How to break this summation

Question

I see least square's slope like this $$a=\frac{\sum_{i=1}^n (x_i-\bar x)(y_i-\bar y)}{\sum_{i=1}^n(x_i-\bar x)^2}$$ But I also see the top part like this $$\sum_{i=1}^n x_iy_i - \frac{\sum_{i=1}^n x_i \sum_{i=1}^n y_i}n$$ My question is how to break $\sum_{i=1}^n (x_i-\bar x)(y_i-\bar y)$ down to $\sum_{i=1}^n x_iy_i - \frac{\sum_{i=1}^n x_i \sum_{i=1}^n y_i}n$ Thanks.

Answer 1

I am not sure about your notation. But I think this is the answers you are looking for:

\begin{eqnarray} \sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y}) &=& \sum_{i=1}^n(x_iy_i - x_i\bar{y}-\bar{x}y_i+\bar{x}\bar{y}) \\ &=& \sum_{i=1}^nx_iy_i - \bar{y}\sum_{i=1}^nx_i - \bar{x}\sum_{i=1}^ny_i+n\bar{x}\bar{y} \\ &=& \sum_{i=1}^nx_iy_i - n\bar{y}\bar{x} - n\bar{x}\bar{y}+n\bar{x}\bar{y} \\ &=& \sum_{i=1}^nx_iy_i - n\bar{y}\bar{x} \\ &=& \sum_{i=1}^nx_iy_i - n\frac{\sum_{i=1}^nx_i}{n}\frac{\sum_{i=1}^ny_i}{n} \\ &=& \sum_{i=1}^nx_iy_i - \frac{\sum_{i=1}^nx_i \sum_{i=1}^ny_i}{n} \end{eqnarray}

Then you got it!