<p>[2] does not imply [1]. [2] and $E(u)=0$ imply $cov(u,x)=0$, which is about <strong>linear independence</strong>. [1] is stronger, as it refers to any type of dependence. </p>

<p>The classic counterexample to show this is $x=u^2$ over a symmetric domain. These are dependent yet linearly independent.</p>

<p>The R code below shows this:</p>

<pre><code>set.seed(1)
u &lt;- runif(100, min = -1, max = 1)
e &lt;- rnorm(100, mean = 0, sd = 0.1)
x &lt;- u^2 + e

plot(u,x)
abline(lm(x ~ u)) # Yields an R^2 of 0.006539

cov(u,x) # Yields 0.01206663
</code></pre>

<p>The plot is, where the black line represents the regression line:</p>

<p><a href="https://i.sstatic.net/xjjKs.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xjjKs.png" alt="enter image description here"></a></p>

<p>Read more about this <a href="https://en.wikipedia.org/wiki/Uncorrelated_random_variables" rel="nofollow noreferrer">here</a>.</p>

CC BY-SA 3.0
2017-10-02T19:00:50.757Z
2017-10-02T19:21:51.293Z
2017-10-02T19:21:51.293Z
luchonacho
<p>[1] is stronger than [2]. By applying the law of total probability expectation and the properties of conditional expectation, we have: $$E(ux)= E(E(ux|x)) =E(xE(u|x)) $$
Thus [1] implies [2] but [2] does not necessarily imply [1]</p>

CC BY-SA 3.0
2017-10-02T21:01:12.717Z
2017-10-02T21:01:12.717Z
HarmlessEcon
<p>I have a question regarding basic econometrics.
Consider the model 
$$y_i=\alpha +\beta x_i +u_i$$</p>

<p>I understand that assumption 4 of the linear regression model states</p>

<p>$$[1] \quad  E(u|x)=0$$</p>

<p>However, I often see this condition written as: </p>

<p>$$[2] \quad E(ux)=0$$</p>

<p>Are these two things equivalent? I see that if [1] and $E(u)=0$ then we get [2]; however I don't understand why [2] would imply [1]. </p>

CC BY-SA 3.0
2017-10-02T19:25:56.673Z
Note that [1] implies [2], *even without the condition $E(u)=0$*. In fact this condition is implied by [1]. Applying the law of iterated expectations to the LHS of [1], we get $E_x[E_u(u\vert x)]=E(u)$. But since $E_u(u\vert x)=0$ by [1], we have $E_x[E_u(u\vert x)]=E_u(u)=E_x(0)=0$
Herr K.
CC BY-SA 3.0
2017-10-03T16:36:49.897Z
I believe this is off topic as it belongs on [Cross Validated](https://stats.stackexchange.com).
Richard Hardy
<econometrics><statistics><regression>
Conditional independence and no correlation
Doon_Bogan
# Conditional independence and no correlation

(Asked by: username_0 on 2017-10-02)

I have a question regarding basic econometrics.
Consider the model
$$y\_i=\alpha +\beta x\_i +u\_i$$

I understand that assumption 4 of the linear regression model states

$$[1] \quad E(u|x)=0$$

However, I often see this condition written as:

$$[2] \quad E(ux)=0$$

Are these two things equivalent? I see that if [1] and $E(u)=0$ then we get [2]; however I don't understand why [2] would imply [1].

**Tags:** econometrics, statistics, regression
> **Comments:**
> **username_1**: Note that [1] implies [2], *even without the condition $E(u)=0$*. In fact this condition is implied by [1]. Applying the law of iterated expectations to the LHS of [1], we get $E_x[E_u(u\vert x)]=E(u)$. But since $E_u(u\vert x)=0$ by [1], we have $E_x[E_u(u\vert x)]=E_u(u)=E_x(0)=0$  
> **username_2**: I believe this is off topic as it belongs on [Cross Validated](https://stats.stackexchange.com).  

---

## 2 answers

### Answer by username_3 on 2017-10-02. Score: 1 | Accepted answer

[2] does not imply [1]. [2] and $E(u)=0$ imply $cov(u,x)=0$, which is about linear independence. [1] is stronger, as it refers to any type of dependence. 
The classic counterexample to show this is $x=u^2$ over a symmetric domain. These are dependent yet linearly independent.
The R code below shows this:
set.seed(1)
u <- runif(100, min = -1, max = 1)
e <- rnorm(100, mean = 0, sd = 0.1)
x <- u^2 + e

plot(u,x)
abline(lm(x ~ u)) # Yields an R^2 of 0.006539

cov(u,x) # Yields 0.01206663

The plot is, where the black line represents the regression line:

Read more about this here.


---

### Answer by username_4 on 2017-10-02. Score: 0

[1] is stronger than [2]. By applying the law of total probability expectation and the properties of conditional expectation, we have: $$E(ux)= E(E(ux|x)) =E(xE(u|x)) $$
Thus [1] implies [2] but [2] does not necessarily imply [1]


---

