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].

  • $\begingroup$ 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 $E_x[E_u(u\vert x)]=E_u(u)=E_x(0)=0$ $\endgroup$ – Herr K. Oct 2 '17 at 19:25
  • $\begingroup$ I believe this is off topic as it belongs on Cross Validated. $\endgroup$ – Richard Hardy Oct 3 '17 at 16:36

[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:

u <- runif(100, min = -1, max = 1)
e <- rnorm(100, mean = 0, sd = 0.1)
x <- u^2 + e

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:

enter image description here

Read more about this here.


[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]


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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