Remaining Orientation Rules

Let us now wrap-up the easier rules. I enumerate the sound variations of FCI’s rules R2-R3 and R6-R8.

Lemma 14. The following variations of FCI’s R2 are sound:

1. R2a: If (1)Oi∗→Oj−∗Ok or Oi−∗Oj∗→Ok, and (2)Oi∗−◦Ok or Oi∗—IOk, then orient

Oi∗−◦Ok or Oi∗—IOk as Oi∗→Ok. 2. R2b: If (1) Oi∗—IOj−∗Ok, Oi∗—Oj −∗Ok, Oi−∗Oj ∗—IOk or Oi−∗Oj∗—Ok and (2) Oi∗−◦Ok or Oi∗—•Ok, then orient Oi∗−◦Ok or Oi ∗—•Ok as Oi∗—IOk. 3. R2c: If (1) Oi∗→Oj •−∗Ok, Oi∗→ Oj−∗Ok, Oi•−∗Oj∗→ Ok or Oi−∗Oj∗→ Ok and (2) Oi∗−◦Ok or Oi∗—•Ok, then orient Oi∗−◦Ok or Oi∗—•Ok as Oi∗—IOk. 4. R2d: If (1) Oi ∗—IOj •−∗Ok, Oi ∗—Oj •−∗Ok, Oi ∗—IOj−∗Ok or Oi ∗—Oj−∗Ok is

contemporaneous with respect to Oi∗→ Oj −∗Ok and (2) Oi ∗−◦Ok or Oi∗—•Ok, then

orient Oi∗−◦Ok or Oi∗—•Ok as Oi∗—IOk.

5. R2e: If (1) Oi•−∗Oj ∗—IOk, Oi−∗Oj ∗—IOk, Oi •−∗Oj ∗—Ok or Oi−∗Oj ∗—Ok is

contemporaneous with respect to Oi −∗Oj∗→ Ok and (2) Oi ∗−◦Ok or Oi∗—•Ok, then

orient Oi∗−◦Ok or Oi∗—•Ok as Oi∗—IOk.

Proof. R2a: Suppose to the contrary that we have an unfilled tail or square at Ok. If

Oi∗→ Oj −∗Ok, then Otj ∈ An(Oti ∪St) for all or some t ∈ S(f(T)), respectively, which

contradicts the arrowhead at Oj. If Oi−∗Oj∗→Ok, then Otk ∈An(Ojt∪St) for all or some

R2b: Suppose to the contrary that we have an unfilled tail at Ok. Then either Ojt ∈

An(Ot

i∪St) orOkt ∈An(Ojt∪St) at allt∈S(f(T)) which contradicts the filled arrowhead

atOj or Ok, respectively.

R2c: Suppose to the contrary that we have an unfilled tail at Ok. Then either Otj ∈

An(Ot

i ∪St) or Okt ∈ An(Ojt ∪St) at some t ∈ S(f(T)) which contradicts the unfilled

arrowhead at Oj orOk, respectively.

R2d: Suppose to the contrary that we have an unfilled tail atOk. ThenOjt ∈An(Oit∪St)

at somet ∈A⊆S(f(T)) which contradicts the filled arrowhead at Oj for t ∈A.

R2e: Suppose to the contrary that we have an unfilled tail atOk. ThenOkt ∈An(Ojt∪St)

at somet ∈A⊆S(f(T)) which contradicts the filled arrowhead at Oj for t ∈A.

We now move onto R3:

Lemma 15. The following variations of FCI’s R3 are sound under mixture faithfulness,

d-separation faithfulness with respect to_PXt,∀t∈S(f(T))as well as parameter faithfulness: 1. R3a: If (1)Oi∗→Oj ←∗Ok, (2)Oi∗—∗Ol∗—∗Ok, (3)Oi ⊥⊥Ok|(W,S)withOl ∈W and

minimal independence set W, (4) f(Oi, Ok|B,S) is stationary for all B ⊆ (W \Ol),

and (5) Ol∗−◦Oj or Ol∗—IOj, then orient Ol∗−◦Oj or Ol∗—IOj as Ol∗→Oj.

2. R3b: If (1) Oi∗—IOj ←∗Ok, Oi∗→OjJ—∗Ok, Oi∗—Oj ←∗Ok or Oi∗→Oj—∗Ok, (2)

Oi∗—∗Ol∗—∗Ok, (3) Oi ⊥⊥Ok|(W,S)with Ol ∈W and minimal independence set W,

(4) f(Oi, Ok|B,S) is stationary for all B⊆(W \Ol), and (5) Ol∗−◦Oj or Ol∗—•Oj,

then orient Ol∗−◦Oj or Ol∗—•Oj as Ol∗—IOj.

3. R3c: If (1) we have the contemporaneous path Oi ∗—IOjJ—∗Ok, Oi ∗—OjJ—∗Ok, Oi∗

—IOj—∗Ok or Oi∗—Oj—∗Ok with respect to Oi∗→Oj ←∗Ok, (2)Oi∗—∗Ol∗—∗Ok, (3)

Oi ⊥⊥Ok|(W,S) with Ol ∈W and minimal independence setW, (4)f(Oi, Ok|B,S)is

stationary for all B ⊆(W \Ol), and (5) Ol∗−◦Oj or Ol∗—•Oj, then orient Ol∗−◦Oj

or Ol∗—•Oj as Ol∗—IOj.

4. R3d: If (1) Oi∗ → Oj ← ∗Ok, (2) Oi ∗—∗ Ol ∗—∗ Ok, (3) Oi ⊥⊥ Ok|(W,S) with

Ol ∈ W and minimal independence set W, (4) f(Oi, Ok|B,S) is non-stationary for

some B⊆(W \Ol), and (5) Ol∗−◦Oj or Ol∗—•Oj, then orient Ol∗−◦Oj or Ol∗—•Oj

Proof. R3a: Observe by (3) and (4) thatOt

l ∈An(Oti ∪Okt ∪St) at all t ∈S(f(T)) due to

Lemma 8. Assume to the contrary that we have an unfilled tail or square at Oj. But then

Ot

j ∈ An(Oit∪Okt ∪St) at some t ∈S(f(T)), which contradicts the unfilled arrowheads at

Oj.

R3b: Observe by (3) and (4) thatO_lt∈An(Ot_i∪O_kt∪St) at allt∈S(f(T)) due to Lemma

8. Assume to the contrary that we have an unfilled tail atOj. But thenOtj ∈An(Oti∪Okt∪St)

at all t∈S(f(T)), which contradicts the arrowheads at Oj.

R3c: Observe by (3) and (4) thatOt

l ∈An(Oit∪Okt∪St) at allt ∈S(f(T)) due to Lemma

8. Assume to the contrary that we have an unfilled tail atOj. But thenOtj ∈An(Oti∪Okt∪St)

at all t∈S(f(T)), which contradicts the contemporaneous filled arrowheads atOj.

R3d: Observe by (3) and (4) that Ot

l ∈ An(Oti ∪Otk∪St) at some t ∈ S(f(T)) due

to Lemma 8. Assume to the contrary that we have an unfilled tail at Oj. But then Otj ∈

An(Ot

i∪Otk∪St) at some t ∈S(f(T)), which contradicts the unfilled arrowheads at Oj.

Lemma 16. The following variations of FCI’s R6 are sound:

1. R6a: IfOi−Oj◦−∗Ok or Oi−Oj•−∗Ok (Oi and Ok may or may not be adjacent), then

orient Oj◦−∗Ok or Oj •−∗Ok as Oj—∗Ok.

2. R6b: If Oi−•Oj◦−∗Ok, Oi−Oj◦−∗Ok, Oi−•OjJ—∗Ok or Oi−OjJ—∗Ok (Oi and Ok

may or may not be adjacent), then orient Oj◦−∗Ok or OjJ—∗Ok as Oj•−∗Ok.

3. R6c: If Oi•−Oj◦−∗Ok, Oi−Oj◦−∗Ok, Oi•−OjJ—∗Ok or Oi−OjJ—∗Ok (Oi and Ok

may or may not be adjacent), then orient Oj◦−∗Ok or OjJ—∗Ok as Oj•−∗Ok.

4. R6d: If Oi •—• Oj◦−∗Ok, Oi—•Oj◦−∗ Ok, Oi •—Oj◦−∗Ok or Oi—Oj◦−∗ Ok is

contemporaneous with respect to Oi−Oj◦−∗Ok (Oi andOk may or may not be adjacent),

then orient Oj◦−∗Ok or OjJ—∗Ok as Oj •−∗Ok.

Proof. For R6a, a filled arrowhead or square at Oj would violate Lemma 11. For R6b

and R6c, an unfilled arrowhead at Oj would also violate Lemma 11. For R6d, an unfilled

arrowhead atOj cannot exist becauseOi, Oj ∈An(St) at somet∈S(f(T)) by the acyclicity

Lemma 17. The following variations of FCI’s R7 are sound under mixture faithfulness, d-separation faithfulness with respect to_PXt,∀t∈S(f(T))as well as parameter faithfulness: 1. R7a: If (1) Oi—∗Oj◦−∗Ok or Oi—∗Oj•−∗Ok, (2) Oi ⊥⊥Ok|(W,S) with Oj ∈W and

minimal independence set W, and f(Oi, Ok|B,S) is stationary for all B ⊆ (W \Oj),

then orient Oj◦−∗Ok or Oj •−∗Ok as Oj—∗Ok.

2. R7b: If (1)Oi•—∗Oj◦−∗Ok, Oi—∗Oj◦−∗Ok, Oi•—∗OjJ—∗Ok or Oi—∗OjJ—∗Ok, (2)

Oi ⊥⊥ Ok|(W,S) with Oj ∈ W and minimal independence set W, and f(Oi, Ok|B,S)

is stationary for all B⊆(W \Oj), then orient Oj◦−∗Ok or OjJ—∗Ok as Oj•−∗Ok.

3. R7c: If (1) Oi—∗Oj◦−∗Ok or Oi—∗OjJ—∗Ok, (2) Oi ⊥⊥ Ok|(W,S) with Oj ∈ W

and minimal independence set W, and f(Oi, Ok|B,S) is non-stationary for some B ⊆

(W \Oj), then orient Oj◦−∗Ok or OjJ—∗Ok as Oj •−∗Ok.

Proof. R7a: Suppose to the contrary that we have a filled arrowhead atOj. Then, we must

have Oi—Oj from Lemma 8 due to the minimal separating set; this however contradicts

Lemma 11.

R7b: Suppose to the contrary that we have an unfilled arrowhead at Oj. Then, we must

have Oi•—Oj from Lemma 8 due to the minimal separating set; this however contradicts

Lemma 11.

R7c: Suppose to the contrary that we have an unfilled arrowhead at Oj. Then, we must

have Oi—•Oj from Lemma 8 due to the minimal separating set; this however contradicts

Lemma 11.

Lemma 18. The following variations of FCI’s R8 are sound:

1. R8a: If (1) Oi −∗Oj −∗Ok, and (2) Oi◦−∗Ok or Oi •−∗Ok, then orient Oi◦−∗Ok or

Oi•−∗Ok as Oi—∗Ok.

2. R8b: If (1)Oi•—∗Oj−∗Ok or Oi—∗Oj−∗Ok, and (2) Oi◦−∗Ok or OiJ—∗Ok, then orient

Oi◦−∗Ok or OiJ—∗Ok as Oi•−∗Ok.

3. R8c: If (1)Oi−∗Oj•—∗Ok or Oi−∗Oj—∗Ok, and (2) Oi◦−∗Ok or OiJ—∗Ok, then orient

4. R8d: If (1) Oi •−∗Oj •−∗Ok, Oi−∗Oj •−∗Ok, Oi•−∗Oj−∗Ok or Oi−∗Oj−∗Ok is

contemporaneous according to Oi−∗Oj−∗Ok, and (2) Oi◦−∗Ok or OiJ—∗Ok, then orient

Oi◦−∗Ok or OiJ—∗Okas Oi•−∗Ok.

Proof. All rules follow due to transitivity of the tail.