Pearson product moment correlation coefficent, coefficient of determination and negative values here is the formula I have:
$$R = \frac {    \frac {1}{n} \sum_{i=1}^n x_i y_i - \bar{x} \bar{y}   }           
{    \sqrt{ \frac {1}{n} \sum_{i=1}^n (x_i - \bar{x})^2   \frac {1}{n} \sum_{i=1}^n (y_i - \bar{y})^2  }}$$
Now assuming we have a set of measurements whose sample correlation coefficient is $R$.
Assuming the relationship (regression curve) : $$ y = \beta_0 + \beta_1 x $$ ( i think we assume the simple linear model), then we have that, given that $\beta_i$ are the MLE of their respective coefficient : $$ R = \hat{\beta}_1 \frac{ \sqrt{ n \sum_{i=1}^n x_i^2 - (\sum_{i=1}^n x_i)^2 }   } {    \sqrt{  n \sum_{i=1}^n y_i^2 - (\sum_{i=1}^n y_i)^2 }} $$
which leads us to: 
$$ R^2 = \frac{\sum_{i=1}^n (y_i - \bar{y} )^2 - \sum_{i=1}^n(y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i)^2 }{\sum_{i=1}^n (y_i - \bar{y})^2} $$
which is I guess is a well-known formula.
all of this can be found page 602 of the book "Introduction to Mathematical Statistics and its applications" by R.J. Larsen and M.L. Marx.

Here is my question. I'm not sure how to tackle this last equation.
  $R^2$ should be positive, does that mean the numerator can't be
  negative, why? Moreover, how do you interpret globally this expression?
  Does it have some meaning? Since we are doing $$ \frac {A + B}{A},
 $$I'm assuming it has some interesting meaning.

 A: $$ R^2 = \frac{\sum^n (y_i - \bar{y} )^2 - \sum^n(y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i)^2 }{\sum^n (y_i - \bar{y})^2} $$
$$=1-\frac{\sum^n(y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i)^2 }{\sum^n (y_i - \bar{y})^2}.$$
I will assume that our regression weights are determined by the method of least squares. Then, the second term is always positive and smaller than $1$. It is positive because it is a fraction of the sum of squares. The value is smaller than $1$ because the regression equation $\hat{y}_i=\hat{\beta}_0+\hat{\beta}_1x_i$ is almost always closer to $y_i$ than the mean $\overline{y}$. Only if our observed outputs have the same constant values the regression equation will give the mean $\overline{y}=\hat{\beta}_0$ as the solution. This means that the deviations in the numerator are almost always smaller than the deviations in the denominator. Which implies that the second term is always smaller than $1$. Or to put it more simply your regression can always be as poor as the constant mean of the outputs $\overline{y}$.
The $R^2$ value can be interpreted as the ratio of the sum of squares for the regression
$$\text{SS}_\text{Reg}=\sum_{n=1}^N\left[\hat{y}_n - \overline{y}\right]^2$$
and the total sum of squares 
$$\text{SS}_\text{Total}=\sum_{n=1}^N\left[y_n-\overline{y} \right]^2.$$
If our regression equation $\hat{y}_n$ is perfectly fitting the observed outputs $y_n$ then the $R^2$ value will have the value $1$. If the regression is only able to fit the mean then $R^2=0$ which indicates a very poor fit. Note, that the $R^2$ value is very liberal when $n$ is not much larger than the number of regression weights.
