# conditional and interventional expectation

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.

• 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. – Noah May 5 '19 at 19:02

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 $$E[Y|do(X_1, X_2)]$$:

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.

• 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? – markowitz Sep 20 '19 at 12:41