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Math: Effective Sample Sizes of Weighted Moving Averages?

Discussion in 'Strategy Development' started by Traden4Alpha, Sep 20, 2002.

Anyone know how to compute the effective sample size for variably-weighted moving averages?

Many statistical analyses take the sample size as a variable. This includes: calculating the standard error, correcting the standard deviation for small sample sizes, doing the z-transform of the correlation coefficient, testing for significance based on the t-distribution, etc.

For a simple moving averages and exponential moving averages, the sample size is just the period of the moving average. But what is the effective sample size for more complex variably-weighted moving averages? I can tell from some preliminary work, that variable weighting tends to reduce the effective sample size by magnifying the variance contributed by highly weighted samples, and reducing the variance reduction provided by lightly weighted samples. It seems like there should be some way to estimate the effective sample size (= variance reduction) associated with more complex sets of weights on the samples. Anyone got any ideas?

Hoping for a large sample size of above-average ideas,

Well, I decided to derive the answer myself. The intermediate results looked pretty ugly until the whole thing simplified down to a ratio of sums. The result is that Neff, the effective sample size of an arbitrarily weighted average, is:

Neff = ( SUM(Weight)^2/( SUM(Weight^2) ) )

Like DUH! Should have know that it was just the square of the sum-of-the-weights divided by the sum-of-the-squares-of-the-weights. But it was fun to derive the answer by using projections of Gaussian ellipsoids on the geometry of hyperplanes -- always neat when geometry and statistics agree.

P.S. www.wilmott.com has a great forum for more math-intensive questions <b>(Thanx vulture!)</b>

kut2k2 likes this.
3. nitro

Paul Wilmott is very kewl...

Thanks for the site...

nitro

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