Conditional expectation $E[Y|X]$ and interventional expectation $E[Y|do(X)]$ are related but conceptually very different things.

I know that if $X$ is a randomly assigned by an experiment, we have that $E[Y|X]=E[Y|do(X)]$ In some other case we can achieve the equivalence by conditioning on proper set of variables $Z$: $E[Y|X,Z]=E[Y|do(X)]$

My question: is possible to consider both $X$ and $Z$ as vector of variables? Usually in an experiment we are focused in just one causal variable ($X$ as scalar), however logically seems me that the generalization is permitted.

  • $\begingroup$ Definitely. There is a literature on causal interaction, which concerns two treatments (i.e., $X$s). VanderWeele (2009) is a decent paper on this issue. There is also a literature on sequential treatments and mediation, which consider the outcomes were one to experimental intervene on a sequence of treatments. VanderWeele (2009) (different paper) is another decent paper on this issue. Note these use potential outcomes, which are similar to do(x) operations. $\endgroup$
    – Noah
    May 5, 2019 at 19:02

1 Answer 1


Yes, you can consider $X$ and $Z$ to be arbitrary vectors of variables. The identification problem of expressions of the type $E[Y|do(X)]$ and $E[Y|do(X), Z]$ for arbitrary vectors of variables $X$ and $Z$ has been solved for nonparametric models using the do-calculus (via the ID-algorithm).

For instance, in the model below, suppose you are interested in identifying $E[Y|do(X_1, X_2)]$:

enter image description here

This is given by (here you can just use the truncated factorization formula):

$$ E[Y|do(X_1, X_2)] = \sum_{Z_1, Z_2} P(Y|X_1, X_2, Z_2) P(Z_2|X_1,Z_1) P(Z_1) $$

Or equivalently, using inverse probability weights:

$$ E[Y|do(X_1, X_2)] = \sum_{Z_1, Z_2} \frac{P(Y, X_1, X_2, Z_1, Z_2)}{P(X_2|X_1, Z_1, Z_2)P(X_1|Z_1)} $$

The R package causaleffect has several of the existing identification algorithms implemented.

  • $\begingroup$ I know that the backdoor criterion permit us to find several sets $Z$ that permit to identify the ACE of $X$ on $Y$; then $E[Y|X,Z]=E[Y|do(X)]$. Now, backdoor criterion is applicable even if $X$ is a vector? $\endgroup$
    – markowitz
    Sep 20, 2019 at 12:41
  • $\begingroup$ If the SCM in your example was a linear system, what regression I have to compute in practice for identify $E[Y|do(X_1,X_2)]$? In other words, what regression is associated with the first equation you wrote if any? $\endgroup$
    – markowitz
    Oct 5, 2021 at 9:03

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