For example, in theorem 3.2 in Causation, Prediction, and Search, we have a result that says that faithfulness holds with probability 1 if we have a linear model with coefficients drawn randomly from distributions with positive densities.
It is not clear to me why we should expect faithfulness to hold in a situation like this, where Z is constructed from other variables with a particular purpose in mind.
Consider the graph Y<-X->Z. If I set Y:=X and Z:=X, we have that X⊥Y|Z, violating faithfulness. How are you sure that you don’t violate faithfulness by constructing Z?
Consider the graph Y<-X->Z. If I set Y:=X and Z:=X
I would say then the graph is reduced to the graph with just one node, namely X. And faithfulness is not violated because we wouldn’t expect X⊥X|X to hold.
In contrast, the graph X-> Y ← Z does not reduce straightforwardly even though Y is deterministic given X and Z, because there are no two variables which are information equivalent.
I’m not completely sure though if it reduces in a different way, because Y and {X, Z} are information equivalent (just like X and {Y,Z}, as well as Z and {X,Y}). And I agree that the conditions for theorem 3.2. aren’t fulfilled in this case. Intuitively I’d still say X-> Y ← Z doesn’t reduce, but I’d probably need to go through this paper more closely in order to be sure.
Finally got around to looking at this. I didn’t read the paper carefully, so I may have missed something, but I could not find anything that makes me more at ease with this conclusion.
Ben has already shown that it is perfectly possible that Y causes X. If this is somehow less likely that X causes Y, this is exactly what needs to be made precise. If faithfulness is the assumption that makes this work, then we need to show that faithfulness is a reasonable assumption in this example. It seems that this work has not been done?
If we can find the precise and reasonable assumptions that exclude that Y causes X, that would be super interesting.
For example, in theorem 3.2 in Causation, Prediction, and Search, we have a result that says that faithfulness holds with probability 1 if we have a linear model with coefficients drawn randomly from distributions with positive densities.
It is not clear to me why we should expect faithfulness to hold in a situation like this, where Z is constructed from other variables with a particular purpose in mind.
Consider the graph Y<-X->Z. If I set Y:=X and Z:=X, we have that X⊥Y|Z, violating faithfulness. How are you sure that you don’t violate faithfulness by constructing Z?
I would say then the graph is reduced to the graph with just one node, namely X. And faithfulness is not violated because we wouldn’t expect X⊥X|X to hold.
In contrast, the graph X-> Y ← Z does not reduce straightforwardly even though Y is deterministic given X and Z, because there are no two variables which are information equivalent.
I’m not completely sure though if it reduces in a different way, because Y and {X, Z} are information equivalent (just like X and {Y,Z}, as well as Z and {X,Y}). And I agree that the conditions for theorem 3.2. aren’t fulfilled in this case. Intuitively I’d still say X-> Y ← Z doesn’t reduce, but I’d probably need to go through this paper more closely in order to be sure.
Finally got around to looking at this. I didn’t read the paper carefully, so I may have missed something, but I could not find anything that makes me more at ease with this conclusion.
Ben has already shown that it is perfectly possible that Y causes X. If this is somehow less likely that X causes Y, this is exactly what needs to be made precise. If faithfulness is the assumption that makes this work, then we need to show that faithfulness is a reasonable assumption in this example. It seems that this work has not been done?
If we can find the precise and reasonable assumptions that exclude that Y causes X, that would be super interesting.