
Re: Simulation for the standard deviation
Posted:
Feb 24, 2013 8:08 PM


On 24/02/2013 21:49, Ray Koopman wrote: > On Feb 24, 8:51 am, Cristiano <cristi...@NSgmail.com> wrote: >> On 22/02/2013 6:15, Ray Koopman wrote: >> >>> 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. >> >> Your limit works, but now I'm trying to find a way which says >> how good is the sample SD (w.r.t. 1/sqrt(12)). >> I thought to calculate the above p for which I get the sample SD; >> for example: >> sd = 0.224508, xmin = 0.4087, xmax = 0.847092 >> your lower bound = 0.595019. >> But (as you know) I can't use that lower bound as pvalue, because >> a good SD will have a "pvalue" around 0.5, while it should be 1. >> >> Is there any way to use your procedure to calculate a pvalue for SD? >> >> Thank you >> Cristiano > > How did you get that lower bound? What were n and p?
Sorry, but I mistakenly wrote that the lower bound is 0.595019, while I find p which gives that lower bound, so n= 3, p= 0.595019 and lower bound = sd = 0.224508.
> I'm not sure what you're really trying to do. Are you trying to use > the sample SD to make inferences about the true SD for samples from a > uniform distribution? Why? If you know you're sampling from a uniform > distribution then you should be using the range, not the SD.
I randomly pick 3 numbers in U(0,1) and I get, for example, sd= .1234, but we know that the expected sd is .288675. How good .1234 is? Is there any way to calculate a pvalue which says how good is .1234?
Cristiano

