Pearl et al. "Causal Inference in Statistics: A Primer" (2016) p. 108 contains the following adjustment formula (based on the backdoor criterion) for probabilities of counterfactuals expressed using observed data: \begin{align} P(Y_x|X=x')=\sum_z P(Y=y|X=x,Z=z) P(Z=z|X=x'). \tag{4.21} \end{align} The derivation is as follows: \begin{align} P(Y_x|X=x') &= \sum_z P(Y_x=y|X=x\color{red}{'},Z=z) P(Z=z|X=x') \\ &= \sum_z P(Y_\color{red}{x}=y|X=x,Z=z) P(Z=z|X=x') \\ &= \sum_z P(Y=y|X=x,Z=z) P(Z=z|X=x'). \end{align} The top equality simply partitions the original $P(Y_x|X=x')$ w.r.t $Z$. Since $Z$ satisfies the backdoor criterion, $(Y_x\perp \!\!\! \perp X)|Z$, so $P(Y_x|X,Z)=P(Y_x|Z)$. This allows replacing $x\color{red}{'}$ by $x$ and yields the middle equality. The bottom equality follows from the consistency rule: if $X=x$ then $Y_\color{red}{x}=Y$.

I am interested in taking things one step further. If $(Y_x\perp \!\!\! \perp X)|Z$, could we not get rid of conditioning on $X=x$ altogether and have \begin{align} P(Y_x|X=x') &= \sum_z P(Y=y|\color{red}{X=x},Z=z) P(Z=z|X=x') \\ &= \sum_z P(Y=y|Z=z) P(Z=z|X=x') \quad ? \end{align}

  • $\begingroup$ Does $(Y_x \perp \!\! \perp X)|Z$ imply $(Y \perp \!\! \perp X)|Z?$ Counterfactuals I find rather tricky and slippery. On the face of it, I wouldn't expect my implication to hold. Isn't $Y_x$ a more specific $Y$ than just $Y?$ $\endgroup$ Apr 14, 2020 at 18:01
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    $\begingroup$ @AdrianKeister, you might very well be right. $Y_x$ is the counterfactual of $Y$, so they are not the same. $\endgroup$ Apr 14, 2020 at 18:55

1 Answer 1


As noted in the comments, this were valid if $Y_x \perp X|Z$ would imply $Y \perp X|Z$. But it does not. The simplest counterexample is when $X$ affects $Y$. Then $Z$ blocks all back-door paths, but there is an open path $X \rightarrow Y$ which is not blocked by $Z$, and so $X$ is informative about $Y$ even conditional on $Z$.

If, however, $X$ was assumed to not affect $Y$, then we would have $Y_x = Y$, and the conclusion would follow.


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