5.2 The Mixture Approach
5.2.4 Fourth Orientation Rule
I will now move on to proving the soundness of the proposed algorithm’s orientation rules. In general, I adopt the following strategy for the orientation rules: for each rule in FCI, (1) eliminate all endpoints in the rule which are not required in the proofs of soundness in [Zhang,
2008], and then (2) combine the minimalist rule with all (logically useful) combinations of stationary and non-stationary densities.
Now orientation rules 1, 4, 5, 9 and 10 require longer arguments than the other rules, so I will present the arguments for rules 1, 4, 5, 9 and 10 in separate subsections. I will order the presentation of the rules according to a logical progression, rather then simply present the rules in numerical order. I will first cover rule 4 in this subsection, then rule 1 in Subsection 5.2.5, then rules 5, 9 and 10 in Subsection 5.2.6, and finally the remaining rules in Subsection5.2.7.
Now the fourth orientation rule requires the following lemma:
Lemma 7. Assume d-separation faithfulness with respect to PXt,∀t∈S(f(T)) and mixture faithfulness. Let πik ={Oi, . . . , Ol, Oj, Ok} be a sequence of at least four vertices which may
satisfy the following:
A1. Oi ⊥⊥Ok|(W,S) with W ⊆O\ {Oi, Ok},
A2. Any two successive verticesOh and Oh+1 on πik are conditionally dependent given (Y \
A3. We have the stationary density f(Oh, Oh+1|((Y \ {Oh, Oh+1})∪S) for all Y ⊆ W for any two successive vertices Oh and Oh+1 on πik,
A4. All vertices Oh between Oi and Oj (not including Oi and Oj) satisfy Oht ∈ An(Okt) and
Ot
h 6∈ An({Oht−1, Oht+1} ∪ St) for all time points t ∈ S(f(T)), where Oh−1 and Oh+1 denote the vertices adjacent to Oh on πik.
The following conclusions hold, ordered from strongest to weakest:
C1. Assume A1-A4 hold exactly. If Oj ∈ W, then (a1) Ojt ∈ An(Okt ∪ St) at all t ∈
S(f(T)) and (a2) Ot
k ∈ An(Ojt ∪ St) at all t ∈ S(f(T)). If Oj 6∈ W, then (a3)
Ot
j 6∈An({Otl, Otk}∪St)at allt ∈S(f(T))and (a4)Okt 6∈An(Otj∪St)at allt∈S(f(T)).
C2. Assume A1-A4 hold except, for A3, suppose there exists one and only one non-stationary densityf(Oh, Oh+1|Y \{Oh, Oh+1}∪S)for someY ⊆W. However, assume that at least one member of {f(Oh|Y \ {Oh, Oh+1} ∪S), f(Oh+1|Y \ {Oh, Oh+1} ∪S)}is stationary. If
Oj ∈W, then (b1) Ojt ∈An(Okt∪St) at some t∈S(f(T))and (b2) Otk6∈An(Otj∪St)
at all t∈S(f(T)). If Oj 6∈W, then (b3) Otj 6∈An({Otl, O t k} ∪S t) at some t ∈S(f(T)) and (b4) Ot k6∈An(Otj∪St) at all t ∈S(f(T)).
C3. Assume A1-A4 hold except, for A4, suppose ∃t0 ∈S(f(T)) where all vertices Oh between
Oi andOj (not includingOi andOj) satisfyOt
0 h ∈An(Ot 0 k)andOt 0 h 6∈An({Ot 0 h−1, Ot 0 h+1}∪ St0). If Oj ∈ W, then (c1) Otj ∈ An(Otk∪S t) at some t ∈ S(f(T)) and (c2) Ot k 6∈ An(Ot j ∪St) at some t ∈ S(f(T)). If Oj 6∈ W, then (c3) Otj 6∈ An({Olt, Okt} ∪St) at
some t ∈S(f(T)) and (c4) Okt 6∈An(Otj∪St) at some t∈S(f(T)).
Proof. We can claim that A1 and A2 hold at all t ∈ S(f(T)) by invoking Lemma 4 for A1 and Lemma 2 with A3 for A2. Observe that the conditions in A4 hold at all t ∈ S(f(T)). By acyclicity of the CMJ-DAG at any one time point, conclusions (a1)-(a4) follow due to Lemma 3.2 in [Colombo et al., 2012] applied at any arbitrary t∈S(f(T)).
The proof for conclusions (c1)-(c4) proceeds similarly except A4 only holds at some
t ∈ S(f(T)). We can therefore invoke Lemma 3.2 in [Colombo et al., 2012] again but only applied at some t∈S(f(T)), where all of the sufficient conditions are satisfied.
Let us now tackle conclusions (b1) and (b3) (but not (b2) and (b4)). Notice that A2 only holds at some t∈S(f(T)) by Lemmas 2and 3, but A1 and A4 hold at allt ∈S(f(T))
in this case. We can therefore invoke Lemma 3.2 in [Colombo et al., 2012] but only applied at somet ∈S(f(T)), where all of the sufficient conditions are satisfied.
We now prove conclusions (b2) and (b4). From A4, we know that Ot
l ∈ An(Okt) and
Olt 6∈ An(St) at all t ∈ S(f(T)). Hence, we must have Okt 6∈ An(St) at all t ∈ S(f(T)). We therefore only need to prove that Ot
k 6∈ An(Otj) at all t ∈ S(f(T)). Assume contrary
to the claim that Okt ∈ An(Otj) at some t ∈ S(f(T)). Recall that Olt ∈ An(Otk) at all
t ∈ S(f(T)) from A4, so Ot
l ∈ An(Ojt) also at some t ∈ S(f(T)). This statement however
contradicts another part of A4, where we must haveOlt6∈An(Ojt) at all t∈S(f(T)). Hence
Ot
k 6∈An(Otj) at all t∈S(f(T)).
We can use the above lemma to apply the following three orientation rules:
1. R4a: Suppose (1) there exists a path πik = {Oi, . . . , Ol, Oj, Ok}, (2) Oi ⊥⊥ Ok|(W,S),
(3) any two successive vertices Oh and Oh+1 on πik are conditionally dependent given
(Y \{Oh, Oh+1})∪S for allY ⊆W, (4) we have the stationary densityf(Oh, Oh+1|((Y \ {Oh, Oh+1})∪S) for all Y ⊆ W for any two successive vertices Oh and Oh+1 on πik,
(5) all vertices Oh between Oi and Oj (not including Oi and Oj) satisfy Oth ∈ An(Okt)
and Oth 6∈An({Oht−1, Oth+1} ∪St) for all time pointst ∈S(f(T)), where Oh−1 and Oh+1 denote the vertices adjacent to Oh onπik. If further Oj ∈ W, then orient Oj◦−∗Ok or
Oj •—∗Ok as Oj → Ok. Otherwise, if Oj 6∈ W, then orient the triple hOl, Oj, Oki as
Ol↔Oj ↔Ok.
2. R4b: Suppose (1)-(5) hold in R4a except, for (4), suppose there exists one and only one non-stationary density f(Oh, Oh+1|(Y \ {Oh, Oh+1})∪S) for some Y ⊆ W. How- ever, assume that at least one member of {f(Oh|(Y \ {Oh, Oh+1})∪S), f(Oh+1|(Y \ {Oh, Oh+1})∪S)}is stationary. Further, ifOj ∈W, then orientOj◦−∗OkorOjJ—∗Okas
Oj•→Ok. Otherwise, if Oj 6∈W, then orient the triple hOl, Oj, Oki as Ol ←IOjJ→Ok.
3. R4c: Suppose (1)-(5) hold in R4a except, for (5), suppose ∃t0 ∈ S(f(T)) where all vertices Oh between Oi and Oj (not including Oi and Oj) satisfy Ot
0 h ∈ An(Ot 0 k) and Ot0 h 6∈An({Ot 0 h−1, Ot 0 h+1} ∪St 0
). Further, if Oj ∈W, then orientOj◦−∗Ok orOjJ—∗Ok as
Let us pay special attention to R4c. Here, we may identify the existence of the time point
t0 when there exists only one filled endpoint on the discriminating path or, more generally, when the discriminating path is contemporaneous:
Definition 5. Consider a path π = hO1, . . . , Oni. Also consider a similar path E =
hO1, . . . , Oni but with only unfilled endpoints. Here, an endpoint on π may have a corre-
sponding unfilled endpoint in E. I say that π is contemporaneous according to E if and only if ∃t0 ∈ S(f(T)) such that, for every arbitrary endpoint on π with a corresponding unfilled endpoint in E, say at Oi on the edge between Oi and Oj, we have (1) Ot
0 i 6∈ An(Ot 0 j ∪St 0 ) when we have a corresponding unfilled arrowhead in E, or (2)Ot0
i ∈An(Ot
0
j ∪St
0
) when we have a corresponding unfilled tail in E.
In other words, π is contemporaneous according to E when the unfilled endpoints in E correspond to endpoints on π with ancestral/non-ancestral relations that exist at the same point in time. For example, suppose we create the v-structure Oi ∗—IOjJ—∗Ok with VSb.
We then have a contemporaneous path Oi∗—IOjJ—∗Ok according to Oi∗→Oj ←∗Ok. The
setE therefore corresponds to an unshielded v-structure with unfilled endpoints in this case. Similarly, the set E corresponds to a discriminating path with unfilled endpoints in R4c.