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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].

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  • $\begingroup$ 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
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[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:

enter image description here

Read more about this here.

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[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]

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