# Cov(e,X) in the population regression

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.

• 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. Oct 15, 2021 at 20:40
• 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.
– FWL
Oct 15, 2021 at 21:08
• 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. Oct 15, 2021 at 21:36
• I agree with @Aksakal: you do not need, and do not use, OLS when you have the entire population. Oct 15, 2021 at 21:37
• How would one go around finding the $\beta$'s then?
– FWL
Oct 15, 2021 at 21:54

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.
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.