
Re: Simulation for the standard deviation
Posted:
Feb 22, 2013 9:12 AM


On 22/02/2013 0:52, Rich Ulrich wrote: > On Thu, 21 Feb 2013 20:54:44 +0100, Cristiano <cristiapi@NSgmail.com> > wrote: > >> To check my simulation, I calculated the CI for the normal distribution >> as explained here: >> http://www.itl.nist.gov/div898/handbook/prc/section2/prc231.htm > > > CAREFUL. You stated a problem about the uniform. You are > pointing now to an answer about the normal? Not relevant.
I know how to exactly calculate the confidence interval for the normal distribution, so I use the exact confidence interval to check my simulation. I'd say that it is relevant! :)
>> For example, for N(0,1), when N= 60 and alpha= 0.01, I get: >> 0.806465 <= sigma <= 1.30263 (twosided) >> and >> 0.822722 <= sigma, 0 <= sigma <= 1.26795 >> >> while the simulation converges to 0.7885 <= sigma. >> >>> Assuming that you are describing a recommended procedure for >>> bootstrapping, that seems like the way to simulate that CI. >> >> I found this link: >> http://en.wikipedia.org/wiki/Bootstrapping_%28statistics%29#Deriving_confidence_intervals_from_the_bootstrap_distribution >> and I think that I'm using the "Percentile Bootstrap". >> >> There is an interesting phrase: "This method can be applied to any >> statistic. It will work well in cases where the bootstrap distribution >> is symmetrical and centered on the observed statistic [...]". >> >> The reason of my wrong result could be that the bootstrap distribution >> of the standard deviation is not symmetrical, I guess. > > When you are looking at one tail only, the matter of symmetry does > not arise. You stated your tentative solution as just one percentile. > The difficulty of asymmetry does not arise for one limit alone.
Then there should be something wrong in my simulation, because I get a totally wrong result for the onesided CI of the SD for the normal distribution: 0.7885 instead of 0.8227.
Cristiano

