el2.cen.EMs {emplik2} | R Documentation |
Computes p-value for a single mean-type hypothesis, based on two independent samples that may contain censored data.
Description
This function uses the EM algorithm to calculate a maximized empirical likelihood ratio for the hypothesis
H_o: E(g(x,y)-mean)=0
where E
indicates expected value; g(x,y)
is a user-defined function of x
and y
; and
mean
is the hypothesized value of E(g(x,y))
. The samples x
and y
are assumed independent.
They may be uncensored, right-censored, left-censored, or left-and-right (“doubly”) censored. A p-value for
H_o
is also calculated, based on the assumption that -2*log(empirical likelihood ratio) is approximately
distributed as chisq(1).
Usage
el2.cen.EMs(x,dx,y,dy,fun=function(x,y){x>=y}, mean=0.5, maxit=25)
Arguments
x |
a vector of the data for the first sample |
dx |
a vector of the censoring indicators for |
y |
a vector of the data for the second sample |
dy |
a vector of the censoring indicators for |
fun |
a user-defined, continuous-weight-function |
mean |
the hypothesized value of |
maxit |
a positive integer used to set the number of iterations of the EM algorithm; default is 25 |
Details
The value of mean
should be chosen between the maximum and minimum values of g(x_i,y_j)
; otherwise
there may be no distributions for x
and y
that will satisfy H_o
. If mean
is inside
this interval, but the convergence is still not satisfactory, then the value of mean
should be moved
closer to the NPMLE for E(g(x,y))
. (The NPMLE itself should always be a feasible value for mean
.)
Value
el2.cen.EMs
returns a list of values as follows:
xd1 |
a vector of the unique, uncensored |
yd1 |
a vector of the unique, uncensored |
temp3 |
a list of values returned by the |
mean |
the hypothesized value of |
funNPMLE |
the non-parametric-maximum-likelihood-estimator of |
logel00 |
the log of the unconstrained empirical likelihood |
logel |
the log of the constrained empirical likelihood |
"-2LLR" |
|
Pval |
the estimated p-value for |
logvec |
the vector of successive values of |
sum_muvec |
sum of the probability jumps for the uncensored |
sum_nuvec |
sum of the probability jumps for the uncensored |
constraint |
the realized value of |
Author(s)
William H. Barton <bbarton@lexmark.com>
References
Barton, W. (2010). Comparison of two samples by a nonparametric likelihood-ratio test. PhD dissertation at University of Kentucky.
Chang, M. and Yang, G. (1987). “Strong Consistency of a Nonparametric Estimator of the Survival Function
with Doubly Censored Data.” Ann. Stat.
,15, pp. 1536-1547.
Dempster, A., Laird, N., and Rubin, D. (1977). “Maximum Likelihood from Incomplete Data via the EM Algorithm.” J. Roy. Statist. Soc.
, Series B, 39, pp.1-38.
Gomez, G., Julia, O., and Utzet, F. (1992). “Survival Analysis for Left-Censored Data.” In Klein, J. and Goel, P. (ed.),
Survival Analysis: State of the Art.
Kluwer Academic Publishers, Boston, pp. 269-288.
Li, G. (1995). “Nonparametric Likelihood Ratio Estimation of Probabilities for Truncated Data.”
J. Amer. Statist. Assoc.
, 90, pp. 997-1003.
Owen, A.B. (2001). Empirical Likelihood
. Chapman and Hall/CRC, Boca Raton, pp.223-227.
Turnbull, B. (1976). “The Empirical Distribution Function with Arbitrarily Grouped, Censored and Truncated Data.”
J. Roy. Statist. Soc.
, Series B, 38, pp. 290-295.
Zhou, M. (2005). “Empirical likelihood ratio with arbitrarily censored/truncated data by EM algorithm.”
J. Comput. Graph. Stat.
, 14, pp. 643-656.
Zhou, M. (2009) emplik
package on CRAN website.
The el2.cen.EMs
function extends el.cen.EM
function from one-sample to two-samples.
Examples
x<-c(10,80,209,273,279,324,391,415,566,785,852,881,895,954,1101,
1133,1337,1393,1408,1444,1513,1585,1669,1823,1941)
dx<-c(1,2,1,1,1,1,1,2,1,1,1,1,1,1,1,0,0,1,0,0,0,0,1,1,0)
y<-c(21,38,39,51,77,185,240,289,524,610,612,677,798,881,899,946,
1010,1074,1147,1154,1199,1269,1329,1484,1493,1559,1602,1684,1900,1952)
dy<-c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,0,0,0,0,0,1,0,0,0)
# Ho1: X is stochastically equal to Y
el2.cen.EMs(x, dx, y, dy, fun=function(x,y){x>=y}, mean=0.5, maxit=25)
# Result: Pval = 0.7090658, so we cannot with 95 percent confidence reject Ho1
# Ho2: mean of X equals mean of Y
el2.cen.EMs(x, dx, y, dy, fun=function(x,y){x-y}, mean=0.5, maxit=25)
# Result: Pval = 0.9695593, so we cannot with 95 percent confidence reject Ho2