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I have code that calculates $R^2$ with summations $$R^2 = \frac{(\sum xy - \frac1n \sum x \sum y)^2}{(\sum x^2 - \frac1n \sum x \sum x) (\sum y^2 - \frac1n \sum y \sum y)},$$ which is equivalent to $$R^2 = \frac{cov(x, y) \cdot cov(x, y)}{var(x) \cdot var(y)}.$$

I know the code is correct by benchmarking, but I have never seen this form. Can someone please explain or provide a reference? Thanks!

FWIW, the code is built for speed. It does rolling regressions and can quickly find each summation by differencing a cumulative sum.

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You've never seen which form? The first, or the second? The first form is a very dangerous way to calculate $R^2$. There are much more numerical stable one-step update methods. – cardinal Oct 15 '11 at 23:24
@cardinal -- I haven't seen either (maybe I shouldn't have converted to var/covar -- I thought it might save an answerer some time). I am more familiar with $R^2 = 1 - SS_{err}/SS_{tot}$ version. Why does the var/covar form work? – richardh Oct 16 '11 at 0:44

1 Answer

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The correlation is the covariance scaled by the SDs, $r=\text{cor}(x,y)=\text{cov}(x,y)/[\text{SD}(x) \; \text{SD}(y)]$. The formula you cite follows immediately. A reference seems unnecessary.

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Is $R^2 = \rho_{x,y} \cdot cov(x,y) / (sd(x) \cdot sd(y))$? – richardh Oct 16 '11 at 0:47
@richardh - $R^2=r^2=[\text{cor}(x,y)]^2$ – Karl Oct 16 '11 at 0:50
:) Thanks. As I look back I comment, it's clear. – richardh Oct 16 '11 at 0:53

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