Please help me to find the reason when applied this program gave me the same
result for
b1,b2,b3 for all (BS,BIAS, RMSE , and SE)
I want to compare between these beta's
tq.
################## Ridge MM ##################
rm(list=ls())
library(MASS)
library(mvoutlier)
library(robustbase)
library(car)
library(quantreg)
n<-100
r<-0.95
R<-10
p<-3
B<-matrix(c(1,1,1,1),p+1,1)
n.out<-round(0.10*n)
SUM.MMR.BS<-c(0,0,0,0) # MM ridge
SUM.MMR.MSE<-c(0,0,0,0)
set.seed(n)
b0=matrix(nrow=10,ncol=1)
b1=matrix(nrow=10,ncol=1)
b2=matrix(nrow=10,ncol=1)
b3=matrix(nrow=10,ncol=1)
for (i in 1:R){
z11<-rnorm(n,0,1)
z22<-rnorm(n,0,1)
z33<-rnorm(n,0,1)
z44<-rnorm(n,0,1)
x1<-(1-r)*z11+sqrt(r)*z44
#x1[1:n.out]<-100
x2<-(1-r)*z22+sqrt(r)*z44
#x2[1:n.out]<-100
x3<-(1-r)*z33+sqrt(r)*z44
#x3[1:n.out]<-100
x<-as.matrix(data.frame(x1,x2,x3))
e<-rnorm(n,0,1)
#e<-rcauchy(n,0,1)
#e<-rt(n,3)
e[1:n.out]<-100
y<-1+x1+x2+x3+e
data<-data.frame(y,x1,x2,x3)
MM<- rlm(y~.,data,psi=psi.bisquare,method="MM",k2=4.685)
beta<-as.matrix(coef(MM))
stand<-function(x){(x-mean(x))/sd(x)*(n-1)^-0.5}
x11<-stand(x1)
x22<-stand(x2)
x33<-stand(x3)
X<-as.matrix(data.frame(1,x11,x22,x33))
y1<-stand(y)
data1<-data.frame(y1,x11,x22,x33)
cell<-data1
y<-cell[,1]
x1<-cell[,2]
x2<-cell[,3]
x3<-cell[,4]
data2<-data.frame(y,x1,x2,x3)}
###################### MM ridge ####################
b.MM<- rlm(y~.,data,psi=psi.bisquare,method="MM",k2=4.685)$coef
eMM<-rlm(y~.,data,psi=psi.bisquare,method="MM",k2=4.685)$residuals
sMM<-sum(eMM^2)/(n-p-1)
B.MM<-as.matrix(b.MM)
kMM<-4*sMM^2/t(b.MM)%*%b.MM
kMM<-c(kMM)
II<-diag(x = 1, nrow=p+1, ncol=p+1)
KMM<-II*kMM
MMR.BS<-solve(t(X)%*%X+KMM)%*%t(X)%*%X%*%b.MM
SUM.MMR.BS<-SUM.MMR.BS+MMR.BS
SUM.MMR.MSE<-SUM.MMR.MSE+(MMR.BS-1)^2
##############################################
#Bias,RMSE and SE for MMR
MMR.AVE<-SUM.MMR.BS/R
MMR.MSE<-SUM.MMR.MSE/R
MMR.BIAS<-MMR.AVE-1
MMR.RMSE<-sqrt(MMR.MSE)
MMR.SE<-abs(MMR.MSE-MMR.BIAS^2)^0.5
MMR.output<-cbind(MMR.BS,MMR.BIAS,MMR.RMSE,MMR.SE)
MMR.output
BS BIAS RMSE SE
x11 9.359232e-06 -0.9999991 0.3162248 0.9486833
x22 9.324826e-06 -0.9999991 0.3162248 0.9486833
x33 9.311783e-06 -0.9999991 0.3162248 0.9486833
Kafi Dano Pati
Ph.D candidate ( mathematics/statistics)
Department of mathematical Science/ faculty of Science
University Technology Malaysia
81310 UTM, Johor Bahru, Johor, Malaysia
IC. NO. 201202F10234
Matric No. PS113113
HP. No. 00601117517559
E-mail: kafi_d...@yahoo.com
supervisor- Assoc. Prof. Robiah Binti Adnan
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