3

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

BCLC
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  • The errors are assumed to be uncorrelated. – dsaxton Mar 19 '16 at 15:13
  • @dsaxton How do you know? It doesn't seem to be part of the assumptions of SLR – BCLC Mar 19 '16 at 15:14
  • Scroll down to assumptions: https://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. – dsaxton Mar 19 '16 at 15:19
  • @dsaxton I know it's uncorrelated, mean independent or independent. I wanted to know which specifically it was. 'Independence of errors. This assumes that the errors of the response variables are uncorrelated with each other.' Should I find it strange that the book doesn't include that? – BCLC Mar 19 '16 at 15:21
  • @dsaxton Wait so is it independent? Or just uncorrelated? – BCLC Mar 19 '16 at 15:21
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    Sometimes independence is stated as an assumption, but lack of correlation should always be. You can find it here as well: https://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 $\text{E}(u_i u_j) = 0$. – dsaxton Mar 19 '16 at 15:29
  • @dsaxton Oh sorry I mean I knew it was uncorrelated or something else based on what I read before, but this particular book doesn't seem to have anything. Might there be a way of deducing uncorrelatedness from another assumption here? – BCLC Mar 19 '16 at 17:39
  • As I commented before, uncorrelatedness doesn't follow from the other assumptions, otherwise it wouldn't need to be stated as an additional assumption as it is. – dsaxton Mar 19 '16 at 23:10
  • @dsaxton So Wooldridge's assumptions are incomplete? – BCLC Mar 20 '16 at 00:05
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    @BCLC independence follows from the random sampling assumption, see answer below – Carlos Cinelli Nov 22 '17 at 23:35

1 Answers1

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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 $E[u_i u_j | x] =E[u_i|x] E[u_j | x] = 0$

Carlos Cinelli
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  • 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? – BCLC Nov 25 '17 at 05:31
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    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. – Carlos Cinelli Nov 25 '17 at 05:54
  • Actually, you have to be a little more precise about what "random sample" actually means. In finite populations, elements of simple random samples are not independent. So you really do have to assume independence, unless "random sampling" has previously been defined as "independent sampling." – BigBendRegion Oct 18 '20 at 17:02
  • @CarlosCinelli, could you clarify the notation in the last paragraph? Is the x without a subscript a vector, i.e., x = (x_i, x_j)? And if other covariates were included, say, z_i, would it be appropriate to write E(u_i, u_j | x_i, x_j, z_i, z_j) = 0? – hendogg87 Jun 11 '21 at 08:27