Proving Linear Estimator (beta) is BLUE?

Question

In the book Statistical Inference pg 570 of pdf, There's a derivation on how a linear estimator can be proven to be BLUE.

I got all the way up to 11.3.18 and then the next part stuck me.

After finding the $d_i$s that satisfy the condition: $\Sigma d_i = 0$ & $\Sigma d_ix_i = 1$, we take 11.31.7 derived from the lemma: $$\sum_{i=1}^nd_ix_i=\sum_{i=1}^nK(x_i-\bar{x})x_i=KS_{xx}$$

Where did they get $S_{xx}$ from? Isn't that supposed to be defined as $\sum_{i=1}^n(x_i-\bar{x})^2$ which does not equal to the equation above.

Furthermore, if I'm on the right tracking then I'm not sure how 11.3.19 is worked out.

I'm so close to figuring it out guys and I would appreciate some guidance.

Answer 1

If you want to prove that the OLS for $\hat{\beta}$ is BLUE (best linear unbiased estimator) you have to prove the following two things: First that $\hat{\beta}$ is unbiased and second that $Var(\hat{\beta})$ is the smallest among all linear unbiased estimators.

Proof that OLS estimator is unbiased can be found here http://economictheoryblog.com/2015/02/19/ols_estimator/

and proof that $Var(\hat{\beta})$ is the smallest among all linear unbiased estimators can be found here http://economictheoryblog.com/2015/02/26/markov_theorem/

Answer 2

$$S_{xx} = \sum_{i = 1}^n (x_i - \bar{x})^2 = \sum_{i = 1}^n (x_i^2 - 2x_i\bar{x} + \bar{x}^2)$$ $$= \left(\sum_{i = 1}^n x_i^2\right) - n\bar{x}^2 = \left(\sum_{i = 1}^n x_i^2\right) - \bar{x}\sum_{i = 1}^nx_i = \sum_{i = 1}^n x_i^2 - \bar{x}x_i = \sum_{i = 1}^n (x_i - \bar{x})x_i$$

Answer 3

It seems you need to show:

$\sum_{i=1}^n(x_i-\bar{x})x_i=\sum_{i=1}^n(x_i-\bar{x})^2$

Try this: Expand the right hand side out into two terms, one of which is the left hand side.

Then just show the other term is zero.

This then shows you the manipulation required (though in reverse order) for your derivation.