What is the difference between strict / strong and weak exogeneity

Question

Let be two variables $y$ and $x$, the latter being expected to be a cause of the latter. If we suppose linearity, we can set up a model:

$$y=\beta_0+\beta_1x+u$$

Where $\beta_0$ and $\beta_1$ are coefficients to be determined, and $u$ is a random noise ($\mathbb{E}[u]=0$) representing all other effects on $y$ not mediated through $x$.

If we make an so-called "exogeneity" hypothesis, we can interpret $\beta_1$ as the effect of $x$ on $y$. (Intuitively, exogeneity means that we can only estimate the effect of $x$ on $y$ if there is no spurious causal pathways, like a third variable $z$ causing both $x$ and $y$, messing with the estimation.) However, there are two versions of exogeneity, weak and strong, and I cannot make my mind around the intuitive difference between the two.

• Weak exogeneity is $\mathbb{E}[ux]=0$ ("null correlation") ;

• Strong or strict exogeneity is $\mathbb{E}[u|x]=0$ ("zero conditionnal mean").

I know that $\mathbb{E}[u|x]=0$ implies $\mathbb{E}[ux]=0$, since: $$\mathbb{E}_{x, u}[u\cdot x]=\mathbb{E}_x[\mathbb{E}_u[u \cdot x|x]]=\mathbb{E}_x[x\cdot\mathbb{E}_u[u|x]]=\mathbb{E}_x[x\cdot 0]=0$$

... but what does it intuitively mean?

Also, if I have a third variable $z$, what do these conditions become? I usually read something like $\mathbb{E}[u|x,z]=0$ for strong exogeneity. Is this stronger than $\mathbb{E}[u|x]=0$ and $\mathbb{E}[u|z]=0$ holding simultaneously or is that equivalent? And week exogeneity? Is it only $\mathbb{E}[ux]=0$ and $\mathbb{E}[uz]=0$ ? Do hypotheses like $\mathbb{E}[uz|x]=0$ make sense? And if it does, what does it intuitively mean? (For context, I am only interested in correctly estimating $\beta_1$ and I actually don't care about $z$ except for its confounding effects on $x$ and $y$.)

Answer 1

As far as I understood, the strict exogeneity implies no peer effects. Exogeneity means that the independent variable belonging to an individual chosen from a sample is orthogonal to that individual's error term in expectation. When it is strict, that error term is ortagonal to every other individual's error term in expectation. If there the element i and j of the sample are correlated in a way that is not contained in the independent variables of the model, then it is called a peer effect and it implies that the strict exogeneity is violated.

Answer 2

$E(xu)=0$, simply rules out the possibility that $x$ and $u$ are correlated. While the conditional zero mean assumption, $E(u|x)=0$, implies, among others, that for any function $g$ (with some restrictions) we have $E(u|g(x))=0$; this is practically of interest, since with this assumption you may include $x^2$, say, as a regressor and still have OLS providing you unbiased estimates.

When $E(xu)=0$, then under OLS assumptions for $y=b_0+b_1x+u$ you obtain unbiased estimates for model parameters, but not necessarily for $y=b_0+b_1x+b_2x^2+u$, say.

Regarding your second question $E(u|x,z)=0$ is not equivalent to $E(u|z)=0$ and $E(u|x)=0$, but both imply zero correlation between $z$ and $u$ and also $x$ and $u$.