I'm trying to find out what are $\hat{β_1}, \hat{β_2}$
$ \left \{ \begin{array}{c @{=} c} \frac{∂S( \hat{β_1}, \hat{β_2})}{∂S \hat{β_1}} =-2\sum(yi − \hat{β_1} − \hat{β_2}xi) = 0, \\ \frac{∂S( \hat{β_1}, \hat{β_2})}{∂S \hat{β_2}}=-2\sum x_i(yi − \hat{β_1} − \hat{β_2}xi) = 0. \\ \end{array} \right. $
- The first equation gives $\hat{β_1}+ \hat{β_2}\sum(x_i)^2=\sum y_i$
which gives the estimator of the y-intercept: $\hat{β_1} = \bar{y} − \hat{β_2}\bar{x}$
- The second equation gives: $\hat{β_1}\sum x_i + \hat{β_2}\sum(x_i)^2=\sum x_iy_i$
replacing $\hat{β_1}$ by the expression we found, we have: $\frac{\sum x_iy_i - \sum x_i\bar{y}}{\sum x_i^2 - \sum x_i\bar{x}}$
And then I received an hint to conclude: "use shrewdly the fact that $\sum x_i-\bar{x}$ to conclude on $\hat{β_2}$"
I have to found that $\hat{β_2} =\frac{\sum (x_i-\bar{x})y_i}{\sum (x_i-\bar{x})^2}$
but I thought we couldn't divide by zero? Is the division by $\sum x_i-\bar{x}$ actually null, then?