
Re: Simulation for the standard deviation
Posted:
Feb 22, 2013 7:02 PM


On Fri, 22 Feb 2013 15:12:03 +0100, Cristiano <cristiapi@NSgmail.com> wrote:
>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! :)
Okay  So you needed to make it evident to me that you were testing the simulation, and not comparing outcomes... since I (still) don't follow your comparison.
> >>> 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
Puzzled out, maybe. The first line has the two 0.5% onesided limits  which is how you create the twosided CI  and the second line has the two 1% onesided limits. I'm not bothering to compute it to find out.
>>> >>> 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. >
Those numbers that you list above are limits around sigma/s rather than s/sigma  You took the formula at .nist and divided through by s. Personally, I was surprised that the formula seemed to be upside down. I was expecting s/sigma. Now I wonder which should be used for what.
However, when I take the reciprocal of 1.26795, it comes out as 0.7887  which makes me think that the only error you see is computation error on someone's CDF generator. Cumulative roundoff in the simulation seems possible, but less likely if you did enough steps to converge.
 Rich Ulrich

