My question is as follows
Y~N(mu2,sigma2^2), i.e. Y has cdf F2
X generates from:
X=F2^{-1}(u)+a*|u-tau|^b*I(u>0.75), where u~U(0,1)
Given tau=0.75, I want to find a and b such that
E(X)-E(Y)=2 and Var(X)/Var(Y)=3
I try optim and grid seaching and get different results, any solution?
Thanks,
Kate
##########
#R code:
mu2=0.4
sigma2=4
tau=0.75
mean.diff=2
var.ratio=3
#Use optim:
parameter<-function(c)
{
a<-c[1]
b<-c[2]
u<-runif(10000)
Y<-qnorm(u,mean=mu2,sd=sigma2)
u<-runif(10000)
X<-qnorm(u,mean=mu2,sd=sigma2)+a*abs(u-tau)^b*(u>tau)
return((abs(mean(X)-mean(Y))-mean.diff)^2+(var(X)/var(Y)-var.ratio)^2)
}
c0<-c(3,1)
cstar<-optim(c0,parameter)$par
astar<-cstar[1] #4.1709
bstar<-cstar[2] #-0.2578
#Use grid seaching (I randomly assign a rage (-10,100)):
parameter.X<-function(a,b)
{
TSE<-matrix(0, length(a),length(b))
u<-runif(10000)
Y<-qnorm(u,mean=mu2,sd=sigma2)
u<-runif(10000)
for(i in 1: length(a))
{
for(j in 1: length(b))
{
X<-qnorm(u,mean=mu2,sd=sigma2)+a[i]*abs(u-tau)^b[j]*(u>tau)
TSE[i,j]<-(abs(mean(X)-mean(Y))-mean.diff)^2+(var(X)/var(Y)-var.ratio)^2
}
}
minTSE<-min(TSE)
a.optimal<-a[which(TSE==min(TSE),arr.ind = TRUE)[1]]
b.optimal<-b[which(TSE==min(TSE),arr.ind = TRUE)[2]]
return(list(a.optimal=a.optimal,b.optimal=b.optimal,minTSE=minTSE))
}
a0<-seq(-10,100,,50)
b0<-seq(-10,100,,50)
tse1<-parameter.X(a0,b0)
astar<-tse1$a.optimal # 84.28571
bstar<-tse1$b.optimal #1.224490
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
