# 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 $$\text{PA}(X)$$.

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. Apr 7, 2020 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). Apr 7, 2020 at 13:10

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?

$$Z$$ satisfy the backdoor criterion because no backdoor paths between $$X$$ and $$Y$$ remain open in the DAG if we condition on $$Z$$.

Considering that we are interested in the (total) causal effect of $$X$$ on $$Y$$ a control set that contain $$Z$$ is a good control set. Moreover even the empty set is good, indeed even it deal with backdoor criterion.

If your concern is about the fairness of the definition you reported above, I suggest:

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 conditioning on the control set $$Z$$ no directed/causal paths are blocked and no spurious/backdoor paths remain open.

The reason that conditioning on the parents of $$X$$, irrespective of what the DAG looks like, always satisfies the backdoor criterion relative to $$(X,Y)$$ is that there is a parent of $$X$$ on each backdoor path and parents of $$X$$ cannot be colliders, by definition of parents of $$X$$ (which implies an arrow from the parent to $$X$$), hence conditioning on the the set of parents of $$X$$ will block all the backdoor paths, not open any spurious paths, and leave all directed paths untouched.

With regards to your specific question on this DAG: $$Z \rightarrow X \rightarrow Y$$: $$Z$$, the parent of $$X$$, does satisfy the backdoor criterion, albeit trivially. There is no backdoor path that remains open once we condition on $$Z$$; all directed paths from $$X$$ to $$Y$$ remain unperturbed; no new spurious paths are created. But, of course, the empty set also satisfies the backdoor criterion in this case.

However, there are at least 3 reasons why when interested in the causal effect of $$X$$ on $$Y$$, we would prefer to reduce the DAG you brought up $$Z \rightarrow X \rightarrow Y$$ to the following $$X \rightarrow Y$$ instead.

1. $$Y$$ and $$Z$$ are independent conditional on X: $$P(Y|X,Z)=P(Y|X)$$. We gain nothing by conditioning on $$Z$$, once we've already conditioned on $$X$$. Put differently, $$Z$$ here is neutral in terms of bias reduction.
2. Controlling for $$Z$$ will reduce the variation in $$X$$ and hence will reduce the precision of the estimate of the average causal effect.
3. To the extent that there are unobserved common causes of $$X$$ and $$Y$$, controlling for Z will amplify the bias (due to the association via U).

See here for more on this.

• Your DAG at the end there is a canonical instrumental variables estimation DAG. :) Jan 21, 2022 at 18:13