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In the context of ordinary least squares model $y_i = \beta_1 + \beta_2 x_i + \varepsilon_i$ some authors assume $E(\varepsilon_i|x_i)=0$ and some authors assume $Cov(\varepsilon_i, x_i)=0$.

Is the assumption $E(\varepsilon_i|x_i)=0$ weaker than $Cov(\varepsilon_i, x_i)=0$?

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    $\begingroup$ Some explanation would be helpful, because in this context there are two possible distinct meanings of "$E[\varepsilon_i\mid x_i]:$" one is a conditional probability and the other is not. The answer might very well rest on that distinction. $\endgroup$
    – whuber
    Commented Nov 2, 2019 at 22:01
  • $\begingroup$ Yes sorry, the Expectation is supposed to represent conditional probability. In written form: The expectation of ei given xi is equal to zero $\endgroup$ Commented Nov 2, 2019 at 22:58
  • $\begingroup$ @whuber what is the other meaning other than conditional expectation? And, why are you and OP referring it as conditional probability, instead of expectation? $\endgroup$
    – gunes
    Commented Nov 3, 2019 at 6:39
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    $\begingroup$ @Gunes You're right: I should have written "conditional expectation" instead of "conditional probability." Thank you for clarifying that. The other interpretation of the notation is that $x_i$ is simply a number, not a random variable. The resulting expression is an expectation, not a conditional expectation. This is the usual model of any designed experiment, for instance, where $x_i$ is determined by the experimenter. $\endgroup$
    – whuber
    Commented Nov 3, 2019 at 13:21

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I'll use $x$ instead of $x_i$ and $u$ instead of $\varepsilon_i$.

No, $E(u|x)=0$ is stronger than $Cov(u, x)=0$. Here is the proof: $$ Cov(u, x) = E(ux) - E(u)E(x) = E(E(ux|x))-E(E(u|x))E(x)= E(xE(u|x)) - 0 = 0 $$ But $E(u|x) = 0$ is weaker than independence of $u$ and $x$. For counterexample consider $u$ that takes values $-1$, $1$, $2$, $-2$ with equal probabilities and $x=u^2$.

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  • $\begingroup$ You seem to equate zero covariance with "independence," but that's not correct. $\endgroup$
    – whuber
    Commented Nov 3, 2019 at 18:32
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    $\begingroup$ @whuber, no :) I just say that: independence of $u$ and $x$ implies $E(u|x)=0$, which in turn, implies $Cov(u, x)=0$. $\endgroup$
    – Roah
    Commented Nov 4, 2019 at 19:08
  • $\begingroup$ The OP (not me) says $E(u|x)=0$ is weaker than $Cov(u,x)=0$ and asks for explanations why. This claim is false, and I give a proof that zero covariance is weaker and follows from zero conditional expected value. $\endgroup$
    – Roah
    Commented Nov 4, 2019 at 19:10
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    $\begingroup$ I appreciate your comments. It would help to include those clarifications in the post itself, because in its current form it is ambiguous. $\endgroup$
    – whuber
    Commented Nov 4, 2019 at 19:27

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