% operators that are separated from the operand by a space
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% Paper specific
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 X a topological space:
P(X) is the space of Borel probability measures on X equipped with the weak* topology.
C(X) is the Banach space of continuous functions X→R with uniform norm.
B(X) is the Borel σ-algebra on X.
U(X) is the σ-algebra of universally measurable sets on X.
Given μ∈P(X), suppμ denotes the support of μ.
Given X and Y measurable spaces, K:Xmk−→Y is a Markov kernel from X to Y. For any x∈X, we have K(x)∈P(Y). Given μ∈P(X), μ⋉K∈P(X×Y) is the semidirect product of μ and K and K∗μ∈P(Y) is the pushforward of μ by K.
Given X, Y Polish spaces, π:X→Y Borel measurable and μ∈P(X), we denote μ∣π the set of Markov kernels K:Ymk−→X s.t. π∗μ⋉K is supported on the graph of π and K∗π∗μ=μ. By the disintegration theorem, μ∣π is always non-empty and any two kernels in μ∣π coincide π∗μ-almost everywhere.
##Results
The way we previously laid out the dominant market formalism, the sequence of observations (represented by the sets {Xi}) was fixed. To study forecasting, we instead need to assume this sequence is sampled from some probability measure (the true environment).
For each n∈N, let On be a compact Polish space. On represents the space of possible observations at time n. Denote
Yn:=∏m<nOm
Given n≤m, πnm:Ym→Yn denotes the projection mapping and πn:=πn,n+1. Denote
X=∞∏n=0On
X is a compact Polish space. For each n∈N we denote πnω:X→Yn the projection mapping. Given y∈Yn, we denote Xy:=π−1nω(y), a closed subspace of X. Given n∈Z and x∈X, we denote x(n):=πmax(n,0)ω(x).
#Definition 1
A market is a sequence of mappings {Mn:Yn→P(X)}n∈N s.t.
Each Mn is measurable w.r.t. U(Yn) and B(P(X)).
For any y∈Yn, suppMn(y)⊆Xy.
As before, we define the space of trading strategies T(X):=C(P(X)×X), but this time we regard it as a Banach space.
#Definition 2
A trader is a sequence of mappings {Tn:Yn×P(X)n→T(X)}n∈N which are measurable w.r.t. U(Yn)⊗B(P(X)n) and B(T(X)).
Given a trader T and a market M, we define the mappings {TMn:Yn→T(X)}n∈N (measurable w.r.t. U(Yn) and B(T(X))) and {¯TMn:Yn→C(X)}n∈N (measurable w.r.t. U(Yn) and B(C(X))) as follows:
The "market maker" lemma now requires some additional work due to the measurability requirement:
#Lemma
Consider any trader T. Then, there is a market M s.t. for all n∈N and y∈Yn
suppMn(y)⊆argmaxXy¯TMn(y)
As before, we have the operator W:T(X)→T(X) defined by
Wτ(μ,x):=τ(μ,x)−Ez∼μ[τ(μ,z)]
We also introduce the notation {W¯TMn:Yn→C(X)}n∈N and {ΣWTMn:Ymax(n−1,0)→C(X)}n∈N which are measurable mappings defined by
W¯TMn(y)=WTMn(y,Mn(y))
ΣWTMn(y)=∑m<nW¯TMm(πmn(y))
#Definition 3
A market M is said to dominate a trader T when for any x∈X, if
infn∈Nminz∈Xx(n)ΣWTMn+1(x(n),z)>−∞
then
supn∈Nmaxz∈Xx(n)ΣWTMn+1(x(n),z)<+∞
#Theorem 1
Given any countable set of traders R, there is a market M s.t. M dominates all T∈R.
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 μ∗∈P(X) s.t. if a market dominates a trader from this class, a certain function of the pricing converges to 0 with μ∗-probability 1. In a future post, we will apply this result to a trader associated with an incomplete models Φ⊆P(X) by observing that the trader is in the class for any μ∗∈Φ.
#Definition 4
A trading metastrategy is a uniformly bounded family of measurable mappings {υn:Yn→T(X)}n∈N. Given μ∗∈P(X), υ is said said to be profitable for μ∗, when there are β>0 and {Kn∈μ∗∣πnω}n∈N s.t. for any n∈N, πnω∗μ∗-almost any y∈Yn and any μ∈P(Xy):
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 υ. We define the trader Tυ and the measurable mappings {Uυn:Yn×P(X)n→C(X)}n∈N recursively as follows:
We can view Z as the graph of a multivalued mapping from Yn×T(X) to P(X). 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 M.
Z1 is closed by Proposition B.7. Z2 is closed by Proposition B.5. Z=Z1∩Z2 by Proposition B.6 and hence closed. In particular, the fiber Zyτ of Z over any (y,τ)∈Y×T(X) is also closed.
For any y∈Y, τ∈T(X), define iy:Y′→X by iy(y′):=(y,y′) and τy∈T(Y′) by τy(ν,y′):=τ(iy∗ν,y,y′). Applying Proposition A.1 to τy we get ν∈P(Y′) s.t.
suppτy(ν)⊆argmaxτy
It follows that (y,τ,iy∗ν)∈Z and hence Zyτ is non-empty.
Consider any U⊆P(X) open. Then, AU:=(Y×T(X)×U)∩Z is locally closed and in particular Fσ. Therefore, the image of AU under the projection to Y×T(X) is also Fσ and in particular Borel.
Applying the Kuratowski-Ryll-Nardzewski measurable selection theorem, we get the desired result.
#Proof of Lemma
For any n∈N, let Mn:Yn×T(X)→P(X) be as in Proposition A.2. We define Mn recursively by:
Mn(y):=Mn(y,TMn(y))
#Proposition A.3
Consider X a probability space, {Fn⊆2X}n∈N a filtration of X, t,α,β>0, {Sn:X→R}n∈N and {Δn:X→[0,t]}n∈N stochastic processes adapted to F. Assume that:
E[|S0|]<∞
∀n′≥n:|Sn′−Sn|≤n′−1∑m=nΔm+α
E[Sn+1−Sn∣Fn]≥βΔn
Then, infnSn>−∞ with probability 1.
The proof will use the following definition:
#Definition A
Consider a sequence {tn∈[0,1]}n∈N. The accumulation times of t are {nk∈N⊔{∞}}k∈N defined recursively by
n0:=0
nk+1={inf{n∈N∣∑n−1m=nktm≥1} if nk<∞∞ if nk=∞
Consider X a probability space and {Δn:X→[0,1]}n∈N a stochastic process. The accumulation times of Δ are {Nk:X→N⊔{∞}}k∈N defined pointwise as above. Clearly, they are stochastic processes and whenever Δ is adapted to a filtration {Fn⊆2X}n∈N, they are stopping times w.r.t. F.
#Proof of Proposition A.3
Without loss of generality, we can assume t=1 (otherwise we can renormalize S, Δ and α by a factor of t−1). Define {S0n:X→R}n∈N by
S0n:=Sn−βn−1∑m=0Δm
By Proposition B.8, S0 is a submartingale. Let {Nn:X→N⊔{∞}}n∈N be the accumulation times of Δ. By proposition N23, {S0min(n,Nk)}n∈N are submartingales for all k. By Proposition\ N24, each of them is uniformly integrable. Using the fact that Nk≤Nk+1 to apply Theorem C, we get
E[S0Nk+1∣FNk]≥S0Nk
Clearly, {S0Nk}k∈N is adapted to {FNk}k∈N. Doob's second martingale convergence theorem implies that E[|S0Nk|]<∞ (S0Nk is the limit of the uniformly integrable submartingale {S0min(n,Nk)}n∈N). We conclude that {S0Nk}k∈N is a submartingale.
By Proposition B.12, |S0Nk+1−S0Nk|≤α+2. Applying the Azuma-Hoeffding inequality, we conclude that for any positive integer k:
It remains to show that if x∈X is s.t. infnSn(x)=−∞ then the condition above fails. Consider any such x∈X. |Sn(x)−S0(x)|≤∑n−1m=0Δn(x)+α, therefore ∑∞n=0Δn(x)=∞. On the other hand, by Proposition B.14, infkSNk(x)(x)=−∞.
#Proposition A.4
Consider X a probability space, {Fn⊆2X}n∈N a filtration of X, t,α,β>0, {S′n:X→R}n∈N and {Δn:X→[0,t]}n∈N stochastic processes adapted to F and {Sn:X→R}n∈N an arbitrary stochastic process. Assume that:
|Sn−S′n|≤α4
E[|S0|]<∞
|Sn+1−Sn|≤Δn
E[Sn+1−Sn∣Fn]≥βΔn
Then, infnSn>−∞ (equivalently infnS′n>−∞) with probability 1.
By Proposition A.3, infnS′′n>−∞ with probability 1. Since |S′′n−Sn|≤α2, we get the desired result.
#Proposition A.5
Consider {On}n∈N, {Yn:=∏m<nOm}n∈N and X:=∏nOn as before. Consider μ∗∈P(X), {Kn∈μ∗∣πnω}n∈N, υ a metastrategy profitable for μ∗ and M a market. Then, for μ∗-almost any x∈X:
infn∈Nminz∈Xx(n)ΣWTMυ,n+1(x(n),z)>−∞
#Proof of Proposition A.5
We regard X as a probability space using the σ-algebra U(X) and the probability measure μ∗. For any n∈N, we define Fn⊆U(X) and Sn,S′n,Δn:X→R by
Clearly, F is a filtration of X, S,S′,Δ are stochastic processes and S′,Δ are adapted to F. υ is uniformly bounded, therefore Tυ is uniformly bounded and so is Δ. Obviously, Δ is also non-negative.
By Proposition B.15, |Sn−S′n| are uniformly bounded. S0 is bounded and in particular E[|S0|]<∞. We have
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 x∈X:
supn∈NSn(x)<∞⟹∞∑n=0Δn(x)<∞
#Proof of Proposition A.7
Define S′′n:=E[Sn∣Fn]. As in the proof of Proposition A.4, S′′ meets the conditions of Proposition A.3 and thus of Proposition A.6 also. By Proposition A.6, for almost all x∈X:
supn∈NS′′n(x)<∞⟹∞∑n=0Δn(x)<∞
As in the proof of Proposition A.4, |S′′n−Sn| is uniformly bounded, giving the desired result.
#Proof of Theorem 2
Let F, S, S′ and Δ be as in the proof of Proposition A.5. Using Proposition A.5 and the assumption that M dominates Tυ, we conclude that for μ∗-almost any x∈X, supnSn(x)<∞. 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 ∑nΔn(x)<∞. By Proposition B.17, it follows that for any n≫0, TMυn(x(n))=υ(x(n)). We get
If X,Y are compact Polish spaces and f:X×Y→R is continuous, then F:X→C(Y) defined by F(x)(y):=f(x,y) is continuous.
We omit the proof of Proposition B.1, since it appeared as "Proposition A.2" before.
#Proposition B.2
Fix X,Y compact Polish spaces. Define e:C(Y×X)×Y→C(X) by e(f,y)(x):=f(y,x). Then, e is continuous. In particular, we can apply this to Y=P(X) in which case e:T(X)×P(X)→C(X).
#Proof of Proposition B.2
Consider fk→f and yk→y. We have
maxx∈X|fk(yk,x)−f(yk,x)|≤∥fk−f∥→0
By Proposition B.1
maxx∈X|f(yk,x)−f(y,x)|→0
Combining, we get
maxx∈X|fk(yk,x)−f(y,x)|→0
#Proposition B.3
Fix Y,Y′ compact Polish spaces and denote X:=Y×Y′. Define F:Y×C(X)→R by
F(y,f):=maxy′∈Y′f(y,y′)
Then, F is continuous.
#Proof of Proposition B.3
Consider yk→y, fk→f. By Proposition B.1, yk→y implies that
limk→∞maxy′∈Y′f(yk,y′)=maxy′∈Y′f(y,y′)
Since fk→f, we get
limk→∞maxy′∈Y′fk(yk,y′)=maxy′∈Y′f(y,y′)
#Proposition B.4
Fix Y,Y′ compact Polish spaces. Denote X:=Y×Y′. Define Z⊆Y×P(X)×C(X) by
Z:={(y,μ,f)∈Y×P(X)×C(X)∣Eμ[f]=maxy′∈Y′f(y,y′)}
Then, Z is closed.
#Proof of Proposition B.4
Consider yk→y, μk→μ, fk→f, (yk,μk,fk)∈Z. By Proposition B.3, we get
Consider yk→y, μk→μ, (yk,μk)∈Z. We have Eμk[dyk]=0. By Proposition B.1, dyk→dy, therefore Eμ[dy]=0. By Proposition B.6, suppμ⊆y×Y′ and hence (y,μ)∈Z.
#Proposition B.8
Consider X a probability space, {Fn⊆2X}n∈N a filtration of X, {Sn:X→R}n∈N and {Δn:X→[0,1]}n∈N stochastic processes adapted to F. Assume that there are α,β>0 s.t.:
Let X be a probability space, {Fn⊆2X}n∈N a filtration on X, {Sn:X→R}n∈N a stochastic process adapted to F and N:X→N⊔{∞} a stopping time (w.r.t. F). Suppose S is submartingale. Then, {Smin(n,N)}n∈N is also submartingale.
#Proof of Proposition B.9
Clearly, {Smin(n,N)}n∈N is adapted to F. We have
|Smin(n,N)|≤∑m≤n|Sm|
E[|Smin(n,N)|]≤E[∑m≤n|Sm|]=∑m≤nE[|Sm|]<∞
For any n∈N, define An⊆X by
An:={x∈X∣N(x)>n}
N is a stopping time, therefore An∈Fn. We have
Smin(n+1,N)−Smin(n,N)=χAn(Sn+1−Sn)
E[Smin(n+1,N)−Smin(n,N)∣Fn]=E[χAn(Sn+1−Sn)∣Fn]
E[Smin(n+1,N)−Smin(n,N)∣Fn]=χAnE[Sn+1−Sn∣Fn]≥0
#Proposition B.10
Consider a sequence {tn∈[0,1]}n∈N. Let {nk∈N⊔{∞}}k∈N be the accumulation times of t. Then:
nk−1∑n=0tn≤2k
#Proof of Proposition B.10
We prove by induction on k. For k=0, the claim is obvious. If nk=∞ then nk+1=∞ and, by the induction hypothesis
nk+1−1∑n=0tn=∞∑n=0tn=nk−1∑n=0tn≤2k≤2(k+1)
Now assume nk<∞. Then nk+1>nk (because for the sum in the definition of accumulation times to be ≥1 it has to be non-empty), therefore
nk+1−1∑n=0tn=nk−1∑n=0tn+nk+1−1∑n=nktn
By the induction hypothesis
nk+1−1∑n=0tn≤2k+nk+1−1∑n=nktn
If nk+1<∞ then
nk+1−1∑n=0tn≤2k+nk+1−2∑n=nktn+tnk+1−1
By definition of accumulation times, the middle term is <1. By definition of t, the last term is ≤1. We get
nk+1−1∑n=0tn<2k+1+1=2(k+1)
Finally, assume nk+1=∞. We have
nk+1−1∑n=0tn≤2k+∞∑n=nktn
By definition of accumulation times, we get that for any m∈N:
m∑n=nktn<1
It follows that
∞∑n=nktn≤1
Combining, we have
nk+1−1∑n=0tn≤2k+1<2(k+1)
#Proposition B.11
Consider X a probability space, {Fn⊆2X}n∈N a filtration of X, {Sn:X→R}n∈N and {Δn:X→[0,1]}n∈N stochastic processes adapted to F. Define {Un:X→R}n∈N by
Un:=n−1∑m=0Δm
Assume that there is α≥0 s.t.
E[|S0|]<∞
|Sn−S0|≤Un+α
Let {Nk:X→N⊔{∞}}k∈N be the accumulation times of Δ. Then, for any k∈N, {Smin(n,Nk)}n∈N and {Umin(n,Nk)}n∈N are uniformly integrable.
#Proof of Proposition B.11
By Proposition B.10, Umin(n,Nk)≤UNk≤2k, so {Umin(n,Nk)}n∈N is uniformly bounded and in particular uniformly integrable. Moreover:
|Smin(n,Nk)|≤|S0|+Umin(n,Nk)≤|S0|+2k
Since E[|S0|]<∞, it follows that {Smin(n,Nk)}n∈N is uniformly integrable.
#Proposition B.12
Consider sequences {sn∈R}n∈N and {tn∈[0,1]}n∈N. Let {nk:X→N⊔{∞}}k∈N be the accumulation times of t. Assume that there is α≥0 s.t.
∀n′≥n:|sn′−sn|≤n′−1∑m=ntm+α
Assume further that either all nk are finite or s converges. Denote snk:=limn→nksn. Then, |snk+1−snk|<α+2.
#Proof of Proposition B.12
Fix k∈N. If nk=∞, the claim is trivial. Consider the case nk+1<∞. We have
Now consider the case nk<∞, nk+1=∞. By definition of accumulation times:
∞∑n=nktn≤1
It follows that |snk+1−snk|=|limn→∞sn−snk|≤α+1.
#Proposition B.13
Consider a sequence {tn∈[0,1]}n∈N. Let {nk∈N⊔{∞}}k∈N be the accumulation times of t. Assume that ∑∞ntn=∞. Then:
nk−1∑n=0tn≥k
#Proof of Proposition B.13
We prove by induction on k. For k=0, the claim is obvious. ∑∞ntn=∞ implies that n0<n1<…<∞, therefore
nk+1−1∑n=0tn=nk−1∑n=0tn+nk+1−1∑n=nktn
The first term is ≥k by induction. The second term is ≥1 by definition of accumulation time.
#Proposition B.14
Consider sequences {sn∈R}n∈N and {tn∈[0,1]}n∈N. Assume that there is α≥0 s.t.
∀n′≥n:|sn′−sn|≤n′−1∑m=ntm+α
Assume further that infnsn=−∞. In particular, ∑∞ntn=∞. Let {nk∈N}k∈N be the accumulation times of t (finite because of the previous observation). Then, infksnk=−∞.
#Proof of Proposition B.14
We need to prove that for any s>0 and k∈N, there is l≥k s.t. snl<−s. We know that there is n≥nk s.t. sn<−(s+α+2). We have n0<n1<…<∞, therefore we can choose l≥k s.t. nl≤n<nl+1. We get
By the assumption, the first two terms total to ≤1, yielding the desired result.
#Proposition B.16
Consider X a compact Polish space, f∈C(X) and {Xn⊆X}n∈N closed s.t. Xn+1⊆Xn and ⋂nXn={x∗} for some x∗∈X. Then
limn→∞(maxXnf−minXnf)=0
#Proof of Proposition B.16
Choose {x+n∈argmaxXnf}n∈N and {x−n∈argminXnf}n∈N. maxXnf−minXnf 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 x+n→x+ω and x−n→x−ω. For each n∈N, the fact that Xn is closed implies that x+ω,x−ω∈Xn. Therefore, x+ω=x−ω=x∗. We get
Taking the sum of the last two inequalities, we conclude that for any n≥n1
maxXx(n)ΣWTMυn(x(n−1))−minXx(n)ΣWTMυn(x(n−1))<1
By definition of Tυ, 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 X be a probability space, {Fn⊆2X}n∈N a filtration on X, {Sn:X→R}n∈N a stochastic process adapted to F and M,N:X→N⊔{∞} stopping times (w.r.t. F) s.t. M≤N. Assume that {Smin(n,N)}n∈N is a uniformly integrable submartingale. Using Doob's martingale convergence theorem, we can define SN:X→R by
SN(x):=limn→∞Smin(n,N)(x)={SN(x)(x) if N(x)<∞limn→∞Sn(x) if N(x)=∞
(the above is well-defined almost everywhere, which is sufficient for our purpose) We define SM:X→R is an analogous way (this time we can't use Doob's martingale convergence theorem, but whenever M(x)=∞ we also have N(x)=∞, therefore the limit almost surely converges). We also define FM⊆2X by
% operators that are separated from the operand by a space
% operators that require brackets
% operators that require parentheses
% Paper specific
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 X a topological space:
P(X) is the space of Borel probability measures on X equipped with the weak* topology.
C(X) is the Banach space of continuous functions X→R with uniform norm.
B(X) is the Borel σ-algebra on X.
U(X) is the σ-algebra of universally measurable sets on X.
Given μ∈P(X), suppμ denotes the support of μ.
Given X and Y measurable spaces, K:Xmk−→ Y is a Markov kernel from X to Y. For any x∈X, we have K(x)∈P(Y). Given μ∈P(X), μ⋉K∈P(X×Y) is the semidirect product of μ and K and K∗μ∈P(Y) is the pushforward of μ by K.
Given X, Y Polish spaces, π:X→Y Borel measurable and μ∈P(X), we denote μ∣π the set of Markov kernels K:Ymk−→X s.t. π∗μ⋉K is supported on the graph of π and K∗π∗μ=μ. By the disintegration theorem, μ∣π is always non-empty and any two kernels in μ∣π coincide π∗μ-almost everywhere.
##Results
The way we previously laid out the dominant market formalism, the sequence of observations (represented by the sets {Xi}) was fixed. To study forecasting, we instead need to assume this sequence is sampled from some probability measure (the true environment).
For each n∈N, let On be a compact Polish space. On represents the space of possible observations at time n. Denote
Yn:=∏m<nOm
Given n≤m, πnm:Ym→Yn denotes the projection mapping and πn:=πn,n+1. Denote
X=∞∏n=0On
X is a compact Polish space. For each n∈N we denote πnω:X→Yn the projection mapping. Given y∈Yn, we denote Xy:=π−1nω(y), a closed subspace of X. Given n∈Z and x∈X, we denote x(n):=πmax(n,0)ω(x).
#Definition 1
A market is a sequence of mappings {Mn:Yn→P(X)}n∈N s.t.
Each Mn is measurable w.r.t. U(Yn) and B(P(X)).
For any y∈Yn, suppMn(y)⊆Xy.
As before, we define the space of trading strategies T(X):=C(P(X)×X), but this time we regard it as a Banach space.
#Definition 2
A trader is a sequence of mappings {Tn:Yn×P(X)n→T(X)}n∈N which are measurable w.r.t. U(Yn)⊗B(P(X)n) and B(T(X)).
Given a trader T and a market M, we define the mappings {TMn:Yn→T(X)}n∈N (measurable w.r.t. U(Yn) and B(T(X))) and {¯TMn:Yn→C(X)}n∈N (measurable w.r.t. U(Yn) and B(C(X))) as follows:
TMn(y):=Tn(y,M0(π0n(y)),M1(π1n(y))…Mn−1(πn−1,n(y)))
¯TMn(y):=TMn(y,Mn(y))
The "market maker" lemma now requires some additional work due to the measurability requirement:
#Lemma
Consider any trader T. Then, there is a market M s.t. for all n∈N and y∈Yn
suppMn(y)⊆argmaxXy¯TMn(y)
As before, we have the operator W:T(X)→T(X) defined by
Wτ(μ,x):=τ(μ,x)−Ez∼μ[τ(μ,z)]
We also introduce the notation {W¯TMn:Yn→C(X)}n∈N and {ΣWTMn:Ymax(n−1,0) →C(X)}n∈N which are measurable mappings defined by
W¯TMn(y)=WTMn(y,Mn(y))
ΣWTMn(y)=∑m<nW¯TMm(πmn(y))
#Definition 3
A market M is said to dominate a trader T when for any x∈X, if
infn∈Nminz∈Xx(n)ΣWTMn+1(x(n),z)>−∞
then
supn∈Nmaxz∈Xx(n)ΣWTMn+1(x(n),z)<+∞
#Theorem 1
Given any countable set of traders R, there is a market M s.t. M dominates all T∈R.
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 μ∗∈P(X) s.t. if a market dominates a trader from this class, a certain function of the pricing converges to 0 with μ∗-probability 1. In a future post, we will apply this result to a trader associated with an incomplete models Φ⊆P(X) by observing that the trader is in the class for any μ∗∈Φ.
#Definition 4
A trading metastrategy is a uniformly bounded family of measurable mappings {υn:Yn→T(X)}n∈N. Given μ∗∈P(X), υ is said said to be profitable for μ∗, when there are β>0 and {Kn∈μ∗∣πnω}n∈N s.t. for any n∈N, πnω∗μ∗-almost any y∈Yn and any μ∈P(Xy):
EKn(y)[υ(y,μ)]−Eμ[υ(y,μ)]≥β(maxXyυ(y,μ)−minXyυ(y,μ))
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 υ. We define the trader Tυ and the measurable mappings {Uυn:Yn×P(X)n→C(X)}n∈N recursively as follows:
U0:=0
Tυ0:=υ0
Uυ,n+1(y,{μm}m≤n):=Uυn(πn(y),{μm}m<n)+Tυn(y,{μm}m≤n)
Tυ,n+1(y,{μm}m≤n):={υn+1(y) if maxXyUυ,n+1(y,{μm}m≤n)−minXyUυ,n+1(y,{μm}m≤n)≤10 otherwise
#Theorem 2
Consider μ∗∈P(X), {Kn∈μ∗∣πnω}n∈N, υ a metastrategy profitable for μ∗ and M a market. Assume M dominates Tυ. Then, for μ∗-almost any x∈X:
limn→∞(EKn(x(n))[υ(x(n),Mn(x(n)))]−EMn(x(n))[υ(x(n),Mn(x(n)))])=0
That is, the market price of the "stock portfolio" traded by υ converges to its true μ∗-expected value.
##Appendix A
#Proposition A.1
Fix X a compact Polish space and τ∈T(X). Then, there exists μ∈P(X) s.t.
suppτ(μ)⊆argmaxτ
#Proof of Proposition A.1
Follows immediately from "Proposition 1" from before and Proposition B.6.
#Proposition A.2
Fix Y,Y′ compact Polish spaces. Denote X:=Y×Y′. Then, there exists M:Y×T(X)→P(X) measurable w.r.t. B(Y×T(X)) and B(P(X)) s.t. for any y∈Y and τ∈T(X):
suppM(y,τ)⊆argmaxy×Y′τ(M(y,τ))
#Proof of Proposition A.2
Define Z1,Z2,Z⊆Y×T(X)×P(X) by
Z1:={(y,τ,μ)∈Y×T(X)×P(X)∣suppμ⊆Xy}
Z2:={(y,τ,μ)∈Y×T(X)×P(X)∣Eμ[τ(μ)]=maxy∈Y′τ(μ,y,y′)}
Z:={(y,τ,μ)∈Y×T(X)×P(X)∣suppμ⊆argmaxy×Y′τ(μ)}
We can view Z as the graph of a multivalued mapping from Yn×T(X) to P(X). 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 M.
Z1 is closed by Proposition B.7. Z2 is closed by Proposition B.5. Z=Z1∩Z2 by Proposition B.6 and hence closed. In particular, the fiber Zyτ of Z over any (y,τ)∈Y×T(X) is also closed.
For any y∈Y, τ∈T(X), define iy:Y′→X by iy(y′):=(y,y′) and τy∈T(Y′) by τy(ν,y′):=τ(iy∗ν,y,y′). Applying Proposition A.1 to τy we get ν∈P(Y′) s.t.
suppτy(ν)⊆argmaxτy
It follows that (y,τ,iy∗ν)∈Z and hence Zyτ is non-empty.
Consider any U⊆P(X) open. Then, AU:=(Y×T(X)×U)∩Z is locally closed and in particular Fσ. Therefore, the image of AU under the projection to Y×T(X) is also Fσ and in particular Borel.
Applying the Kuratowski-Ryll-Nardzewski measurable selection theorem, we get the desired result.
#Proof of Lemma
For any n∈N, let Mn:Yn×T(X)→P(X) be as in Proposition A.2. We define Mn recursively by:
Mn(y):=Mn(y,TMn(y))
#Proposition A.3
Consider X a probability space, {Fn⊆2X}n∈N a filtration of X, t,α,β>0, {Sn:X→R}n∈N and {Δn:X→[0,t]}n∈N stochastic processes adapted to F. Assume that:
E[|S0|]<∞
∀n′≥n:|Sn′−Sn|≤n′−1∑m=nΔm+α
E[Sn+1−Sn∣Fn]≥βΔn
Then, infnSn>−∞ with probability 1.
The proof will use the following definition:
#Definition A
Consider a sequence {tn∈[0,1]}n∈N. The accumulation times of t are {nk∈N⊔{∞}}k∈N defined recursively by
n0:=0
nk+1={inf{n∈N∣∑n−1m=nktm≥1} if nk<∞∞ if nk=∞
Consider X a probability space and {Δn:X→[0,1]}n∈N a stochastic process. The accumulation times of Δ are {Nk:X→N⊔{∞}}k∈N defined pointwise as above. Clearly, they are stochastic processes and whenever Δ is adapted to a filtration {Fn⊆2X}n∈N, they are stopping times w.r.t. F.
#Proof of Proposition A.3
Without loss of generality, we can assume t=1 (otherwise we can renormalize S, Δ and α by a factor of t−1). Define {S0n:X→R}n∈N by
S0n:=Sn−βn−1∑m=0Δm
By Proposition B.8, S0 is a submartingale. Let {Nn:X→N⊔{∞}}n∈N be the accumulation times of Δ. By proposition N23, {S0min(n,Nk)}n∈N are submartingales for all k. By Proposition\ N24, each of them is uniformly integrable. Using the fact that Nk≤Nk+1 to apply Theorem C, we get
E[S0Nk+1∣FNk]≥S0Nk
Clearly, {S0Nk}k∈N is adapted to {FNk}k∈N. Doob's second martingale convergence theorem implies that E[|S0Nk|]<∞ (S0Nk is the limit of the uniformly integrable submartingale {S0min(n,Nk)}n∈N). We conclude that {S0Nk}k∈N is a submartingale.
By Proposition B.12, |S0Nk+1−S0Nk|≤α+2. Applying the Azuma-Hoeffding inequality, we conclude that for any positive integer k:
Pr[S0Nk−S0<−βk]≤exp(−(βk)22(α+2)2k)=exp(−β2k2(α+2)2)
Since ∑kexp(−β2k2(α+2)2)<∞, it follows that
Pr[∃k∈N∀l>k:S0Nl−S0≥−βl]=1
Pr[∃k∈N∀l>k:SNl−S0≥β(Nl−1∑n=0Δn−l)]=1
By Proposition B.13
Pr[∞∑n=0Δn=∞⟹∃k∈N∀l>k:SNl−S0≥0]=1
It remains to show that if x∈X is s.t. infnSn(x)=−∞ then the condition above fails. Consider any such x∈X. |Sn(x)−S0(x)|≤∑n−1m=0Δn(x)+α, therefore ∑∞n=0Δn(x)=∞. On the other hand, by Proposition B.14, infkSNk(x)(x)=−∞.
#Proposition A.4
Consider X a probability space, {Fn⊆2X}n∈N a filtration of X, t,α,β>0, {S′n:X→R}n∈N and {Δn:X→[0,t]}n∈N stochastic processes adapted to F and {Sn:X→R}n∈N an arbitrary stochastic process. Assume that:
|Sn−S′n|≤α4
E[|S0|]<∞
|Sn+1−Sn|≤Δn
E[Sn+1−Sn∣Fn]≥βΔn
Then, infnSn>−∞ (equivalently infnS′n>−∞) with probability 1.
#Proof of Proposition A.4
Define S′′n:=E[Sn∣Fn]. We have
E[|S′′0|]=E[|E[S0∣F0]|]≤E[E[|S0|∣F0]]=E[|S0|]<∞
|S′′n−S′n|=|E[Sn∣Fn]−S′n|=|E[Sn−S′n∣Fn]|≤α4
|S′′n−Sn|≤|S′′n−S′n|+|S′n−Sn|≤α2
∀n′≥n:|S′′n′−S′′n|≤|Sn′−Sn|+α≤n′−1∑m=nΔm+α
E[S′′n+1−S′′n∣Fn]=E[E[Sn+1∣Fn+1]−E[Sn∣Fn]∣Fn]=E[Sn+1−Sn∣Fn]≥βΔn
By Proposition A.3, infnS′′n>−∞ with probability 1. Since |S′′n−Sn|≤α2, we get the desired result.
#Proposition A.5
Consider {On}n∈N, {Yn:=∏m<nOm}n∈N and X:=∏nOn as before. Consider μ∗∈P(X), {Kn∈μ∗∣πnω}n∈N, υ a metastrategy profitable for μ∗ and M a market. Then, for μ∗-almost any x∈X:
infn∈Nminz∈Xx(n)ΣWTMυ,n+1(x(n),z)>−∞
#Proof of Proposition A.5
We regard X as a probability space using the σ-algebra U(X) and the probability measure μ∗. For any n∈N, we define Fn⊆U(X) and Sn,S′n,Δn:X→R by
Fn:=π−1nω(U(Yn))
Sn(x):=ΣWTMυn(x(n−1),x)
S′n(x):=minz∈Xx(n−1)ΣWTMυn(x(n−1),z)
Δn(x):=maxz∈Xx(n)¯TMυn(x(n),z)−minz∈Xx(n)¯TMυn(x(n),z)
Clearly, F is a filtration of X, S,S′,Δ are stochastic processes and S′,Δ are adapted to F. υ is uniformly bounded, therefore Tυ is uniformly bounded and so is Δ. Obviously, Δ is also non-negative.
By Proposition B.15, |Sn−S′n| are uniformly bounded. S0 is bounded and in particular E[|S0|]<∞. We have
|Sn+1(x)−Sn(x)|=|W¯TMυn(x(n),x)|≤Δn(x)
Let β>0 and Kn∈μ∗∣πnω be as in Definition 4.
E[Sn+1−Sn∣Fn]=Ez∼Kn(x(n))[W¯TMυn(x(n),z)]
E[Sn+1−Sn∣Fn]=Ez∼Kn(x(n))[¯TMυn(x(n),z)]−Ez∼Mn(x(n))[¯TMυn(x(n),z)]
By definition of Tυ, ¯TMυn(x(n)) is equal to either υn(x(n),Mn(x(n))) or 0. In either case, we get (almost everywhere)
E[Sn+1−Sn∣Fn]≥β(maxXx(n)¯TMυn(x(n))−minXx(n)¯TMυn(x(n)))=βΔn
Applying Proposition A.4, we get the desired result.
#Proposition A.6
Consider the setting of Proposition A.3. Then, for almost all x∈X:
supn∈NSn(x)<∞⟹∞∑n=0Δn(x)<∞
#Proof of Proposition A.6
Define {S0n:X→R}n∈N by
S0n:=Sn−βn−1∑m=0Δm
Let {Nn}n∈N be the accumulation times of Δ. Consider any x∈X s.t. supnSn(x)=s(x)<∞ but ∑nΔn(x)=∞. Proposition B.13 implies that
S0Nk(x)(x)≤SNk(x)(x)−βk≤s(x)−βk
As in the proof of Proposition A.3, we can apply the Azuma-Hoeffding inequality to S0N and get that for any positive integer k
Pr[S0Nk−S0<−k34]≤exp(−k322(α+2)2k)=exp(−k122(α+2)2)
It follows that
∞∑k=1Pr[S0Nk−S0<−k34]<∞
Pr[∃k∈N∀l>k:S0Nl−S0<−l34]=0
Pr[∃m∈N∀k∈N:S0Nk≤m−βk]=0
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 x∈X:
supn∈NSn(x)<∞⟹∞∑n=0Δn(x)<∞
#Proof of Proposition A.7
Define S′′n:=E[Sn∣Fn]. As in the proof of Proposition A.4, S′′ meets the conditions of Proposition A.3 and thus of Proposition A.6 also. By Proposition A.6, for almost all x∈X:
supn∈NS′′n(x)<∞⟹∞∑n=0Δn(x)<∞
As in the proof of Proposition A.4, |S′′n−Sn| is uniformly bounded, giving the desired result.
#Proof of Theorem 2
Let F, S, S′ and Δ be as in the proof of Proposition A.5. Using Proposition A.5 and the assumption that M dominates Tυ, we conclude that for μ∗-almost any x∈X, supnSn(x)<∞. 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 ∑nΔn(x)<∞. By Proposition B.17, it follows that for any n≫0, TMυn(x(n))=υ(x(n)). We get
EKn(x(n))[υ(x(n),Mn(x(n)))]−EMn(x(n))[υ(x(n),Mn(x(n)))]=EKn(x(n))[¯TMυn(x(n))]−EMn(x(n))[¯TMυn(x(n))])
EKn(x(n))[υ(x(n),Mn(x(n)))]−EMn(x(n))[υ(x(n),Mn(x(n)))]≤Δn(x)
limn→∞(EKn(x(n))[υ(x(n),Mn(x(n)))]−EMn(x(n))[υ(x(n),Mn(x(n)))])=0
##Appendix B
#Proposition B.1
If X,Y are compact Polish spaces and f:X×Y→R is continuous, then F:X→C(Y) defined by F(x)(y):=f(x,y) is continuous.
We omit the proof of Proposition B.1, since it appeared as "Proposition A.2" before.
#Proposition B.2
Fix X,Y compact Polish spaces. Define e:C(Y×X)×Y→C(X) by e(f,y)(x):=f(y,x). Then, e is continuous. In particular, we can apply this to Y=P(X) in which case e:T(X)×P(X)→C(X).
#Proof of Proposition B.2
Consider fk→f and yk→y. We have
maxx∈X|fk(yk,x)−f(yk,x)|≤∥fk−f∥→0
By Proposition B.1
maxx∈X|f(yk,x)−f(y,x)|→0
Combining, we get
maxx∈X|fk(yk,x)−f(y,x)|→0
#Proposition B.3
Fix Y,Y′ compact Polish spaces and denote X:=Y×Y′. Define F:Y×C(X)→R by
F(y,f):=maxy′∈Y′f(y,y′)
Then, F is continuous.
#Proof of Proposition B.3
Consider yk→y, fk→f. By Proposition B.1, yk→y implies that
limk→∞maxy′∈Y′f(yk,y′)=maxy′∈Y′f(y,y′)
Since fk→f, we get
limk→∞maxy′∈Y′fk(yk,y′)=maxy′∈Y′f(y,y′)
#Proposition B.4
Fix Y,Y′ compact Polish spaces. Denote X:=Y×Y′. Define Z⊆Y×P(X)×C(X) by
Z:={(y,μ,f)∈Y×P(X)×C(X)∣Eμ[f]=maxy′∈Y′f(y,y′)}
Then, Z is closed.
#Proof of Proposition B.4
Consider yk→y, μk→μ, fk→f, (yk,μk,fk)∈Z. By Proposition B.3, we get
maxy′∈Y′f(y,y′)=limk→∞maxy′∈Y′fk(yk,y′)=limk→∞Eμk[fk]=Eμ[f]
Hence, (y,μ,f)∈Z.
#Proposition B.5
Fix Y,Y′ compact Polish spaces. Denote X:=Y×Y′. Define Z⊆Y×P(X)×T(X) by
Z:={(y,μ,τ)∈Y×P(X)×T(X)∣Eμ[τ(μ)]=maxy′∈Y′τ(μ,y,y′)}
Then, Z is closed.
#Proof of Proposition B.5
By Proposition B.2, Z is the continuous inverse image of a subset of Y×P(X)×C(X) which is closed by Proposition B.4.
#Proposition B.6
Fix X a compact Polish space. Consider f∈C(X) and μ∈P(X) and denote M:=maxf. Then, suppμ⊆f−1(M) iff Eμ[f]=M.
#Proof of Proposition B.6
If suppμ⊆f−1(M) then Prx∼μ[f(x)≠M]=0 and therefore Eμ[f]=M.
Now, assume Eμ[f]=M. For any k∈N, Markov's inequality yields
Prx∼μ[M−f(x)≥1k]≤kEx∼μ[M−f(x)]=0
Taking k→∞, we get Prx∼μ[M>f(x)]=0 and hence suppμ⊆f−1(M).
#Proposition B.7
Consider Y,Y′ compact Polish spaces. Denote X:=Y×Y′. Define Z⊆Y×P(X) by
Z:={(y,μ)∈Y×P(X)∣suppμ⊆y×Y′}
Then, Z is closed.
#Proof of Proposition B.7
We fix metrizations for Y and Y′ and metrize X by
dX((y1,y′1),(y2,y′2)):=max(dY(y1,y2),dY′(y′1,y′2))
For each y∈Y, denote dy:=dy×Y′.
Consider yk→y, μk→μ, (yk,μk)∈Z. We have Eμk[dyk]=0. By Proposition B.1, dyk→dy, therefore Eμ[dy]=0. By Proposition B.6, suppμ⊆y×Y′ and hence (y,μ)∈Z.
#Proposition B.8
Consider X a probability space, {Fn⊆2X}n∈N a filtration of X, {Sn:X→R}n∈N and {Δn:X→[0,1]}n∈N stochastic processes adapted to F. Assume that there are α,β>0 s.t.:
E[|S0|]<∞
|Sn−S0|≤n−1∑m=0Δm+α
E[Sn+1−Sn∣Fn]≥βΔn
Define {S0n:X→R}n∈N by
S0n:=Sn−βn−1∑m=0Δm
Then, S0 is a submartingale.
#Proof of Proposition B.8
Obviously, S0 is adapted to F. We have
E[|S0n|]≤E[|Sn|]+βn≤E[|S0|]+E[|Sn−S0|]+βn≤E[|S0|]+2βn+α<∞
E[S0n+1∣Fn]=E[Sn+1∣Fn]−βn∑m=0Δm
E[S0n+1∣Fn]≥E[Sn∣Fn]+βΔn−βn∑m=0Δm
E[S0n+1∣Fn]≥E[Sn∣Fn]−βn−1∑m=0Δm
E[S0n+1∣Fn]≥E[S0n∣Fn]
#Proposition B.9
Let X be a probability space, {Fn⊆2X}n∈N a filtration on X, {Sn:X→R}n∈N a stochastic process adapted to F and N:X→N⊔{∞} a stopping time (w.r.t. F). Suppose S is submartingale. Then, {Smin(n,N)}n∈N is also submartingale.
#Proof of Proposition B.9
Clearly, {Smin(n,N)}n∈N is adapted to F. We have
|Smin(n,N)|≤∑m≤n|Sm|
E[|Smin(n,N)|]≤E[∑m≤n|Sm|]=∑m≤nE[|Sm|]<∞
For any n∈N, define An⊆X by
An:={x∈X∣N(x) >n}
N is a stopping time, therefore An∈Fn. We have
Smin(n+1,N)−Smin(n,N)=χAn(Sn+1−Sn)
E[Smin(n+1,N)−Smin(n,N)∣Fn]=E[χAn(Sn+1−Sn)∣Fn]
E[Smin(n+1,N)−Smin(n,N)∣Fn]=χAnE[Sn+1−Sn∣Fn]≥0
#Proposition B.10
Consider a sequence {tn∈[0,1]}n∈N. Let {nk∈N⊔{∞}}k∈N be the accumulation times of t. Then:
nk−1∑n=0tn≤2k
#Proof of Proposition B.10
We prove by induction on k. For k=0, the claim is obvious. If nk=∞ then nk+1=∞ and, by the induction hypothesis
nk+1−1∑n=0tn=∞∑n=0tn=nk−1∑n=0tn≤2k≤2(k+1)
Now assume nk<∞. Then nk+1>nk (because for the sum in the definition of accumulation times to be ≥1 it has to be non-empty), therefore
nk+1−1∑n=0tn=nk−1∑n=0tn+nk+1−1∑n=nktn
By the induction hypothesis
nk+1−1∑n=0tn≤2k+nk+1−1∑n=nktn
If nk+1<∞ then
nk+1−1∑n=0tn≤2k+nk+1−2∑n=nktn+tnk+1−1
By definition of accumulation times, the middle term is <1. By definition of t, the last term is ≤1. We get
nk+1−1∑n=0tn<2k+1+1=2(k+1)
Finally, assume nk+1=∞. We have
nk+1−1∑n=0tn≤2k+∞∑n=nktn
By definition of accumulation times, we get that for any m∈N:
m∑n=nktn<1
It follows that
∞∑n=nktn≤1
Combining, we have
nk+1−1∑n=0tn≤2k+1< 2(k+1)
#Proposition B.11
Consider X a probability space, {Fn⊆2X}n∈N a filtration of X, {Sn:X→R}n∈N and {Δn:X→[0,1]}n∈N stochastic processes adapted to F. Define {Un:X→R}n∈N by
Un:=n−1∑m=0Δm
Assume that there is α≥0 s.t.
E[|S0|]<∞
|Sn−S0|≤Un+α
Let {Nk:X→N⊔{∞}}k∈N be the accumulation times of Δ. Then, for any k∈N, {Smin(n,Nk)}n∈N and {Umin(n,Nk)}n∈N are uniformly integrable.
#Proof of Proposition B.11
By Proposition B.10, Umin(n,Nk)≤UNk≤2k, so {Umin(n,Nk)}n∈N is uniformly bounded and in particular uniformly integrable. Moreover:
|Smin(n,Nk)|≤|S0|+Umin(n,Nk)≤|S0|+2k
Since E[|S0|]<∞, it follows that {Smin(n,Nk)}n∈N is uniformly integrable.
#Proposition B.12
Consider sequences {sn∈R}n∈N and {tn∈[0,1]}n∈N. Let {nk:X→N⊔{∞}}k∈N be the accumulation times of t. Assume that there is α≥0 s.t.
∀n′≥n:|sn′−sn|≤n′−1∑m=ntm+α
Assume further that either all nk are finite or s converges. Denote snk:=limn→nksn. Then, |snk+1−snk|<α+2.
#Proof of Proposition B.12
Fix k∈N. If nk=∞, the claim is trivial. Consider the case nk+1<∞. We have
|snk+1−snk|≤nk+1−1∑n=nktn+α=nk+1−2∑n=nktn+tnk+1−1+α<1+1+α=α+2
Now consider the case nk<∞, nk+1=∞. By definition of accumulation times:
∞∑n=nktn≤1
It follows that |snk+1−snk|=|limn→∞sn−snk|≤α+1.
#Proposition B.13
Consider a sequence {tn∈[0,1]}n∈N. Let {nk∈N⊔{∞}}k∈N be the accumulation times of t. Assume that ∑∞ntn=∞. Then:
nk−1∑n=0tn≥k
#Proof of Proposition B.13
We prove by induction on k. For k=0, the claim is obvious. ∑∞ntn=∞ implies that n0<n1<…<∞, therefore
nk+1−1∑n=0tn=nk−1∑n=0tn+nk+1−1∑n=nktn
The first term is ≥k by induction. The second term is ≥1 by definition of accumulation time.
#Proposition B.14
Consider sequences {sn∈R}n∈N and {tn∈[0,1]}n∈N. Assume that there is α≥0 s.t.
∀n′≥n:|sn′−sn|≤n′−1∑m=ntm+α
Assume further that infnsn=−∞. In particular, ∑∞ntn=∞. Let {nk∈N}k∈N be the accumulation times of t (finite because of the previous observation). Then, infksnk=−∞.
#Proof of Proposition B.14
We need to prove that for any s>0 and k∈N, there is l≥k s.t. snl<−s. We know that there is n≥nk s.t. sn < −(s+α+2). We have n0<n1<…<∞, therefore we can choose l≥k s.t. nl≤n<nl+1. We get
|sn−snl|≤n−1∑m=nltm+α≤nl+1−1∑m=nltm+α=nl+1−2∑m=nltm+tnl+1−1+α<1+1+α=α+2
Therefore, snl<−s.
#Proposition B.15
Consider {On}n∈N, {Yn:=∏m<nOm}n∈N and X:=∏nOn as before. Consider υ a metastrategy and M a market. Define Sn,S′n:X→R by
Sn(x):=ΣWTMυn(x(n−1),x)
S′n(x):=minz∈Xx(n−1)ΣWTMυn(x(n−1),z)
Then:
|Sn−S′n|≤2supm∈Nsupy∈Ym∥υ(y)∥+1
#Proof of Proposition B.15
We prove by induction. For the basis, S0=S′0=0. Consider any n∈N and x∈X. First, assume that
maxz∈Xx(n)ΣWTMυn(x(n−1),z)−minz∈Xx(n)ΣWTMυn(x(n−1),z)>1
Then, by definition of Tυ, TMυn(x(n))≡0 and therefore
ΣWTMυ,n+1(x(n))=ΣWTMυn(x(n−1))
It follows that
|Sn+1(x)−S′n+1(x)|=|ΣWTMυ,n+1(x(n),x)−minz∈Xx(n)ΣWTMυ,n+1(x(n),z)|
|Sn+1(x)−S′n+1(x)|=|ΣWTMυn(x(n−1),x)−minz∈Xx(n)ΣWTMυn(x(n−1),z)|
|Sn+1(x)−S′n+1(x)|=|ΣWTMυn(x(n−1),x)−minz∈Xx(n−1)ΣWTMυn(x(n−1),z)|
|Sn+1(x)−S′n+1(x)|≤|Sn(x)−S′n(x)|
Using the induction hypothesis, we conclude
|Sn+1(x)−S′n+1(x)|≤2supm∈Nsupy∈Ym∥υ(y)∥+1
Now, assume that
maxz∈Xx(n)ΣWTMυn(x(n−1),z)−minz∈Xx(n)ΣWTMυn(x(n−1),z)≤1
Then, TMυn(x(n))=υn(x(n)) and therefore
ΣWTMυ,n+1(x(n))=ΣWTMυn(x(n−1))+υn(x(n),Mn(x(n)))
We get
|Sn+1(x)−S′n+1(x)|≤maxz∈Xx(n)ΣWTMυn(x(n−1),z)−minz∈Xx(n)ΣWTMυn(x(n−1),z)+2∥υn(x(n))∥
By the assumption, the first two terms total to ≤1, yielding the desired result.
#Proposition B.16
Consider X a compact Polish space, f∈C(X) and {Xn⊆X}n∈N closed s.t. Xn+1⊆Xn and ⋂nXn={x∗} for some x∗∈X. Then
limn→∞(maxXnf−minXnf)=0
#Proof of Proposition B.16
Choose {x+n∈argmaxXnf}n∈N and {x−n∈argminXnf}n∈N. maxXnf−minXnf 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 x+n→x+ω and x−n→x−ω. For each n∈N, the fact that Xn is closed implies that x+ω,x−ω∈Xn. Therefore, x+ω=x−ω=x∗. We get
limn→∞(maxXnf−minXnf)=limn→∞(f(x+n)−f(x−n))=f(x+ω)−f(x−ω)=f(x∗)−f(x∗)=0
#Proposition B.17
Consider υ a trading metastrategy, M a market and x∈X. Assume that
∞∑n=0(maxXx(n)¯TMυn(x(n))−minXx(n)¯TMυn(x(n)))<∞
Then, for any n≫0, TMυn(x(n))=υ(x(n)).
#Proof of Proposition B.17
Choose n0∈N s.t.
∞∑n=n0(maxXx(n)¯TMυn(x(n))−minXx(n)¯TMυn(x(n)))<12
Since ¯TMυn(x(n)),W¯TMυn(x(n))∈C(X) differ by a constant function, we have
∞∑n=n0(maxXx(n)W¯TMυn(x(n))−minXx(n)W¯TMυn(x(n)))<12
In particular, for any n≥n0
n−1∑m=n0(maxXx(n)W¯TMυm(x(m))−minXx(n)W¯TMυm(x(m)))≤n−1∑m=n0(maxXx(m)W¯TMυm(x(m))−minXx(m)W¯TMυm(x(m)))<12
By Proposition B.16, there is n1≥n0 s.t. for any n≥n1
maxXx(n)ΣWTMυn0(x(n0−1))−minXx(n)ΣWTMυn0(x(n0−1))<12
Taking the sum of the last two inequalities, we conclude that for any n≥n1
maxXx(n)ΣWTMυn(x(n−1))−minXx(n)ΣWTMυn(x(n−1))<1
By definition of Tυ, 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 X be a probability space, {Fn⊆2X}n∈N a filtration on X, {Sn:X→R}n∈N a stochastic process adapted to F and M,N:X→N⊔{∞} stopping times (w.r.t. F) s.t. M≤N. Assume that {Smin(n,N)}n∈N is a uniformly integrable submartingale. Using Doob's martingale convergence theorem, we can define SN:X→R by
SN(x):=limn→∞Smin(n,N)(x)={SN(x)(x) if N(x)<∞limn→∞Sn(x) if N(x)=∞
(the above is well-defined almost everywhere, which is sufficient for our purpose) We define SM:X→R is an analogous way (this time we can't use Doob's martingale convergence theorem, but whenever M(x)=∞ we also have N(x)=∞, therefore the limit almost surely converges). We also define FM⊆2X by
FM:={A⊆X measurable∣∀n∈N:A∩M−1(n)∈Fn}
Then:
E[SN∣FM]≥SM