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We generalize the formalism of dominant markets to account for stochastic "deductive processes," and prove a theorem regarding the asymptotic behavior of such markets. In a following post, we will show how to use these tools to formalize the ideas outlined here.

Appendix A contains the key proofs. Appendix B contains the proofs of technical propositions used in Appendix A, which are mostly straightforward. Appendix C contains the statement of a version of the optional stopping theorem from Durret.

##Notation

Given a topological space:

• is the space of Borel probability measures on equipped with the weak* topology.

• is the Banach space of continuous functions with uniform norm.

• is the Borel -algebra on .

• is the -algebra of universally measurable sets on .

• Given , denotes the support of .

Given and measurable spaces, is a Markov kernel from to . For any , we have . Given , is the semidirect product of and and is the pushforward of by .

Given , Polish spaces, Borel measurable and , we denote the set of Markov kernels s.t. is supported on the graph of and . By the disintegration theorem, is always non-empty and any two kernels in coincide -almost everywhere.

##Results

The way we previously laid out the dominant market formalism, the sequence of observations (represented by the sets ) was fixed. To study forecasting, we instead need to assume this sequence is sampled from some probability measure (the true environment).

For each , let be a compact Polish space. represents the space of possible observations at time . Denote

Given , denotes the projection mapping and . Denote

is a compact Polish space. For each we denote the projection mapping. Given , we denote , a closed subspace of . Given and , we denote .

#Definition 1

A market is a sequence of mappings s.t.

• Each is measurable w.r.t. and .

• For any , .

As before, we define the space of trading strategies , but this time we regard it as a Banach space.

#Definition 2

A trader is a sequence of mappings which are measurable w.r.t. and .

Given a trader and a market , we define the mappings (measurable w.r.t. and ) and (measurable w.r.t. and ) as follows:

The "market maker" lemma now requires some additional work due to the measurability requirement:

#Lemma

Consider any trader . Then, there is a market s.t. for all and

As before, we have the operator defined by

We also introduce the notation and which are measurable mappings defined by

#Definition 3

A market is said to dominate a trader when for any , if

then

#Theorem 1

Given any countable set of traders , there is a market s.t. dominates all .

Theorem 1 is proved exactly as before (modulo Lemma), and we omit the details.

We now describe a class of traders associated with a fixed environment s.t. if a market dominates a trader from this class, a certain function of the pricing converges to 0 with -probability 1. In a future post, we will apply this result to a trader associated with an incomplete models by observing that the trader is in the class for any .

#Definition 4

A trading metastrategy is a uniformly bounded family of measurable mappings . Given , is said said to be profitable for , when there are and s.t. for any , -almost any and any :

Even if a metastrategy is profitable, it doesn't mean that a smart trader should use this metastrategy all the time: in order to avoid running out of budget, a trader shouldn't place too many bets simultaneously. The following construction defines a trader that employs a metastrategy only when all previous bets are closed to being resolved.

#Definition 5

Fix a metastrategy . We define the trader and the measurable mappings recursively as follows:

#Theorem 2

Consider , , a metastrategy profitable for and a market. Assume dominates . Then, for -almost any :

That is, the market price of the "stock portfolio" traded by converges to its true -expected value.

##Appendix A

#Proposition A.1

Fix a compact Polish space and . Then, there exists s.t.

#Proof of Proposition A.1

Follows immediately from "Proposition 1" from before and Proposition B.6.

#Proposition A.2

Fix compact Polish spaces. Denote . Then, there exists measurable w.r.t. and s.t. for any and :

#Proof of Proposition A.2

Define by

We can view as the graph of a multivalued mapping from to . We will now show this multivalued mapping has a selection, i.e. a single-valued measurable mapping whose graph is a subset. Obviously, the selection is the desired .

is closed by Proposition B.7. is closed by Proposition B.5. by Proposition B.6 and hence closed. In particular, the fiber of over any is also closed.

For any , , define by and by . Applying Proposition A.1 to we get s.t.

It follows that and hence is non-empty.

Consider any open. Then, is locally closed and in particular . Therefore, the image of under the projection to is also and in particular Borel.

Applying the Kuratowski-Ryll-Nardzewski measurable selection theorem, we get the desired result.

#Proof of Lemma

For any , let be as in Proposition A.2. We define recursively by:

#Proposition A.3

Consider a probability space, a filtration of , , and stochastic processes adapted to . Assume that:

Then, with probability 1.

The proof will use the following definition:

#Definition A

Consider a sequence . The accumulation times of are defined recursively by

Consider a probability space and a stochastic process. The accumulation times of are defined pointwise as above. Clearly, they are stochastic processes and whenever is adapted to a filtration , they are stopping times w.r.t. .

#Proof of Proposition A.3

Without loss of generality, we can assume (otherwise we can renormalize , and by a factor of ). Define by

By Proposition B.8, is a submartingale. Let be the accumulation times of . By proposition N23, are submartingales for all . By Proposition\ N24, each of them is uniformly integrable. Using the fact that to apply Theorem C, we get

Clearly, is adapted to . Doob's second martingale convergence theorem implies that ( is the limit of the uniformly integrable submartingale ). We conclude that is a submartingale.

By Proposition B.12, . Applying the Azuma-Hoeffding inequality, we conclude that for any positive integer :

Since , it follows that

By Proposition B.13

It remains to show that if is s.t. then the condition above fails. Consider any such . , therefore . On the other hand, by Proposition B.14, .

#Proposition A.4

Consider a probability space, a filtration of , , and stochastic processes adapted to and an arbitrary stochastic process. Assume that:

Then, (equivalently ) with probability 1.

#Proof of Proposition A.4

Define . We have

By Proposition A.3, with probability 1. Since , we get the desired result.

#Proposition A.5

Consider , and as before. Consider , , a metastrategy profitable for and a market. Then, for -almost any :

#Proof of Proposition A.5

We regard as a probability space using the -algebra and the probability measure . For any , we define and by

Clearly, is a filtration of , are stochastic processes and are adapted to . is uniformly bounded, therefore is uniformly bounded and so is . Obviously, is also non-negative.

By Proposition B.15, are uniformly bounded. is bounded and in particular . We have

Let and be as in Definition 4.

By definition of , is equal to either or 0. In either case, we get (almost everywhere)

Applying Proposition A.4, we get the desired result.

#Proposition A.6

Consider the setting of Proposition A.3. Then, for almost all :

#Proof of Proposition A.6

Define by

Let be the accumulation times of . Consider any s.t. but . Proposition B.13 implies that

As in the proof of Proposition A.3, we can apply the Azuma-Hoeffding inequality to and get that for any positive integer

It follows that

Comparing with the inequality from before, we reach the desired conclusion.

#Proposition A.7

Consider the setting of Proposition A.4. Then, for almost all :

#Proof of Proposition A.7

Define . As in the proof of Proposition A.4, meets the conditions of Proposition A.3 and thus of Proposition A.6 also. By Proposition A.6, for almost all :

As in the proof of Proposition A.4, is uniformly bounded, giving the desired result.

#Proof of Theorem 2

Let , , and be as in the proof of Proposition A.5. Using Proposition A.5 and the assumption that dominates , we conclude that for -almost any , . As in the proof of Proposition A.5, the conditions of Proposition A.4 are satisfied, and therefore the conditions of Proposition A.7 are also satisfied. Applying Proposition A.7, we conclude that . By Proposition B.17, it follows that for any , . We get

##Appendix B

#Proposition B.1

If are compact Polish spaces and is continuous, then defined by is continuous.

We omit the proof of Proposition B.1, since it appeared as "Proposition A.2" before.

#Proposition B.2

Fix compact Polish spaces. Define by . Then, is continuous. In particular, we can apply this to in which case .

#Proof of Proposition B.2

Consider and . We have

By Proposition B.1

Combining, we get

#Proposition B.3

Fix compact Polish spaces and denote . Define by

Then, is continuous.

#Proof of Proposition B.3

Consider , . By Proposition B.1, implies that

Since , we get

#Proposition B.4

Fix compact Polish spaces. Denote . Define by

Then, is closed.

#Proof of Proposition B.4

Consider , , , . By Proposition B.3, we get

Hence, .

#Proposition B.5

Fix compact Polish spaces. Denote . Define by

Then, is closed.

#Proof of Proposition B.5

By Proposition B.2, is the continuous inverse image of a subset of which is closed by Proposition B.4.

#Proposition B.6

Fix a compact Polish space. Consider and and denote . Then, iff .

#Proof of Proposition B.6

If then and therefore .

Now, assume . For any , Markov's inequality yields

Taking , we get and hence .

#Proposition B.7

Consider compact Polish spaces. Denote . Define by

Then, is closed.

#Proof of Proposition B.7

We fix metrizations for and and metrize by

For each , denote .

Consider , , . We have . By Proposition B.1, , therefore . By Proposition B.6, and hence .

#Proposition B.8

Consider a probability space, a filtration of , and stochastic processes adapted to . Assume that there are s.t.:

Define by

Then, is a submartingale.

#Proof of Proposition B.8

Obviously, is adapted to . We have