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Re: Chebyshev Inequality for Sample Variance
Posted:
Oct 2, 2010 5:50 AM
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On 2 Ekim, 12:05, C Hanck <chrha...@aol.com> wrote:
> On Oct 2, 2:09 am, Cagdas Ozgenc <cagdas.ozg...@gmail.com> wrote:
>
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>
>
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> > On 1 Ekim, 20:15, Ludovicus <luir...@yahoo.com> wrote:
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> > > On Sep 24, 9:42 am, Cagdas Ozgenc <cagdas.ozg...@gmail.com> wrote:
>
> > > > How do you adjust Chebyshev Inequality for Sample Variance when
> > > > Population Variance is not known?
>
> > > That's impossible because Chebyshev Inequality is an arithmetic
> > > theorem applied to Probability based in the sample variance.
> > > Chebyshev Theorem:
> > > "Given any set of of numbers with Standard deviation s, the fraction
> > > that deviates more than k.s from the mean is always less than 1/k^2
> > > Ludovicus
>
> > Of course possible. I found a paper regarding this matter,
> > unfortunately I don't have access to it.
>
> >http://www.jstor.org/pss/2683249
>
> If you know that it is a random sample and the sufficiently high
> moments exist, then
>
> P(|S^2-\sigma^2|>eps)<Var(S^2)/eps^2->0
>
> as n->oo because the variance of S^2 will then tend to zero.
>
> Also the claim about Chebychev's inequality being about the sample
> mean is wrong, this is just the most prominent application. In
> general, it says that the probability that some r.v. is more than an
> epsilon away from another in absolute value is bounded from above by
> the expected value of the squared deviation divided by epsilon^2
S^2 is sample variance here?
I don't see how this is related to my question. I just would like to
determine some loose confidence bounds from data.
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