Difference between $Cov(x, y)$ and $Cov(x_i, y_i)$

Question

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?

Answer 1

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.