Re: Variance estimation by Permutation Samples
- From: "A.G.McDowell" <mcdowella@xxxxxxxxxxxx>
- Date: Fri, 30 Dec 2005 07:50:37 +0000
In article <1825138.1135848292218.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxx
g>, Luis A. Afonso <licas_@xxxxxxxxxxx> writes
>(a Project)
>
>
>It was nice that someone could check the following strategy I´ll try to put in
>action:
>
>Algorithm:
>To obtain a generic variance estimate*, var, then its Confidence Interval by:
>
>First step: to find MM by the permutation X´, (MM denoting the modified mean)
>__MM = [j*x´j]/c__
>__ c = n*(1+n)/2
>Second step: to get a new data Permutation X´´ and
>_ssd = [(x´´j - MM)^2]
>Third step
>_var = ssd/(n-1)
>
>* repeated a larger number of times to be possible a CI estimation, which I
>suppose (I hope) narrower than the conventional algorithm:
>___ [xj - xbar]^2 /(n-1).
>___xbar = [xj] / n
>
>_licas (Luís A. Afonso)
Here's a test example for resampling-based methods. Consider n i.i.d.
samples which are 0 with probability p and 1 with probability 1-p. The
variance of each sample is therefore p(1-p). With probability 2^-(n-1)
all the samples you see will be the same. Consider a method for
constructing confidence intervals that is supposed to fail to provide an
interval containing the true value with probability less than 2^-(n-1).
Feed it n identical values. Does the resulting interval contain the true
variance?
--
A.G.McDowell
.
- References:
- Variance estimation by Permutation Samples
- From: \"Luis A. Afonso\"
- Variance estimation by Permutation Samples
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