Is conditional mean independence, $E(e_i | x_i)=0$, a different assumption from $E(x e) = 0$? (from Hansen) This question comes out of Hansen's Econometrics ((https://www.ssc.wisc.edu/~bhansen/econometrics/Econometrics.pdf))
(Following Hansen's notation of using $e$ to denote errors, $\hat{e}$ to denote residuals)
In section 2.18, we only impose the assumptions of finite variance and $Q_{xx}$ being positive definite, and then derive the linear projection coefficient $\beta = E(\textbf{x} \textbf{x}^\prime)^{-1} E(\textbf{x} y)$ as the minimizer of the expected squared error of the linear projection model $y = \textbf{x}^\prime {\beta} + e$. This leads to the implication (NOT the assumption) that $E(\textbf{x} e) = \textbf{0}$. 
Then in section 4.4, under assumption 4.2, we have: $E(e_i | \textbf{x}_i)=0$.
My question is: is $E(e_i | \textbf{x}_i)=0$ a newly-imposed assumption in chapter 4? Or is it equivalent to the condition $E(\textbf{x} e) = \textbf{0}$ which we reached in chapter 2, not by assumption but by implication?
I know that $E(\textbf{x} e) = E(E(\textbf{x} e|\textbf{x})) = E(\textbf{x} E(e|\textbf{x}))$
 (by law of iterated expectations and conditioning theorem, respectively), so imposing the assumption that $E(e|\textbf{x}) = \textbf{0}$ yields the implication that $E(\textbf{x} e) = \textbf{0}$. But is the reverse true? Does $E(e_i | \textbf{x}_i)=0$ follow from $E(\textbf{x} e) = \textbf{0}$, or is $E(e_i | \textbf{x}_i)=0$ a new assumption that we're imposing?
 A: We have
$$
\mathbb{E}(X \mid Y) = 0 \Rightarrow \mathbb{E}(XY) = 0 
$$
since
$$
\mathbb{E} (XY) = \mathbb{E} \big (   Y \mathbb{E}(X\mid Y) \big) = 0
$$
But we do not have
$$
\mathbb{E}(XY) = 0 \Rightarrow \mathbb{E}(X \mid Y) = 0 
$$
Take for example $X \perp Y$ with $\mathbb{E}(X) > 0$ and $\mathbb{E}(Y) =0$.
Then 
\begin{align*}
\mathbb{E}(X Y) &= \mathbb{E}(X )\mathbb{E}(Y) \\
&=0
\end{align*}
but
\begin{align*}
\mathbb{E}(X \mid Y) &= \mathbb{E}(X) \\
&>0
\end{align*}
A: It's not a new assumption.
On page 21 (Section 2.8) of the book you linked the author goes on to show $E(e|x) = 0$ and that this implies $E(e)=0$. From there, all you have to do is apply it to individual data points and errors: $E(e_i | x_i) = 0$. 
It's essentially the same thing as saying $E(x_i) = E(x)=$ μ. 
However, you cannot get $E(e_i|x_i)=0$ from $E(xe) = 0$. 
To do this, you need to show that $x$ and $e$ are independent. You need $E(e|x)=0$ and $E(e)=0$ to do this. 
TLDR; not a new assumption for chapter 4, but it does come from someplace different. 
