<p><span class="math-container">$E(\mathbf{x}e)=0$</span> does not imply <span class="math-container">$E(e|\mathbf{x})=0$</span>, to see that, suppose that x is a variable that always takes the value of zero, but <span class="math-container">$e$</span> is independent of <span class="math-container">$\mathbf x$</span> and has a mean of, say, <span class="math-container">$3$</span>. Then the expected value of the product is necessarily <span class="math-container">$0$</span>, but the conditional expectation will be equal to 3 (given independence, the conditional mean is equal to the unconditional mean).</p>

CC BY-SA 4.0
2019-10-23T18:48:41.063Z
2019-10-23T18:48:41.063Z
Regio
<p>This question comes out of Hansen's Econometrics ((<a href="https://www.ssc.wisc.edu/~bhansen/econometrics/Econometrics.pdf" rel="nofollow noreferrer">https://www.ssc.wisc.edu/~bhansen/econometrics/Econometrics.pdf</a>))</p>

<p>In section 2.18, we only impose the assumptions of finite variance and <span class="math-container">$Q_{xx}$</span> being positive definite, and then derive the linear projection coefficient <span class="math-container">$\beta = E(\textbf{x} \textbf{x}^\prime)^{-1} E(\textbf{x} y)$</span> as the minimizer of the expected squared error of the linear projection model <span class="math-container">$y = \textbf{x}^\prime {\beta} + e$</span>. This leads to the implication (NOT the assumption) that <span class="math-container">$E(\textbf{x} e) = \textbf{0}$</span>. </p>

<p>Then in section 4.4, under assumption 4.2, we have: <span class="math-container">$E(e_i | \textbf{x}_i)=0$</span>.</p>

<p>My question is: is <span class="math-container">$E(e_i | \textbf{x}_i)=0$</span> a newly-imposed assumption in chapter 4? Or is it equivalent to the condition <span class="math-container">$E(\textbf{x} e) = \textbf{0}$</span> which we reached in chapter 2, not by assumption but by implication?</p>

<p>I know that <span class="math-container">$E(\textbf{x} e) = E(E(\textbf{x} e|\textbf{x})) = E(\textbf{x} E(e|\textbf{x}))$</span>
 (by law of iterated expectations and conditioning theorem, respectively), so imposing the assumption that <span class="math-container">$E(e|\textbf{x}) = \textbf{0}$</span> yields the implication that <span class="math-container">$E(\textbf{x} e) = \textbf{0}$</span>. But is the reverse true? Does <span class="math-container">$E(e_i | \textbf{x}_i)=0$</span> follow from <span class="math-container">$E(\textbf{x} e) = \textbf{0}$</span>, or is <span class="math-container">$E(e_i | \textbf{x}_i)=0$</span> a new assumption that we're imposing?</p>

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2019-10-23T11:19:54.013Z
maybe this is better placed at https://stats.stackexchange.com/
E. Sommer
<regression>
Is conditional mean independence, $E(e_i | x_i)=0$, a different assumption from $E(x e) = 0$? (from Hansen)
user24465
# Is conditional mean independence, $E(e_i | x_i)=0$, a different assumption from $E(x e) = 0$? (from Hansen)

(Asked by: username_0 on 2019-10-23)

This question comes out of Hansen's Econometrics ((<https://www.ssc.wisc.edu/~bhansen/econometrics/Econometrics.pdf>))

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?

**Tags:** regression

> **Comments:**
> **username_1**: maybe this is better placed at https://stats.stackexchange.com/  

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## 1 answers

### Answer by username_2 on 2019-10-23. Score: 1

$E(\mathbf{x}e)=0$ does not imply $E(e|\mathbf{x})=0$, to see that, suppose that x is a variable that always takes the value of zero, but $e$ is independent of $\mathbf x$ and has a mean of, say, $3$. Then the expected value of the product is necessarily $0$, but the conditional expectation will be equal to 3 (given independence, the conditional mean is equal to the unconditional mean).


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