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I have seen the Gauss Markov assumption of uncorrelated dependent variables and error term presented in three different ways. I want to make sure that I am correctly interpreting the underlying mathematics.

The three different presentations:

$$cov(x_i, U)=0$$

$$cov(x_i, u_i)=0$$

and

$$cov(X, U)=0$$

As usual x is the explanatory variable, and u the error term. Just to be clear, the "i" on the equations must denote variables and not observations, as it usually denotes (correct?). Covariance between single observations does not make any sense as a concept. So for multiple equation system with multiple explanatory variables we have $cov(x_i, u_i)=0$, which is the same as $cov(X, u_i)=0 = cov(X,U)$. In other words the conditions are exactly the same. Is my understanding here correct?

kjetil b halvorsen
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Dole
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    Re "covariance between single observations does not make any sense," please see the discussion at http://stats.stackexchange.com/questions/179346/covariance-of-non-random-vectors-equal-to-zero/181275#181275 . – whuber Mar 10 '16 at 22:19
  • @whuber Interesting, but if it indeed is the observation, how come the different definitions? – Dole Mar 11 '16 at 09:30
  • I'm unsure what you are referring to: different definitions of what? – whuber Mar 11 '16 at 14:03
  • @whuber Apologies, I have observed the three different formulations of Gauss Markov assumptions in different sources. As explained, since there is a difference between $cov()_ii$(cov for each observation) and $cov()$(cov for all observations), is there a mistake on some papers I am reading?

    (This also relates to a question I asked before): http://stats.stackexchange.com/questions/198638/are-gauss-markov-assumptions-ee-ix-i-0-and-covx-i-e-j-0-equivalent

     – Dole Mar 11 '16 at 14:36
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    I doubt there is a mistake: you need to read each paper on its own terms, using its own definitions and explanations of its notation. Expositions may differ in whether they view the regressors $X$ as being fixed or being random variables. – whuber Mar 11 '16 at 15:21
  • Assuming that $U$ is a vector of the $u_{i}$ and that $X$ is the vector of the $x_{i}$ ( where $u_{i}$ and $x_{i}$ are assumed to be scalars ) then the RHS of the first presentation is a zero vector, the second is a zero scalar and the third one is a zero matrix. But, the third one is really just the most general representation so that you don't need to say "for every i = 1 to whatever". So, they're not the same thing but close to it AFAIK. – mlofton Sep 04 '18 at 08:10

1 Answers1

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This is a matter of non-fully standardized notation. So one has to pay attention for how the notation is used in each study.

E.g.

Suppose that the authors are strict in using uppercase letters for the random variables, lowercase letters for the realization of the random variables. In that case, $X$ denotes the r.v. in abstract, $X_i$ denotes the r.v. indexed (for some reason) by $i$, $x$ denotes a value in the support of the r.v. $X$, and $x_i$ denotes a realization of $X$ indexed by $i$. Then, any covariance expression that includes lowercase letters is by construction zero $$ {\rm Cov}(x_i,U) = {\rm Cov}(x,U) = 0$$ because fixed values/realizations, are, exactly, fixed and not varying.

But is is also common to see studies using lowercase for the random variable itself, in which case ${\rm Cov}(x_i,u_i)$ would represent the covariance between two random variables, and then it is not necessarily zero, it will be an assumption or a result.

Alecos Papadopoulos
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