I’m not quite convinced by this response. Would it be possible to formalize “set of probability distributions in which Y causes X is a null set, i.e. it has measure zero.”?
It is true that if the graph was (Y->X, X->Z, Y->Z), then we would violate faithfulness. There are results that show that under some assumptions, faithfulness is only violated with probability 0. But those assumptions do not seem to hold in this example.
Would it be possible to formalize “set of probability distributions in which Y causes X is a null set, i.e. it has measure zero.”?
We are looking at the space of conditional probability table (CPT) parametrizations in which the indepencies of our given joint probability distribution (JPD) hold.
If Y causes X, the independencies of our JPD only hold for a specific combination of conditional probabilities. Namely those in which P(X,Z) = P(X)P(Z). The set of CPT parametrizations with P(X,Z) = P(X)P(Z) has measure zero (it is lower-dimensional than the space of all CPT parametrizations).
In contrast, any CPT parametrization of graph 1 would fulfill the independencies of our given JPD. So the set of CPT parametrizations in which our independencies hold has a non-zero measure.
There are results that show that under some assumptions, faithfulness is only violated with probability 0. But those assumptions do not seem to hold in this example.
Could you say what these assumptions are that don’t hold here?
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.
I’m not quite convinced by this response. Would it be possible to formalize “set of probability distributions in which Y causes X is a null set, i.e. it has measure zero.”?
It is true that if the graph was (Y->X, X->Z, Y->Z), then we would violate faithfulness. There are results that show that under some assumptions, faithfulness is only violated with probability 0. But those assumptions do not seem to hold in this example.
We are looking at the space of conditional probability table (CPT) parametrizations in which the indepencies of our given joint probability distribution (JPD) hold.
If Y causes X, the independencies of our JPD only hold for a specific combination of conditional probabilities. Namely those in which P(X,Z) = P(X)P(Z). The set of CPT parametrizations with P(X,Z) = P(X)P(Z) has measure zero (it is lower-dimensional than the space of all CPT parametrizations).
In contrast, any CPT parametrization of graph 1 would fulfill the independencies of our given JPD. So the set of CPT parametrizations in which our independencies hold has a non-zero measure.
Could you say what these assumptions are that don’t hold here?
I’m pretty sure that you can prove that finite factored sets have this property directly, actually!
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.