In the book Causal Inference In Statistics by Pearl, page 63, while referring to the below DAG, it says -

Thus to compute the $w$-specific causal effect, written $P(y|do(x),w)$, we adjust for $T$, and obtain

$P(Y=y|do(X=x),W=w)$ $=$ $\sum_t {P(Y=y|X=x,W=w,T=t)P(T=t|X=x,W=w)}$ (3.11)

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I have the following queries -

  1. Why does it say - "to compute the $w$-specific causal effect, written $P(y|do(x),w)$"? Given the definition of $do(x)$ presented here, it cannot be guaranteed that $P(y|do(x),w)$ calculates the respective causal effect, when conditioning on $w$ opens up a non-causal path (highlighted in pink in the figure). Am I understanding the definition of $do(x)$ incorrectly here?
  2. In the equation if the summation on the right-hand side is performed, $\sum_t {P(Y=y|X=x,W=w,T=t)P(T=t|X=x,W=w)}$
    $=\sum_t {P(Y=y,T=t|X=x,W=w)}$
    which should not be the causal-effect as it seems to be including the association rising from the non-causal path. What am I missing here?

1 Answer 1


The $w$-specific causal effect of $X$ on $Y$ is quite distinct from the causal effect of $X$ on $Y.$ The causal effect of $X$ on $Y$ is just $P(y|do(x)).$ The $w$-specific causal effect of $X$ on $Y$ is as you've written: $P(y|do(x),w).$ Essentially, you are stratifying the causal effect of $X$ on $Y$ by values of $w.$ Now, in the particular case in question, there are backdoor or other non-causal paths opened up by conditioning on $w,$ so it is necessary to stop those up to get the true $w$-specific causal effect (just condition on $Z$ or $T$ as well to stop up the undesired path - hence the first formula in your question). I don't think you're necessarily misunderstanding the $do$ operator.

In your second question, you need to understand what's meant by "including the association rising from the non-causal path". You could include that association in more than one way. In this example, we're "including" that association by adjusting for it so that it does not bias our results.

  • $\begingroup$ In $P(y|do(x),w)$ is $W=w$ measured post-intervention? And in the final expression of the second query - $P(Y=y|X=x,W=w)$ - is $w$ here a measure from observational/pre-intervention data? I think it should be because we are only theoretically simulating the intervention, and hence, do not have actual access to post-intervention data. $\endgroup$ Commented Jan 24, 2023 at 22:16
  • 1
    $\begingroup$ $W$ is post-intervention because the arrow goes from $X$ to $W.$ $W$ is measured, so it should be in the data. It's not possible to adjust for unmeasured variables, unless you do something like the front-door adjustment. $\endgroup$ Commented Jan 25, 2023 at 0:43
  • $\begingroup$ Given the Rule 2 in this query, should it be $P(T=t|X=x,W=w)$ or $P(T=t|W=w)$ in equation 3.11? This is similar to the this query, where you supported the later expression. According to this source it should be $P(T=t)$. Any idea which should be the correct one, and some solid source for it? $\endgroup$ Commented Jan 25, 2023 at 2:11
  • $\begingroup$ Found the following reliable errata page from Pearl's own UCLA website. The errata here changes the figure to replace $W \leftarrow X$ to $W \leftarrow V \rightarrow X$ and also suggests $P(T=t|W=w)$. In fact an earlier version of this same errata page introduced $X=x$ in the first place. @AdrianKeister, do I need to edit the question accordingly? $\endgroup$ Commented Jan 25, 2023 at 2:39
  • $\begingroup$ Those are good questions. I'm not sure I can answer them fully - we probably need Carlos Cinelli for that (probably CrossValidated's best expert on causality). Off the top of my head, I'd say that some of those expressions are probably equivalent if the DAG has certain characteristics. For example, if $X$ and $Y$ are completely independent in the DAG, then you can say $P(Y|X)=P(Y).$ That sort of thing. But it requires detailed thinking to tease out all the possibilities. $\endgroup$ Commented Jan 26, 2023 at 14:36

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