Halpern (2015) provides three variants of the Halpern-Pearl definitions of actual causation. I’m trying to get my head around the formalism, which is elegant, concise, and precise, but tedious to use in practice so I wrote an R script to do the sums. This blog post shows two worked examples, primarily for my own benefit; however, they may help you too if you’re also struggling with the examples in the paper!
The second (“updated”) definition of an actual cause asserts that \(\vec{A} = \vec{a}\) is a cause of \(\varphi\) in \((M,\vec{u})\) iff the following conditions hold:
AC1 \((M,\vec{u}) \models (\vec{A} =\vec{a}) \land \varphi\).
This says, if \(\vec{A} = \vec{a}\) is an actual cause of \(\varphi\) then they both hold in the actual world, \((M,\vec{u})\). Note, for this condition, we are just having a look at the model and not doing anything to it.
AC2 There is a partition of the endogenous variables in \(M\) into \(\vec{Z} \supseteq \vec{X}\) and \(\vec{W}\) and there are settings \(\vec{x’}\) and \(\vec{w}\) such that
(a) \((M,\vec{u}) \models [ \vec{X} \leftarrow \vec{x’}, \vec{W} \leftarrow \vec{w}] \neg \varphi\).
So, we’re trying to show that undoing the cause, i.e., setting \(\vec{X}\) to \(\vec{x’} \ne \vec{x}\), prevents the effect. We are allowed to modify \(\vec{W}\) however we want to show this, whilst leaving \(\vec{Z}-\vec{X}\) free to do whatever the model tells these variables to do.
(b) If \((M,\vec{u}) \models \vec{Z} = \vec{z^{\star}}\), for some \(\vec{z^{\star}}\), then for all \(\vec{W’} \subseteq \vec{W}\) and \(\vec{Z’} \subseteq \vec{Z}-\vec{X}\),
\((M,\vec{u}) \models [ \vec{X} \leftarrow \vec{x}, \vec{W’} \leftarrow \vec{w’}, \vec{Z’} \leftarrow \vec{z^{\star}}] \varphi\).
This says, trigger the cause (unlike AC1, we aren’t just looking to see if it holds) and check whether it leads to the effect under all subsets of \(\vec{Z}\) (as per actual world) that aren’t \(\vec{X}\) and all subsets of the modified \(\vec{W}\) that we found for AC2(a). Note how we are setting \(\vec{Z}\) for those subsets, rather than just observing it.
AC3 There is no \(\vec{A’} \subset \vec{A}\) such that \(\vec{A’} = \vec{a’}\) satisfies AC1 and AC2.
This says, there’s no superfluous stuff in \(\vec{A}\). You taking a painkiller and waving a magic wand doesn’t cause your headache to disappear, under AC3, if the painkiller works without the wand.
Example 1: an (actual) actual cause
Let’s give it a go with an overdetermined scenario (lightly edited from Halpern) that Alice and Bob both lob bricks at a glasshouse and smash the glass. Define
\(\mathit{AliceThrow} = 1\)
\(\mathit{BobThrow} = 1\)
\(\mathit{GlassBreaks} = \mathit{max}(\mathit{AliceThrow},\mathit{BobThrow})\)
So, if either Alice or Bob (or both) hit the glasshouse, then the glass breaks. Strictly speaking, I should have setup one or more exogenous variables, \(\vec{u}\), that define the context and then defined \(\mathit{AliceThrow}\) and \(\mathit{BobThrow}\) in terms of \(\vec{u}\), but it works fine to skip that step as I have here since I’m holding \(\vec{u}\) constant anyway.
Is \(\mathit{AliceThrow} = 1\) an actual cause of \(\mathit{GlassBreaks} = 1\)?
AC1 holds since \((M,\vec{u}) \models \mathit{AliceThrow} = 1 \land \mathit{GlassBreaks} = 1\). The first conjunct comes directly from one of the model equations. Spelling out the second conjunct,
\(\mathit{GlassBreaks} = \mathit{max}(\mathit{AliceThrow},\mathit{BobThrow})\)
\(= \mathit{max}(1, 1)\)
\(= 1\)
For AC2, we need to find a partition of the endogenous variables such that AC2(a) and AC2(b) hold. Try \(\vec{Z} = \{ \mathit{AliceThrow}, \mathit{GlassBreaks} \}\) and \(\vec{W}= \{ \mathit{BobThrow} \}\).
AC2(a) holds since \((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 0, \mathit{BobThrow} \leftarrow 0] \mathit{GlassBreaks} = 0\).
For AC2(b), we begin with \(\vec{Z} = \{ \mathit{AliceThrow}, \mathit{GlassBreaks} \}\) and the settings as per the unchanged model, so
\((M,\vec{u}) \models \mathit{AliceThrow} = 1 \land \mathit{GlassBreaks} = 1\).
We need to check that for all \(\vec{W’} \subseteq \vec{W}\) and \(\vec{Z’} \subseteq \vec{Z}-\vec{X}\),
\((M,\vec{u}) \models [ \vec{X} \leftarrow \vec{x}, \vec{W’} \leftarrow \vec{w’}, \vec{Z’} \leftarrow \vec{z^{\star}}] \varphi\).
Here are the combinations and \(\varphi \equiv \mathit{GlassBreaks} = 1\) holds for all of them:
\((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 1, \mathit{GlassBreaks} \leftarrow 1, \mathit{BobThrow} \leftarrow 0 ] \varphi\)
\((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 1, \mathit{BobThrow} \leftarrow 0 ] \varphi\)
\((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 1, \mathit{GlassBreaks} \leftarrow 1 ] \varphi\)
\((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 1 ] \varphi \)
AC3 is easy since the cause only has one variable, so there’s nothing superfluous.
Example 2: not an actual cause
Now let’s try an example that isn’t an actual cause: the glass breaking causes Alice to throw the brick. It’s obviously false; however, it wasn’t clear to me exactly where it would fail until I worked through this…
AC1 holds since in the actual world, \(\mathit{GlassBreaks} = 1\) and \(\mathit{AliceThrow} = 1\) hold.
Examining the function defintions, they don’t provide a way to link \(\mathit{AliceThrow}\) to a change in \(\mathit{GlassBreaks}\), so the only apparent way to do so is through \(\vec{W}\). Therefore, use the partition \(\vec{W} = \{\mathit{AliceThrow}\}\) and \(\vec{Z} = \{\mathit{GlassBreaks}, \mathit{BobThrow}\}\).
Now for AC2(a), we can easily get \(\mathit{AliceThrow} = 0\) as required, since we can do what we like with \(\vec{W}\). It doesn’t help when we move onto AC2(b) since we have to hold \(\mathit{AliceThrow} = 0\), which is the negation of what we want. The same is the case for the other partition including \(\mathit{AliceThrow}\) in \(\vec{W}\), i.e., \(\vec{W} = \{ \mathit{AliceThrow}, \mathit{BobThrow} \}\).
So, the broken glass does not cause Alice to throw a brick. The setup we needed to get through AC2(a) set us up to fail AC2(b).