Coefficient of determination

Remarks

Example: take the points $(x_i, y_i) = (0,0), (1,1), (2,3)$ and fit $y = a + bx$ by ordinary least squares. The fit is $b = 3/2$, $a = -1/6$, with fitted values $-1/6$, $4/3$, $17/6$. Then $SS_{\mathrm{res}} = 1/36 + 4/36 + 1/36 = 1/6$ and $SS_{\mathrm{tot}} = 16/9 + 1/9 + 25/9 = 14/3$, so $R^2 = 1 - 1/28 = 27/28$.

When the fit is by ordinary least squares with an intercept, $SS_{\mathrm{tot}} = SS_{\mathrm{reg}} + SS_{\mathrm{res}}$ with $SS_{\mathrm{reg}} = \sum_i (\hat y_i - \bar y)^2$. So $R^2 = SS_{\mathrm{reg}} / SS_{\mathrm{tot}}$ lies in $[0, 1]$. In that case $R^2$ is the square of the sample correlation between $y$ and $\hat y$. For a straight line fit it is the square of the sample correlation between $x$ and $y$.

Non-example: if the fitted values do not come from least squares with an intercept, the formula can give a negative number. With the same data and the constant prediction $\hat y_i = 3$, one gets $SS_{\mathrm{res}} = 13 > SS_{\mathrm{tot}}$, so $R^2 < 0$.

The adjusted $R^2$ for a model with $p$ coefficients, including the intercept, is $1 - (1 - R^2)(n - 1)/(n - p)$. It penalises extra regressors.