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Re: Simulation for the standard deviation
Posted:
Feb 22, 2013 7:02 PM
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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 (two-sided)
>>> and
>>> 0.822722 <= sigma, 0 <= sigma <= 1.26795
Puzzled out, maybe. The first line has the two 0.5% one-sided
limits -- which is how you create the two-sided CI -- and the
second line has the two 1% one-sided 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
>>>> 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.
>
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
round-off in the simulation seems possible, but less likely if you
did enough steps to converge.
--
Rich Ulrich
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