# How come parents of $X$ always satisfy the backdoor criterion relative to $(X,Y)$?

Pearl et al. "Causal Inference in Statistics: A Primer" (2016) p. 61 presents the backdoor criterion:

Definition 3.3.1 (The Backdoor Criterion) Given an ordered pair of variables $$(X,Y)$$ in a directed acyclic graph $$G$$, a set of variables $$Z$$ satisfies the backdoor criterion relative to $$(X,Y)$$ if no node in $$Z$$ is a descendant of $$X$$, and $$Z$$ blocks every path between $$X$$ and $$Y$$ that contains an arrow into $$X$$. If a set of variables $$Z$$ satisfies the backdoor criterion for $$X$$ and $$Y$$, then the causal effect of $$X$$ on $$Y$$ is given by the formula $$P(Y=y|do(X=x))=\sum_{z} P(Y=y|X=x,Z=z)P(Z=z)$$ just as when we adjust for $$\text{PA}(X)$$.

It then adds in parentheses:

Note that $$\text{PA}(X)$$ always satisfies the backdoor criterion.

The latter note is not obvious to me. Consider a DAG that is a simple chain $$Z \rightarrow X \rightarrow Y$$ Here, $$\text{PA}(X)=Z$$. At the same time, there is no backdoor path between $$X$$ and $$Y$$, so $$Z$$ does not block any. Question: How come $$Z$$ satisfies the backdoor criterion then?

• I suppose it's simply a matter of reading $Z$ blocks every path between $X$ and $Y$ that contains an arrow into $X$ if any exists. – Tim Mak Apr 7 '20 at 7:14
• @TimMak Exactly right. The empty set could serve as satisfying the backdoor criterion here, but adding that particular $Z$ doesn't hurt anything (that is, it doesn't open up previously closed paths via a collider). – Adrian Keister Apr 7 '20 at 13:10