If we had a model:

$y=x\beta +\eta$

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?'


1 Answer 1


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".

  • 1
    $\begingroup$ 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? $\endgroup$
    – Steve
    Aug 21, 2020 at 2:14
  • 1
    $\begingroup$ 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). $\endgroup$ Aug 21, 2020 at 6:54
  • $\begingroup$ @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/… $\endgroup$
    – markowitz
    Aug 26, 2020 at 5:52
  • $\begingroup$ @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)? $\endgroup$
    – wut
    Nov 15, 2022 at 6:24
  • $\begingroup$ 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 $E(Y(x)|X] = E[Y(x)]$. $\endgroup$ Jan 18 at 9:41

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