
Re: Simulation for the standard deviation
Posted:
Feb 23, 2013 2:25 PM


On Feb 23, 4:58 am, Cristiano <cristi...@NSgmail.com> wrote: > On 22/02/2013 21:06, Ray Koopman wrote: >> On Feb 22, 6:05 am, Cristiano <cristi...@NSgmail.com> wrote: >>> On 22/02/2013 6:15, Ray Koopman wrote: >>>> On Feb 20, 3:54 am, Cristiano <cristi...@NSgmail.com> wrote: >>>> >>>>> Short question: does anybody know how to calculate the confidence >>>>> interval of the standard deviation for the uniform distribution? >>>> >>>> For n iid samples from a continuous uniform distribution, >>>> Pr(r/R <= x) = F(x) = n*x^(n1)  (n1)*x^n, where >>>> r is the sample range, R is the true range, and 0 <= x <= 1. >>>> A 100p% confidence interval for R is R >= r/x, where F(x) = p. >>>> Divide that by sqrt(12) to get a lower bound for the SD. >>> >>> Suppose I randomly pick 0.1, 0.4 and 0.2 (n = 3); >>> what should I write to calculate a 99% confidence interval? >> >> F(x) = 3 x^2  2 x^3 = p >> >> F(.941097) = .99 >> >> SD >= (.4  .1)/(.941097 * sqrt(12)) > > That lower bound doesn't work. > According to the practical definition of CI given here: > http://www.itl.nist.gov/div898/handbook/eda/section3/eda352.htm > I wrote a simulation which counts how many times the sample SD > exceeds the calculated lower bound. > Using 10^6 trials and a confidence level of .9, I see that the > calculated SD is greater than your lower bound 10^6 times, while > the calculated SD is greater than my lower bound 900221 times > (I find the confidence limits as explained in my original post).
To check a lowerbound estimation procedure, at each trial you calculate the bound and compare it to the true value of the parameter, not to the sample value of a point estimate of the parameter. The bound should be less than the true value 100p% of the time.

