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While reviewing the basics of ordinary least squares (OLS) regression in a matrix context, I stumbled upon the variance covariance matrix of the residuals. In an OLS context, the population model is usually assumed to be (for a cross-section of data):

$ \begin{aligned} & y_i = \textbf{x}_i' \beta + u_i. \end{aligned} $

$y$ is then an n$\times$1 vector with the dependent variable, $\textbf{X}$ is an n$\times$k matrix of exogenous variables and $u$ is the n$\times$1 vector of residuals. In order to derive the distribution of the OLS estimator, the following proposition about the residuals is made by Cameron & Trivedi:

Proposition: Data are independent over $i$ with $\text{E}\left({\textbf{u}|\textbf{X}}\right) = \textbf{0} $ and $\text{E}\left({\textbf{uu'}|\textbf{X}}\right) = \bf{\Omega}$$ = $ $\text {Diag}\left({\sigma_i^2}\right)$
Cameron, Trivedi (2005): p. 73

So far, I only ever thought about homoskedasticity, which implies that $\text{E}(u^2) = \sigma^2$. But Cameron & Trivedi make their proposition more general, in that they also allow for heteroskedastic errors with variance $\sigma_i^2$ for different $i$. I knew the concepts of homoskedasticity and heteroskedasticity before, but the way that Cameron & Trivedi frame this proposition makes you think about the distribution of the errors in a matrix way. Clearly, $\bf{\Omega}$ is an $n \times n$ matrix, but when I thought about it, I had troubles understanding the single entries in this variance covariance matrix.

Looking at the diagonal elements of $\bf{\Omega}$: How does $u_1$, for instance, have a variance? We are in a cross-sectional setting, so $u_1$ is essentially one observation point. So how do I obtain the variance for a variable with one observation? The same goes for the off-diagonal elements: How come there exists a covariance for 2 single observations?


Note:

The source is:
Cameron, Trivedi (2005): Microeconometris: Methods and Applications.

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How does (a point) $u_{i=1,...,n}$ have a variance?

For sure this can be really confusing at first.

But actually, what you need to understand is that $u_1$ is likely to have many other possible values, and thus a variance. And I am not talking about other components of the $n \times 1$ vector $\boldsymbol{u}$, but really about the variance of each of its $n$ components.

If you look at the definition of the covariance matrix, you will see that the covariance between two components of $\boldsymbol{u}$, say the $i$th and the $j$th ones, is

$\mathrm{E}(\boldsymbol{u}\boldsymbol{u}') = [\mathrm{cov}(u_i,u_j)] = [\mathrm{E}(u_i u_j) - \mathrm{E}(u_i) \mathrm{E}(u_j)]$

where $\mathrm{E}(u_i)$ stands for the average of all the values that $u_i$ can have, classically equal to $0$. Idem for $\mathrm{E}(u_j)$. And $\mathrm{E}(u_i u_j)$ stands for the average of all the values that their product can have ! Incidentally, the variance is a particular case of covariance in which $i=j$.

Thus, to conclude, for sure you only get point values of $u_i \forall i\in [1,n]$, but each of this point is actually randomly picked up given an underlying distribution, be it empirically observed/computed (computed using bootstrapping or Bayesian technics) or theoretically derived/assumed.

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  • $\begingroup$ Thanks very much! After reading your answer, I realized that my confusion was caused by an incomplete understanding of the basics of OLS regression. So the question should actually be more general and not refer to spatial econometrics in particular, but to OLS. I will edit it accordingly. $\endgroup$ – Tartan Leaves Jul 9 '17 at 18:56
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    $\begingroup$ I changed it now. I think your answer should still fit the question perfecty fine. $\endgroup$ – Tartan Leaves Jul 9 '17 at 20:15

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