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I cant find any where information or algorithm how to apply in steps the bootstrap procedure to estimate the 95% confidence intervals for the 95% quantile from a random sample. Does anyone knows how to do it and can you write it? Thanks in advance.

vasil
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  • Needs clarification. Do you have a given sample? What size? Small samples will be almost useless. Parametric or nonparametric bootstrap? Title says gamma distribution. Do you know its parameters? If not, what role does knowing it's gamma play? If estimating parameters, MME or MLE? – BruceET Jul 29 '15 at 07:36

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The simplest possible approach is a nonparametric bootstrap as follows:

You have a large sample from a process, but no idea what the underlying distribution may be (no family such as gamma, no parameters). Maybe it is the length of time until a particular item fails when operated at too high a temperature. You can easily make a histogram of the SAMPLE and see it has a long tail to the right. And that it's 95th percentile is at 92.62 days. Knowing the 95th percentile of this PROCESS is important, and you want a 95% CI for that.

 # Generate fake data (for example using Gamma(5, .1)
 set.seed(1234)  # so you can reproduce same dataset
 n = 1000;  sh = 5;  rt = .1
 x = rgamma(n, sh, rt)
 q95.obs = quantile(x, .95)
 ##      95% 
 ## 92.62036 
 summary(x)
 ##  Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 ## 8.064  33.830  47.180  50.500  62.440 171.900

Here is one method to find a 95% CI based on the observed 95th percentile of the sample programmed in R. (There are many styles of nonparametric bootstrap. I have no idea which ones you may have encountered. This one uses the 'quantile method'.)

 B = 10000;  q95 = numeric(B)
 for (i in 1:B) {
   x.b = sample(x,length(x),repl=T)  # bootstrap re-sample
   q95[i] = quantile(x.b, .95) }
 2*q95.obs - quantile(q95, c(.975,.025))
 ##  88.25883 98.23680 
 length(unique(q95))
 ## 265

So based on your sample, a bootstrap 95% CI for the 95th percentile of the process is $(88.3, 98.2).$ The final statement says that, among 10,000 bootstrap iterations, we encountered 're-samples' that produced 265 uniquely distinct 95th percentiles, which is enough for a reasonably good CI. I suspect that using a very small sample of size for such a bootstrap would not yield a useful CI.

Because pseudorandom simulation is involved in getting a bootstrap CI, you may get a slightly different result if you run the same program again. One additional run gave slightly different result that agrees with the one above when rounded to one decimal place.

I don't know whether this is a drill exercise in a class or a real problem from your work. If you have some kind of parametric bootstrap method in mind, please provide a more specific description of the situation.

BruceET
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  • If it's not a problem, could you explain this line: 2*q95.obs - quantile(q95, c(.975,.025))? – Milos Jun 24 '18 at 15:50
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    Short version: it's bias correction. With a fresh run, bootstrapping q95 and using quantile(q.95, c(,025, .975) gives Ci (87.03, 96.98). whereas 2*q95.obs - quantile(.95, c(.975,.025)) as in my Answ gives bias-corrected CI (88.26, 98.22). For rationale, see this Q & A. – BruceET Jun 24 '18 at 17:46