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?


2 Answers 2


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.

  • $\begingroup$ 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
    Commented Oct 23, 2019 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
    Commented Oct 23, 2019 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
    Commented Oct 23, 2019 at 14:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.