Consider $$ \hat{\theta_2}=\frac{n\sum_{i=1}^{n}x_iy_i-\sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i}{n\sum_{i=1}^{n}x_i^2-(\sum_{i=1}^{n}x_i)^2} $$ Show that $Var(\hat{\theta_2})=\frac{\sigma^2}{\sum_{i=1}^{n}(x_i-\overline{x})^2}.$
Here we consider the linear modell $$ y_i=\theta_1+\theta_2x_i+u_i, Var(u_i)=\sigma^2 $$ where the $x_i$ are not arbitrary but the $y_i$ and the $y_i$ are indepedent.
So I tried to calculate it:
$$ Var(\hat{\theta_2})=\frac{1}{(n\sum x_i^2-(\sum x_i)^2)^2}Var(n\sum x_iy_i-\sum x_i\sum y_i)\\ =\frac{1}{(n\sum x_i^2-(\sum x_i)^2)^2}Var(\sum nxy_iy_i)-Var(\sum x_i\sum y_i)\\ =\frac{1}{(n\sum x_i^2-(\sum x_i)^2)^2}n^2\sigma^2\sum x_i^2-n\sigma^2 (\sum x_i)^2\\ =\frac{1}{(n\sum x_i^2-(\sum x_i)^2)^2}\sigma^2 n(n(\sum x_i^2-(\sum x_i)^2)\\ =\frac{\sigma^2}{\sum x_i^2-\frac{1}{n}(\sum x_i)^2} $$
But thats not what I wanted to show.
Can anybody please help me?
What you've done is perfectly okay, because $$\sum (x_i-\bar{x})^2=\sum x_i^2-2\sum x_i \bar{x}+\sum\bar{x}^2=\sum x_i^2-2n\bar{x}^2+n\bar{x}^2=\sum x_i^2-n\bar{x}^2\\=\sum x_i^2-\frac{1}{n}\left(\sum x_i\right)^2$$