Ray Koopman <firstname.lastname@example.org> 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).