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Re: Chebyshev Inequality for Sample Variance
Posted:
Oct 2, 2010 5:50 AM


On 2 Ekim, 12:05, C Hanck <chrha...@aol.com> wrote: > On Oct 2, 2:09 am, Cagdas Ozgenc <cagdas.ozg...@gmail.com> wrote: > > > > > > > On 1 Ekim, 20:15, Ludovicus <luir...@yahoo.com> wrote: > > > > 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.



