5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '19 at 19:02
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '19 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.