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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?

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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*}

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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.

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  • $\begingroup$ But section 2.8 is about the CEF, and in that section $E(e|x)=0$ is a statement of the CEF error, not the linear projection error. Does he mention the $E(e|x)=0$ condition for the linear projection error at any point before 4.4? $\endgroup$ – user24465 Oct 23 '19 at 14:03
  • $\begingroup$ @user24465, on p.28 he says that linearity is a special case of CEF, and gives the equation in your question, together with $E(e|x)=0$. $\endgroup$ – GridAlien Oct 23 '19 at 14:09
  • $\begingroup$ He says a linear CEF is a special case of a CEF, but is still talking about the CEF error in the section you reference. I'm asking about the linear projection error, for the general case (with an almost certainly non-linear CEF) $\endgroup$ – user24465 Oct 23 '19 at 14:59

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