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All of FjolleJagt's Comments + Replies

Thanks, that makes sense! Could you say a little about why the weak union axiom holds? I've been struggling to prove that from your definitions. I was hoping that $h^{F} (X | z, w) \subseteq h^{F} (X | z)$ would hold, but I don't think that $h^{F} (X | z)$ satisfies the second condition in the definition of conditional history for $h^{F} (X | z, w)$ .

3Scott Garrabrant4y

When I prove it, I prove and use (a slight notational variation on) these two lemmas. 1. If hF(X|E)∩hF(Y|E)={}, then hF(X|E)=hF(X|(y∩E)) for all y∈Y. 2. hF((X∨SY)|E)=hF(X|E)∪⋃x∈XhF(Y|x∩E). (These are also the two lemmas that I have said elsewhere in the comments look suspiciously like entropy.) These are not trivial to prove, but they might help.

Finite Factored Sets

FjolleJagt4y10

I'm confused by the definition of conditional history, because it doesn't seem to be a generalisation of history. I would expect $h^{F} (X | \emptyset) = h^{F} (X)$ , but both of the conditions in the definition of $h^{F} (X | \emptyset)$ are vacuously true if $E = \emptyset$ . This is independent of what $H$ is, so $h^{F} (X | \emptyset) = \emptyset$ . Am I missing something?

3Scott Garrabrant4y

E is the event you are conditioning on, so the thing you should expect is that hF(X|S)=hF(X), which does indeed hold.