robcbi (version 1.1-4)

robcbi-package: Robust Fit for Discrete Generalized Linear Model

Description

Conditionally unbiased bounded influence estimates as described in Kuensch et al. (1989) in three special cases of the Generalized Linear Model: Bernoulli, Binomial, and Poisson distributed responses.

Author

A. Marazzi <[e-mail]>

Maintainer: A. Randriamiharisoa <[e-mail]>

Details

Package:	cubinf
#Version:	1.0
#Date:	2013-07-04
License:	GPL (>= 2)

References

Kuensch, H.R., Stefanski L.A., Carroll R.J. (1989). Conditionally unbiased bounded-influence estimation in general regression models, with application to generalized linear models. Journal of the American Statistical Association, 84, 460-466.

Marazzi, A. (1993). Algorithms, Routines, and S-functions for robust Statistics. Chapman and Hall, New York.

Examples

 library(robcbi)
# First example
 data(Finney)
 Vol <- Finney$Vol; Rate <- Finney$Rate; Resp <- Finney$Resp
if (FALSE) {
 plot(Vol,Rate,type="n")
 points(Vol[Resp==0],Rate[Resp==0],pch=5, cex=1.2)
 points(Vol[Resp==1],Rate[Resp==1],pch=16,cex=1.2)
}
 lVol <-log(Vol); lRate <- log(Rate)
 z.glm <- glm(Resp~lVol+lRate,family=binomial)
 summary(z.glm)
 z.cub <- glm(Resp~lVol+lRate,family=binomial,method="cubinf", ufact=3.2)
 summary(z.cub)
 weights(z.cub)
if (FALSE) {
 plot(z.cub, smooth=TRUE, ask=TRUE)
}
 comp <- fits.compare(z.glm,z.cub)
 comp
if (FALSE) {
 plot(comp)
}
# Second example
 data(Breslow)
if (FALSE) { 
 help(Breslow)
}
 y  = Breslow$sumY
 x1 = Breslow$Age10
 x2 = Breslow$Base4
 x3 = rep(0,length(y))
 x3[Breslow$Trt=="progabide"] = 1
 x4 = x2*x3
 CBA = glm(y~x1+x2+x3+x4,family=poisson,method=cubinf,ufact=3.2)
if (FALSE) {
 plot(CBA,num=5)
}
 weights(CBA)
#
# compute the $R_n^2$ statistic (Section 2.5) to compare CBA 
# with a reduced model with three variables:
#
 CBA.red = update(CBA, .~.-x3-x4)
 np = 5       # number of parameters of the full model
 nq = 3       # number of parameters of the reduced model
 CVR = covar(CBA)
 CFF = coef(CBA)
 K22 = CVR[(nq+1):np,(nq+1):np]
 cff = as.matrix(CFF[(nq+1):np])
 Rn2 = t(cff)%*%solve(K22)%*%cff
 Rn2