How to find the deviated form of beta 1 in OLS
Y=β1+β2X+u
estimated β1=(ΣX^2ΣY-ΣXΣXY)/(nΣX^2-(ΣX)^2)
I do not know how to turn this part (ΣX^2ΣY-ΣXΣXY)into deviated form.
I found that estimated β2=(nΣXY-ΣXΣY)/(nΣX^2-(ΣX)^2)=Σxy/Σx^2
x & y in deviated forms, ΣX=n(mean of X).
The expression of $\beta_1$ is
$\beta_1=\Large{\frac{\sum_{i=1}^n x_i^2\sum_{i=1}^ny_i-\sum_{i=1}^n x_i\sum_{i=1}^nx_iy_i}{\sum_{i=1}^n x_i^2-\left( \sum_{i=1}^n x_i \right)^2}}$
$x=X_i-\overline X$ and $y=Y_i-\overline Y$
Only the numerator of $\beta'_1$:
$\sum_{i=1}^n (X_i-\overline X)^2\sum_{i=1}^n Y_i-\overline Y-\sum_{i=1}^n (X_i-\overline X)\sum_{i=1}^n (X_i-\overline X)(Y_i-\overline Y)$
$\sum_{i=1}^n Y_i=n\cdot \overline Y$ and $\sum_{i=1}^n \overline Y=n\cdot \overline Y \Rightarrow \sum_{i=1}^n Y_i-\overline Y=0$
Thus $\sum_{i=1}^n (X_i-\overline X)^2\sum_{i=1}^n Y_i-\overline Y=\sum_{i=1}^n (X_i-\overline X)^2\cdot 0 =0$.
$\sum_{i=1}^n X_i=n \cdot \overline X$ and $\sum_{i=1}^n \overline X=n\cdot \overline X$
It follows, that $\sum_{i=1}^n (X_i-\overline X)=0$.
Thus $\sum_{i=1}^n (X_i-\overline X)\sum_{i=1}^n (X_i-\overline X)(Y_i-\overline Y)=0$
In total $\beta'_1=0$