0
$\begingroup$

Causal Inference In Statistics by Pearl, section 3.5, page 70 clearly mentions that -

This effect, written $P(Y=y|do(X=x),Z=z)$, measures the distribution of $Y$ in a subset of the population for which $Z$ achieves the value $z$ after the intervention.

In that context Rule 2 says -

$P(Y=y|do(X=x),Z=z)$
$= \sum_s {P(Y=y|X=x,S=s,Z=z)P(S=s|Z=z)}$

where $S \cup Z$ satisfies the backdoor criterion.

Query: Is $Z=z$ in the second expression pre- or post-intervention measure?

Note: In subsequent section where the effect of conditional intervention is evaluated, it says -

$P(Y=y|do(X=g(Z)))$
$=\sum_z{P(Y=y|do(X=g(Z)),Z=z)P(Z=z|do(X=g(Z)))}$
$=\sum_z{P(Y=y|do(X=g(z)),Z=z)P(Z=z)}$

The equality $P(Z=z|do(X=g(Z)))=P(Z=z)$ stems, of course, from the fact that $Z$ occurs before $X$; hence any control exerted on $X$ can have no effect on the distribution of $Z$.

Here it seems that $Z=z$ is a pre-intervention measure. Hence, in the above sections, when is $Z-z$ a pre-intervention measure and when is it a post-intervention measure?

$\endgroup$
3
  • 1
    $\begingroup$ It's pre-intervention if either $Z\to X$ or there are a series of arrows from $Z$ to $X.$ It's post-intervention if either $X\to Z$ or there are a series of arrows from $X$ to $Z.$ $\endgroup$ Jan 23, 2023 at 13:45
  • $\begingroup$ @AdrianKeister could you please provide some reference for the above claim? Besides, in $P(Y=y|do(X=x),Z=z)$, $Z=z$ may be post-intervention measure. But in the RHS expression $\sum_s {P(Y=y|X=x,S=s,Z=z)P(S=s|Z=z)}$, $Z=z$ needs to be observational/pre-intervention correct (as it's just a theoretically simulated intervention and we do not have access to post-intervention data)? $\endgroup$ Jan 24, 2023 at 22:25
  • 1
    $\begingroup$ Alas, I don't have a reference for it. I heard this concept in a presentation that was utilizing causality. But this idea makes sense: for something to be pre-intervention, it must be before $X;$ and for it to have any interest, it must be causally related to $X.$ Hence, the arrows go into $X$. You see it in passing here: stats.stackexchange.com/questions/598107/… $\endgroup$ Jan 25, 2023 at 0:50

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.