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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

#Proposition B.9

Let be a probability space, a filtration on , a stochastic process adapted to and a stopping time (w.r.t. ). Suppose is submartingale. Then, is also submartingale.

#Proof of Proposition B.9

Clearly, is adapted to . We have

For any , define by

is a stopping time, therefore . We have

#Proposition B.10

Consider a sequence . Let be the accumulation times of . Then:

#Proof of Proposition B.10

We prove by induction on . For , the claim is obvious. If then and, by the induction hypothesis

Now assume . Then (because for the sum in the definition of accumulation times to be it has to be non-empty), therefore

By the induction hypothesis

If then

By definition of accumulation times, the middle term is . By definition of , the last term is . We get

Finally, assume . We have

By definition of accumulation times, we get that for any :

It follows that

Combining, we have

#Proposition B.11

Consider a probability space, a filtration of , and stochastic processes adapted to . Define by

Assume that there is s.t.

Let be the accumulation times of . Then, for any , and are uniformly integrable.

#Proof of Proposition B.11

By Proposition B.10, , so is uniformly bounded and in particular uniformly integrable. Moreover:

Since , it follows that is uniformly integrable.

#Proposition B.12

Consider sequences and . Let be the accumulation times of . Assume that there is s.t.

Assume further that either all are finite or converges. Denote . Then, .

#Proof of Proposition B.12

Fix . If , the claim is trivial. Consider the case . We have

Now consider the case , . By definition of accumulation times:

It follows that .

#Proposition B.13

Consider a sequence . Let be the accumulation times of . Assume that . Then:

#Proof of Proposition B.13

We prove by induction on . For , the claim is obvious. implies that , therefore

The first term is by induction. The second term is by definition of accumulation time.

#Proposition B.14

Consider sequences and . Assume that there is s.t.

Assume further that . In particular, . Let be the accumulation times of (finite because of the previous observation). Then, .

#Proof of Proposition B.14

We need to prove that for any and , there is s.t. . We know that there is s.t. . We have , therefore we can choose s.t. . We get

Therefore, .

#Proposition B.15

Consider , and as before. Consider a metastrategy and a market. Define by

Then:

#Proof of Proposition B.15

We prove by induction. For the basis, . Consider any and . First, assume that

Then, by definition of , and therefore

It follows that

Using the induction hypothesis, we conclude

Now, assume that

Then, and therefore

We get

By the assumption, the first two terms total to , yielding the desired result.

#Proposition B.16

Consider a compact Polish space, and closed s.t. and for some . Then

#Proof of Proposition B.16

Choose and . is a non-increasing sequence, therefore it is sufficient to prove that a subsequence converges to 0. Therefore, we can assume without loss of generality that and . For each , the fact that is closed implies that . Therefore, . We get

#Proposition B.17

Consider a trading metastrategy, a market and . Assume that

Then, for any , .

#Proof of Proposition B.17

Choose s.t.

Since differ by a constant function, we have

In particular, for any

By Proposition B.16, there is s.t. for any

Taking the sum of the last two inequalities, we conclude that for any

By definition of , we get the desired result.

##Appendix C

The following version of the optional stopping theorem appears in Durret as Theorem 5.4.7:

#Theorem C

Let be a probability space, a filtration on , a stochastic process adapted to and stopping times (w.r.t. ) s.t. . Assume that is a uniformly integrable submartingale. Using Doob's martingale convergence theorem, we can define by

(the above is well-defined almost everywhere, which is sufficient for our purpose) We define is an analogous way (this time we can't use Doob's martingale convergence theorem, but whenever we also have , therefore the limit almost surely converges). We also define by

Then:

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