Conditional Expectation, Orthogonality, and Correlation I know that if $\epsilon$ and $x$ are independent, then $E[\epsilon|x]=E[\epsilon]$ and Cov$(\epsilon,x)=0$. However, $E[\epsilon|x]=E[\epsilon]=0$ implies Cov$(\epsilon,x)=0$ iff $\epsilon$ and $x$ are orthogonal. In pg 45 of Greene's book (Econometric Analysis, 2008), he says: "Assumption…states that $\epsilon$ and $x$ are orthogonal," that is, "$E[\epsilon|x]=E[\epsilon]=0$." I don't see why that conditional expectation means the same thing as orthogonality. Could someone help me here? 
Thanks.
 A: 
The assumption that $x$ and $\epsilon$ are orthogonal, that is, that $\mathrm{Cov}(\epsilon,x)=0$, does not imply that $E(\epsilon|x)=E(\epsilon)$. Thus the quote in your question is indeed faulty (even assuming that $E(\epsilon)=0$).

Example: let $x$ uniform on $(-1,1)$ and $\epsilon=1-3x^2$, then $E(x)=E(\epsilon)=E(x\epsilon)=0$ hence $\mathrm{Cov}(\epsilon,x)=0$ while $E(\epsilon|x)=\epsilon\ne E(\epsilon)$.
A: Reediting my poor answer, to a hopefully improved one. 
I think your understanding the text of the theorem wrong, or tell me if I'm the one wrong.
Usually in econometrics, the reasonning goes like this:
$E(\epsilon_i|X)=0 \Rightarrow E(\epsilon_i)=0$
To be orthogonal, $\mathbf{X}'\mathbf{\epsilon}=0\Leftrightarrow \sum \mathbf{x}_i\epsilon_i=0\Rightarrow E(\mathbf{x}_i\epsilon_i)=0$, where we assume the terms in the summation respect the conditions of the WLLN. Having $E(\mathbf{x}_i\epsilon_i)=0$, when $E(\epsilon_i)=0$, is equivalent to having $Cov(x_{ik},\epsilon_i)=0=Cov(x_{i},\epsilon)$
The second equality is true, if they are identically distributed.
Check Fumio Hayashi Economtrics book.
P.S.: You may have better luck with an answer at https://stats.stackexchange.com/ for this type of statistical questions.
