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I hope I'm at right place, as I asked this question at stackoverflow.com where the question was closed and suggested I go ask here. I'll try to be more comprehensive also, looking at couple of replies I got there.

The question is simple for those who know how algorithms work, and as I'm physics student I'm not sure about the answer. I had exam in Statistics where professor suggested that formula I used (red outlined) in determining "Least-Squares Regression Line" was computationally inferior to the one derived below (green outline). My simple thinking was that first one uses n squares while derived (n+1) squares, but supposedly that's not enough. Can someone point why could derived formula be computationally more effective?

variance

Thanks

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up vote 3 down vote accepted

You're comparing the two variance formulae

$$s_x^2=\frac1{n}\sum_{i=1}^n \left(x_i-\bar{x}\right)^2$$

and

$$s_x^2=\frac1{n}\left(\sum_{i=1}^n x_i^2\right)-\bar{x}^2$$

With respect to the computational effort needed, the second formula looks better: it only requires one loop to pass through your data, as opposed to the two needed for the first one (first for computing the mean, and then for adding up the squared deviations).

The truth of the matter, however, is that the first formula makes much more computational sense than the second one. In practice, the two terms being subtracted in the second form can be very large; if the variance is small, you lose a lot of significant figures in the subtraction of two nearly-equal quantities (subtractive cancellation). Worst comes to worst, you may even end up with a negative variance with the second formula!

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Thank you for your reply, that even backs up my simple reasoning on subject and provides more into. In the meantime one of stackoverflow.com members replied with this (although the question is closed there): [quote]The first formula requires two passes (one to compute sample mean, another to calculate sum(squared-error), and you have to remember all the Xi's). The second formula can be applied in one pass, keeping running totals of sum(X^2) and sum(X) without having to store the whole sequence[/quote] Does this make sense, as I can't follow it? –  romor Oct 1 '10 at 6:31
    
Meh, there is even wiki about the problem: en.wikipedia.org/wiki/Algorithms_for_calculating_variance –  romor Oct 1 '10 at 6:34
    
The short version of what that commenter said is that it takes more effort to compute the variance with the first formula. It is however more accurate than the second formula; the second formula may be faster, but it is more prone to losing significant figures. –  J. M. Oct 1 '10 at 6:49
    
Here is an article you might want to peruse regarding the matter. –  J. M. Oct 1 '10 at 6:51
    
I finally understand - you don't need explicit formula for sample mean (as in formula 1.2 in your kindly provided link), which is considered during calculation. So second formula is faster, but prone to errors while first is suggested as more sane. Basically what answer sais, but I need time to comprehend it. Huh, thanks :) –  romor Oct 1 '10 at 7:35
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