P(W|X,Y) is deterministic, because X and Y already fully describe our sample space. This means P(W|X,Y) is either 0 or 1
Z and W are independent
X and W are dependent
Y and W are dependent
I think your arguments in Section 3 to rule out Graph 3 can’t be correct if you accept Graph 2.
To see this, note that there is a symmetry between X and Z. Namely, if we use FFS temporal inference, then we know that X and Z are both before Y (and ¬Y ).(here we even have P(X=x,Z=z)=P(X=z,Z=x), so they are entirely exchangeable).
Therefore, if you accept Graph 2 then we can clearly switch X and Z in Graph 2 and obtain a solution for Graph 3. Also, note that in these solutions X=¬W or Z=¬W, so if we see variables as their information content, as in FFS, this is Graph 1 in disguise.
Also in Graph 2 there is a typo P(W=0|Z=0) instead of P(Z=0|W=0)
I think your arguments in Section 3 to rule out Graph 3 can’t be correct if you accept Graph 2.
To see this, note that there is a symmetry between X and Z. Namely, if we use FFS temporal inference, then we know that X and Z are both before Y (and ¬Y ).(here we even have P(X=x,Z=z)=P(X=z,Z=x), so they are entirely exchangeable).
Therefore, if you accept Graph 2 then we can clearly switch X and Z in Graph 2 and obtain a solution for Graph 3. Also, note that in these solutions X=¬W or Z=¬W, so if we see variables as their information content, as in FFS, this is Graph 1 in disguise.
Also in Graph 2 there is a typo P(W=0|Z=0) instead of P(Z=0|W=0)
Good point, you are right! I just edited the post and now say that graph 2, 3, and 4 only have instantiations that are equivalent to graph 1.
fixed, thanks!