Problem in applying conditional looping
You can use something like this to keep running your algorithm
# repeat until all values are positive
repeat{
x <- rnorm(10)
if (all(x >= 0)) break # conditions met; quit
}
On Thu, Jul 3, 2008 at 5:25 AM, Nadeem Shafique
<nadeemshafique at gmail.com> wrote:
Respected All,
I hope you are enjoying good health, I am tring to write a program in R but
could not be very sucessful. My program draws random sample form bivariate
normal distribution and then compute a variable PIJ. For certian samples
some entries of variable PIJ is apearing as negative, which result
in negative variance estimator. I want to introduce a loop in my program
that verify the each value of PIJ and if any value in PIJ is negitive return
back and draw another sample until a sample is selected that produces all
values of PIJ positive.
library(mvtnorm)
library(combinat)
yg <- function(n,m,s) {
n<-10
m<-c(1,2)
s<- matrix(c(4,.2,.2,3), ncol=2)
x <- rmvnorm(n=n, mean=m, sigma=s)
pi <- x[,2]/sum(x[,2])
P <- pi*((1-2*pi)/(1-pi)+sum(pi/(1-pi)))
Pi<-t(combn(pi,2))
y<-t(combn(x[,1],2))
PI<-t(combn(P,2))
yht <- y[,1]/PI[,1]+ y[,2]/PI[,2]
*PIJ <- (Pi[,1]*Pi[,2]*(2-Pi[,1]-Pi[,2]))/((1-Pi[,1])*(1-Pi[,2]))*
Eyht <- sum(yht*PIJ)
E2yht <- sum(yht^2*PIJ)
var1 <- E2yht - Eyht^2
var1
}
I want to impose the condition on PIJ(which is a vector). If any value in
the PIJ is negative then go back and take another sample until a sample is
selected which produces the PIJ positive(all values in PIJ).
I hope now you can understand.
Best Regards,
Nadeem Shafique Butt
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