Are you sure the “arrow forgetting” operation is correct?
You say there should be an arrow when
for all probability distributions P over the elements of Ni, the approximation of Nj over P is no worse than the approximation of Ni over P
wherein [Ni]=[A→B←C] here, and [Nj]=[A−B−C]. If we take a distribution in which A⊥/C|B , then Nj is actually a worse approximation, because A⊥C|B must hold in any distribution which can be expressed by a graph in [Nj] right?
Also, I really appreciate the diagrams and visualizations throughout the post!
Thanks for the feedback. There’s a condition which I assumed when writing this which I have realized is much stronger than I originally thought, and I think I should’ve devoted more time to thinking about its implications.
When I mentioned “no information being lost”, what I meant is that in the interaction A→B, each value b∈B (where B is the domain of PB) corresponds to only one value of a∈A. In terms of FFS, this means that each variable must be the maximally fine partition of the base set which is possible with that variable’s set of factors.
Under these conditions, I am pretty sure that A⊥C⟹A⊥C|B
Are you sure the “arrow forgetting” operation is correct?
You say there should be an arrow when
wherein [Ni]=[A→B←C] here, and [Nj]=[A−B−C]. If we take a distribution in which A⊥/C|B , then Nj is actually a worse approximation, because A⊥C|B must hold in any distribution which can be expressed by a graph in [Nj] right?
Also, I really appreciate the diagrams and visualizations throughout the post!
Thanks for the feedback. There’s a condition which I assumed when writing this which I have realized is much stronger than I originally thought, and I think I should’ve devoted more time to thinking about its implications.
When I mentioned “no information being lost”, what I meant is that in the interaction A→B, each value b∈B (where B is the domain of PB) corresponds to only one value of a∈A. In terms of FFS, this means that each variable must be the maximally fine partition of the base set which is possible with that variable’s set of factors.
Under these conditions, I am pretty sure that A⊥C⟹A⊥C|B