Theorem 6: Causal IPOMDP's: Any infra-POMDP with transition kernel K:S×Aik→O×S which fulfills the niceness conditions and always produces crisp infradistributions, and starting infradistribution ψ∈□S, produces a causal belief function via Θ(π):=pr(A×O)ω∗(ψ⋉Kπ:∞).
So, first up, we can appeal to Theorem 4 on Pseudocausal IPOMDP's, to conclude that all the Kπ:∞ are well-defined, and that this Θ we defined fulfills all belief function conditions except maybe normalization.
The only things to check are that the resulting Θ is normalized, and that the resulting Θ is causal. The second is far more difficult and makes up the bulk of the proof, so we'll just dispose of the first one.
T6.1 First, K, since it's crisp by assumption, always has K(s,a)(c)=c for all constants c. The same property extends to all the Kπn (they're all crisp too) and then from there to Kπ:∞ by induction. Then we can just use the following argument, where π is arbitrary.
The last two inequalities were by Kπ:∞ mapping constants to the same constant regardless of π, and by ψ being normalized since it's an infradistribution. This same proof works if you swap out the 1 with a 0, so we have normalization for Θ.
T6.2 All that remains is checking the causality condition. We'll be using the alternate rephrasing of causality. We have Θ defined as
and want to derive our rephrasing of causality, that for any U and π′ and ζ and family Uπ, we have
Accordingly, we get to fix a U,π′,ζ, and family of functions Uπ, and assume
Our proof target is
Applying our reexpression of Θ in terms of ψ and Kπ:∞ and projection, we get
Undoing the projections on both sides, we get
Unpacking the semidirect product, we get
Now, we can notice something interesting here. By concavity for infradistributions, we have
So, our new proof target becomes
Because if we hit that, we can stitch the inequalities together. The obvious move to show this new result is to apply monotonicity of infradistributions. If the inner function on the left hand side was always above the inner function on the right hand side, we'd have our result by infradistribution monotonicity. So, our new proof target becomes
At this point, now that we've gotten rid of ψ, we can use the full power of crisp infradistributions. Crisp infradistributions are just a set of probability distributions! Remember that the semidirect product, ψ⋉K, can be viewed as taking a probability distribution from ψ, and for each x, picking a conditional probability distribution from K(x), and all the probability distributions of that form make up ψ⋉K.
In particular, the infinitely iterated semidirect product here can be written as a repeated process where, for each history of the form s0,aos1:n,a, Murphy gets to select a probability distribution over the next observation and state from K(sn,a) (as a set of probability distributions) to fill in the next part of the action-observation-state tree. Accordingly every single probability distribution in Kπ:∞(s0) can be written as π interacting with an expanded probabilistic environment eΔ+ of type (A×O×S)<ω×A→Δ(O×S) (which gives Murphy's choice wherever it needs to pick the next observation and state), subject to the constraints that
We'll abbreviate this property that eΔ+ always picks stuff from the appropriate K set and starts off picking from K(s0,a) as eΔ+∼K,s0.
Our old proof goal of
can be rephrased as
and then rephrased as
And now, by our earlier discussion, we can rephrase this statement in terms of picking expanded probabilistic environments.
Now, on the right side, since the set we're selecting the expanded environments from is the same each time and doesn't depend on π, we can freely move the inf outside of the expectation. It is unconditionally true that
So now, our proof goal shifts to
because if that were true, we could pair it up with the always-true inequality to hit our old proof goal. And we'd be able to show this proof goal if we showed the following, more ambitious result.
Note that we're not restricting to compatibility with K and s0, we're trying to show it for any expanded probabilistic environment of type (A×O×S)<ω×A→Δ(O×S). We can rephrase this new proof goal slightly, viewing ourselves as selecting an action and observation history.
Now, for any expanded probabilistic environment eΔ+:(A×O×S)<ω×A→Δ(O×S), there is a unique corresponding probabilistic environment eΔ of type (A×O)<ω×A→ΔO where, for any policy, pr(A×O)ω∗(π⋅eΔ+)=π⋅eΔ. This is just the classical result that any POMDP can be viewed as a MDP with a state space consisting of finite histories, it still behaves the same. Using this swap, our new proof goal is:
Where eΔ is an arbitrary probabilistic environment. Now, you can take any probabilistic environment eΔ and view it as a mixture of deterministic environments e. So we can rephrase eΔ as, at the start of time, picking a e from ξ∈ΔE for some particular ξ. This is our mixture of deterministic environments that makes up eΔ. This mixture isn't necessarily unique, but there's always some mixture of deterministic enivironments that works to make eΔ where eΔ is arbitrary. So, our new proof goal becomes:
Since π⋅e is just a dirac-delta distribution on a particular history, we can substitute that in to get the equivalent
Now we just swap the two expectations.
And then we remember that our starting assumption was
So we've managed to hit our proof target and we're done, this propagates back to show causality for Θ.
Theorem 7: Pseudocausal to Causal Translation: The following diagram commutes when the bottom belief function Θ is pseudocausal. The upwards arrows are the causal translation procedures, and the top arrows are the usual morphisms between the type signatures. The new belief function Θc is causal. Also, if U:(A×O)ω→[0,1], and UN:(A×(O+N))ω→[0,1] is defined as 1 for any history containing Nirvana, and U(h) for any nirvana-free history, then ∀π:Θc(π)(UN)=Θ(π)(U).
T7.0 Preliminaries. Let's restate what the translations are. The notation d(e,h) is that e"defends" h. e defends h if it responds to any prefix of h which ends in an action with either the next observation in h, or Nirvana, and responds to any non-prefixes of h with Nirvana.
Also, we'll write hN∼π′,h if hN is "compatible" with π′ and h. The criterion for this is that hN∼π′, and that any prefix of hN which ends in an action and contains no Nirvana observations must either be a prefix of h, or only deviate from h in the last action. An alternate way to state that is that hN is capable of being produced by a copolicy which defends h, interacting with π′.
With these two notes, we introduce the infrakernels Kd:(A×O)ωik→E, and Kπ′:(A×O)ωik→(A×(O+N))ω.
Kd(h) is total uncertainty over e such that e defends h. Kd(h)(f):=infe:d(e,h)f(e)
And Kπ′(h) is total uncertainty over hN such that hN is compatible with π′ and h. Kπ′(h)(UN):=infhN∼π′,hUN(hN)
Now, to briefly recap the three translations.For first-person static translation, we have
Where Θ is the original belief function.
For third-person static translation, we have
Where D is the original infradistribution over (A×O)ω.
For first-person dynamic translation, the new infrakernel has type signature
Ie, the state space is destinies plus a single Nirvana state, and the observation state is the old space of observations plus a Nirvana observation. N is used for the Nirvana observation, and N is used for the Nirvana state. The transition dynamics are:
and, if a′≠a, then →N(aoh,a′):=δN,N
and otherwise, →N(aoh,a):=δo,h∨δN,N
The initial infradistribution is i∗(ψF), the obvious injection of ψF from (A×O)ω to (A×O)ω+N.
The obvious way to show this theorem is that we can assume the bottom part of the diagram commutes and ψF,Θ,D are all pseudocausal, and then show that all three paths to Θc produce the same result, and it's causal. Then we just wrap up that last part about the utility functions.
Fortunately, assuming all three paths produce the same Θc belief function, it's trivial to show that it's causal. i∗(ψF) is an infradistribution since ψF is, and →N is a crisp infrakernel, so we can just invoke Theorem 6 that crisp infrakernels produce causal belief functions to show causality for Θc. So we only have to check that all three paths produce the same result, and show the result about the utility functions. Rephrasing one of the proofs is one phase, and then there are two more phases showing that the other two phases hit the same target, and the last phase showing our result about the utility functions.
We want to show that
T7.1 First, let's try to rephrase Θc(π′)(UN) into a more handy form for future use. Invoking our definition of Θc(π′), we have:
Then, we undo the projection
And unpack the semidirect product
And unpack what ⊤Π means
And then apply the definition of Kπ′(h)
This will be our final rephrasing of Θc(π′)(UN). Our job is to derive this result for the other two paths.
T7.2 First up, the third-person static translation. We can start with definitions.
And then, we can recall that since D and Θ are assumed to commute, we can define D in terms of Θ via D=pr(A×O)ω∗(⊤Π⋉Θ). Making this substitution, we get
We undo the first pushforward
And then the projection
And remove the projection
and unpack the semidirect product and what ⊤Π is
Now, we recall that Kd(h) is all the e that defend h, to get
At this point we can realize that for any e which defends h, when e interacts with π′, it makes a hN that is compatible with π′ and h. Further, all hN compatible with π′ and h have an e which produces them. If hN has its nirvana-free prefix not being a prefix of h, then the defending e which doesn't end h early, just sends deviations to Nirvana, is capable of producing it. If hN looks like h but ended early with Nirvana, you can just have a defending e which ends with Nirvana at the same time. So we can rewrite this as
And this is exactly the form we got for Θ(π′)(UN).
T7.3 Time for the third and final translation. We want to show that
is what we want. Because the pseudocausal square commutes, we have that ψF=pr(A×O)ω∗(⊤Π⋉Θ), so we can make that substitution to yield
Remove the projection, to get
For this notation, we're using sN to refer to an element of (A×O)ω+N, the state space. aosN1:∞ is an element of (A×(O+N)×((A×O)ω+N))ω, a full unrolling. aoN1:∞ is the projection of this to (A×(O+N))ω.
We unpack the semidirect product to yield
and rewrite the injection pushforward
and remove the projection
And unpack the semidirect product along with ⊤Π.
Now we rewrite this as a projection to yield
And do a little bit of reindexing for convenience, swapping out the symbol aoN1:∞ for hN, getting
And now, we can ask what pr(A×(O+N))ω∗(→π′N:∞(h)) is. Basically, the starting state is the destiny h, and then π′ interacts with it through the kernel →N. Due to how →N is defined, it can do one of three things. First, it could just unroll h all the way to completion, if h is the sort of history that's compatible with π′. Second, it could partially unroll h and deviate to Nirvana early, even if π′ is compatible with more, because of how →N(aoh,a) is defined. Finally, π′ could deviate from h and then →N would have to respond with Nirvana. And of course, once you're in a Nirvana state, you're in there for good. Hm, you have total uncertainty over histories hN compatible with π′ and h, any of them at all could be made by pr(A×(O+N))ω∗(→π′N:∞(h)). So, this projection of the infrakernel unrolling h is just Kπ′(h). We get
And then we use what Kπ′(h) does to functions, to get
Which is exactly the form we need. So, all three translation procedures produce identical belief functions.
T7.4 The only thing which remains to wrap this result up is guaranteeing that, if UN:(A×(O+N))ω→[0,1] is 1 if the history contains Nirvana, (or infinity for the R type signature), and U otherwise, then Θc(π′)(UN)=Θ(π′)(U).
We rephrased Θc(π′)(UN) as infπΘ(π)(λh.infhN∼π′,hUN(hN))
We can notice something interesting. If h is the sort of history that π′ is capable of making, ie h∼π′, then all of the hN∼π′,h are going to end up in Nirvana except h itself, because all the hN are like "follow π′, any deviations of π′ from h go to Nirvana, and also Nirvana can show up early". In this case, since any history ending with Nirvana is maximum utility, and otherwise U is copied, we have infhN∼π′,hUN(hN)=U(h). However, if h is incompatible with π′, then no matter what, a history hN which follows π′ is going to hit Nirvana and be assigned maximum utility. So, we can rewrite
Which can be written as an update, yielding
And then, we remember that regardless of π′ and π, for pseudocausal belief functions, which Θ is, we have that u1∼π′(Θ(π))(U)≥Θ(π′)(U), so the update assigns higher expectation values than Θ(π′) does. Also, u1∼π′(Θ(π′))=Θ(π′). Thus, the infinimum is attained by π′, and we get
and we're done.
Theorem 8: Acausal to Pseudocausal Translation: The following diagram commutes when the bottom belief function Θ is acausal. The upwards arrows are the pseudocausal translation procedures, and the top arrows are the usual morphisms between the type signatures. The new belief function Θp will be pseudocausal.
T8.0 First, preliminaries. The two translation procedures are, to go from an infradistribution over policy-tagged destinies D, to an infradistribution over destinies Dp, we just project.
And, to go from an acausal belief function Θ to a pseudocausal one Θp, we do the translation
We assume that D and Θ commute, and try to show that these translation procedures result in a Θp and Dp which are pseudocausal and commute with each other. So, our proof goals are that
Once we've done that, it's easy to get that Θp is pseudocausal. This occurs because the projection of any infradistribution is an infradistribution, so we can conclude that Dp is an infradistribution over (A×O)ω. Also, back in the Pseudocausal Commutative Square theorem, we were able to derive that any infradistribution Dp over (A×O)ω, when translated to a Θp, makes a pseudocausal belief function, no special properties were used in that except being an infradistribution.
T8.1 Let's address the first one, trying to show that
We'll take the left side, plug in some function, and keep unpacking and rearranging until we get the right side with some function plugged in. First up is applying definitions.
by how Θp was defined. Now, we start unpacking. First, the projection.
Then unpack the semidirect product and ⊤Π, for
Then unpack the update, to get
Then pack up the infπ
And pack up the semidirect product
And since D=⊤Π⋉Θ for the original acausal belief function, we can rewrite this as
And pack this up in a projection.
And then, since we defined Dp:=pr(A×O)ω∗(D), we can pack this up as
And then pack up the update
So, that's one direction done.
T8.2 Now for the second result, showing that
Again, we'll take the more complicated form, and write it as the first one, for arbitrary functions U.bWe begin with
and then unpack what Θp(π′) is to get
Undo the projection
Unpack the semidirect product and ⊤Π.
Remove the projection
Unpack the semidirect product and ⊤Π again.
Let's fold the two infs into one inf, this doesn't change anything, just makes things a bit more concise.
and unpack the update.
Now, we can note something interesting here. If π′ is selected to be equal to π, that inner function is just U, because Θ(π) is only supported over histories compatible with π. If π′ is unequal to π, that inner function is U with 1's for some inputs, possibly. So, our choice of π′ is optimal for minimizing when its equal to π, so we can rewrite as
and write this as ⊤Π
And pack up as a semidirect product
and then because D=⊤Π⋉Θ, we substitute to get
and then write as a projection.
Now, this is just Dp, yielding
Proposition 2: The two formulations of x-pseudocausality are equivalent for crisp acausal belief functions.
One formulation of x-pseudocausality is that
∀π,π′:u1∼π′(Θ(π))⊆Θx(π′) where Θx(π′)(U):=Θ(π′)(inf(x,U))
The second formulation is,
This proof will be slightly informal, it can be fully formalized without too many conceptual difficulties. We need to establish some basic probability theory results to work on this. The proof outline is we establish/recap four probability-theory/inframeasure-theory results, and then show the two implication directions of the iff statement.
P2.1 Our first claim is that, given any two probability distributions μ and ν, you can shatter them into measures m∗,mμ, and mν such that m∗+mμ=μ and m∗+mν=ν, and mμ(1)=mν(1)=dTV(μ,ν). What this decomposition is basically doing is finding the largest chunk of measure that μ and ν agree on (that's m∗), and then subtracting that out from μ and ν yields you your mμ and mν. These measures have no overlap, because if there was overlap, you could put that overlap into the m∗ chunk of measure. And, the mass of the "unique to μ" piece must equal the mass of the "unique to ν" piece because they must add up to probability distributions, and this quantity is exactly the total variation distance of μ,ν.
P2.2 Our second claim is that
Why is this? Well, there's an alternate formulation of total variation distance where
As intuition for this, we can pick a function U which is 1 on the support of mμ, and 0 elsewhere. In such a case, we'd get
and then remember that mμ and mν have disjoint supports, so we get
And any other function in the [0,1] range wouldn't do as well at attaining different expectation values on mμ and mν. So that's where the identity comes from. If we try to do the optimal separation but instead restrict U to be in [0,x], we'd get