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

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 $$Cov(\varepsilon_i, x_i)=0$$?

• 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.
– whuber
Commented Nov 2, 2019 at 22:01
• Yes sorry, the Expectation is supposed to represent conditional probability. In written form: The expectation of ei given xi is equal to zero Commented Nov 2, 2019 at 22:58
• @whuber what is the other meaning other than conditional expectation? And, why are you and OP referring it as conditional probability, instead of expectation? Commented Nov 3, 2019 at 6:39
• @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.
– whuber
Commented Nov 3, 2019 at 13:21

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 $$x=u^2$$.

• You seem to equate zero covariance with "independence," but that's not correct.
– whuber
Commented Nov 3, 2019 at 18:32
• @whuber, no :) I just say that: independence of $u$ and $x$ implies $E(u|x)=0$, which in turn, implies $Cov(u, x)=0$.
– Roah
Commented Nov 4, 2019 at 19:08
• 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.
– Roah
Commented Nov 4, 2019 at 19:10
• I appreciate your comments. It would help to include those clarifications in the post itself, because in its current form it is ambiguous.
– whuber
Commented Nov 4, 2019 at 19:27