Is it true that (in all prior joint distributions where A is independent of B, but A is evidence of C, and B is evidence of
C) A is none-independent of B, given C is held constant?
No, but I think it’s true if A,B,C are binary. In general, if a distribution p is Markov relative to a graph G, then if something is d-separated in G, then there is a corresponding independence in p. But, importantly, the implication does not always go the other way. Distributions in which the implication always goes the other way are very special and are called faithful.
“Markov” is used in the standard memoryless sense. By definition, the graph G represents any distribution p where each variable on the graph is independent of its past given its parents. This is the Markov property.
Ilya is discussing probability distributions p that may or may not be represented by graph G. If every variable in p is independent of its past given its parents in G, then you can use d-separation in G to reason about independences in p.
No, but I think it’s true if A,B,C are binary. In general, if a distribution p is Markov relative to a graph G, then if something is d-separated in G, then there is a corresponding independence in p. But, importantly, the implication does not always go the other way. Distributions in which the implication always goes the other way are very special and are called faithful.
What is Markov relative?
“Markov” is used in the standard memoryless sense. By definition, the graph G represents any distribution p where each variable on the graph is independent of its past given its parents. This is the Markov property.
Ilya is discussing probability distributions p that may or may not be represented by graph G. If every variable in p is independent of its past given its parents in G, then you can use d-separation in G to reason about independences in p.