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I have been trying to get my head around using do-calculus in causal inference, and I've run into a little toy problem that's confused me and my classmates.

Imagine you have a DAG for some causal model that is defined by $(X \rightarrow M, M \rightarrow Y, U \rightarrow X, U \rightarrow Y)$, where $M$ is a mediator, $X$ is a treatment, $Y$ is an outcome and $U$ is unobserved confounders. I want to calculate $P(Y|do(M=m))$.

As I understand it, this amounts to cutting the link $ X \rightarrow M$, so the new DAG lacks that chain. Then there are no backdoor paths from $M \rightarrow Y$, so you can get the effect of the intervention trivially easily with $P(Y|do(M=m)) = P(Y|M=m)$. My classmate thinks that you still need to account for values of $X$ though, so you need to adjust for those and it should actually be $P(Y|do(M=m))=\Sigma_{X'} P(Y | M=m, X')P(X')$.

Which of us is right? My answer seems too simple, but I just don't understand why you need to condition for $X$ if there are no backdoor paths from $M$ to $Y$ via $X$?

V. Vancak
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Griffin Farrow
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1 Answers1

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Your classmate is correct. I'll use the potential outcomes notations, but it carry the same idea. In order to identify the causal effect of $M$ on $Y$, i.e., $$ P(Y=y|do(M=m)) = P(Y_m=y), $$ you need to block every back door path. Particularly, $ Y\xleftarrow{} U\to X \to M$. In order to block it you need to adjust either for $X$ or for $U$ (or both). Since $U$ is unobserved, then your only choice is $X$. Formally, you have conditinioal exhangibility of $Y_m$ and $M$ given $X$, i.e., $$ Y_m \perp M | X. $$ Formally, $$ P(Y_m = y) = \sum P(Y_m = y|X=x)P(X=x) $$ which is always true (the law of total probability). Now, you can use the conditional independence, i.e., $$ P(Y_m = y) = \sum P(Y_m = y|M=m, X=x)P(X=x) $$ now use the causal consistency, i.e., $Y_m = Y$ when $ M=m$ $$ P(Y_m = y) = \sum P(Y = y|M, X=x)P(X=x), $$ which completes the proof.

V. Vancak
  • 17,021