# 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?

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*}

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.

• 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? – user24465 Oct 23 '19 at 14:03
• @user24465, on p.28 he says that linearity is a special case of CEF, and gives the equation in your question, together with $E(e|x)=0$. – GridAlien Oct 23 '19 at 14:09
• 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) – user24465 Oct 23 '19 at 14:59