# In covariate-specific effect - $P(Y=y|do(X=x),Z=z)$ - is $Z=z$ pre- or post-treatment measure?

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?

• 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 $Z.$ Jan 23, 2023 at 13:45
• @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)? Jan 24, 2023 at 22:25
• 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/… Jan 25, 2023 at 0:50