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We often square coefficients like the R coefficient in simple/multiple linear regression or standardized factor loadings to get a percentage of variance accounted for by predictor variables.

Can the same principle be applied to a Pearson correlation coefficient? Wikipedia suggests that

When an intercept is included, then r2 is simply the square of the sample correlation coefficient (i.e., r) between the observed outcomes and the observed predictor values

But in a simple Pearson correlation coefficient, intercepts are not included. So is squaring still appropriate to determine the percent of shared variance between variables?

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    $\begingroup$ You can always square a correlation $r$ (between $x$ and $y$, say) and the result is equal to the $r^2$ or $R^2$ (notation varies, but for two variables the difference is unimportant) you would get if you did either regression, $y$ on $x$ or $x$ on $y$. But shared variance is not good wording: $y$ and $x$ will have often have quite different units and so their variances won't be comparable at all. Rather, $r^2$ or $R^2$ is the fraction of the variance of either variable associated with (or "explained by", although many dislike such expressions) regression on the other. $\endgroup$ – Nick Cox Oct 23 '18 at 9:21
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    $\begingroup$ Otherwise put, every correlation implies two regressions, each with an intercept. Whether you carry out either regression is another matter. $\endgroup$ – Nick Cox Oct 23 '18 at 9:22
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    $\begingroup$ I think you have misunderstood Wikipedia. The Pearson correlation removes mean from dependent and independent variables(centers).in a regression model with non centred dependent and independent variables you need intercept term. Put another way you need intercept for uncentred data, the Pearson correlation centres the data $\endgroup$ – seanv507 Oct 23 '18 at 15:14
  • $\begingroup$ Wow - all brilliant and enlightening comments. Many thanks. I'm just wondering why neither of you 'answered' the question. $\endgroup$ – PyjamaNinja Oct 23 '18 at 17:22

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