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Re: Simulation for the standard deviation
Posted:
Feb 22, 2013 9:12 AM
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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 (two-sided)
>> 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
>>> boot-strapping, 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 one-sided CI of the SD for the normal
distribution: 0.7885 instead of 0.8227.
Cristiano
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