Here is a more concrete example of me using FFS the way I intend them to be used outside of the inference problem. (This is one specific application, but maybe it shows how I intend the concepts to be manipulated).
I can give an example of embedded observation maybe, but it will have to come after a more formal definition of observation (This is observation of a variable, rather than the observation of an event above):
Definition: Given a FFS F=(S,B), and A, W, X, which are partitions of S, where X={x1,…,xn}, we say A observes X relative to W if: 1) A⊥X,
2) A can be expressed in the form A=A0∨S⋯∨SAn, and
3) Ai⊥W|(S∖xi).
(This should all be interpreted combinatorially, not probabilistically.)
The intuition of what is going on here is that to observe an event, you are being promised that you 1) do not change whether the event holds, and 3) do not change anything that matters in the case where that event does not hold. Then, to observe a variable, you can basically 2) split yourself up into different fragments of your policy, where each policy fragment observes a different value of that variable. (This whole thing is very updateless.)
Example 1: (non-observation) An agent A={L,R} does not observe a coinflip X={H,T}, and chooses to raise either his left or right hand. Our FFS F=(S,B) is given by S=A×X, and B={A,X}. (I am abusing notation here slightly by conflating A with the partition you get on A×X by projecting onto the A coordinate.) Then W is the discrete partition on A×X.
In this example, we do not have observation. Proof: A only has two parts, so if we express A as a common refinement of 2 partitions, at least one of these two partitions must be equal to A. However, A is not orthogonal to W given H and A is not orthogonal to W given T. (hF(A|H)=hF(W|H)=hF(A|T)=hF(W|T)={A}). Thus we must violate condition 3.
Example 2: (observation)
An agent A={LL,LR,RL,RR} does observe a coinflip X={H,T}, and chooses to raise either his left or right hand. We can think of A as actually choosing a policy that is a function from X to {L,R}, where the two character string in the parts in A are the result of H followed by the result of T.
Our FFS F=(S,B) is given by S=X×AH×AT, and B={X,AH,AT}, where AH={LH,RH} represents what the agent would do seeing heads, and AT={LT,RT} represents what the agent word do given seeing tails. A=AH∨SAT. We also have a partition representing what the agent actually does Y={L,R}, where L and R are each four element sets in the obvious way. We will then say W=X∨SY, so W does not get to see what A would have done, it only gets to see the coin flip and what Aactually did.
Now I will prove that A observes X relative to W in this example. First, hF(A)={AH,AT}, and hF(X)={X}, so we get the first condition, A⊥X. We will break up A in the obvious way set up in the problem for condition 2, so it suffices now to show that AH⊥W|T, (and it will follow symmetrically that AT⊥W|H.)
Im not going to go through the details, but hF(AH|T)={AH}, while hF(W|T)={AT}, which are disjoint. The important thing here is that W doesn’t care about AH in worlds in which T holds.
Discussion:
So largely I am sharing this to give an example for how you can manipulate FFS combinatorially, and how you can use this to say things that you might otherwise want to say using graphs, Granted, you could also say the above things using graphs, but now you can say more things, because you are not restricted to the nodes you choose, you can ask the same combinatorial question about any of the other partitions, The benefit is largely about not being dependent on our choice of variables.
It is interesting to try to translate this definition of observation to transparent Newcomb or counterfactual mugging, and see how some of the orthogonalities are violated, and thus it does not count as an observation.
Here is a more concrete example of me using FFS the way I intend them to be used outside of the inference problem. (This is one specific application, but maybe it shows how I intend the concepts to be manipulated).
I can give an example of embedded observation maybe, but it will have to come after a more formal definition of observation (This is observation of a variable, rather than the observation of an event above):
Definition: Given a FFS F=(S,B), and A, W, X, which are partitions of S, where X={x1,…,xn}, we say A observes X relative to W if:
1) A⊥X,
2) A can be expressed in the form A=A0∨S⋯∨SAn, and
3) Ai⊥W|(S∖xi).
(This should all be interpreted combinatorially, not probabilistically.)
The intuition of what is going on here is that to observe an event, you are being promised that you 1) do not change whether the event holds, and 3) do not change anything that matters in the case where that event does not hold. Then, to observe a variable, you can basically 2) split yourself up into different fragments of your policy, where each policy fragment observes a different value of that variable. (This whole thing is very updateless.)
Example 1: (non-observation)
An agent A={L,R} does not observe a coinflip X={H,T}, and chooses to raise either his left or right hand. Our FFS F=(S,B) is given by S=A×X, and B={A,X}. (I am abusing notation here slightly by conflating A with the partition you get on A×X by projecting onto the A coordinate.) Then W is the discrete partition on A×X.
In this example, we do not have observation. Proof: A only has two parts, so if we express A as a common refinement of 2 partitions, at least one of these two partitions must be equal to A. However, A is not orthogonal to W given H and A is not orthogonal to W given T. (hF(A|H)=hF(W|H)=hF(A|T)=hF(W|T)={A}). Thus we must violate condition 3.
Example 2: (observation)
An agent A={LL,LR,RL,RR} does observe a coinflip X={H,T}, and chooses to raise either his left or right hand. We can think of A as actually choosing a policy that is a function from X to {L,R}, where the two character string in the parts in A are the result of H followed by the result of T.
Our FFS F=(S,B) is given by S=X×AH×AT, and B={X,AH,AT}, where AH={LH,RH} represents what the agent would do seeing heads, and AT={LT,RT} represents what the agent word do given seeing tails. A=AH∨SAT. We also have a partition representing what the agent actually does Y={L,R}, where L and R are each four element sets in the obvious way. We will then say W=X∨SY, so W does not get to see what A would have done, it only gets to see the coin flip and what Aactually did.
Now I will prove that A observes X relative to W in this example. First, hF(A)={AH,AT}, and hF(X)={X}, so we get the first condition, A⊥X. We will break up A in the obvious way set up in the problem for condition 2, so it suffices now to show that AH⊥W|T, (and it will follow symmetrically that AT⊥W|H.)
Im not going to go through the details, but hF(AH|T)={AH}, while hF(W|T)={AT}, which are disjoint. The important thing here is that W doesn’t care about AH in worlds in which T holds.
Discussion:
So largely I am sharing this to give an example for how you can manipulate FFS combinatorially, and how you can use this to say things that you might otherwise want to say using graphs, Granted, you could also say the above things using graphs, but now you can say more things, because you are not restricted to the nodes you choose, you can ask the same combinatorial question about any of the other partitions, The benefit is largely about not being dependent on our choice of variables.
It is interesting to try to translate this definition of observation to transparent Newcomb or counterfactual mugging, and see how some of the orthogonalities are violated, and thus it does not count as an observation.