight) function used to calculate
the mean as in H_0.

mu 
a real number used in the constraint, the mean value of f(X). 
maxit 
an optional integer, used to control maximum number of iterations. 
error 
an optional positive real number specifying the tolerance of iteration error. This is the bound of the L_1 norm of the difference of two successive weights. 
... 
additional arguments, if any, to pass to fun. 
Details
This implementation is all in R and have several forloops in it. A faster version would use C to do the forloop part. But this version seems faster enough and is easier to port to Splus.
We return the log likelihood all the time. Sometimes, (for right censored and no censor 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.
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), similar for the left censored, smallest observation. μ is a given constant. When the given constants μ is too far away from the NPMLE, there will be no distribution satisfy the constraint. In this case the computation will stop. The 2 Log empirical likelihood ratio should be infinite.
The constant mu
must be inside
( \min f(x_i) , \max f(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 —
∑_{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
.
Difference to the function el.cen.EM2
: here duplicate (input) observations
are collapsed (with weight 2, 3, ... etc.) but those
will stay separate by default in the el.cen.EM2
. This will lead to
a different loglik
value. But the 2LLR
value should be same
in either version.
Value
A list with the following components:
loglik 
the maximized empirical log likelihood under the constraint. 
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 chisquare distributed under Ho. 
Pval 
The Pvalue of the test, using chisquare approximation. 
lam 
The Lagrange multiplier. Added 5/2007. 
Author(s)
Mai Zhou
References
Zhou, M. (2005). Empirical likelihood ratio with arbitrary censored/truncated data by EM algorithm. Journal of Computational and Graphical Statistics, 643656.
Murphy, S. and van der Vaart (1997) Semiparametric likelihood ratio inference. Ann. Statist. 25, 14711509.
Examples
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22  ## 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 < (xtheta)*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 xvector 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.
