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Re: Simulation for the standard deviation
Posted:
Feb 23, 2013 2:25 PM
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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^(n-1) - (n-1)*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 lower-bound 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.
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