Causality as a theoretical concept: explanatory warrant and empirical content of the theory of causal nets

Abstract

We start this paper by arguing that causality should, in analogy with force in Newtonian physics, be understood as a theoretical concept that is not explicated by a single definition, but by the axioms of a theory. Such an understanding of causality implicitly underlies the well-known theory of causal (Bayes) nets (TCN) and has been explicitly promoted by Glymour (Br J Philos Sci 55:779–790, 2004). In this paper we investigate the explanatory warrant and empirical content of TCN. We sketch how the assumption of directed cause–effect relations can be philosophically justified by an inference to the best explanation. We then ask whether the explanations provided by TCN are merely post-facto or have independently testable empirical content. To answer this question we develop a fine-grained axiomatization of TCN, including a distinction of different kinds of faithfulness. A number of theorems show that although the core axioms of TCN are empirically empty, extended versions of TCN have successively increasing empirical content.

Appendix: Proofs of lemmata and theorems

Proof of Lemma 1

For (1.1): Assume $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}})$ is acyclic and $\hbox {X}\rightarrow \hbox {Y}$ in ${{\varvec{\mathcal {E}}}}$. For reductio, assume ${\uppi }$ is a path in ${{\varvec{\mathcal {E}}}}{-}\{\hbox {X}{\rightarrow }\hbox {Y}\}$ that connects X with Y and is activated by $\hbox {par(Y)}{-}\{\hbox {X}\}$. So ${\uppi }$ must have the form $\hbox {X}--\,\hbox {Z}\leftarrow \hbox {Y}$. Thus ${\uppi }$ must carry at least one common effect Z*; otherwise ${\uppi }$ would have the form $\hbox {X}\leftarrow \leftarrow \hbox {Y}$ and $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}})$ would be cyclic. But since $\hbox {Z}^{*}\notin \hbox {par(Y)}{-}\{\hbox {X}\}, \,{\uppi }$ is blocked by $\hbox {par(Y)}{-}\{\hbox {X}\}$, contradicting our assumption.

For (1.2): Assume $\mathbf{U} \supset \hbox {par(Y)}{-}\{\hbox {X}\}$. Both $\hbox {par(Y)}{-}\{\hbox {X}\}$ and U d-separate X from Y in $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{-}\{\hbox {X}\rightarrow \hbox {Y}\})$ (by assumption and lemma 1.1, respectively). Direction $\Leftarrow$ is trivial. For direction $\Rightarrow$ we have to prove that $\hbox {INDEP}(\hbox {X,Y}{\vert }\hbox {par(Y)}{-}\{\hbox {X}\})$ implies $\hbox {INDEP}(\hbox {X,Y}{\vert }\mathbf{U})$. We proceed by induction on the number of elements in $\mathbf{U}-(\hbox {par(Y)}{-}\{\hbox {X}\})$. Assume the claim has been proved for some $\mathbf{U} \supseteq \hbox {par(Y)}{-}\{\hbox {X}\}$ and let $\mathbf{U}^{\prime } = \mathbf{U}\cup \{\hbox {Z}\}$, where U and U $^{\prime }$ d-separate X from Y in $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{-}\{\hbox {X}\rightarrow \hbox {Y}\})$. We show that

(*) Either Z is d-separated from X by $\mathbf{U}\cup \{\hbox {Y}\}$, or Z is d-separated from Y by $\mathbf{U}\cup \{\hbox {X}\}$.

For reductio, assume (*) does not hold. We distinguish two cases. Case (A): Z is d-connected with X given $\mathbf{U}\cup \{\hbox {Y}\}$, and with Y given $\mathbf{U}\cup \{\hbox {X}\}$, by two paths ${\uppi }_{\mathrm{X}}$: Z $-$ X and ${\uppi }_{\mathrm{Y}}{:}\;\hbox {Z}-\hbox {Y}$, respectively, which both do not carry X $\rightarrow$ Y. In this case, Z is d-connected with X and with Y by the respective paths given U alone. Let ${\uppi }$ be the concatenation of ${\uppi }_{\mathrm{X}}$ and ${\uppi }_{\mathrm{Y}}$. If Z is a common effect on ${\uppi }$, then U $^{\prime }$ would activate a new X–Y-connecting path, which is excluded, and if Z is not a common effect on ${\uppi }$, then ${\uppi }$ would d-connect X and Y given U, which is excluded. So case (A) is impossible. The other possible case is (B): Z is d-connected with Y and with X only by paths ${\uppi }$ that contain $\hbox {X}\rightarrow \hbox {Y}$ as a subpath; these paths must either have the form (i) $\hbox {X}\rightarrow \hbox {Y}-\hbox {Z}$ or (ii) $\hbox {Z}-\hbox {X}\rightarrow \hbox {Y}$, but not both (else we would have case (A)). But this is impossible since in case (i) X is d-separated from Z by $\mathbf{U}\cup \{\hbox {Y}\}$ (since if $\hbox {Y}\leftarrow \hbox {Y}^{\prime }$ is on ${\uppi }$, then $\hbox {Y}^{\prime } \in \mathbf{U}$), and in case (ii) Z is d-separated from Y by $\mathbf{U}\cup \{\hbox {X}\}$.

Dawid’s axioms of probabilistic independence (cf. Pearl 1988, p. 84) include the following (probabilistically valid) axioms:

• Contraction: $\hbox {INDEP}(\hbox {X,Y}{\vert }\{\hbox {Z}\}\cup \mathbf{U})\wedge \hbox {INDEP}(\hbox {X,Z}{\vert }\mathbf{U}) \Rightarrow \hbox {INDEP}(\hbox {X},\{\hbox {Y,Z}\}{\vert }\mathbf{U})$.

• Decomposition: $\hbox {INDEP}(\hbox {X},\{\hbox {Y,Z}\}{\vert }\mathbf{U}) \Rightarrow \hbox {INDEP}(\hbox {X,Y}{\vert }\mathbf{U})\wedge \hbox {INDEP}(\hbox {X,Z}{\vert }\mathbf{U})$.

• Weak union: $\hbox {INDEP}(\hbox {X},\{\hbox {Y,Z}\}{\vert }\mathbf{U}) \Rightarrow \hbox {INDEP}(\hbox {X,Y}{\vert }\{\hbox {Z}\}\cup \mathbf{U})$.

Assume by (*) that Z is d-separated from X by $\mathbf{U}\cup \{\hbox {Y}\}$. (The other possibility is that Z is d-separated from Y by $\mathbf{U}\cup \{\hbox {X}\}$; the proof proceeds in exactly the same way.) Since $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{,}\hbox {P})$ satisfies (C), (a) $\hbox {INDEP}(\hbox {X,Z}{\vert }\mathbf{U}\cup \{\hbox {Y}\})$ follows. From the induction hypothesis $\hbox {INDEP}(\hbox {X,Y}{\vert }\mathbf{U})$ and (a) $\hbox {INDEP}(\hbox {X,Z}{\vert }\mathbf{U}\cup \{\hbox {Y}\})$ we get (b) $\hbox {INDEP}(\hbox {X},\{\hbox {Y,Z}\}{\vert }\mathbf{U})$ by contraction, and from (b) we get $\hbox {INDEP}(\hbox {X,Y}{\vert }\mathbf{U}\cup \{\hbox {Z}\})$, i.e. $\hbox {INDEP}(\hbox {X,Y}{\vert }\mathbf{U}^{\prime })$ by weak union.$\square$

Proof of Theorem 2

Proof of $(P) \Rightarrow (Min)$: Assume $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{,}\hbox {P})$ is not minimal. So there exists an $\hbox {X}{\rightarrow }\hbox {Y}$ in ${{\varvec{\mathcal {E}}}}$ such that $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}^{-},\hbox {P})$ satisfies (C), where ${{\varvec{\mathcal {E}}}}^{-} := {{\varvec{\mathcal {E}}}}{-}\{\hbox {X}{\rightarrow }\hbox {Y}\}$. Since $\hbox {par(Y)}{-}\{\hbox {X}\}$ d-separates X from Y in $({\varvec{\mathcal {V}}},{\varvec{\mathcal {E}}}^{-})$ (by lemma 1.1), $\hbox {INDEP}(\hbox {X,Y}{\vert }\hbox {par(Y)}{-}\{\hbox {X}\})$ holds because of (C). So (P) is violated.

Proof of (Min) $\Rightarrow (P)$: Assume that $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{,}\hbox {P})$ satisfies (Min), which means that there is no $\hbox {X,Y}\in {{\varvec{\mathcal {V}}}}$ with $\hbox {X}{\rightarrow }\hbox {Y}\in {{\varvec{\mathcal {E}}}}$ such that $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{-}\{\hbox {X}{\rightarrow }\hbox {Y}\},\hbox {P})$ still satisfies (C). The latter is the case iff

(*) the parent set par(Y) of every $\hbox {Y}\in {{\varvec{\mathcal {V}}}}$ (with $\hbox {par(Y)} \ne \varnothing$) is minimal in the sense that removing one of Y’s parents X from par(Y) would make a difference for Y, meaning that $\hbox {P}(\hbox {y}{\vert }\hbox {x,par(Y)}{-}\{\hbox {X}\}) \ne \hbox {P}(\hbox {y}{\vert }\hbox {par(Y)}{-}\{\hbox {X}\})$ holds for some X-value x, Y-value y, and some instantiations of $\hbox {par(Y)}{-}\{\hbox {X}\}$.

For otherwise P would admit the Markov factorization according to (8.2) both relative to $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{,}\hbox {P})$ and relative to $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{-}\{\hbox {X}{\rightarrow }\hbox {Y}\},\hbox {P})$. This implies by Theorem 1 that $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{,}\hbox {P})$ and $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{-}\{\hbox {X}{\rightarrow }\hbox {Y}\},\hbox {P})$ satisfy (C), i.e. $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{,}\hbox {P})$ is not minimal, which contradicts our assumption. Now, (*) entails that $\hbox {Dep}(\hbox {X,Y}{\vert }\hbox {par(Y)}{-}\{\hbox {X}\})$ holds for all $\hbox {X,Y}\in \mathbf{V}$ with $\hbox {X}{\rightarrow }\hbox {Y}$, i.e., that $({{\varvec{\mathcal {V}}}}{,}{{\varvec{\mathcal {E}}}}{,}\hbox {P})$ satisfies (P). $\square$

Proof of Theorem 5

For (5.1): Recall Dawid’s axioms of probabilistic independence from the proof of lemma (1.2). By switching Y with Z and setting $\mathbf{U} = \varnothing$, the contraposed forms of decomposition and contraction give us $\hbox {DEP(X,Z)} \Rightarrow \hbox {DEP}(\hbox {X}{,}\{\hbox {Y,Z}\})$ and $\hbox {DEP}(\hbox {X}{,}\{\hbox {Y,Z}\}) \wedge \hbox {INDEP}(\hbox {X,Z}{\vert }\hbox {Y}) \Rightarrow \hbox {DEP}(\hbox {X,Y})$. In the same way, switching X with Y and setting $\mathbf{U} = \varnothing$ gives us $\hbox {DEP(Y,Z)} \Rightarrow \hbox {DEP}(\hbox {Y}{,}\{\hbox {X,Z}\})$ and $\hbox {DEP}(\hbox {Y}{,}\{\hbox {X,Z}\}) \wedge \hbox { INDEP}(\hbox {X,Y}{\vert }\hbox {Z}) \Rightarrow \hbox {DEP}(\hbox {Y,Z})$.

By these considerations, the assumptions DEP(X,Z) and $\hbox {INDEP}(\hbox {X,Z}{\vert }\hbox {Y})$ of Theorem 5.1 entail DEP(X,Y) and DEP(Y,Z). So (by (C)) X and Y as well as Y and Z are d-connected given $\varnothing$ by two paths $\hbox {X} --\,\hbox {Y}$ and $\hbox {Y}--\,\hbox {Z}$, whence (a) these two paths don’t carry a common effect. (F) implies that X and Z are d-separated by Y which together with (a) implies (b) that $\hbox {X}-\cdots \rightarrow \hbox {Y}\leftarrow \cdots -\hbox {Z}$ is impossible. (a) + (b) entail that either $\hbox {X}\leftarrow \leftarrow \hbox {Y}$ or $\hbox {Y}\rightarrow \rightarrow \hbox {Z}$. But both possibilities are excluded by condition (T).

For (5.2): Because of (F) and INDEP(X,Z), X and Z are d-separated (by $\varnothing$). Thus and because of (C) and $\hbox {DEP}(\hbox {X,Z}{\vert }\hbox {Y})$, X and Z are d-connected by a path ${\uppi }$ that carries a common effect Y$^{\prime }$ that is either identical with Y or has Y as an effect. So ${\uppi }{:}\;\hbox {X}-{\cdots }\rightarrow \hbox {Y}^{\prime } \leftarrow {\cdots }-\hbox {Z}$, where ${\uppi }$ contains no colliders except Y$^{\prime }$. By condition (T) and assumption (*), this path cannot carry a common cause of Y$^{\prime }$ and X and one of Y$^{\prime }$ and Z. So either (a) $\hbox {X}\rightarrow \rightarrow \hbox {Y}^{\prime }$ or (b) $\hbox {Y}^{\prime } \leftarrow \leftarrow \hbox {Z}$ must be a subpath of ${\uppi }$. Both cases are excluded by (T): In case (a), t(Y) $<$ t(X) holds by assumption (*), whence also t(Y$^{\prime }$) $<$ t(X) must hold (since either Y$^{\prime }$ = Y or $\hbox {Y}^{\prime } \rightarrow \rightarrow \hbox {Y}$, which implies by (T) that $\hbox {t}(\hbox {Y}^{\prime }) < \hbox {t(Y)}$). The same argument applies to case (b). $\square$

Proof of Theorem 6

For (6.1): By DEP(X,Y) and (C), X and Y are d-connected (by $\varnothing$), and by (T) and t(X) = t(Y), X and Y can only be d-connected by common causes Z lying in their past. By (C), conditionalization on the set U of all such common causes must screen off X from Y.

For (6.2): Assume Z is in the future of X and Y and screens off X from Y, whence DEP(X,Y) and $\hbox {INDEP}(\hbox {X,Y}{\vert }\hbox {Z})$. It follows from the axioms of decomposition and contraction (similar as in the proof of Theorem 5.1) that DEP(X,Z) and DEP(Y,Z) hold. Thus, X and Y, X and Z, and also Y and Z must be d-connected given $\varnothing$. From this together with (C), (T) and $\hbox {t(Z)} > \hbox {t(X)} = \hbox {t(Y)}$ it follows that X and Y must be d-connected by a common cause path ${\uppi }_{\mathbf{U}}{:}\; \hbox {X}\leftarrow \leftarrow \mathbf{U}\rightarrow \rightarrow \hbox {Y}$ (where U is the set of all common causes of X and Y), and that X and Z must be d-connected by a cause–effect or a common cause path; the same holds for Y instead of X. So there will also be a path $\hbox {X}-{\cdots }\rightarrow \hbox {Z}^{\prime } \leftarrow {\cdots }-\hbox {Y}$, where Z$^{\prime }$ is the only collider on this path and either (i) Z$^{\prime }$ = Z or (ii) $\hbox {Z}^{\prime } \rightarrow \rightarrow \hbox {Z}$ holds. In what follows we write Z for $Z^{\prime }.$

Let $\mathbf{P} = \hbox {par(X)}\cup \hbox {par(Y)}$ be the set of all parents of X and of Y. By the Markov-condition (M) and (T), P is a past screen-off set for {X,Y} (though a redundant one: par(X) or par(Y) alone is one, too); so $\hbox {INDEP}(\hbox {X,Y}{\vert }\mathbf{P})$ holds. Note also that for some p, DEP(X,p) and DEP(Y,p) must hold, since otherwise, by the proof in footnote 10, the path $\pi _{\mathbf{U}}$ that d-connects X and Y could not transmit dependence between X and Y.

There are two possible cases: either (A) $\hbox {INDEP}(\hbox {X,Y}{\vert }\mathbf{P}\cup \{\hbox {Z}\})$ or (B) $\hbox {DEP}(\hbox {X,Y}{\vert }\mathbf{P}\cup \{\hbox {Z}\})$. Assume (B) is the case. Then we have $\hbox {INDEP}(\hbox {X,Y}{\vert }{Z})$ and $\hbox {DEP}(\hbox {X,Y}{\vert }\mathbf{P}\cup \{\hbox {Z}\})$, i.e. X and Y are linked up and thus d-connected given Z conditional on P, but not unconditionally. In other words, conditionalizing on P isolates the common effect path $\hbox {X}-\cdots \rightarrow \hbox {Z}\leftarrow \cdots -\hbox {Y}$ between X and Y which was exactly canceled by $\pi _\mathbf{U}$ before. This would be a case of cancelation unfaithfulness, which is, according to (RF), highly improbable.

Now assume case (A), $\hbox {INDEP}(\hbox {X,Y}{\vert }\mathbf{P}\cup \{\hbox {Z}\})$. Since X and Y are d-connected given $\mathbf{P}\cup \{\hbox {Z}\}$, this constitutes, again, a case of unfaithfulness. We will show that it is (a) either one of cancelation and, hence, made improbable by axiom (RF), or (b) a case of deterministic dependence of some X- or P-value on every $\hbox {z}\in \hbox {Val(Z)}$. Let us assume that (b) is false, i.e.:

(*) $\forall \mathbf{p}\forall \hbox {x}\exists \hbox {z}{:}\;\hbox {P}(\mathbf{p},\!\hbox {x,z}) \ne 0,1$.

Without restricting the assumption of our proof we can assume that the prior probabilities of all values of P, X, and Z are positive; simply by removing all values with zero-probability from the value-space.

We compute, attaching indices to the identity signs for easy reference:

Identity “$=_{3}$” holds by assumption $\hbox {INDEP}(\hbox {X,Y}{\vert }\hbox {Z})$, given (*). (Note that by the definition of “$\hbox {INDEP}(\hbox {X,Y}{\vert }\hbox {Z})$” $\forall \hbox {x,y,z}{:}\;\hbox {P}(\hbox {y}{\vert }\hbox {x,z}) = \hbox {P}(\hbox {y}{\vert }\hbox {z})$ or $\hbox {P}(\hbox {x}{\vert }\hbox {z}) = 0$ or $\hbox {P(z)} = 0$ holds; the latter case is excluded by our assumption of positive prior probabilities.) The identities “$=_{1}$” and “$=_{5}$” hold by probability theory, and “$=_{2}$” follows from case (A) (plus assumption (*) and positive priors). This gives identity “$=_{4}$”. The identity “$=_{3}$” can only hold in two cases:

Case (A.1): $\hbox {DEP}(\hbox {Y},\!\mathbf{P}{\vert }\hbox {Z})$, i.e. there exist at least two distinct values $\hbox {P}(\hbox {Y}{\vert }\mathbf{p}_{1},\!\hbox {Z})\ne \hbox {P}(\hbox {Y}{\vert }\mathbf{p}_{2},\!\hbox {Z})$ in the above sums at the right hand side. Assume $\hbox {P}(\mathbf{p}{\vert }\hbox {X,Z})$ differs from $\hbox {P}(\mathbf{p}{\vert }\hbox {Z})$ for some P-values p. The values of $\hbox {P}(\mathbf{p}{\vert }\hbox {X,Z})$ and $\hbox {P}(\mathbf{p}{\vert }\hbox {Z})$ are weights whose sums always add up to one. Since the differences in these weights do not change the resulting sums $(\Sigma _{\mathrm{p}}\hbox {P}(\hbox {Y}{\vert }\mathbf{p},\hbox {Z}) \cdot \hbox { weight}_{\mathrm{p}})$, this can only be because these differences are exactly canceled by the differences in the $\hbox {P}(\hbox {Y}{\vert }\mathbf{p}{,}\hbox {Z})$-values. This would constitute a case of unfaithfulness due to internal canceling paths (in the sense of Näger, see Sect. 3.3), which is made improbable by axiom (RF). So we infer $\hbox {P}(\mathbf{p}{\vert }\hbox {X,Z}) = \hbox {P}(\mathbf{p}{\vert }\hbox {Z})$ for all P-values p, i.e. $\hbox {INDEP}(\hbox {X},\!\mathbf{P}{\vert }\hbox {Z})$.

• Case (A.2): $\hbox {INDEP}(\hbox {Y},\!\mathbf{P}{\vert }\hbox {Z})$. From the two cases (A.1+2) we conclude:

• Case (A*): $\hbox {INDEP}(\hbox {X},\!\mathbf{P}{\vert }\hbox {Z}) \vee \hbox {INDEP}(\hbox {Y},\!\mathbf{P}{\vert }\hbox {Z})$, i.e. Z screens off either X or Y from P.

For the rest of the proof we assume

• (a) $\hbox {INDEP}(\hbox {X},\!\mathbf{P}{\vert }\hbox {Z})$.

If $\hbox {INDEP}(\hbox {Y},\!\mathbf{P}{\vert }Z)$, X and Y change their roles and the proof works in exactly the same way.

By the remarks above, the causal structure between X, P and Z has one of the following three forms:

Note that the path(s) from P to Z in (i) and (ii) may or may not include a common cause path (that may go through Y); this is indicated by “$- \rightarrow$”.

Assume causal structure (i): We assume (+) $\hbox {DEP}(\hbox {X,Z}{\vert }\mathbf{P})$—otherwise the case is treated exactly as in the proof for structure (ii), which rests on condition $\hbox {INDEP}(\hbox {X,Z}{\vert }\mathbf{P})$. Likewise we assume (++) $\hbox {DEP}(\mathbf{P},\!\hbox {Z}{\vert }\hbox {X})$—otherwise the case is treated exactly as in the proof for structure (iii), which rests on $\hbox {INDEP}(\mathbf{P},\!\hbox {Z}{\vert }\hbox {X})$.

From (+) and (++) it follows that DEP(Z,P) must hold (otherwise, INDEP(P,Z) and $\hbox {INDEP}(\mathbf{P}{,}\hbox {X}{\vert }\hbox {Z})$ would imply by the axioms of contraction and decomposition INDEP(P,X), which contradicts DEP(P,X)). This means that we have a case of cancelation unfaithfulness: though both paths $\mathbf{P}\rightarrow \hbox {X}$ and $\mathbf{P}-\cdots \rightarrow \hbox {Z}\leftarrow \hbox {X}$ transmit probabilistic P–X-dependence when conditionalizing on Z, they exactly compensate each other. This case is made improbable by axiom (RF).

Assume causal structure (ii): Here we have $\hbox {INDEP}(\hbox {X,Z}{\vert }\mathbf{P})$ by the causal Markov condition (M).¹⁸ From $\hbox {INDEP}(\hbox {X,Z}{\vert }\mathbf{P}), \hbox {INDEP}(\hbox {X},\!\mathbf{P}{\vert }\hbox {Z})$ and our assumption of positive priors we get:

• (b) $\forall \hbox {x,z},\!\mathbf{p}{:}\;\hbox {P}(\hbox {x}{\vert }\hbox {z},\!\mathbf{p}) = \hbox {P}(\hbox {x}{\vert }\hbox {z})\vee \hbox {P}(\mathbf{p}{\vert }\hbox {z}) = 0$, and

• (c) $\forall \hbox {x,z},\!\mathbf{p}{:}\;\hbox {P}(\hbox {x}{\vert }\hbox {z},\!\mathbf{p}) = \hbox {P}(\hbox {x}{\vert }\mathbf{p}) \vee \hbox {P}(\mathbf{p}{\vert }\hbox {z}) = 0$.

• (d) By DEP(X,P) there exist $\mathbf{p}_{1},\!\mathbf{p}_{2}$ such that $\hbox {P}(\hbox {x}{\vert }\mathbf{p}_{1}) \ne \hbox {P}(\hbox {x}{\vert }\mathbf{p}_{2})$.

Thus either $\forall \hbox {z}{:}\;\hbox {P}(\mathbf{p}_{1}{\vert }\hbox {z}) = 0 \,\vee \, \hbox {P}(\mathbf{p}_{2}{\vert }\hbox {z}) = 0$; or for some z: $\hbox {P}(\hbox {x}{\vert }\mathbf{p}_{\mathrm{i}}) = \hbox {P}(\hbox {x}{\vert }\hbox {z})$ holds by (b) + (c) for i = 1 and i = 2, which contradicts (d). So $\forall \hbox {z}{:}\;\hbox {P}(\mathbf{p}_{1}{\vert }\mathrm{z}) = 0\, \vee \,\hbox {P}(\mathbf{p}_{2}{\vert }\hbox {z}) = 0)$, i.e. every Z-value has some P-value that depends deterministically on it.

Finally assume causal structure (iii): Here we have $\hbox {INDEP}(\mathbf{P},\!\hbox {Z}{\vert }\hbox {X})$ since P and Z are d-separated by X. $\hbox {INDEP}(\mathbf{P}{,}\hbox {Z}{\vert }\hbox {X})$ and $\hbox {INDEP}(\mathbf{P}{,}\hbox {X}{\vert }\hbox {Z})$ plus positive priors give us (similarly as in case (ii))

• (b${^\prime }$)$\,\,\forall \hbox {x,z},\!\mathbf{p}{:}\;\hbox {P}(\mathbf{p}{\vert }\hbox {z,x}) = \hbox {P}(\mathbf{p}{\vert }\hbox {x}) \vee \hbox { P}(\hbox {x}{\vert }\hbox {z}) = 0$, and

• (c${^\prime }$) $\forall \hbox {x,z},\!\mathbf{p}{:}\;\hbox {P}(\mathbf{p}{\vert }\hbox {z,x}) = \hbox {P}(\mathbf{p}{\vert }\hbox {z}) \vee \hbox { P}(\hbox {x}{\vert }\hbox {z}) = 0$.

• (d${^\prime }$) By DEP(X,P) exist $\hbox {x}_{1},\hbox {x}_{2}$ such that $\hbox {P}(\mathbf{p}{\vert }\hbox {x}_{1}) \ne \hbox {P}(\mathbf{p}{\vert }\hbox {x}_{2})$.

Thus either $\forall \hbox {z}{:}\;\hbox {P}(\hbox {x}_{1}{\vert }\hbox {z}) = 0 \vee \hbox {P}(\hbox {x}_{2}{\vert }\hbox {z}) = 0$; or for some z: $\hbox {P}(\mathbf{p}{\vert }\hbox {x}_{\mathrm{i}}) = \hbox {P}(\mathbf{p}{\vert }\hbox {z})$ holds by (b$^{\prime }$) + (c$^{\prime }$) for i = 1 and i = 2, which contradicts (d$^{\prime }$). So every Z-value has some X-value that depends deterministically on it.

Thus, assumption (*) must be false, which concludes our proof.$\square$