Prove $\sum_i \frac{\bar{x}(x_i-\bar{x})}{nS_{xx}} =0$

Question

Prove

$$\sum_i \frac{\bar{x}(x_i-\bar{x})}{nS_{xx}} =0$$

That is one detail in a proof of the variance of the intercept $\alpha$ in the simple linear regression $Y_i=\alpha+\beta x_i$.

Answer 1

The term $$\frac{\bar x}{nS_{xx}}$$ is independent of $i$ and can be taken out of the summation. Thus $$\sum_{i}\frac{\bar{x}(x_i-\bar{x})}{nS_{xx}}=\frac{\bar x}{nS_{xx}}\sum_{i}\left(x_i-\bar{x}\right)=$$ where the latter term $\sum_i \left(x_i-\bar{x}\right)$ is known to be equal to zero. But, for the sake of completeness $$\sum_i \left(x_i-\bar{x}\right)=\sum_i x_i - \sum_i \bar x=n \cdot \underbrace{\left(\frac{1}{n} \sum_i x_i\right)}_{=\bar x} -n \bar x= n \bar x - n \bar x =0$$

Answer 2

You have $$ \sum_{i=1}^n \frac{\bar{x}(x_i-\bar{x})}{nS_{xx}} =\frac{\bar{x}}{nS_{xx}}\sum_{i=1}^n x_i -\frac{\bar{x}}{nS_{xx}}\sum_{i=1}^n \bar{x} =\frac{\bar{x}^2}{S_{xx}} -\frac{\bar{x}^2}{S_{xx}}=0, $$ since $$ \bar{x} = \frac{1}{n}\sum_{i=1}^n x_i $$ and $$ \bar{x} = \frac{1}{n}\sum_{i=1}^n \bar{x}. $$