David Winsemius <[EMAIL PROTECTED]> wrote in news:[EMAIL PROTECTED]:
> "Jacques Wagnor" <[EMAIL PROTECTED]> wrote in > news:[EMAIL PROTECTED]: > >> I have a model as follows: >> >> x <- replicate(100, sum(rlnorm(rpois(1,5), 0,1))) >> y <- quantile(x, 0.99)) >> >> How would one go about estimating the boundaries of a 95% >> confidence interval for y? >> >> Any pointers would be greatly appreciated. > > I'm not a statistician, so giving the answer in terms of extreme > value statistics is beyond me, but the R Team gives us a (sharp) > tool. > > quantile(x,99) is returning the midpoint of the 99th and 100th > elements of the sorted 100 element vector you created. > > If you repeat that process 1000 times, sort again, and pick the 25th > and the 975th points, you can pull the 0.025 and 97.5 percentile > points from the simulated distribution. Obviously an estimate and > will vary depending on the seed. > > Here's what I got after that process: >> sort(y1000.df$midpt)[25] > [1] 20.8424 >> sort(y1000.df$midpt)[1000-25] > [1] 47.47615 > My apologies to Ivan Frohne and Rob J Hyndman, the authors of stats:::quantile. After looking further at the type definition for the default interpolation algorithm (type = 7), I do not think my description offered above is accurate. Jacques, you should not worry if your simulations produce more narrow bounds for the 99th percentile. When I use quantile(), rather than my imagined behavior of it, and do 10,000 iterations, I get results that are somewhat different: + ))[c(250,9750)] 99% 99% 19.11080 38.56948 -- David Winsemius ______________________________________________ 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.
