A theory of structural determination

Abstract

While structural equations modeling is increasingly used in philosophical theorizing about causation, it remains unclear what it takes for a particular structural equations model to be correct. To the extent that this issue has been addressed, the consensus appears to be that it takes a certain family of causal counterfactuals being true. I argue that this account faces difficulties in securing the independent manipulability of the structural determination relations represented in a correct structural equations model. I then offer an alternate understanding of structural determination, and I demonstrate that this theory guarantees that structural determination relations are independently manipulable. The account provides a straightforward way of understanding hypothetical interventions, as well as a criterion for distinguishing hypothetical changes in the values of variables which constitute interventions from those which do not. It additionally affords a semantics for causal counterfactual conditionals which is able to yield a clean solution to a problem case for the standard ‘closest possible world’ semantics.

Appendix

Define the rank of a variable $V \in {\mathcal {U}}\cup {\mathcal {V}}$ recursively as follows:

$$\begin{aligned} rank(V) = 0 \quad&\iff \quad V \in {\mathcal {U}}\\ rank(V) = k+1 \quad&\iff \quad \max \{ rank(P) : P \in {\mathbf {PA}}(V)\} = k \end{aligned}$$

Graphically, a variable’s rank is the largest number of edges lying between that variable and an exogenous variable along a directed path. Let ‘${\mathbf {Rank}}(i)$’ denote the set of all variables of rank $i$, and let ‘${\mathbf {Rank}}(i, j, \ldots , k)$’ denote the union ${\mathbf {Rank}}(i) \cup {\mathbf {Rank}}(j) \cup \dots \cup {\mathbf {Rank}}(k)$.

Lemma 1

Given $({\mathcal {V}}1)$, $({\mathcal {U}}1)$, and $(f1)$, for all $V \in {\mathcal {V}}$, all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, all $\mathbf {x}$, and all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w),\,{\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$, where ${\mathbf {PA}}(V)_{{\mathbf {X}} = {\mathbf {x}}}$ assigns the values to ${\mathbf {PA}}(V)$ determined by the structural equations in ${\mathcal {E}}- \bigcup _i (\phi _{X_i})$, for every endogenous $X_i \in \mathbf {X}$, and $({\mathcal {U}}- {\mathbf {X}})_w \cup \mathbf {x}$.

Proof

By induction on the rank of the variables in ${\mathcal {V}}$. $\square$

Base case

For all $V \in {\mathbf {Rank}}(1)$, all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, all $\mathbf {x}$, and all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w),\,{\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{{\mathbf {X}} = \mathbf {x}}$.

Proof

If $rank(V) = 1$, then every $P \in {\mathbf {PA}}(V)$ is exogenous. Without loss of generality, consider one $P \in {\mathbf {PA}}(V)$. If $P \in \mathbf {X}$, then $f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, P = P_{\mathbf {X}=\mathbf {x}}$ (the value assigned to $P$ by $\mathbf {x}$), by $(f1)$. If $P \notin \mathbf {X}$, then $f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, P = P_w$, by $({\mathcal {U}}1)$. In either case, $P$ takes on the value assigned to it by ${\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$. $\square$

Inductive step

If for all $V \in {\mathbf {Rank}}(1, 2, \dots , k)$, it is true that, for all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, all $\mathbf {x}$, and all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w),\,{\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$, then for all $V \in {\mathbf {Rank}}(k+1)$, it will be true that, for all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, all $\mathbf {x}$, and all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w),\,{\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$.

Proof

Without loss of generality, consider one $V \in {\mathbf {Rank}}(k+1)$, one ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, one $\mathbf {x}$, and one $P \in {\mathbf {PA}}(V)$. Either $P \in \mathbf {X}$ or $P \notin \mathbf {X}$. Suppose that $P \in \mathbf {X}$. Then, $f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, P = P_{\mathbf {X}=\mathbf {x}}$, by $(f1)$. If $P \notin \mathbf {X}$, then, since $rank(P) \leqslant k, {\mathbf {PA}}(P)_{w'} = {\mathbf {PA}}(P)_{\mathbf {X} =\mathbf {x}}$, for all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w)$, by the inductive hypothesis (since $P \notin {\mathbf {X}}, {\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- P)$). Then, $({\mathcal {V}}1)$ guarantees that $f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, P = \phi _P({\mathbf {PA}}(P)_{\mathbf {X} = \mathbf {x}}$). So, whether $P \in \mathbf {X}$ or $P \notin {\mathbf {X}}, P$ takes on the value $P_{\mathbf {X}=\mathbf {x}}$ at every $w' \in f({\mathbf {X}} = {\mathbf {x}}, w)$. Since $P, V, \mathbf {X}$, and $\mathbf {x}$ were arbitrary, for all $V \in {\mathbf {Rank}}(k+1)$, all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, and all ${\mathbf {x}}, {\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$ for every $w' \in f({\mathbf {X}} = {\mathbf {x}}, w)$. $\square$

Lemma 2

Given $({\mathcal {V}}3)$, $({\mathcal {U}}1)$, and $(f1)$, for all $V \in {\mathcal {V}}$, all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, all $\mathbf {x}$, and all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w),\,{\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$, where ${\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$ assigns the values to ${\mathbf {PA}}(V)$ determined by the assignment of values $({\mathcal {U}}- {\mathbf {X}})_w \cup \mathbf {x}$ and the structural equations in ${\mathcal {E}}- \bigcup _i (\phi _{X_i})$, for every endogenous $X_i \in \mathbf {X}$.

Proof

By induction on the rank of the variables in ${\mathcal {V}}$. $\square$

Base case

For all $V \in {\mathbf {Rank}}(1)$, all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, all $\mathbf {x}$, and all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w),\,{\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$.

Proof

Consider, without loss of generality, a variable $V \in {\mathbf {Rank}}(1)$. Since $V$’s rank is 1, every $P \in {\mathbf {PA}}(V)$ is exogenous. If $P \in \mathbf {X}$, then $f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, P = P_{\mathbf {X}=\mathbf {x}}$ (the value assigned to $X$ by $\mathbf {x}$), by $(f1)$. If $P \notin \mathbf {X}$, then $f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, P = P_w$, by $({\mathcal {U}}1)$. In either case, $P$ takes on the value assigned to it by ${\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$. $\square$

Inductive step

If for all $V \in {\mathbf {Rank}}(1, 2, \ldots , k)$, it is true that, for all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, all $\mathbf {x}$, and all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w),\,{\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$, then for all $V \in {\mathbf {Rank}}(k+1)$, it will be true that, for all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, all $\mathbf {x}$, and all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w),\,{\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$.

Proof

Without loss of generality, consider one $V \in {\mathbf {Rank}}(k+1)$, one ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, one $\mathbf {x}$, and one $P \in {\mathbf {PA}}(V)$. Either $P \in \mathbf {X}$ or $P \notin \mathbf {X}$. Suppose that $P \in \mathbf {X}$. Then, $f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, P = P_{\mathbf {X}=\mathbf {x}}$, by $(f1)$. If $P \notin \mathbf {X}$, then, since $rank(P) \leqslant k, {\mathbf {PA}}(P)_{w'} = {\mathbf {PA}}(P)_{\mathbf {X} = \mathbf {x}}$, for all $w' \in f({\mathbf {X}} = {\mathbf {x}}, w)$, by the inductive hypothesis (since $P \notin \mathbf {X}, \mathbf {X} \subseteq {\mathcal {U}}\cup ( {\mathcal {V}}- P)$). Then, $({\mathcal {V}}3)$ and $({\mathfrak {F}}1)$ guarantee that $f({\mathbf {X}} = {\mathbf {x}}, w) \subset {\mathfrak {F}}_P$ and ${\mathfrak {F}}_P \,\models \, P = \phi _P({\mathbf {PA}}(P))$. So $f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, P = \phi _P({\mathbf {PA}}(P)_{\mathbf {X} = \mathbf {x}}).$ So, whether $P \in \mathbf {X}$ or $P \notin {\mathbf {X}}, P$ takes on the value $P_{\mathbf {X}=\mathbf {x}}$ at every $w' \in f({\mathbf {X}} = {\mathbf {x}}, w)$. Since $P, V, \mathbf {X}$, and $\mathbf {x}$ were arbitrary, for all $V \in {\mathbf {Rank}}(k+1)$, all $\mathbf {X} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, and all ${\mathbf {x}}, {\mathbf {PA}}(V)_{w'} = {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}$ for every $w' \in f(\mathbf {X} = \mathbf {x}, w)$. $\square$

Theorem 1

Given (f1), in a causal model ${\mathcal {M}}= ( {\mathcal {U}}, {\mathcal {V}}, {\mathcal {E}})$, if ${\mathcal {U}}$ satisfies $({\mathcal {U}}1)$, then ${\mathcal {V}}$ satisfies $({\mathcal {V}}3)$ iff ${\mathcal {V}}$ satisfies $({\mathcal {V}}1)$.

Proof

First assume that ${\mathcal {V}}$ satisfies $({\mathcal {V}}1)$. Then, we know that for all $V \in {\mathcal {V}}$, all ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, and all assignments $\mathbf {x}$ to $\mathbf {X}$,

$$\begin{aligned} f({\mathbf {X}} = {\mathbf {x}}, w) \,\models \, V = \phi _V( {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}}) \end{aligned}$$

By Lemma 1, it then follows that

$$\begin{aligned} \forall w' \in f({\mathbf {X}} = {\mathbf {x}}, w), V_{w'} = \phi _V( {\mathbf {PA}}(V)_{w'}) \end{aligned}$$

So

$$\begin{aligned} f({\mathbf {X}} ={ \mathbf {x}}, w) \,\models \, V = \phi _V( {\mathbf {PA}}(V)) \end{aligned}$$

So, if for every $V$, every $\mathbf {X} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, and every $\mathbf {x}$, we include every $w' \in f({\mathbf {X}} = {\mathbf {x}}, w)$ in ${\mathfrak {F}}_V$, then we will have a set ${\mathfrak {F}}_V$ which satisfies $({\mathfrak {F}}1)$ and which entails that $V = \phi _V( {\mathbf {PA}}(V))$. So every $V \in {\mathcal {V}}$ will satisfy $({\mathcal {V}}3)$.

To establish the other direction, assume that ${\mathcal {V}}$ satisfies $({\mathcal {V}}3)$. Then, for every $V \in {\mathcal {V}}$, every ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ({\mathcal {V}}- V)$, and every ${\mathbf {x}}, f({\mathbf {X}} ={\mathbf {x}}, w) \in {\mathfrak {F}}_V$ and ${\mathfrak {F}}_V \,\models \, V = \phi _V({\mathbf {PA}}(V))$. By Lemma 2, it then follows that, for all $V, \mathbf {X}$, and $\mathbf {x}$,

$$\begin{aligned} f({\mathbf {X}} ={\mathbf {x}}, w) \,\models \, V = \phi _V( {\mathbf {PA}}(V)_{\mathbf {X} = \mathbf {x}} ) \end{aligned}$$

So ${\mathcal {V}}$ must satisfy $({\mathcal {V}}1)$ as well. $\square$

Theorem 2

Given (f3), if $\phi _V$ belongs to a causal model satisfying $({\mathcal {V}}4)$ at a world $w_0$, then $\phi _V$ will continue to belong to a causal model satisfying $({\mathcal {V}}4)$ after any number of consecutive hypothetical interventions to set the values of any ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ( {\mathcal {V}}- V)$.

Proof

By induction on the number of interventions. $\square$

Inductive step

If $\phi _V$ belongs to a causal model satisfying $({\mathcal {V}}4)$ at world $w_k$ after $k$ interventions to set the values of any ${\mathbf {X}} \subseteq {\mathcal {U}}\cup ( {\mathcal {V}}- V)$, then $\phi _V$ will belong to a causal model satisfying $({\mathcal {V}}4)$ at the world $w_{k+1}$ where there is a $k+1^{st}$ intervention to set the values of any $\mathbf {X} \subseteq {\mathcal {U}}\cup ( {\mathcal {V}}- V)$.

Proof

By the inductive hypothesis, $\phi _V$ belongs to a causal model satisfying $({\mathcal {V}}4)$ at $w_k$. This means that there must exist a set of worlds ${\mathfrak {F}}_V$ which satisfies $({\mathfrak {F}}2)$ and which contains $w_k$. By $(f3)$, an intervention setting the value of some $\mathbf {X} \subseteq {\mathcal {U}}\cup ( {\mathcal {V}}- V)$ to $\mathbf {x}$ must take us to a world $w_{k+1} \in f({\mathbf {X}} = {\mathbf {x}}, w_k)$. Since $w_k \in {\mathfrak {F}}_V, ({\mathfrak {F}}2)$ guarantees that $f({\mathbf {X}} = {\mathbf {x}}, w_k) \subseteq {\mathfrak {F}}_V$, so $w_{k+1} \in {\mathfrak {F}}_V$ as well. And, by assumption, ${\mathfrak {F}}_V \,\models \, V = \phi _V({\mathbf {PA}}(V))$. So there is a ${\mathfrak {F}}_V \ni w_{k+1}$ such that ${\mathfrak {F}}_V$ satisfies $({\mathfrak {F}}2)$ and ${\mathfrak {F}}_V \,\models \, V = \phi _V({\mathbf {PA}}(V))$. As $\phi _V$ was arbitrary, the same holds for every $V' \notin \mathbf {X}$. So, $({\mathcal {V}}4)$ will hold at $w_{k+1}$. So $\phi _V$ will belong to a causal model satisfying $({\mathcal {V}}4)$ at $w_{k+1}$.

Setting $k = 0$ in the proof of the inductive step establishes the base case. ${\square}$

Theorem 3

Given (f1), if a causal model ${\mathcal {M}}= ( {\mathcal {U}}, {\mathcal {V}}, {\mathcal {E}})$ is correct according to $({\mathcal {M}}2)$, then ${\mathfrak {F}}_{\mathcal {M}}$ $\mathop {=}\limits ^\mathrm{{{def}}}$ $\bigcap _{V \in {\mathcal {V}}} {\mathfrak {F}}_{V}$ contains all and only allowed assignment of values to the variables $V \in {\mathcal {U}}\cup {\mathcal {V}}$, where an assignment is allowed just in case it is a solution to the equations in ${\mathcal {E}}$.

Proof

The proof proceeds by induction on the rank of the variables in ${\mathcal {V}}$. $\square$

Base case

${\mathfrak {F}}_{\mathcal {M}}$ contains all and only allowed assignment of values to the variables in ${\mathbf {Rank}}(0)$.

Proof

For every $V \in {\mathcal {V}}, {\mathfrak {F}}_V$ contains $f({\mathcal {U}}= {\mathbf {u}}, w)$, for every assignment $\mathbf {u}$ to ${\mathcal {U}}$, and every $w \in {\mathfrak {F}}_V$. So ${\mathfrak {F}}_{\mathcal {M}}$ contains $f({\mathcal {U}}= {\mathbf {u}}, w)$, for every assignment $\mathbf {u}$ to ${\mathcal {U}}$ and every $w \in {\mathfrak {F}}_{\mathcal {M}}$. If $U \in {\mathbf {Rank}}(0)$, then $U$ is exogenous, $U \in {\mathcal {U}}$. Every assignment of values to the exogenous variables is allowed. So ${\mathfrak {F}}_{\mathcal {M}}$ contains all and only allowed assignments to the variables in ${\mathbf {Rank}}(0)$. $\square$

Inductive step

If ${\mathfrak {F}}_{\mathcal {M}}$ contains all and only allowed assignment of values to the variables in ${\mathbf {Rank}}(0, 1, \ldots , k)$, then it contains all and only allowed assignment of values to the variables in ${\mathbf {Rank}}(0, 1, \ldots , k, k+1)$.

Proof

Take an arbitrary $V \in {\mathbf {Rank}}(k+1)$. Since ${\mathbf {PA}}(V)\subseteq {\mathbf {Rank}}(0, 1, \dots , k)$, the inductive hypothesis gets us that every and only the allowed assignment of values to ${\mathbf {PA}}(V)$ are realized in ${\mathfrak {F}}_{\mathcal {M}}$. And because ${\mathcal {M}}$ is correct, $V_w = \phi ({\mathbf {PA}}(V)_w)$, for every $w \in {\mathfrak {F}}_{V}$. Since ${\mathfrak {F}}_{\mathcal {M}}\subseteq {\mathfrak {F}}_{V}$, this means that $V_w = \phi ({\mathbf {PA}}(V)_w)$ for every $w \in {\mathfrak {F}}_{\mathcal {M}}$ as well. So ${\mathfrak {F}}_{\mathcal {M}}$ contains all and only the allowed values of $V$. Since $V$ was arbitrary, the above holds for every $V \in {\mathbf {Rank}}(k+1)$. $\square$

Theorem 4

Given $({\mathcal {M}}2)$ and (f3), for any $\mathbf {V} \,\subsetneq \,{\mathcal {V}}$,

$$\begin{aligned} \bigcap _{W \notin {\mathbf {V}}} {\mathfrak {F}}_{W} - \bigcup _{V \in \mathbf {V}} {\mathfrak {F}}_{V} \end{aligned}$$

is non-empty and contains every assignment of values to the variables in ${\mathbf{V}} \cup {\mathcal {U}}$.

Proof

Take an arbitrary $\mathbf {V} \,\subsetneq\, {\mathcal {V}}$, an arbitrary assignment of values $\mathbf {v}$ to $\mathbf {V}$, an arbitrary $W \notin \mathbf {V}$, and an arbitrary assignment $\mathbf {u}$ to ${\mathcal {U}}$. Then, ${\mathfrak {F}}_W$ contains worlds at which $\mathbf {V} \cup {\mathcal {U}}$ is set to $\mathbf {v} \cup \mathbf {u}$ by an intervention, by $({\mathfrak {F}}2)$ and $(f3)$. These worlds are not in $\bigcup _{V \in \mathbf {V}} {\mathfrak {F}}_V$, by the definition of an intervention. Since $W, \mathbf {V}, \mathbf {v}, \mathbf {u}$, and $V$ were arbitrary, for every $W \notin \mathbf {V}$, every $\mathbf {u}$, and every $V \in \mathbf {V}$, there are worlds in ${\mathfrak {F}}_W$ which are not in ${\mathfrak {F}}_V$ and at which the value of $V \cup {\mathcal {U}}$ is set to any value $\mathbf {v} \cup \mathbf {u}$. Thus, $\bigcap _{W \notin \mathbf {V}} {\mathfrak {F}}_W - \bigcup _{V \in \mathbf {V}} {\mathfrak {F}}_V$ is non-empty and contains every assignment of values to the variables in $\mathbf {V} \cup {\mathcal {U}}$. $\square$