Learning incomplete models using dominant markets

[Vanessa Kosoy]

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This post is formal treatment of the idea outlined here.

Given a countable set of incomplete models, we define a forecasting function that converges in the Kantorovich-Rubinstein metric with probability 1 to every one of the models which is satisfied by the true environment. This is analogous to Blackwell-Dubins merging of opinions for complete models, except that Kantorovich-Rubinstein convergence is weaker than convergence in total variation. The forecasting function is a dominant stochastic market for a suitably constructed set of traders.

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Appendix A contains the proofs. Appendix B restates the theorems about dominant stochastic markets for ease of reference. Appendix C states a variant of the Michael selection theorem due to Yannelis and Prabhakar which is used in Appendix A.

##Notation

Given $X$ a metric space, $x\in X$ and $r\in R$, $B_r\left(x\right):=\{y\in X\mid d(x,y)<r\}$.

Given $X$ a topological space:

• $P\left(X\right)$ is the space of Borel probability measures on $X$ equipped with the weak* topology.

• $C\left(X\right)$ is the Banach space of continuous functions $X\to R$ with uniform norm.

• $T\left(X\right):=C\left(P\right(X)\times X)$

• $B\left(X\right)$ is the Borel $\sigma$-algebra on $X$.

• $U\left(X\right)$ is the $\sigma$-algebra of universally measurable sets on $X$.

• Given $\mu \in P\left(X\right)$, $\mathrm{supp}\mu$ denotes the support of $\mu$.

Given $X$ and $Y$ measurable spaces, $K:X\stackrel{\text{mk}}{\to }Y$ is a Markov kernel from $X$ to $Y$. For any $x\in X$, we have $K\left(x\right)\in P\left(Y\right)$. Given $\mu \in P\left(X\right)$, $\mu ⋉K\in P(X\times Y)$ is the semidirect product of $\mu$ and $K$ and $K_{\ast }\mu \in P\left(Y\right)$ is the pushforward of $\mu$ by $K$.

Given $X$, $Y$ Polish spaces, $\pi :X\to Y$ Borel measurable and $\mu \in P\left(X\right)$, we denote $\mu \mid \pi$ the set of Markov kernels $K:Y\stackrel{\text{mk}}{\to }X$ s.t.${\pi }_{\ast }\mu ⋉K$ is supported on the graph of $\pi$ and $K_{\ast }{\pi }_{\ast }\mu =\mu$. By the disintegration theorem, $\mu \mid \pi$ is always non-empty and any two kernels in $\mu \mid \pi$ coincide ${\pi }_{\ast }\mu$-almost everywhere.

##Results

Consider $X$ any compact Polish metric space and $d:X\times X\to R$ its metric. We denote $\mathrm{Lip}\left(X\right)$ the Banach space of Lipschitz continuous functions $X\to R$ with the norm

$$\parallel f{\parallel }_{\mathrm{Lip}}:=\underset{x}{max}\left|f\right(x\left)\right|+\underset{x\ne y}{sup}\frac{\left|f\right(x)-f(y\left)\right|}{d(x,y)}$$

$P\left(X\right)$ (as before, with the weak topology) can be regarded as a compact subset of $\mathrm{Lip}(X{)}^{\prime }$ (with the strong topology), yielding a metrization of $P\left(X\right)$ which we will call the Kantorovich-Rubinstein metric $d_{\text{KR}}$:

$$d_{\text{KR}}(\mu ,\nu ):=\underset{\parallel f{\parallel }_{\mathrm{Lip}}\le 1}{sup}\left|E_{\mu }\right[f]-E_{\nu }[f\left]\right|$$

In fact, the above differs from the standard definition of the Kantorovich-Rubinstein metric (a.k.a. 1st Wasserstein metric, a.k.a. earth mover’s metric), but this abuse of terminology is mild since the two are strongly equivalent.

Now consider $\Phi \subseteq P\left(X\right)$ convex. We will describe a class of trading strategies that are designed to exploit any $\mu \in \Phi$.

#Definition 1

Let $\alpha >0$. $\tau \in T\left(X\right)$ is said to be an $\alpha$-representative of $\Phi$ when for all $\mu \in P\left(X\right)$, $\nu \in \Phi$ we have

$$\parallel \tau \left(\mu \right)\parallel \le max\left(d_{\text{KR}}\right(\mu ,\Phi )-\alpha ,0)$$

$$E_{\nu }\left[\tau \right(\mu \left)\right]\ge E_{\mu }\left[\tau \right(\mu \left)\right]+\frac{1}{2}\left(d_{\text{KR}}\right(\mu ,\Phi )-\alpha )d_{\text{KR}}(\mu ,\Phi )$$

#Lemma 1

Consider $Y$ another compact Polish space and $\Phi \subseteq Y\times P\left(X\right)$ closed. For any $y\in Y$, define ${\Phi }_y\subseteq P\left(X\right)$ by

$${\Phi }_y:=\{\mu \in P(X)\mid (y,\mu )\in \Phi \}$$

Assume that for any $y\in Y$, ${\Phi }_y$ is convex. Then, for any $\alpha >0$, there exists ${\upsilon }_{\alpha }:Y\to T\left(X\right)$ measurable w.r.t.$U\left(Y\right)$ and $B\left(T\right(X\left)\right)$ s.t. for all $y\in Y$, ${\upsilon }_{\alpha }\left(y\right)$ is an $\alpha$-representative of ${\Phi }_y$.

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For each $n\in N$, let $O_n$ be a compact subset of $R^{D\left(n\right)}$ ($D:N\to N$ is arbitrary). Denote

$$Y_n:=\prod _{m<n}{O}_m$$

$Y_n$ is a compact subset of $R^{{\sum }_{m<n}D\left(m\right)}$. We will regard it as equipped with the Euclidean metric. Denote

$$X=\prod _{n=0}^{\infty }{O}_n$$

$X$ is a compact Polish space. For each $n\in N$ we denote ${\pi }_n:X\to Y_n$ the projection mapping. Given $n\in N$ and $x\in X$, we have $x\left(n\right):={\pi }_n\left(x\right)\in Y_n$ and $x_n\in O_n$.

The following definition provides a notion of updating an incomplete model by observations:

#Definition 2

Consider $\Phi \subseteq P\left(X\right)$. For any $n\in N$ and $y\in Y_n$, we define ${\Phi }_y^{\prime \prime }\subseteq P\left(X\right)$ by

$${\Phi }_y^{\prime \prime }:=\left\{\underset{r\to 0}{lim}\right(\mu \mid {\pi }_n^{-1}\left(B_r\right(y\left)\right))\mid \mu \in \Phi \}$$

Note that the limit in the definition above need not exist for every $\mu \in \Phi$.

Denote ${\Phi }_y^{\prime }\subseteq P\left(X\right)$ the convex hull of ${\Phi }_y^{\prime \prime }$ and define ${\Phi }_n^{\prime }\subseteq Y_n\times P\left(X\right)$ by

$${\Phi }_n^{\prime }:=\left\{\right(y,\mu )\in Y_n\times P(X)\mid \mu \in {\Phi }_y^{\prime }\}$$

Finally, we define ${\Phi }_n\subseteq Y_n\times P\left(X\right)$ to be the closure of ${\Phi }_n^{\prime }$. Given $y\in Y_n$, we define ${\Phi }_y\subseteq P\left(X\right)$ by

$${\Phi }_y:=\{\mu \in P(X)\mid (y,\mu )\in {\Phi }_n\}$$

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Note that the above definition uses the Euclidean metric on $Y_n$. This is the only place through which the assumption that $O_n$ is a compact subset of $R^{D\left(n\right)}$ (rather than an arbitrary compact Polish space) enters. This is used the in the proof of Lemma 2 below, via the Lebesgue differentiation theorem.

Fix a family of metrizations of $X$: $\{d_n:X\times X\to R{\}}_{n\in N}$. The reason we need a family rather than a single metric is that convergence in $d_{\text{KR}}$ is trivial unless we renormalize the metric for each $n$. On the other hand, completely arbitrary renormalization is allowed. For example, assuming the diameters of $O_n$ are uniformly bounded, we can take

$$d_n(x^1,x^2)=\underset{m\in N}{max}{c}_{nm}d(x_m^1,x_m^2)$$

Here, $\{c_{nm}>0{\}}_{n,m\in N}$ are required to satisfy ${lim}_{m\to \infty }{c}_{nm}=0$. To get a non-trivial result, we need the $c_{nm}$ to fall slower with $m$ as $n$ increases. The stronger we make this trend, the stronger conclusion we get (although it always remains weaker than convergence in total variation).

#Lemma 2

Consider $\Phi \subseteq P\left(X\right)$ and $\alpha >0$. Let $\upsilon$ be a metastrategy s.t. for each $n\in N$ and $y\in Y_n$, ${\upsilon }_n\left(y\right)$ is an $\alpha$-representative of ${\Phi }_y$ (relatively to $d_n$). Then, $\upsilon$ is profitable for any ${\mu }^{\ast }\in \Phi$.

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Combining Theorem B.2 and Lemma 2, it is easy to get the following:

#Corollary 1

Consider $\Phi \subseteq P\left(X\right)$, $\alpha >0$ and $\upsilon$ a metastrategy s.t. for each $n\in N$ and $y\in Y_n$, ${\upsilon }_n\left(y\right)$ is an $\alpha$-representative of ${\Phi }_y$. Let $M$ be a market which dominates $T_{\upsilon }$. For every $n\in N$, denote $d_{\text{KR}}^n$ the Kantorovich-Rubinstein metric associated with $d_n$. Then, for any ${\mu }^{\ast }\in \Phi$ and ${\mu }^{\ast }$-almost any $x\in X$:

$$\underset{n\to \infty }{limsup}{d}_{\text{KR}}^n\left(M_n\right(x\left(n\right)),{\Phi }_{x\left(n\right)})\le \alpha$$

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Thus, if we construct a set of traders $T_{{\upsilon }_k}$ as in Corollary 1 where ${\upsilon }_k$ corresponds to $\alpha =\frac{1}{k}$, then a market dominating all of these traders would have to converge to ${\Phi }_{y_n}$ with ${\mu }^{\ast }$-probability 1. Putting this together with Lemma 1 (so that ${\upsilon }_k$ as above actually exists) and Theorem B.1 we finally get

#Theorem 1

Consider any $H\subseteq 2^{P\left(X\right)}$ countable. Then, there exists a market $M^H$ s.t. for any $\Phi \in H$, ${\mu }^{\ast }\in \Phi$ and ${\mu }^{\ast }$-almost any $x\in X$:

$$\underset{n\to \infty }{lim}{d}_{\text{KR}}^n\left(M_n^H\right(x\left(n\right)),{\Phi }_{x\left(n\right)})=0$$

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This leaves an interesting open question, namely whether the counterpart of Theorem 1 with $d_{\text{KR}}$ replaced by total variation distance is true.

##Appendix A

#Proposition A.1

Let $X$ be a compact Polish space, $\phi \in \mathrm{Lip}(X{)}^{\prime \prime }$ and $ϵ>0$. Then, there exists $f_{ϵ}\in \mathrm{Lip}\left(X\right)$ s.t.$\parallel f_{ϵ}{\parallel }_{\mathrm{Lip}}\le \parallel \phi \parallel$ and for all $\mu \in P\left(X\right)$, $\left|E_{\mu }\right[f_{ϵ}]-\phi (\mu \left)\right|<ϵ$.

#Proof of Proposition A.1

Without loss of generality, assume that $\parallel \phi \parallel =1$ (if $\parallel \phi \parallel =0$ the theorem is trivial, otherwise we can rescale everything by a scalar). Using the compactness of $P\left(X\right)$, choose a finite set $A_{ϵ}\subseteq P\left(X\right)$ s.t.

$$\underset{\mu \in P\left(X\right)}{max}\underset{\nu \in A_{ϵ}}{min}{d}_{\text{KR}}(\mu ,\nu )<\frac{ϵ}{4}$$

Using the Goldstine theorem, choose $f_{ϵ}\in \mathrm{Lip}\left(X\right)$ s.t.$\parallel f_{ϵ}{\parallel }_{\mathrm{Lip}}\le 1$ and for any $\nu \in A_{ϵ}$

$$\left|E_{\nu }\right[f_{ϵ}]-\phi (\nu \left)\right|<\frac{ϵ}{2}$$

Consider any $\mu \in P\left(X\right)$. Choose $\nu \in A_{ϵ}$ s.t. $d_{\text{KR}}(\mu ,\nu )<\frac{ϵ}{4}$. We get

$$\left|E_{\mu }\right[f_{ϵ}]-\phi (\mu \left)\right|\le \left|E_{\mu }\right[f_{ϵ}]-E_{\nu }[f_{ϵ}\left]\right|+\left|\phi \right(\mu )-\phi (\nu \left)\right|+\left|E_{\nu }\right[f_{ϵ}]-\phi (\nu \left)\right|<\frac{ϵ}{4}+\frac{ϵ}{4}+\frac{ϵ}{2}=ϵ$$

#Proposition A.2

Let $X$ be a compact Polish space and $\Phi \subseteq P\left(X\right)$ convex. Consider any $\alpha >0$ and ${\mu }_0\in P\left(X\right)$ s.t. $r:=d_{\text{KR}}({\mu }_0,\Phi )>\alpha$. Then, there exists $f\in \mathrm{Lip}\left(X\right)$ s.t.

$$\parallel f{\parallel }_{\mathrm{Lip}}<r-\alpha$$

$$\underset{\nu \in \Phi }{inf}{E}_{\nu }\left[f\right]>E_{{\mu }_0}\left[f\right]+\frac{1}{2}(r-\alpha )r$$

#Proof of Proposition A.2

Define ${\Phi }^r\subseteq \mathrm{Lip}(X{)}^{\prime }$ by

$${\Phi }^r:=\{\mu \in \mathrm{Lip}(X{)}^{\prime }\mid \underset{\nu \in \Phi }{inf}\parallel \mu -\nu \parallel <r\}$$

By the Hahn-Banach separation theorem, there is $\phi \in \mathrm{Lip}(X{)}^{\prime \prime }$ s.t. for all $\nu \in {\Phi }^r$, $\phi \left(\nu \right)>\phi \left({\mu }_0\right)$. Multiplying by a scalar, we can make sure that $\parallel \phi \parallel =\frac{3}{4}(r-\alpha )$. For any $\delta >0$, we can choose $\zeta \in \mathrm{Lip}(X{)}^{\prime }$ s.t.$\parallel \zeta \parallel <1$ and $\phi \left(\zeta \right)>\frac{3}{4}(r-\alpha )-\delta$. For any $\nu \in \Phi$, $\nu -r\zeta \in {\Phi }^r$ and therefore

$$\phi (\nu -r\zeta )>\phi \left({\mu }_0\right)$$

$$\phi \left(\nu \right)>\phi \left({\mu }_0\right)+r\phi \left(\zeta \right)>\phi \left({\mu }_0\right)+r\left(\frac{3}{4}\right(r-\alpha )-\delta )$$

Taking $\delta$ to 0 we get $\phi \left(\nu \right)\ge \phi \left({\mu }_0\right)+\frac{3}{4}r(r-\alpha )$. Applying Proposition A.1 for $ϵ=\frac{1}{16}r(r-\alpha )$, we get $f\in \mathrm{Lip}\left(X\right)$ s.t.

$$\parallel f\parallel \le \frac{3}{4}(r-\alpha )<r-\alpha$$

$$E_{\nu }\left[f\right]\ge \phi \left(\nu \right)-\frac{1}{16}r(r-\alpha )\ge \phi \left({\mu }_0\right)+\frac{11}{16}r(r-\alpha )\ge E_{{\mu }_0}\left[f\right]+\frac{5}{8}r(r-\alpha )>E_{{\mu }_0}\left[f\right]+\frac{1}{2}r(r-\alpha )$$

#Proposition A.3

Let $X$ be a compact Polish space, $\Phi \subseteq P\left(X\right)$ convex and $\alpha >0$. Let $U\subseteq P\left(X\right)$ (an open set) be defined by

$$U:=\{\mu \in P(X)\mid d_{\text{KR}}(\mu ,\Phi )>\alpha \}$$

Then, there exists ${\tau }^{\prime }:U\to \mathrm{Lip}\left(X\right)$ continuous s.t. for all $\mu \in U$

$$\parallel {\tau }^{\prime }\left(\mu \right){\parallel }_{\mathrm{Lip}}<d_{\text{KR}}(\mu ,\Phi )-\alpha$$

$$\underset{\nu \in \Phi }{inf}{E}_{\nu }\left[{\tau }^{\prime }\right(\mu \left)\right]>E_{\mu }\left[{\tau }^{\prime }\right(\mu \left)\right]+\frac{1}{2}\left(d_{\text{KR}}\right(\mu ,\Phi )-\alpha )d_{\text{KR}}(\mu ,\Phi )$$

#Proof of Proposition A.3

Define $F:U\to 2^{\mathrm{Lip}\left(X\right)}$ by

$$F\left(\mu \right):=\{f\in \mathrm{Lip}(X)\mid \parallel f{\parallel }_{\mathrm{Lip}}<d_{\text{KR}}(\mu ,\Phi )-\alpha ,\underset{\nu \in \Phi }{inf}{E}_{\nu }[f]>E_{\mu }[f]+(d_{\text{KR}}(\mu ,\Phi )-\alpha \left)d_{\text{KR}}\right(\mu ,\Phi \left)\right\}$$

By Proposition A.2, for any $\mu \in U$, $F\left(\mu \right)\ne \varnothing$. Clearly, $F\left(\mu \right)$ is convex. Fix $f\in \mathrm{Lip}\left(X\right)$ and consider $F^{-1}\left(f\right):=\{\mu \in U\mid f\in F(\mu \left)\right\}$. Consider any ${\mu }_0\in F^{-1}\left(f\right)$, and take $ϵ>0$ s.t.

$$\parallel f{\parallel }_{\mathrm{Lip}}<d_{\text{KR}}({\mu }_0,\Phi )-ϵ-\alpha$$

$$\underset{\nu \in \Phi }{inf}{E}_{\nu }\left[f\right]>E_{{\mu }_0}\left[f\right]+ϵ+\left(d_{\text{KR}}\right({\mu }_0,\Phi )+ϵ-\alpha )\left(d_{\text{KR}}\right({\mu }_0,\Phi )+ϵ)$$

Define $V:=\{\mu \in U\mid d_{\text{KR}}(\mu ,{\mu }_0)<ϵ,E_{\mu }[f]<E_{{\mu }_0}[f]+ϵ\}$. Obviously $V$ is open and $V\subseteq F^{-1}\left(f\right)$, hence $F^{-1}\left(f\right)$ is open. Applying Theorem C, we get the desired result.

#Proposition A.4

Let $X$ be a compact Polish space, $\Phi \subseteq P\left(X\right)$ convex and $\alpha >0$. Then, there exists $\tau \in T\left(X\right)$ an $\alpha$-representative of $\Phi$.

#Proof of Proposition A.4

Let $U\subseteq P\left(X\right)$ be defined by

$$U:=\{\mu \in P(X)\mid d_{\text{KR}}(\mu ,\Phi )>\alpha \}$$

Use Proposition A.3 to obtain ${\tau }^{\prime }:U\to \mathrm{Lip}\left(X\right)$. Define $\tau :P\left(X\right)\to \mathrm{Lip}\left(X\right)$ by

$$\tau \left(\mu \right):=\{\begin{array}{c}{\tau }^{\prime }\left(\mu \right)\text{if}\mu \in U\\ 0\text{if}\mu \notin U\end{array}$$ Obviously $\tau$ is continuous in $U$. Consider any $\mu \notin U$ and $\{{\mu }_n\in P(X){\}}_{n\in N}$ s.t. ${\mu }_n\to \mu$. We have $d_{\text{KR}}({\mu }_n,\Phi )\to d_{\text{KR}}(\mu ,\Phi )\le \alpha$ and hence $\parallel \tau \left({\mu }_n\right){\parallel }_{\mathrm{Lip}}\le max\left(d_{\text{KR}}\right({\mu }_n,\Phi )-\alpha ,0)\to 0$. Therefore, $\tau$ is continuous everywhere.

#Proof of Lemma 1

Define $Z_{1,2,3}\subseteq Y\times T\left(X\right)\times P(X{)}^4$ by

$$Z_1:=\left\{\right(y,\tau ,\mu ,\nu ,\xi ,\zeta )\in Y\times T(X)\times P(X{)}^4\mid (y,\zeta )\in \Phi \}$$

$$Z_2:=\left\{\right(y,\tau ,\mu ,\nu ,\xi ,\zeta )\in Y\times T(X)\times P(X{)}^4\mid \parallel \tau \left(\mu \right)\parallel \le max\left(d_{\text{KR}}\right(\mu ,\xi )-\alpha ,0)\}$$

$$Z_3:=\left\{\right(y,\tau ,\mu ,\nu ,\xi ,\zeta )\in Y\times T(X)\times P(X{)}^4\mid E_{\nu }\left[\tau \right(\mu \left)\right]\ge E_{\mu }\left[\tau \right(\mu \left)\right]+\frac{1}{2}max\left(d_{\text{KR}}\right(\mu ,\zeta )-\alpha ,0)d_{\text{KR}}(\mu ,\zeta )\}$$

Define $Z:=Z_1\cap Z_2\cap Z_3$. It is easy to see that $Z_{1,2,3}$ are closed and therefore $Z$ also. Define $Z^{\prime }\subseteq Y\times T\left(X\right)\times P(X{)}^3$ by

$$Z^{\prime }:=\left\{\right(y,\tau ,\mu ,\nu ,\xi )\in Y\times T(X)\times P(X{)}^3\mid \exists \zeta \in P\left(X\right):(y,\tau ,\mu ,\nu ,\xi ,\zeta )\in Z\}$$

$Z^{\prime }$ is closed since it is the projection of $Z$ and $P\left(X\right)$ is compact. The ${\Phi }_y$ are compact, therefore $d_{\text{KR}}(\mu ,{\Phi }_y)=d_{\text{KR}}(\mu ,\zeta )$ for some $\zeta \in {\Phi }_y$, and hence $(y,\tau ,\mu ,\nu ,\xi )\in Z^{\prime }$ if and only if the following two conditions hold:

$$\parallel \tau \left(\mu \right)\parallel \le max\left(d_{\text{KR}}\right(\mu ,\xi )-\alpha ,0)$$

$$E_{\nu }\left[\tau \right(\mu \left)\right]\ge E_{\mu }\left[\tau \right(\mu \left)\right]+max\left(d_{\text{KR}}\right(\mu ,{\Phi }_y)-\alpha ,0)d_{\text{KR}}(\mu ,{\Phi }_y)$$

Define $W\subseteq Y\times T\left(X\right)\times P(X{)}^3$ by

$$W:=\left\{\right(y,\tau ,\mu ,\nu ,\xi )\in Y\times T(X)\times P(X{)}^3\mid \nu ,\xi \in {\Phi }_y\wedge (y,\tau ,\mu ,\nu ,\xi )\notin Z^{\prime }\}$$

$W$ is locally closed and in particular $F_{\sigma }$. Define $W^{\prime }\subseteq Y\times T\left(X\right)$ by

$$W^{\prime }:=\left\{\right(y,\tau )\in Y\times T(X)\mid \exists \mu ,\nu ,\xi \in P(X):(y,\tau ,\mu ,\nu ,\xi )\in W\}$$

$W^{\prime }$ is the projection of $W$ and $P(X{)}^3$ is compact, therefore $W^{\prime }$ is $F_{\sigma }$. Let $A\subseteq Y\times T\left(X\right)$ be the complement of $W^{\prime }$. $A$ is $G_{\delta }$ and in particular Borel. As easy to see, $(y,\tau )\in A$ if and only if $\tau$ is an $\alpha$-representative of ${\Phi }_y$.

For any $y\in Y$, $A_y:=\{\tau \in T(X)\mid (y,\tau )\in A\}$ is closed since

$$A_y=\bigcap _{\mu \in P\left(X\right)}\bigcap _{\nu ,\xi \in {\Phi }_y}\{\tau \in T(X)\mid (y,\tau ,\mu ,\nu ,\xi )\in Z^{\prime }\}$$

Moreover, $A_y$ is non-empty by Proposition A.4.

Consider any $U\subseteq T\left(X\right)$ open. Then, $A\cap (Y\times U)$ is Borel and therefore its image under the projection to $Y$ is analytic and in particular universally measurable. Applying the Kuratorwsi-Ryll-Nardzewski measurable selection theorem, we get the desired result.

#Proposition A.5

Consider $X$ a compact Polish space, $D\in N$, $Y$ a compact subset of $R^D$, $\pi :X\to Y$ continuous, $\mu \in P\left(X\right)$ and $K\in \mu \mid \pi$. Then, for ${\pi }_{\ast }\mu$-almost any $y\in Y$

$$\underset{r\to 0}{lim}\mu \mid {\pi }^{-1}\left(B_r\right(y\left)\right)=K\left(y\right)$$

#Proof of Proposition A.5

Let $\{f_k\in C(X){\}}_{k\in N}$ be dense in $C\left(X\right)$. For any $y\in Y$ and $r>0$ we denote ${\chi }_{yr}:Y\to \{0,1\}$ the characteristic function of $B_r\left(y\right)$ and ${\chi }_{yr}^{\pi }:X\to \{0,1\}$ the characteristic function of ${\pi }^{-1}\left(B_r\right(y\left)\right)$. For any $k\in N$ and $y_0\in \mathrm{supp}{\pi }_{\ast }\mu$ we have $\mu \left({\pi }^{-1}\right(B_r\left(y_0\right)\left)\right)={\pi }_{\ast }\mu \left(B_r\right(y_0\left)\right)>0$ and

$$E_{\mu }[f_k\mid {\pi }^{-1}(B_r\left(y_0\right)\left)\right]=\frac{E_{\mu }\left[{\chi }_{y_{0}r}^{\pi }{f}_k\right]}{\mu \left({\pi }^{-1}\right(B_r\left(y_0\right)\left)\right)}=\frac{E_{y\sim {\pi }_{\ast }\mu }\left[E_{K\left(y\right)}\right[{\chi }_{y_{0}r}^{\pi }{f}_k\left]\right]}{{\pi }_{\ast }\mu \left(B_r\right(y_0\left)\right)}=\frac{E_{y\sim {\pi }_{\ast }\mu }\left[{\chi }_{y_{0}r}{E}_{K\left(y\right)}\right[f_k\left]\right]}{{\pi }_{\ast }\mu \left(B_r\right(y_0\left)\right)}$$

In particular, the above holds for ${\pi }_{\ast }\mu$-almost any $y_0\in Y$. Applying the Lebesgue differentiation theorem, we conclude that for ${\pi }_{\ast }\mu$-almost any $y_0\in Y$

$$\underset{r\to 0}{lim}{E}_{\mu }[f_k\mid {\pi }^{-1}(B_r\left(y_0\right)\left)\right]=E_{K\left(y_0\right)}\left[f_k\right]$$

Now consider any $f\in C\left(X\right)$. For any $ϵ>0$, there is $k\in N$ s.t.$\parallel f-f_k\parallel <ϵ$ and therefore

$$\underset{r\to 0}{limsup}{E}_{\mu }[f\mid {\pi }^{-1}(B_r\left(y_0\right)\left)\right]\le \underset{r\to 0}{limsup}{E}_{\mu }[f_k\mid {\pi }^{-1}(B_r\left(y_0\right)\left)\right]+ϵ=E_{K\left(y_0\right)}\left[f_k\right]+ϵ\le E_{K\left(y_0\right)}\left[f\right]+2ϵ$$

Similarly

$$\underset{r\to 0}{liminf}{E}_{\mu }[f\mid {\pi }^{-1}(B_r\left(y_0\right)\left)\right]\ge E_{K\left(y_0\right)}\left[f\right]-2ϵ$$

Taking $ϵ$ to 0, we conclude that

$$\underset{r\to 0}{lim}{E}_{\mu }[f\mid {\pi }^{-1}(B_r\left(y_0\right)\left)\right]=E_{K\left(y_0\right)}\left[f\right]$$

$$\underset{r\to 0}{lim}\mu \mid {\pi }^{-1}\left(B_r\right(y_0\left)\right)=K\left(y_0\right)$$

#Proof of Lemma 2

Fix any ${\mu }^{\ast }\in \Phi$ and $\{K_n\in {\mu }^{\ast }\mid {\pi }_n{\}}_{n\in N}$. By Proposition A.5, for ${\pi }_{n\ast }\mu$-almost any $y\in Y$

$$\underset{r\to 0}{lim}{\mu }^{\ast }\mid B_r\left(y\right)=K_n\left(y\right)$$

Therefore, $K_n\left(y\right)\in {\Phi }_y^{\prime \prime }\subseteq {\Phi }_y$. Since ${\upsilon }_n\left(y\right)$ is an $\alpha$-representative of ${\Phi }_y$, for any $\mu \in P\left(X\right)$

$$E_{K_n\left(y\right)}\left[\upsilon \right(y,\mu \left)\right]-E_{\mu }\left[\upsilon \right(y,\mu \left)\right]\ge \frac{1}{2}max\left(d_{\text{KR}}^n\right(\mu ,{\Phi }_y)-\alpha ,0)d_{\text{KR}}^n(\mu ,{\Phi }_y)\ge \frac{1}{2}\parallel \upsilon (y,\mu )\parallel d_{\text{KR}}^n(\mu ,{\Phi }_y)$$

When $d_{\text{KR}}^n(\mu ,{\Phi }_y)\le \alpha$, $\upsilon (y,\mu )=0$, hence for any $\mu \in P\left(X\right)$, $\parallel \upsilon (y,\mu )\parallel d_{\text{KR}}^n(\mu ,{\Phi }_y)\ge \parallel \upsilon (y,\mu )\parallel \alpha$. We get

$$E_{K_n\left(y\right)}\left[\upsilon \right(y,\mu \left)\right]-E_{\mu }\left[\upsilon \right(y,\mu \left)\right]\ge \frac{\alpha }{2}\parallel \upsilon (y,\mu )\parallel$$

#Proof of Corollary 1

By Lemma 2, $\nu$ is profitable for ${\mu }^{\ast }$. By Theorem B.2, for ${\mu }^{\ast }$-almost any $x\in X$

$$\underset{n\to \infty }{lim}\left(E_{K_n\left(x\right(n\left)\right)}\right[\upsilon \left(x\right(n),M_n(x\left(n\right)\left)\right)]-E_{M_n\left(x\right(n\left)\right)}[\upsilon \left(x\right(n),M_n(x\left(n\right)\left)\right)\left]\right)=0$$

By Proposition A.5, $K_n\left(x\right(n\left)\right)\in {\Phi }_{x\left(n\right)}$ and therefore

$$E_{K_n\left(x\right(n\left)\right)}\left[\upsilon \right(x\left(n\right),M_n\left(x\right(n\left)\right)\left)\right]-E_{M_n\left(x\right(n\left)\right)}\left[\upsilon \right(x\left(n\right),M_n\left(x\right(n\left)\right)\left)\right]\ge \frac{1}{2}\left(d_{\text{KR}}^n\right(M_n\left(x\right(n\left)\right),{\Phi }_{x\left(n\right)})-\alpha )d_{\text{KR}}^n\left(M_n\right(x\left(n\right)),{\Phi }_{x\left(n\right)})$$

We get

$$\underset{n\to \infty }{limsup}\left(d_{\text{KR}}^n\right(M_n\left(x\right(n\left)\right),{\Phi }_{x\left(n\right)})-\alpha )d_{\text{KR}}^n\left(M_n\right(x\left(n\right)),{\Phi }_{x\left(n\right)})\le 0$$

$$\underset{n\to \infty }{limsup}{d}_{\text{KR}}^n\left(M_n\right(x\left(n\right)),{\Phi }_{x\left(n\right)})\le \alpha$$

#Proof of Theorem 1

Consider any $\Phi \in H$ and a positive integer $k$. By Lemma 1, for any $n\in N$, there exists ${\upsilon }_{kn}^{\Phi }:Y_n\to T\left(X\right)$ measurable s.t. for all $y\in Y_n$, ${\upsilon }_{kn}^{\Phi }\left(y\right)$ is a $\frac{1}{k}$-representative of $\Phi$. We fix such a ${\upsilon }_{kn}^{\Phi }\left(y\right)$ for each $\Phi$ and $k$. It is easy to see that the $d_{\text{KR}}^n$-diameter of $P\left(X\right)$ is at most 2, therefore $\parallel {\upsilon }_{kn}^{\Phi }\left(y\right)\parallel \le 2$ and $\{{\upsilon }_{kn}^{\Phi }{\}}_{n\in N}$ is a trading metastrategy. By Theorem B.1, there is a stochastic market $M^H$ that dominates $T_{{\upsilon }_k^{\Phi }}$ for all $\Phi$ and $k$. By Corollary 1, for any $\Phi \in H$, ${\mu }^{\ast }\in \Phi$, positive integer $k$ and ${\mu }^{\ast }$-almost any $x\in X$

$$\underset{n\to \infty }{limsup}{d}_{\text{KR}}^n\left(M_n^H\right(x\left(n\right)),{\Phi }_{x\left(n\right)})\le \frac{1}{k}$$

It follows that

$$\underset{n\to \infty }{lim}{d}_{\text{KR}}^n\left(M_n^H\right(x\left(n\right)),{\Phi }_{x\left(n\right)})=0$$

##Appendix B

The following are the theorems about dominant stochastic markets proven before.

#Theorem B.1

Given any countable set of traders $R$, there is a market $M$ s.t.$M$ dominates all $T\in R$.

#Theorem B.2

Consider ${\mu }^{\ast }\in P\left(X\right)$, $\{K_n\in {\mu }^{\ast }\mid {\pi }_n{\}}_{n\in N}$, $\upsilon$ a trading metastrategy profitable for ${\mu }^{\ast }$ and $M$ a market. Assume $M$ dominates $T_{\upsilon }$. Then, for ${\mu }^{\ast }$-almost any $x\in X$:

$$\underset{n\to \infty }{lim}\left(E_{K_n\left(x\right(n\left)\right)}\right[\upsilon \left(x\right(n),M_n(x\left(n\right)\left)\right)]-E_{M_n\left(x\right(n\left)\right)}[\upsilon \left(x\right(n),M_n(x\left(n\right)\left)\right)\left]\right)=0$$

##Appendix C

The following variant of the Michael selection theorem appears in Yannelis and Prabhakar (1983) as Theorem 3.1.

#Theorem C

Consider $X$ a paracompact Hausdorff topological space and $Y$ a topological vector space. Suppose $F:X\to 2^Y$ is s.t.

(i) For each $x\in X$, $F\left(x\right)\ne \varnothing$.

(ii) For each $x\in X$, $F\left(x\right)$ is convex.

(iii) For each $y\in Y$, $\{x\in X\mid y\in F(x\left)\right\}$ is open.

Then, there exists $f:X\to Y$ continuous s.t. for all $x\in X$, $f\left(x\right)\in F\left(x\right)$.