# Minimax forecasting

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This post continues the research programme of attacking the grain of truth problem by departure from the Bayesian paradigm. In the previous post, I suggested using Savage’s minimax regret decision rule, but here I fall back to the simple minimax decision rule. This is because the mathematics is considerably simpler, and minimax should be sufficient to get IUD play in general games and Nash equilibrium in zero-sum two-player games. I hope to build on these results to get analogous results for minimax regret in the future.

We consider “semi-Bayesian” agents following the minimax expected utility decision rule, in oblivious environments with full monitoring (a setting that we will refer to as “forecasting”). This setting is considered in order to avoid the need to enforce exploration, as a preparation for analysis of general environments. We show that such agents satisfy a certain asymptotic optimality theorem. Intuitively, this theorem means that whenever the environment satisfies an incomplete model that is included in the prior, the agent will eventually learn this model i.e. extract at least as much utility as can be guaranteed for this model.

Appendix A contains the proofs of all results. Appendix B contains some theorems from the literature together with a corollary used in Appendix A.

##Notation

Given a topological space, will denote the space of Borel probability measures on . We regard it as a topological space using the weak topology. Given , is defined by . Given , stands for the total variation distance between and . We denote the set of non-empty convex compact subsets of .

Given a finite set, denotes the set of finite strings in alphabet , i.e. . denotes the set of infinite strings in alphabet . Given , is the length of string. Given , is the -th symbol in . Given , is the prefix of of length . Given , the notations , , and mean ” is a proper prefix of ”, ” is a prefix of ” and their negations. is the empty string and .

##Elementary properties of minimax

Fix and compact Polish spaces, representing the agent’s pure policies and representing the pure environments. Let be a continuous utility function.

#Proposition 1

Consider . There exists

Moreover, let be the closure of the convex hull of . Then, the same satisfies

We will say that such a is a minimax policy for .

#Proposition 2

Consider and . Then, there is s.t. any minimax policy for is a minimax policy for .

In particular, Proposition 2 implies that a minimax policy for is the optimal policy for some .

Now we ask what policy is implemented by “subagents” created by a minimax policy. Suppose , where represents the pure policies of the subagent. Assume that there is a Borel set (the condition for the creation of the subagent), a finite set (the actions taken before the creation of the subagent), a Borel measurable mapping and continuous functions and s.t.

Consider and minimax for . Denote the projection mappings. Define by

Define by

If for some we have , then can be arbitrary.

#Proposition 3

It should also be possible to make do without and the associated decomposition of by instead decomposing into and a Markov kernel from to .

##Asymptotic optimality

Fix finite sets (actions) and (observations). We now assume that and with the product topology. We fix a time discount function satisfying and a bounded reward function. Given and , we define by

The utility function is then given by

We will need a notation for a combination of two policies where the agent switches from one to another when observing some . Given and , we define by

Consider any and . We define as follows. For any Borel measurable :

It is easy to see the above indeed defines a probability measure.

Note that changing the policy after a certain event cannot change utility more than times the probability of the event times the corresponding time discount factor. That is, we have the following:

#Proposition 4

We are now ready to formulate the optimality theorem. Consider and . Denote . We think of as the prior, as one of the models in the prior (assigned probability ) and as the convex combination of all other models. Let be a minimax policy for . Define by

That is, is the expected utility that can be guaranteed by changing the policy following , assuming that the true environment is in . We claim that if the true environment is in , then after sufficient time will achieve at least as much utility as can be guaranteed for (guaranteed under the constraint of following until the point of making the current observations). That is, can only be greater than by a quantity that goes to 0 with probability 1, even after normalizing according to Proposition~5:

#Theorem 1

##Appendix A

We will repeatedly use the standard fact that, given a compact Polish space, is also a compact Polish space (it is metrizable by the Levy-Prokhorov metric, compact as a consequence of the Banach-Alaoglu theorem and the fact that any probability measure on a Polish space is Radon, and separability is also not hard to prove).

#Proposition A.1

Consider compact Polish spaces, continuous, and a sequence . Define by and by . Then, converges to uniformly.

#Proof of Proposition A.1

Assume to the contrary that there is and a subsequence of s.t.

This implies we can choose s.t. . Let be an increasing sequence s.t. for some . We get , which is a contradiction because and .

#Proposition A.2

Consider compact Polish spaces and continuous. Define by

Then, is continuous.

#Proof of Proposition A.2

Consider any , and sequences , . We have

On the right hand side, the first term goes to 0 by Proposition A.1 and the second term goes to 0 by definition of the weak* topology.

#Proposition A.3

Consider any compact Polish and continuous. Define by

Then, is continuous.

#Proof of Proposition A.3

Follows by applying Fubini’s theorem and using Proposition A.2 twice.

#Proposition A.4

Consider any compact Polish, and continuous. Define by . Then, is continuous.

#Proof of Proposition A.4

Consider any and a sequence

By Proposition A.1, the right hand converges to 0, therefore the left hand side also converges to 0.

In the following, we will use the notation , defined by . By Proposition A.3, is continuous.

#Proof of Proposition 1

Define by . By Proposition A.4, is continuous and therefore, exists.

is continuous and affine in the 2nd argument (i.e. it sends convex linear combinations to convex linear combinations), and therefore , implying the second part of the proposition.

#Proof of Proposition 2

Define by . Denote . By Proposition A.4, is continuous and therefore, there exists

Choose , s.t. .

and are compact convex sets in the spaces of finite signed Borel measures on and respectively. Therefore, we can apply the minimax theorem to get

Denote the above number . We have

On the other hand

Combining the inequalities, we get the desired result.

#Definition A.1

In the setting of Proposition 3, consider any and . We define as follows. For any Borel measurable :

It is easy to see the above indeed defines a probability measure.

#Proposition A.5

Consider any and and Borel measurable and bounded. Then:

#Proof of Proposition A.5

#Proposition A.6

Given , and

#Proof of Proposition A.6

Applying Proposition A.5:

#Proposition A.7

Consider any . Define and by

If for some we have , then can be arbitrary.

Then,

#Proof of Proposition A.7

Applying Proposition A.6, we get the desired result.

#Proof of Proposition 3

Consider any . By definition of

Applying Proposition A.7 to the right hand side

Applying Proposition A.6 to both sides, we get the desired result.

Given any , we denote and .

#Proof of Proposition 4

We have

Denote , . We have

The desired result now follows from .

#Definition A.2

A splitting of is s.t. , is Borel, is a finite set, is Borel measurable, and are continuous and

#Proposition A.8

Consider , and , and denote . Suppose that is a splitting of . Assume . Fix . Let satisfy

Denote . Then, for any s.t.

#Proof of Proposition A.8

Let satisfy

Denote . Consider any and denote . We have

Applying Proposition A.6 to the right hand side

Also, we have

Applying Proposition A.6 again

Combining the inequalities

The above inequality holds for any , therefore

By definition of , it follows that

Denote and . We have

Applying Proposition A.6 to the right hand side

On the other hand, Proposition A.6 implies that

Substituting in the inequality, we get

Observing that we can combine the inequalities, we get