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^2
Multiply 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

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.