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 is not self-contained – you will need to read the original paper for an introduction to the model. However, it works through two examples, which may help if you’re also struggling with 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 and none of the functions change it. 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:
\((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 1, \mathit{GlassBreaks} \leftarrow 1, \mathit{BobThrow} \leftarrow 0 ] \varphi\)
\((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 1, \mathit{BobThrow} \leftarrow 0 ] \varphi\)
\((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 1, \mathit{GlassBreaks} \leftarrow 1 ] \varphi\)
\((M,\vec{u}) \models [ \mathit{AliceThrow} \leftarrow 1 ] \varphi \)
(The third was rather trivially true; however, as far as I understand, has to be checked given the definition.)
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).
References
Halpern, J. Y. (2015). A Modification of the Halpern-Pearl Definition of Causality. Proceedings of the Twenty-Fourth International Joint Conference on Artificial Intelligence (IJCAI 2015), 3022–3033.
See also this companion blog post.