# What's the MAG of this underlying DAG?

I am studying causal discovery, with an interest on constraint-based algorithms like FCI (Fast Causal Inference).

I want to know what's the Maximal Ancestral Graph (MAG) of this underlying DAG (example taken from Fig-1 in Jiji Zhang et,al) In this DAG, $$\textbf{O}=\{A,Ef,R\}$$ is the set of observed variables, $$\textbf{L}=\{H\}$$ is the set of unobserved variables, and $$\textbf{S}=\{Sel\}$$ is the set of unobserved variables that are mistakenly conditioned on (i.e. the selection variables).

I use the rules given by Sect 2.3 in Jiji Zhang et,al and compute the MAG over $$\textbf{O}$$ of this underlying DAG as:

$$A-E$$, $$E\rightarrow R$$, $$A \rightarrow R$$

My computing process is:

• $$A$$ is adjacent with $$R$$ because there is an inducing path $$\langle A,E,H,R\rangle$$ w.r.t. $$\textbf{L},\textbf{S}$$ between $$A$$ and $$R$$.
• $$A \rightarrow R$$ because $$A$$ is an ancestor of $$\textbf{S}$$, but $$R$$ is not an ancestor of $$A$$ nor $$\textbf{S}$$.
• $$A-E$$ because $$A$$ is an ancestor of $$E$$ (and $$\textbf{S}$$), $$E$$ is also an ancestor of $$\textbf{S}$$.
• $$E\rightarrow R$$ because there is an inducing path $$\langle E,H,R\rangle$$ between $$E$$ and $$R$$, and $$E$$ is an ancestor of $$\textbf{S}$$, but $$R$$ is not an ancestor of $$E$$ nor $$\textbf{S}$$.

Since the MAG of this underlying DAG is not given in Jiji Zhang's paper, I am worrying whether my result is correct. Especially about the orientation between $$A$$ and $$E$$ (it looks very strange to me because $$A$$ is actually a cause of $$E$$ in the underlying DAG).

Am I giving the right results? If not, can anyone point out which step or rule is mistakenly used?

• How is it possible to condition on an unobserved variable? Mar 16, 2022 at 14:30
• In reading some of the linked paper, I think I would say that the Sel variable is measured. The paper says, "The selection variable (Sel) records whether or not a patient remains in the study, thus for all those remaining in the study Sel = StayIn." Mar 16, 2022 at 14:38
• By the way, this is a really great question. Alas, I don't know MAGs well enough to answer it. I'm very curious to see any answers! Mar 16, 2022 at 15:47
• The selection variable (Sel) is indeed unobserved. Refer to Sect. 2.3 of the paper, 'S denotes a set of unobserved selection variables to be conditioned upon'. Jun 8, 2022 at 12:55
• Yes, I believe MAG is for the following scenario: when we collected data, we unconsciously conditioned on some variable and caused a selection bias in the data. In these scenarios, the selection variables are unobserved. Jun 8, 2022 at 13:26

The MAG looks like this: The steps:

The edge between $$Ef$$ and $$R$$: There is an inducing path between $$Ef$$ and $$R$$ relative to $$\langle\{H\},\emptyset\rangle$$, so there must be an edge. The edge is bidirected because of rule (2)(c) in the paper: $$Ef\notin\bf{An}_\mathcal{G}(\{R\}\cup\emptyset)$$ and $$R\notin\bf{An}_\mathcal{G}(\{Ef\}\cup\emptyset)$$. Thus $$Ef\leftrightarrow R.$$

The edge between $$A$$ and $$Ef$$: There is an inducing path between $$A$$ and $$Ef$$ relative to $$\langle\emptyset, \emptyset\rangle$$, so there must be an edge. The edge is directed from $$A$$ to $$Ef$$ because of rule (2)(a) in the paper: $$A\in\bf{An}_\mathcal{G}(\{Ef\}\cup\emptyset)$$ and $$Ef\notin\bf{An}_\mathcal{G}(\{A\}\cup\emptyset)$$. Thus: $$A\to Ef.$$

The edge between $$A$$ and $$R$$: There is an inducing path between $$A$$ and $$R$$ relative to $$\langle\{H\}, \{Sel\}\rangle$$, so there must be an edge. The edge is directed from $$A$$ to $$R$$ because of rule (2)(a) in the paper: $$A\in\bf{An}_\mathcal{G}(\{R\}\cup\{Sel\})$$ and $$R\notin\bf{An}_\mathcal{G}(\{A\}\cup\{Sel\})$$. Thus: $$A\to R.$$

• Hi, my major confusion is that what the notation $A\in An_G(\{R\}\cup \{Sel\})$ means. Does it mean (1) $A\in An_G(\{R\})$ or $A\in An_G(\{Sel\})$ (2) $A\in An_G(\{R\})$ and $A\in An_G(\{Sel\})$ ? Apr 2, 2022 at 6:23
• The set $An_G(S)$ of ancestors of the set $S$ in the graph $G$ is the set of nodes that are ancestors in $G$ of at least one of the members $s$ of $S$. Thus (1) is correct and (2) is not. Apr 2, 2022 at 6:42
• @ frank Hi, I have a question about your MAG. In Sect. 2.3 of Zhang's paper, he said in the MAG, two vertices A and B are adjacent if and only if there is an inducing path with respect to <L,S>, and L,S are respectively latent variables and latent selection variables in the underlying DAG. So, how can you decide the edge (and its orientation) between A and Ef by an inducing path w.r.t. <\emptyset,\emptyset>? Jun 8, 2022 at 12:59
• If the L must be {H}, S must be {Ef}, I think my original solution make sense. Jun 8, 2022 at 13:01

After checking Zhang's paper, I believe my original solution is correct.

The MAG of the causal DAG is indeed $$A - E$$, $$A \to R$$, $$E \to R$$.

This is because when deciding adjacency and edges' orientation, the set L and S must be respectively the set of latent variables (in our case $$\{H\}$$) and the set of selection variables (in our case $$\{S\}$$).

This MAG actually correspond with Zhang's explanation in Sect. 2.3, for example:

• $$A - E$$ because $$A$$ is the cause of some selection variable $$Sel$$, $$E$$ is the cause of some selection variable $$Sel$$.
• $$A \to E$$ is wrong because $$A \to E$$ requires $$E$$ is not the cause of $$A$$ and not the cause of any selection variable.

Indeed, the MAG of the DAG looks quite strange. This is also illustrated by Zhang in his paper, "The positive causal information is admittedly less informative than one would wish, when the possibility of selection bias is allowed."