Hi there, I wonder if anyone kind can tell me how to prove this sample variance formulae? My main concern is how to get from the 2nd to the 3rd line.
s^2 = [1/(n-1)] * {SUM from i=1 to n) (yi - mean of y)^2 = {n * {SUM from i=1 to n) yi^2 - [(SUM from i=1 to n) yi ]^2} / (n-1)
I have paid very fine attention to the bracketing. Thanks.
Please advise asap.
s^2 is the sample variance n is the finite/infinite variable term to be considered i is the running variable
Remember the definition of 'mean' as '[SUM from i=1 to n](x) / n'. So we can rewrite this... = [1/(n-1)] * ({SUM from i=1 to n} (yi - mean(y))^2) = [1/(n-1)] * n * mean((yi - mean(y))^2) = [n/(n-1)] * mean((yi - mean(y))^2)
Some rules about means...
1)When finding the mean, if all the numbers contain a common factor, this number can be factored out before finding the mean. mean(k*y) = k*mean(y)2)Also, the mean of two addends is equal to the sum of the mean of each addend mean(y+z) = mean(y) + mean(z)
3)And the mean of a constant equals the constant mean(k^2) = k^2Multiply out the term inside the ^2 and you get = [n/(n-1)] * mean((yi - mean(y))^2) = [n/(n-1)] * mean(yi^2 - 2*yi*mean(y) + mean(y)^2)
Applying the above rules and remembering that 2, mean(y) and mean(y)^2 are all constant for a given problem, and that mean(yi) = mean(y)
= [n/(n-1)] * (mean(yi^2) - mean(2*yi*mean(y)) + mean(mean(y)^2)) = [n/(n-1)] * (mean(yi^2) - 2*mean(y)*mean(yi) + mean(y)^2) = [n/(n-1)] * (mean(yi^2) - mean(y)^2) = (n*mean(yi^2) - n*mean(y)^2) / (n-1) = ((SUM from i=1 to n)(yi^2) - [(SUM from i=1 to n)(y)]*mean(y)) / (n-1)
Not exactly your results, but perhaps this helps?
daestrom
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