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Topic: standard deviation of overlapping sums
Replies: 2   Last Post: Aug 19, 2011 10:05 AM

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root

Posts: 50
Registered: 8/26/09
Re: standard deviation of overlapping sums
Posted: Aug 19, 2011 10:05 AM
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Ray Koopman <koopman@sfu.ca> wrote:
> On Aug 18, 1:53 pm, Beliavsky <beliav...@aol.com> wrote:
>> If I have N independent normal variates and create (N-M+1) sums of M
>> consecutive variates, what is the standard deviation of those sums?
>> For M << N the standard deviation of the sums is slightly less than
>> sqrt(M), but as M approaches N, the overlap makes the standard
>> deviation appreciably less than sqrt(M).

>
> I'm not 100% sure about this, but it's in the right ballpark.
>
> The sums have the same marginal variances, m, but are correlated.
> Their covariance matrix has m down its main diagonal, m-1 in the
> first offdiagonals, m-2 in the second offdiagonals, etc.
>
> Let k = n+1-m. If k >= m then the sum of the covariances is
> t = k*m + 2((k-1)*(m-1) + (k-2)(m-2) + ... + (k-m+1)*(1))
> = k*m + m(m-1)(2k-1+k-m)/3.
>
> If k < m then interchange k and m in the expression for t.
>
> Using the alternate definition of a variance as half the expected
> squared difference between two independently-chosen sums gives
> m - t/k^2 (i.e., their average variance minus their average
> covariance) as the variance of the sums.
>
> The other alternate definition of a variance, as half the expected
> squared difference between two sums chosen without replacement, gives
> m - (t - k*m)/(k(k-1)).


The OP's problem is related to a moving average. As a good estimate
of the observed reduction of freedom you can take the number of
points as N/(autocorrelation length).



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