Simple regression coefficient manipulation.

Question

Show that

$$\hat{\beta_1}=\frac{\sum_{i=1}^nx_i(y_i-\bar{y})}{\sum_{i=1}^nx_i(x_i-\bar{x})}=\frac{\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n(x_i-\bar{x})^2}$$

I am learning simple linear regression and the text wants me to prove the above as an exercise. I am really just not sure where to begin. I don't have a solid grasp on how to simplify $\bar{x}$ and $\hat{x}$. Any help is appreciated!

Answer 1

Hint:

\begin{align}\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})&=\sum_{i=1}^nx_i(y_i-\bar{y})-\sum_{i=1}^n\bar{x}(y_i-\bar{y}) \\ &= \sum_{i=1}^nx_i(y_i-\bar{y})-\bar{x}\sum_{i=1}^n(y_i-\bar{y}) \\ &= \sum_{i=1}^nx_i(y_i-\bar{y})-\bar{x}(0) \\&= \sum_{i=1}^nx_i(y_i-\bar{y})\end{align}

Answer 2

For $\sum (x_i - \bar x)^2$ note, that:

$\sum (x_i- \bar x)^2= \sum x_i^2 -2 \sum x_i \bar x + \sum \bar x^2 = \sum x_i^2 - 2 \bar x n \bar x + n \bar x^2 = \sum x_i^2 -n \bar x^2$

and

$\sum x_i(x_i-\bar x) = \sum (x_i^2-x_i \bar x) = \sum x_i^2 - \bar x \sum x_i = \sum x_i^2 - n \bar x^2$