# Why do we need $R^2$?

In linear regression, the $R^2$ value is the square of the correlation between predicted values and observed values. But why do we need the $R^2$ value? Why not just use the correlation coefficient? Just like the correlation coefficient, $R^2$ is scale-less (i.e. values are always between 0 and 1), so I can't see why there's a need for $R^2$. I would imagine it is something to do with the fact that the correlation coefficient can be negative, but don't really see why this would be a problem.

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One advantage of the R-squared is its nice objective interpretation as "Proportion of variance (of the response) explained by (differences in) predictors", which follows directly of its definition $$1 - \frac{Var(e)}{Var(Y)},$$ where $e$ is the vector of residuals, $Y$ the vector of observed values of the response and $Var$ is the sample variance. The numerator can be called "Unexplained variance", thus the interpretation above.
Similar "nice objective interpretations" in terms of other measures of variation would seem to be available, too, such as $1-SD(e)/SD(Y)$. You might therefore want to explain why variance ratios are special or important. –  whuber Aug 29 '14 at 14:48
For univariate regression, it is difficult to think of any real advantage of using $R^2$, because--as the OP notes--that actually loses information about the direction of the relationship afforded by the sign of $r$. Maybe this question should be approached by expanding consideration to the case of multiple regression, where $r$ has no clear analog but $R^2$ remains interpretable. Another way to address these issues is in terms of ANOVA, which seeks to decompose measures of variation into additive components: that practically forces $R^2$ upon us and justifies your answer. –  whuber Aug 29 '14 at 15:18
Just a clarification of @Michael's answer. The quantity $1-Var(e)/Var(Y)$ is the adjusted $R^2$. The ordinary $R^2$ is the same thing with the sums of squares: $R^2=1 - SS(\mbox{error})/SS(\mbox{total})$. –  rvl Aug 29 '14 at 16:38
@RussLenth: For "Var" being the ordinary sample variance with $(n-1)$ in the denominator, "my" definition is equivalent to yours (just divide the two SS by $n-1$). –  Michael M Aug 30 '14 at 13:59
@MichaelMayer: But that's not how people compute the variance of the residuals. They use $Var(e)=MS_E=SS_E/(n-p)$, where $p$ is the number of parameters in the regression equation. –  rvl Aug 30 '14 at 14:07