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Say the population regression function is: $$ Y_i = \beta_0 + \beta_1 X_i + \varepsilon_i $$

(In the econometrics context) While I can't just assume that $E[\varepsilon_i | X_i] = 0$, can I not say that $Cov(\varepsilon_i, X_i) = 0$ just as an algebraic consequence of OLS?

Of course, OLS only that says $\widehat{Cov}(\varepsilon_i,X_i) = 0$, but assuming I'm running OLS on the entire population dataset, then this "sample" covariance is just the population covariance, right?

What am I missing here? Is it that we assume that there's a data generating process and thus I can't just say that $\varepsilon_i$ is the residual from OLS on the population dataset? Or is the error term in the population regression function not a residual from OLS but some structural error term of this model?

But if I run OLS on the entire population dataset, it is the residual.

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    $\begingroup$ The assumption regarding the errors $\varepsilon$, not residuals $\hat\varepsilon$, is just that, an assumption. You may or may not make it. It's either true or not. $\endgroup$
    – Aksakal
    Oct 15, 2021 at 20:40
  • $\begingroup$ Right, but are the errors $\varepsilon$ not defined as the residuals of OLS but on the population dataset (as opposed to a sample)? Then it should inherit all the properties of the OLS, including $Cov(\varepsilon_i,X_i) = 0$, without having to claim it as an assumption. $\endgroup$
    – FWL
    Oct 15, 2021 at 21:08
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    $\begingroup$ If you have the population then don’t assume or guess but measure and calculate. This way you’ll establish whether covariance is zero or not. $\endgroup$
    – Aksakal
    Oct 15, 2021 at 21:36
  • $\begingroup$ I agree with @Aksakal: you do not need, and do not use, OLS when you have the entire population. $\endgroup$
    – Sergio
    Oct 15, 2021 at 21:37
  • $\begingroup$ How would one go around finding the $\beta$'s then? $\endgroup$
    – FWL
    Oct 15, 2021 at 21:54

2 Answers 2

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The assumption that $Cov(X,\epsilon)=0$ is not even needed. If you replace it with the more intuitive and practically relevant assumption that $E(Y|X=x) = \beta_0+\beta_1 x$, then the covariance condition is automatically true. So instead of worrying about the covariance assumption, you can instead worry about whether the conditional mean function is truly linear.

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The assumption $\mathrm{cov}[X_i,\epsilon]=0$, like the assumption $E[\epsilon_i=0]$ is needed to identify the parameter being targeted by OLS with the structural parameters in the model.

Let's separate the two notationally to make it easier to compare. Suppose $X$ and epsilon are provided and Y generated by a process $Y\gets \gamma_0+\gamma_1 X+\epsilon$. If $E[\epsilon]=0$ and $\mathrm{cov}[X,\epsilon]=0$ then the OLS $\hat\beta_0$ estimates $\gamma_0$ and the OLS $\hat\beta_1$ estimates $\gamma_1$.

On the other hand, if $\epsilon= 1-X/10$, then $$Y=\gamma_0+\gamma_1X+\epsilon= (\gamma_0+1) + (\gamma_1-1/10)X$$ and the OLS estimates will be consistent (and unbiased conditional on $X$) for $\beta_0=\gamma_0+1$ and $\beta_1=\gamma_1-1/10$.

The constraints are on the border between assumptions and specification choices. In scenarios where we think of $\epsilon$ merely as the difference between $Y$ and $\hat Y$, then $E[\epsilon]=0$ and $\mathrm{cov}[X,\epsilon]=0$ are the choices that we make to identify $\beta_0$ and $\beta_1$. In the (less common) scenarios where $\epsilon$ is genuinely some sort of measurement error in $Y$ and there is a real reason to believe in linearity of the relationship, the constraints on $\epsilon$ are genuinely falsifiable assumptions about the measurement error process.

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