Magdalena Wache comments on Finite Factored Sets in Pictures

[Magdalena Wache]

I see! You are right, then my argument wasn’t correct! I edited the post partially based on your argument above. New version:

Can we also infer that X causes Y?

Let’s concretize the above graphs by adding the conditional probabilities. Graph 1 then looks like this:

Graph 2 is somewhat trickier, because W is not uniquely determined. But one possibility is like this:

Note that W is just the negation of Z here ($W=\neg Z$). Thus, W and Z are information equivalent, and that means graph 2 is actually just graph 1.

Can we find a different variable W such that graph 2 does not reduce to graph 1? I.e. can we find a variable W such that Z is not deterministic given W?

No, we can’t. To see that, consider the distribution $P(Z|X,Y,W)$. By definition of Z, we know that

$P(Z|X,Y,W)=\{\begin{array}{c}1\text{if}Z=X\text{XOR}Y,\\ 0\text{otherwise.}\end{array}$.

In other words, $P(Z|X,Y,W)$ is deterministic.

We also know that W d-separates Z from X,Y in graph 2:

This d-separation implies that Z is independent from X, Z given W:

$P\left(Z|W\right)=P(Z|W,X,Y)$

As $P(Z|X,Y,W)$ is deterministic, $P\left(Z|W\right)$ also has to be deterministic.

So, graph 2 always reduces to graph 1, no matter how we choose W. Analogously, graph 3 and graph 4 also reduce to graph 1, and we know that our causal structure is graph 1:

Which means we also know that X causes Y.