3
$\begingroup$

I know there's a similar post about this, but I believe my question is a bit different.

In my textbook the author rewrites

$-2(\hat{\beta}_1-\beta_1)\sum u_i (x_i-\bar{x})$

into

$-2(\hat{\beta}_1-\beta_1)^2\sum (x_i-\bar{x})^2$

He doesn't use any expectation or variance operator.

If you don't want to go through my whole calculation, then you can check my final result:

$-2(\hat{\beta}_1-\beta_1)((\hat{\beta}_1-\beta_1)\sum(x_i-\bar{x})^2 +\sum\hat{u}_i(x_i-\bar{x}))$

How can I get rid of the last term?

I started the following:

use

$u_i=y_i-\beta_0-\beta_1x_i$

and substitute into first equation:

$-2(\hat{\beta}_1-\beta_1)\sum (y_i-\beta_0-\beta_1x_i) (x_i-\bar{x})$

use

$y_i=\hat{y}_i+\hat{u}_i$ -> $y_i=\hat{\beta}_0+ \hat{\beta}_1 x_i+\hat{u}_i$

and substitute into the following equation

$-2(\hat{\beta}_1-\beta_1)\sum (y_i-\beta_0-\beta_1x_i) (x_i-\bar{x})$

->

$-2(\hat{\beta}_1-\beta_1)\sum ((\hat{\beta}_0+ \hat{\beta}_1 x_i+\hat{u}_i)-\beta_0-\beta_1x_i) (x_i-\bar{x})$

simplifying the equation above gives

$-2(\hat{\beta}_1-\beta_1)\sum (\hat{\beta}_0+ \hat{\beta}_1 x_i+\hat{u}_i-\beta_0-\beta_1x_i) (x_i-\bar{x})$

multiply with $(x_i-\bar{x})$

$-2(\hat{\beta}_1-\beta_1)\sum (\hat{\beta}_0(x_i-\bar{x})+ \hat{\beta}_1 x_i(x_i-\bar{x})+\hat{u}_i(x_i-\bar{x})-\beta_0(x_i-\bar{x})-\beta_1x_i(x_i-\bar{x}))$

set brackets before summation operator

$-2(\hat{\beta}_1-\beta_1)(\sum (\hat{\beta}_0(x_i-\bar{x})+ \hat{\beta}_1 x_i(x_i-\bar{x})+\hat{u}_i(x_i-\bar{x})-\beta_0(x_i-\bar{x})-\beta_1x_i(x_i-\bar{x})))$

simplify further

$-2(\hat{\beta}_1-\beta_1)((\hat{\beta}_0\sum(x_i-\bar{x})+ \hat{\beta}_1 \sum x_i(x_i-\bar{x})+\sum\hat{u}_i(x_i-\bar{x})-\beta_0\sum(x_i-\bar{x})-\beta_1\sum x_i(x_i-\bar{x})))$

$\sum(x_i-\bar{x})$ equals zero, therefore

$-2(\hat{\beta}_1-\beta_1)(\hat{\beta}_1 \sum x_i(x_i-\bar{x})+\sum\hat{u}_i(x_i-\bar{x})-\beta_1\sum x_i(x_i-\bar{x}))$

use $\sum x_i(x_i-\bar{x}) = \sum(x_i-\bar{x})^2$

$-2(\hat{\beta}_1-\beta_1)(\hat{\beta}_1 \sum(x_i-\bar{x})^2 +\sum\hat{u}_i(x_i-\bar{x})-\beta_1\sum(x_i-\bar{x})^2))$

factorize $\sum(x_i-\bar{x})^2$

$-2(\hat{\beta}_1-\beta_1)((\hat{\beta}_1-\beta_1)\sum(x_i-\bar{x})^2 +\sum\hat{u}_i(x_i-\bar{x}))$

How do I get rid now of the last term?

$\endgroup$
4
$\begingroup$

We want to show that $\sum_{i=1}^{n} (y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i) (x_i - \bar{x}) = 0$, which is the same as $\sum_{i=1}^{n} (\hat{\beta}_0 + \hat{\beta}_1 x_i ) (x_i - \bar{x}) = \sum_{i=1}^{n} y_i (x_i - \bar{x})$. This roughly says that the weighted average of fitted $y$ values equals the weighted average of actual $y$ values, using weights $x_i - \bar{x}$. For this we just need to do some algebra and remember the definitions $\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x}$ and $\hat{\beta}_1 = \sum_{i=1}^{n} (y_i - \bar{y}) (x_i - \bar{x}) / \sum_{i=1}^{n} (x_i - \bar{x})^2$. Let's start with the left hand side

\begin{align} \sum_{i=1}^{n} (\hat{\beta}_0 + \hat{\beta}_1 x_i ) (x_i - \bar{x}) &= \sum_{i=1}^{n} (\bar{y} - \hat{\beta}_1 \bar{x} + \hat{\beta}_1 x_i ) (x_i - \bar{x}) \\ &= \bar{y} \sum_{i=1}^{n} (x_i - \bar{x}) + \hat{\beta}_1 \sum_{i=1}^{n} (x_i - \bar{x})^2 . \end{align}

We know the first term is zero, and the sum of squares $\sum_{i=1}^{n} (x_i - \bar{x})^2$ cancels with the denominator of $\hat{\beta}_1$ leaving us with just $\sum_{i=1}^{n} (y_i - \bar{y}) (x_i - \bar{x}) = \sum_{i=1}^{n} y_i (x_i - \bar{x})$, which is what we wanted to show.

$\endgroup$
0
$\begingroup$

I believe I found another way to show that the last term equals zero.

Multiply the last term with $\hat{u}_i$:

$\sum \hat{u}_ix_i-\hat{u}_i\bar{x}$

The first term equals zero because the $Cov(\hat{u}_i,x_i)$ is zero according to the first order condition of $\hat{\beta}_1$ when deriving the OLS estimates.

The second term simplified with the summation operator gives

$-\bar{x}\sum \hat{u}_i$

According to first order condition of $\hat{\beta}_0$ the sum of the residuals equals zero.

$\endgroup$
  • 2
    $\begingroup$ This seems fairly circular to me. If we're allowing ourselves to simply assume $x_i$ and $u_i$ are uncorrelated then there's nothing to prove. The whole point is to show that they're uncorrelated. $\endgroup$ – dsaxton Sep 24 '16 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.