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:

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


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

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


$$ = 0 $$

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


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

  • $\begingroup$ The errors are assumed to be uncorrelated. $\endgroup$
    – dsaxton
    Commented Mar 19, 2016 at 15:13
  • $\begingroup$ @dsaxton How do you know? It doesn't seem to be part of the assumptions of SLR $\endgroup$
    – BCLC
    Commented Mar 19, 2016 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
    Commented Mar 19, 2016 at 15:19
  • 1
    $\begingroup$ 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
    Commented Mar 19, 2016 at 15:29
  • 1
    $\begingroup$ @BCLC independence follows from the random sampling assumption, see answer below $\endgroup$ Commented Nov 22, 2017 at 23:35

1 Answer 1


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

  • $\begingroup$ 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
    Commented Nov 25, 2017 at 5:31
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 25, 2017 at 5:54
  • $\begingroup$ 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." $\endgroup$ Commented Oct 18, 2020 at 17:02
  • $\begingroup$ @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? $\endgroup$
    – hendogg87
    Commented Jun 11, 2021 at 8:27

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