Summatory problem | Ordinary least square estimator

Question

How I can transform the first expression in the second?

\begin{align} \hat{\beta}_{1} & =\frac{n\sum X_{i}Y_{i}-\sum X_{i}\sum Y_{i}}{n\sum X_{i}^{2}-\left(\sum X_{i}\right)^{2}} \\ & = \frac{\sum(X_{i}-\bar{X})(Y_{i}-\bar{Y})}{\sum (X_{i}-\bar{X})^{2}} \end{align}

It is a simple question about summations, I take a while trying to develop terms but did not equal. Thanks in advance.

Answer 1

$\sum_iX_i=n\bar X$, so

$$\begin{align*} \sum_i(X_i-\bar X)^2&=\sum_iX_i^2-2\bar X\sum_iX_i+\sum_i\bar X^2\\ &=\sum_iX_i^2-2\bar X(n\bar X)+n\bar X^2\\ &=\sum_iX_i^2-n\bar X^2\;. \end{align*}$$

Moreover,

$$\begin{align*} \sum_i(X_i-\bar X)(Y_i-\bar Y)&=\sum_iX_iY_i-\bar X\sum_iY_i-\bar Y\sum_iX_i+\sum_i\bar X\bar Y\\ &=\sum_iX_iY_i-\bar X(n\bar Y)-\bar Y(n\bar X)+n\bar X\bar Y\\ &=\sum_iX_iY_i-n\bar X\bar Y\;. \end{align*}$$

Thus,

$$\begin{align*} \frac{\sum_i(X_i-\bar X)(Y_i-\bar Y)}{\sum_i(X_i-\bar X)^2}&=\frac{\sum_iX_iY_i-n\bar X\bar Y}{\sum_iX_i^2-n\bar X^2}\\\\ &=\frac{n\sum_iX_iY_i-n^2\bar X\bar Y}{n\sum_iX_i^2-n^2\bar X^2}\\\\ &=\frac{n\sum_iX_iY_i-\sum_iX_i\sum_iY_i}{n\sum_iX_i^2-\left(\sum_iX_i\right)^2}\;. \end{align*}$$