Hi Solafah, You are right that two commands are equivalent when p= pnorm(a). You can check the results by following codes. n <- 5 a <- -1 set.seed(123456) qnorm(runif(n,0,pnorm(a))) p <- pnorm(a) set.seed(123456) qnorm(p*runif(n))
Anyway, the elements of the lower tail are not chosen equally by this method. I may try another method. Such like: s1 <- rnorm(10000) n <- 5 a <- -1 sample(s1[s1<a],n) ----- A R learner. -- View this message in context: http://r.789695.n4.nabble.com/sampling-from-normal-tp3003016p3003164.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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.
