Fatou closedness under model uncertainty

Abstract

We provide a characterization in terms of Fatou closedness for weakly closed monotone convex sets in the space of ${\mathcal P}$-quasisure bounded random variables, where ${\mathcal P}$ is a (possibly non-dominated) class of probability measures. Applications of our results lie within robust versions the Fundamental Theorem of Asset Pricing or dual representation of convex risk measures.

Appendices

Auxiliary results for Theorem 3.1

Recall the set ${\mathcal Z}$ defined in (3.2).

Proposition A.1

If ${\mathcal Z}=\emptyset$, then there exists a countable subset ${\widetilde{{\mathcal P}}}\subset {\mathcal P}$ such that ${\widetilde{{\mathcal P}}}\thickapprox {\mathcal P}$. The latter implies that there is a probability measure $Q\in {\mathcal M}_1$ such that $\{Q\}\thickapprox {\mathcal P}$.

Proof

We claim that for each $\varepsilon >0$, there exists $P_1,\ldots , P_n\in {\mathcal P}$ and $\delta >0$ such that $P_i(A)<\delta$ for all $i=1,\ldots , n$ implies that for all $P\in {\mathcal P}$ we have $P(A)<\varepsilon$. Suppose this is not the case. Then there exists $\varepsilon >0$ such that for any $P_1\in {\mathcal P}$ there is $A_1\in {\mathcal F}$ and $P_2\in {\mathcal P}$ satisfying

$$\begin{aligned} P_1(A_1)<1/2 \quad \text{ and }\quad P_2(A_1)\ge \varepsilon . \end{aligned}$$

Then there also exists $A_2\in {\mathcal F}$ and $P_3\in {\mathcal P}$ such that

$$\begin{aligned} P_1(A_2)<1/4,\, P_2(A_2)<1/4 \quad \text{ while }\quad P_3(A_2)\ge \varepsilon . \end{aligned}$$

Continuing this procedure we find sequences $(A_n)_{n\in {\mathbb N}}\subset {\mathcal F}$ and $(P_{n})_{n\in {\mathbb N}}\in {\mathcal P}$ such that

$$\begin{aligned} P_i(A_n)<\frac{1}{2^n}, \, i=1,\ldots , n, \quad \text{ and }\quad P_{n+1}(A_n)\ge \varepsilon . \end{aligned}$$

Consider $N:=\bigcap _{n\in {\mathbb N}}\bigcup _{k\ge n} A_k$. Then $P_i(N)=0$ for each $i\in {\mathbb N}$, because for all $n>(i-1)$

$$\begin{aligned} P_i(N)\le \sum _{k=n}^\infty P_i(A_k)\le \frac{1}{2^{n-1}}. \end{aligned}$$

Hence, replacing the above sequence $A_n$ by $B_n:=A_n\setminus N$, $n\in {\mathbb N}$, we still have

$$\begin{aligned} P_i(B_n)<\frac{1}{2^n}, \, i=1,\ldots , n, \quad \text{ and }\quad P_{n+1}(B_n)\ge \varepsilon . \end{aligned}$$

Now let $E_n:=\bigcup _{k\ge n}B_k$, $n\in {\mathbb N}$. It follows that $E_n\downarrow \emptyset$. However, for each $n\in {\mathbb N}$

$$\begin{aligned} c(E_n)\ge P_{n+1}(E_n)\ge P_{n+1}(B_n)\ge \varepsilon \end{aligned}$$

which contradicts ${\mathcal Z}=\emptyset$.

Now let $\delta _n>0$ and let $P^{(n)}_1,\ldots , P^{(n)}_{m(n)}\in {\mathcal P}$ be such that for all $P \in {\mathcal P}$ it holds $P(A)<1/n$ whenever $P^{(n)}_i(A)<\delta _n$ for all $i=1,\ldots , m(n)$. Define

$$\begin{aligned} \mu :=\sum _{n=1}^\infty \sum _{i=1}^{m(n)}\frac{1}{2^n}\frac{1}{2^i}P_i^{(n)}. \end{aligned}$$

Then $\mu \in ca_+$, and $\mu (A)=0$ implies that $P_i^{(n)}(A)=0$ for all $i=1,\ldots , m(n)$ and $n\in {\mathbb N}$. Eventually this implies that for all $P\in {\mathcal P}$ we have $P(A)<1/n$ for all $n\in {\mathbb N}$, hence $P(A)=0$. Thus

$$\begin{aligned} {\widetilde{{\mathcal P}}}:=\{P_i^{(n)}\mid i\in \{1,\ldots , m(n)\}, n\in {\mathbb N}\}\quad \text{ and }\quad Q:=\frac{1}{\mu (\Omega )}\mu \end{aligned}$$

satisfy the assertion. $\square$

Proposition A.2

Let $(B,\Vert \cdot \Vert )$ be a Banach lattice of (equivalence classes of) random variables on $(\Omega ,{\mathcal F})$ containing all simple random variables such that the order $\le$ on B satisfies $0\le 1_A\le 1_{A'}$ whenever $A\subset A'$ for $A,A'\in {\mathcal F}$. If $B^*\subset ca$, in the sense that every $l\in B^*$ is of type

$$\begin{aligned} l(X)=\int X\,d\mu , \quad X\in B, \end{aligned}$$

for some $\mu \in ca$, then $\Vert 1_{A_n}\Vert \rightarrow 0$ $(n\rightarrow \infty )$ for all $(A_n)_{n\in {\mathbb N}}\subset {\mathcal F}$ such that $A_n\downarrow \emptyset$.

Conversely, if $\Vert 1_{A_n}\Vert \rightarrow 0$ $(n\rightarrow \infty )$ for all $(A_n)_{n\in {\mathbb N}}\subset {\mathcal F}$ such that $A_n\downarrow \emptyset$, then for every $l\in B^*$ there is a $\mu \in ca$ such that $l(Y)=\int Y\, d\mu$ for all simple random variables Y.

Proof

Suppose that $B^*\subset ca$ and let $(A_n)_{n\in {\mathbb N}}\subset {\mathcal F}$ such that $A_n\downarrow \emptyset$. Then $1_{A_n}\rightarrow 0$ with respect to $\sigma (B,B^*)$ since every element in $B^*$ corresponds to a $\sigma$-additive measure. Hence,

$$\begin{aligned} 0\in \overline{co\{1_{A_n}\mid n\in {\mathbb N}\}} \end{aligned}$$

where the closure is taken in the $\sigma (B,B^*)$-topology. As the closed convex set in the $\sigma (B,B^*)$-topology and in the norm topology coincide, we have that there is a sequence of convex combinations

$$\begin{aligned} c_k:=\sum _{i=1}^{m(k)}a_i(k)1_{A_{n_i(k)}}, \quad k\in {\mathbb N}, \end{aligned}$$

where $a_i(k)\in {\mathbb R}$ and $n_1(k)\le n_2(k)\le \ldots \le n_{m(k)}(k)$ for all $k\in {\mathbb N}$ such that $\Vert c_k\Vert \rightarrow 0$ for $k\rightarrow \infty$. Moreover, since $0\in \overline{co\{1_{A_n}\mid n\ge N\}}$ for any $N\in {\mathbb N}$, we may assume that $n_1(k)\le n_1(k+1)$ for all $k\in {\mathbb N}$. However, $c_k\ge 1_{A_k}$ where $A_k=A_{n_{m(k)}(k)}$, because $A_n\supset A_{n+1}$ for all $n\in {\mathbb N}$. Thus, as $\Vert \cdot \Vert$ is a lattice norm, the subsequence $1_{A_ k}$ converges to 0 in norm and hence also $1_{A_{n}}$ converges to 0 in the norm topology (again due to $A_n\supset A_{n+1}$ for all $n\in {\mathbb N}$).

Finally suppose that $\Vert 1_{A_n}\Vert \rightarrow 0$ $(n\rightarrow \infty )$ for all $(A_n)_{n\in {\mathbb N}}\subset {\mathcal F}$ such that $A_n\downarrow \emptyset$. Then for any $l\in B^*$, the set function

$$\begin{aligned} \mu (A):=l(1_A), \quad A\in {\mathcal F}, \end{aligned}$$

is $\sigma$-additive. By linearity of l we deduce that $l(X)=\int X\, d\mu$ for all simple random variables X. $\square$

Penot–Volle duality theorem

Theorem B.1

(see e.g. [20, Theorem 1.1]) Let L be a locally convex topological vector space, $L^{\prime }$ be its dual space and $f:L\rightarrow \overline{\mathbb {R}}:= \mathbb {R}\cup \left\{ -\infty \right\} \cup \left\{ \infty \right\}$ be quasiconvex and lower semicontinuous. Then

$$\begin{aligned} f(X)=\sup _{X^{\prime }\in L^{\prime }}R(X^{\prime }(X),X^{\prime }) \end{aligned}$$

(B.1)

where $R:\mathbb {R\times }L^{\prime }\rightarrow \overline{\mathbb {R}}$ is defined by

(B.2)