Indicative and counterfactual conditionals: a causal-modeling semantics

Abstract

We construct a causal-modeling semantics for both indicative and counterfactual conditionals. As regards counterfactuals, we adopt the orthodox view that a counterfactual conditional is true in a causal model M just in case its consequent is true in the submodel $M^*$, generated by intervening in M, in which its antecedent is true. We supplement the orthodox semantics by introducing a new manipulation called extrapolation. We argue that an indicative conditional is true in a causal model M just in case its consequent is true in certain submodels $M^*$, generated by extrapolating M, in which its antecedent is true. We show that the proposed semantics can account for some important minimal pairs nicely and naturally. We also prove a theorem showing under what conditions intervention and extrapolation will yield the same result, and thus explain how counterfactual and indicative conditionals would behave in a causal-modeling semantics.

Appendices

Appendix 1: Proof that $\le$ is a total preorder on ${\mathbb {M}}^{[\varphi ]}$

Recall our definition:

Definition

Let $M=\langle U, V, F, A\rangle$ be a causal model, and ${M_{1}}=\langle U, V, F, A_{1}\rangle$ and ${M_{2}}=\langle U, V, F, A_{2}\rangle$ be submodels in ${\mathbb {M}}^{[\varphi ]}$. Define $M_1\sqsubseteq M_2$ if and only if either of the two conditions holds:

• (i) for every $X\in U\cup V$, $A_1\approx _X A_2$, or

• (ii) for some $X\in U\cup V$, $A_1<_X A_2$, and for any $Y\in U\cup V$ such that $X\prec Y$, $A_1\approx _Y A_2$.

We then define $\le$ to be the transitive closure of $\sqsubseteq$.

Proof

It suffices to show that $\sqsubseteq$ is a total relation on ${\mathbb {M}}^{[\varphi ]}$. To show this, consider any submodels $M_1=\langle U, V, F, A_1\rangle$ and $M_2=\langle U, V, F, A_2\rangle$ in ${\mathbb {M}}^{[\varphi ]}$. If condition (i) holds, then clearly both $M_1\sqsubseteq M_2$ and $M_2\sqsubseteq M_1$ hold. Now, suppose condition (i) does not hold. So, for some $X\in U\cup V$, $A_1\not \approx _X A_2$, and hence if we define $N=\{X\in U\cup V\mid A_1\not \approx _X A_2\}$ then N is non-empty. Now, since $\preccurlyeq$ is a total preorder on a finite set $U\cup V$, there is some X in N such that for any $Y\in N$, $Y\preccurlyeq X$. Hence, for any $Y\in U\cup V$ such that $X\prec Y$, we have $Y\notin N$, i.e., $A_1\approx _Y A_2$. But since $X\in N$, $A_1\not \approx _X A_2$, and so either $A_1 <_X A_2$ or $A_2 <_X A_1$ (by the definition of $<_X$). Hence, either $M_1\sqsubseteq M_2$ or $M_2\sqsubseteq M_1$. $\square$

Remark

Notice that it is necessary to take the transitive closure, for the relation $\sqsubseteq$ as defined by (i) and (ii) above is not guaranteed to be transitive. This can be shown by a simple example. Consider a causal model M which contains only three variables, X, Y, and Z, with the following value assignments: $A(X)=A(Y)=A(Z)=0$. Suppose our priority relation is given as follows:

$$\begin{aligned} X\prec Y\approx Z \end{aligned}$$

Suppose that ${\mathbb {M}}^{[\varphi ]}$ contains three submodels, $M_1$, $M_2$, and $M_3$, such that

$$\begin{aligned}&A_1(X)=1, A_1(Y)=1, A_1(Z)=0 \\&A_2(X)=0, A_2(Y)=0, A_2(Z)=1 \\&A_3(X)=0, A_3(Y)=1, A_3(Z)=0 \end{aligned}$$

It is easy to check that according to our definition, the relation $\sqsubseteq$ applies to any pair of the three submodels except $\langle M_1, M_3\rangle$. That is to say, we have $M_1\sqsubseteq M_2\sqsubseteq M_3$, but not $M_1\sqsubseteq M_3$. Hence $\sqsubseteq$ may fail to be transitive.

Appendix 2: Proof of the Theorem

Lemma 1

Let $M=\langle U, V, F, A\rangle$ be a causal model, and $M_{(\varphi )}=\langle U_{(\varphi )}, V_{(\varphi )}, F_{(\varphi )}, A_{(\varphi )}\rangle$ be its intervention-generated submodel. Then for any variable X, if $X\notin {\mathcal {D}}{\mathcal {E}}(V\varphi )$, then $A_{(\varphi )}(X)=A(X)$.

Proof

Since we assume that M is recursive, the value of each variable X can be uniquely traced back to the values of those exogenous variables in ${\mathcal {A}}{\mathcal {N}}(X)$. But if $X\notin {\mathcal {D}}{\mathcal {E}}(V\varphi )$, then ${\mathcal {A}}{\mathcal {N}}(X)$ should contain no variable in $V\varphi$, and hence $A_{(\varphi )}$ should assign the same value to each variable in $U_{(\varphi )}\cap {\mathcal {A}}{\mathcal {N}}(X)$ as A does. Hence $A_{(\varphi )}(X)=A(X)$. $\square$

Lemma 2

Let $M=\langle U, V, F, A\rangle$ be a causal model such that ${\mathbb {M}}^{[\varphi ]}\ne \emptyset$. Then there is a submodel $M_0=\langle U, V, F, A_0\rangle \in {\mathbb {M}}^{[\varphi ]}$ such that for any $X\notin {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$ we have $A_0(X) = A(X)$.

Proof

Since ${\mathbb {M}}^{[\varphi ]}\ne \emptyset$, there is some submodel $M^*=\langle U, V, F, A^*\rangle \in {\mathbb {M}}^{[\varphi ]}$. Now, let $M_0=\langle U, V, F, A_0\rangle$ be the submodel such that $A_0(X)=A^*(X)$ for any $X\in U \cap {\mathcal {A}}{\mathcal {N}}(V\varphi )$, and $A_0(X)=A(X)$ for any $X\in U \setminus {\mathcal {A}}{\mathcal {N}}(V\varphi )$. Now, since M is recursive, it follows that $A_0(X)=A^*(X)$ for any $X\in V\varphi$, and $A_0(X) = A(X)$ for any $X\notin {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$. Hence $M_0$ is a submodel in ${\mathbb {M}}^{[\varphi ]}$ with the desired property. $\square$

Theorem

Let $M=\langle U, V, F, A\rangle$ be a causal model such that ${\mathbb {M}}^{[\varphi ]}\ne \emptyset$. Suppose there is a set $S\subseteq U\cup V$ such that (i) $S\cap {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))=\emptyset$, (ii) $S\cup V\varphi$ is a screen to $V\psi$, i.e., for each initial chain C that leads to a variable in $V\psi$, we have $C\cap (S\cup V\varphi )\ne \emptyset$, and (iii) for any $X\in {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$ and any $Y\in S$, we have $X\prec Y$. Then $\varphi >\psi$ is true in M if and only if $\varphi \rightarrow \psi$ is true in M.

Proof

Since ${\mathbb {M}}^{[\varphi ]}\ne \emptyset$, by Lemma 2 there is an $M_0=\langle U, V, F, A_0\rangle \in {\mathbb {M}}^{[\varphi ]}$ such that for any $X\notin {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$ we have $A_0(X) = A(X)$. Now, we claim that every minimal submodel in ${\mathbb {M}}^{[\varphi ]}$ agrees with $M_0$ on the truth value of $\psi$. For suppose the claim is false, then there is a minimal submodel $M^*=\langle U, V, F, A^*\rangle \in {\mathbb {M}}^{[\varphi ]}$ and some variable $X\in V\psi$ such that $A^*(X) \ne A_0(X)$. Now, since $S\cup V\varphi$ is a screen to X, any two submodels of M that agree on the values of all variables in $S\cup V\varphi$ should also agree on the value of X. But $A^*(X) \ne A_0(X)$, it follows that for some $Y\in S\cup V\varphi$, we have $A^*(Y) \ne A_0(Y)$. Clearly, for any $Z\in V\varphi$, we have $A_0(Z) = A^*(Z) = A_{(\varphi )}(Z)$, because $\varphi$ is true in $M_0$ and in $M^*$ and in $M_{(\varphi )}$. So, $Y\notin V\varphi$, and hence $Y\in S$. But $S\cap {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))=\emptyset$, so $Y\in S$ implies $Y\notin {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$, and hence we have $A_0(Y) = A(Y)$. So, we have $A_0 <_Y A^*$. Now, since for any $Z\in {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$ and $Y\in S$, we have $Z\prec Y$, it follows that for any Z such that $Y\preccurlyeq Z$, we have $Z\notin {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$, and hence $A_0(Z) = A(Z)$. This implies that for any Z such that $Y\preccurlyeq Z$, $A_0 \lessapprox _Y A^*$, and consequently, we have $M_0 < M^*$, which contradicts the assumption that $M^*$ is a minimal submodel.

According to (CC), $\varphi > \psi$ is true in M iff $\psi$ is true in $M_{(\varphi )}$. Now, we claim that $\psi$ is true in $M_{(\varphi )}$ iff $\psi$ is true in $M_0$. To show this, first observe that for any $X\in S$, we have $X\notin {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$, and hence $A_0(X) = A(X)$. But since ${\mathcal {D}}{\mathcal {E}}(V\varphi )\subseteq {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$, it follows that $X\notin {\mathcal {D}}{\mathcal {E}}({\mathcal {A}}{\mathcal {N}}(V\varphi ))$ implies $X\notin {\mathcal {D}}{\mathcal {E}}(V\varphi )$, and hence by Lemma 1, we have $A_{(\varphi )}(X) = A(X)$. So, $M_{(\varphi )}$ and $M_0$ agree on the values of all variables in S. Now, since $S\cup V\varphi$ is a screen to $V\psi$, any two submodels of M that agree on the values of all variables in $S\cup V\varphi$ should also agree on the value of any variable in $V\psi$. But certainly, $M_{(\varphi )}$ and $M_0$ agree on the values of all variables in $V\varphi$ (since $\varphi$ is true in both), and we have shown that they also agree on the values of all variables in S. Consequently, $M_{(\varphi )}$ and $M_0$ should also agree on the truth value of $\psi$. Since we have shown that every minimal submodel in ${\mathbb {M}}^{[\varphi ]}$ agrees with $M_0$ on the truth value of $\psi$, and according to (IC), $\varphi \rightarrow \psi$ is true in M iff $\psi$ is true in every minimal submodel in ${\mathbb {M}}^{[\varphi ]}$, it follows that $\varphi \rightarrow \psi$ is true in M iff $\psi$ is true in $M_0$. Hence, we can conclude that $\varphi > \psi$ is true in M iff $\varphi \rightarrow \psi$ is true in M. $\square$