# How are standard exogeneity assumptions and indepent of potential outcomes concepts linked?

$$y=x\beta +\eta$$

and assumed exogeneity, so $$E[\eta|x]$$=0, is the fact that x or treatment intensity is now uncorrelated with $$\eta$$ equivalent to saying that x is 'independent of potential outcomes?'

Yes, if we call the model

$$Y = X\beta + \eta$$

"structural" or "causal", we can use it to define potential outcomes as

$$Y(x) = x\beta + \eta.$$

(I'm using upper-case letters $$Y$$ and $$X$$ for random variables, and lower-case $$x$$ for a realization or fixed constant).

Then assuming $$E[\eta|X] = 0$$, it follows that

$$E[Y(x)|X] = x\beta + E[\eta|X] = x\beta,$$

which does not depend on $$X$$, so mean ignorability $$E[Y(x)|X] = E[Y(x)]$$ holds.

This "structural definition of counterfactuals" was proposed by Judea Pearl, see for example his book "Causality", or his book with Jewell and Glymour, "Causality: A Primer".

• Thanks! just one quick follow up, when people use the terminology 'structural' , that is synonymous with saying it just some true theoretical population relationship, correct? Aug 21, 2020 at 2:14
• Not quite, the conditional mean function E[Y|X] is also a population concept, but not necessarily structural. One definition of a structural model is that the model describes how $Y$ changes when $X$ is controlled physically. This entails that the value of the structural $\beta$ is "invariant" to changes in the distribution of $X$, while a regression coefficient would not be (if the regression coefficient is equal to $cov(X, Y)/var(X)$, this will change when the distribution of $X$ changes). More generally, "structural" is almost synonymous with "theoretical" (especially in economics). Aug 21, 2020 at 6:54
• @Steve Infact the exogeneity condition must be related to the structural equation and not to the regression one. If population or only a sample are available do not matters. Frequently this point is misunderstood. Read here can help: stats.stackexchange.com/questions/262609/… Aug 26, 2020 at 5:52
• @JulianSchuessler Hello, according to your derivation, E(Y(x)|X) = E(Y(x)) = Y(x). Is this right? And how does this have any relationship with ignorability, which I understand is that Y(x) is independent of X given some confounders (eta)?
– wut
Nov 15, 2022 at 6:24
• Yes, E[Y(x)|X] = E[Y(x)] as shown, but $E(Y(x)) = x\beta$ whereas $Y(x) = x\beta + \eta$, so these are not the same. Ignorability mathematically is Y(x) \indep X, which implies $E(Y(x)|X] = E[Y(x)]$. Jan 18 at 9:41