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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.