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All I know is that we assume zero conditional mean (and hence zero mean) and conditional homoscedasticity (and hence homoscedasticity).

When trying to prove that $E[(\hat{\beta_1} - \beta_1)\bar{u}] = 0$, where $\beta_1$ is the slope in the linear regression model, $\hat{\beta_1}$ is its estimate and $\bar{u}$ is the average of the errors in the linear regression model (not the residuals!), I encountered:

$$E[(\hat{\beta_1} - \beta_1)\bar{u}|x]$$

$$\vdots$$

$$ = \frac{1}{n}\sum_{i=1}^{n} \frac{(x_i - \bar{x})}{SST_x} \color{red}{[\sum_{j=1}^{n} E[(u_i)u_j|x]]}$$

$$ = \frac{1}{n}\sum_{i=1}^{n} \frac{(x_i - \bar{x})}{SST_x} \color{red}{\sigma^2}$$

$$\vdots$$

$$ = 0 $$

$$\to E[(\hat{\beta_1} - \beta_1)\bar{u}] = 0$$

QED


What is the justification for that part? I tried:

For $i \ne j$, we have $E[(u_i)u_j|x] = Cov[u_i,u_j|x] + E[(u_i)|x]E[u_j|x] \stackrel{(*)}{=} 0 + (0)(0) = 0$

For $i = j$, we have $E[(u_i)u_j|x] = E[(u_i^2)|x] = Var[u_i|x] = \sigma^2$

Is $(*)$ right?

If so, what is the justification?

If not, how does one show that $E[u_i u_j | x] = 0$?


From Wooldridge:


enter image description here


This is from $(ii)$ of this exercise:


enter image description here

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  • $\begingroup$ The errors are assumed to be uncorrelated. $\endgroup$ – dsaxton Mar 19 '16 at 15:13
  • $\begingroup$ @dsaxton How do you know? It doesn't seem to be part of the assumptions of SLR $\endgroup$ – BCLC Mar 19 '16 at 15:14
  • $\begingroup$ Scroll down to assumptions: en.wikipedia.org/wiki/Linear_regression. If you didn't assume the errors were uncorrelated then how else would you conclude this? You can easily imagine a model satisfying all the other conditions where the errors are correlated. $\endgroup$ – dsaxton Mar 19 '16 at 15:19
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    $\begingroup$ Sometimes independence is stated as an assumption, but lack of correlation should always be. You can find it here as well: en.wikipedia.org/wiki/Gauss%E2%80%93Markov_theorem. If you already knew this I'm not sure why you'd ask for help in showing $\text{E}(u_i u_j) = 0$. $\endgroup$ – dsaxton Mar 19 '16 at 15:29
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    $\begingroup$ @BCLC independence follows from the random sampling assumption, see answer below $\endgroup$ – Carlos Cinelli Nov 22 '17 at 23:35
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The key thing here is that Wooldridge makes the assumption of random sampling.

Notice that since we have a random sample this means $(x_i, y_i) \perp (x_j, y_j)$ for $i \neq j$ which means that the single components of the pair are also independent, in particular $x_i \perp x_j$ (to see that just notice the joint is $p(x_i, y_i, x_j, y_j) = p(x_i, y_i)p (x_j, y_j)$ and marginalize over $y$).

This further implies $y_i \perp y_j |x_i, x_j$, since:

$$ p(y_i, y_j|x_i, x_j) = \frac{p(x_i, y_i, x_j, y_j)}{p(x_i, x_j)} = \frac{p(x_i, y_i)p (x_j, y_j)}{p(x_i)p(x_j)} = p(y_i|x_i)p(y_j|x_j) $$

But $y_i|x_i$ is nothing more than the disturbance $u_i$ plus a constant. Hence $u_i \perp u_j |x$ when assuming a random sample and $E[u_i u_j | x] =E[u_i|x] E[u_j | x] = 0$

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  • $\begingroup$ Oh I think I get it we just write $u_m=y_m-b0-b1x_m$ and then apply all that you said to show the penultimate equation you have? Also are introductory econometric students expected to get this? $\endgroup$ – BCLC Nov 25 '17 at 5:31
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    $\begingroup$ That’s one way to think about it. Regarding introductory students, not really, my experience is that people usually memorize the error terms are independent/uncorrelated with random sampling. $\endgroup$ – Carlos Cinelli Nov 25 '17 at 5:54

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