No-arbitrage and optimal investment with possibly non-concave utilities: a measure theoretical approach

Abstract

We consider a discrete-time financial market model with finite time horizon and investors with utility functions defined on the non-negative half-line. We allow these functions to be random, non-concave and non-smooth. We use a dynamic programming framework together with measurable selection arguments to establish both the characterisation of the no-arbitrage property for such markets and the existence of an optimal portfolio strategy for such investors.

Appendix

1.1 Generalised integral and Fubini’s Theorem

For ease of the reader we provide some well-known results on measure theory, stochastic kernels and integrals. The first lemma provides a version of the Fubini Theorem for non-negative functions [see for instance to Theorem 10.7.2 in Bogachev (2007)]. We then present our definition of generalised integral and provide another version of the Fubini Theorem for generalised integral (see Proposition 5.3), which is essential throughout the paper.

Let $(H, {\mathcal {H}})$ and $(K,\mathcal {K})$ be two measurable spaces, p be a probabilty measure on $(H,\mathcal {H})$ and q a stochastic kernel on $(K,\mathcal {K})$ given $(H, {\mathcal {H}})$ , i.e. such that for any $h \in H$, $q(\cdot |h)$ is a probability measure on $(K,\mathcal {K})$ and for any $C \in \mathcal {K}$, $q(C|\cdot )$ is ${\mathcal {H}}$-measurable. Furthermore, for any $A \in \mathcal {H} \otimes \mathcal {K}$ and any $h \in H$, the section of A along h is defined by $\left( A\right) _{h}:=\left\{ k \in K,\; (h,k) \in A\right\} .$

Lemma 5.1

Let $A \in \mathcal {H} \otimes \mathcal {K}$ be fixed. For any $h \in H$ we have $\left( A\right) _{h} \in \mathcal {K}$ and we define P by

$$\begin{aligned} P(A)&:= \int _{H}\int _{K} 1_{A}(h,k) q(dk|h)p(dh)= \int _{H} q(\left( A\right) _{h}|h) p(dh). \end{aligned}$$

(34)

Then P is a probability measure on $(H\times K, \mathcal {H}\otimes \mathcal {H}).$ If f is non-negative, $\mathcal {H} \otimes \mathcal {K}$-measurable with value in $\mathbb {R}_{+} \cup \{\infty \}$ then $h \in H \rightarrow \int _{K} f(h,k)q(dk|h)$ is $\mathcal {H}$-measurable with value in $\mathbb {R}_{+} \cup \{\infty \}$ and we have

$$\begin{aligned} \int _{H\times K} f dP:=\int _{H\times K} f(h,k) P(dh,dk) = \int _{H} \int _{K} f(h,k)q(dk|h)p(dh). \end{aligned}$$

(35)

Proof

Let $h \in H$ be fixed. Let ${{\mathcal {T}}}=\{A \in \mathcal {H}\otimes \mathcal {K} \,| \, \left( A\right) _{h}\in \mathcal {K}\}$. It is easy to see that ${{\mathcal {T}}}$ is a sigma-algebra on $H\times K$ and is included in $\mathcal {H}\otimes \mathcal {K}$. Let $A=B \times C \in \mathcal {H}\times \mathcal {K}$ then $\left( A\right) _{h}=\emptyset$ if $h \notin B$ and $\left( A\right) _{h}=C$ if $h \in B$. Thus $\left( A\right) _{h} \in \mathcal {K}$ and $\mathcal {H}\times \mathcal {K} \subset {{\mathcal {T}}}$. As $\mathcal {T}$ is a sigma-algebra, $\mathcal {H}\otimes \mathcal {K} \subset \mathcal {T}$ and ${{\mathcal {T}}}= \mathcal {H}\otimes \mathcal {K}$ follows.

We show now that $h \rightarrow \int _{K} 1_{A}(h,k)q(dk|h)=\int _{K} 1_{\left( A\right) _{h}}(k)q(d k|h)=q\left( \left( A\right) _{h}|h\right)$ is $\mathcal {H}$-measurable for any $A \in \mathcal {H}\otimes \mathcal {K}$. Let ${{\mathcal {E}}}=\left\{ A \in \mathcal {H}\otimes \mathcal {K} \,| \, h \rightarrow q \left( \left( A\right) _{h}|h\right) \text{ is } \mathcal {H}\text{-measurable } \right\}$. It is easy to see that ${{\mathcal {E}}}$ is a sigma-algebra on $H\times K$ and is included in $\mathcal {H}\otimes \mathcal {K}$. Let $A=B \times C \in \mathcal {H}\times \mathcal {K}$ then $q\left( \left( A\right) _{h})|h\right)$ equals to 0 if $h \notin B$ and to q(C|h) if $h \in B$. So by definition of $q(\cdot |\cdot )$, $\mathcal {H}\times \mathcal {K} \subset {{\mathcal {E}}}$. As $\mathcal {E}$ is a sigma-algebra, $\mathcal {H}\otimes \mathcal {K} \subset \mathcal {E}$ and ${{\mathcal {E}}}= \mathcal {H}\otimes \mathcal {K}$ follows. Thus the last integral in (34) is well-defined. We verify that P defines a probability measure on $(H\times K, \mathcal {H}\otimes \mathcal {H}).$ It is clear that $P(\emptyset )=0$ and $P(H\times K)=1$. The sigma-additivity property follows from the monotone convergence theorem.

Now we prove (35). When $f=1_A$ with $A\in \mathcal {H}\otimes \mathcal {K}$, it is just (34). Then using linear combinations, it is true for step functions and finally for any arbitrary non-negative measurable f using the monotone convergence theorem. We prove now that for $f: H\times K \rightarrow \mathbb {R}_{+} \cup \{+\infty \}$ non-negative and $\mathcal {H} \otimes \mathcal {K}$-measurable, $h \in H \rightarrow \int _{K} f(h,k)q(dk|h)$ is $\mathcal {H}$-measurable and (35) holds true. If $f=1_A$ for $A\in \mathcal {H}\otimes \mathcal {K}$ the claim is proved. By taking linear combinations, it is proved for $\mathcal {H}\otimes \mathcal {K}$-measurable step functions. Then if $f: H\times K \rightarrow \mathbb {R} \cup \{+\infty \}$ is non-negative and $\mathcal {H}\otimes \mathcal {K}$-measurable, then there exists some increasing sequence $(f_{n})_{n \ge 1}$ such that $f_{n}: H\times K \rightarrow \mathbb {R}$ is a $\mathcal {H}\otimes \mathcal {K}$-measurable step function and $(f_{n})_{n \ge 1}$ converge to f. Using the monotone convergence theorem and (35) for steps functions, we conclude that (35) holds true for f.   $\square$

Definition 5.2

Let $f: H\times K \rightarrow \mathbb {R} \cup \{\pm \infty \}$ be a $\mathcal {H}\otimes \mathcal {K}$-measurable function. If $\int _{H\times K} f^{+} dP< \infty$ or $\int _{H\times K} f^{-}dP< \infty$, we define the generalised integral of f by $\int _{H\times K} f dP := \int _{H\times K} f^{+}dP - \int _{H\times K} f^{-}dP.$

Proposition 5.3

Let $f: H\times K \rightarrow \mathbb {R} \cup \{\pm \infty \}$ be a $\mathcal {H}\otimes \mathcal {K}$-measurable function such that $\int _{H\times K} f^{+}dP< \infty$. Then, we have

$$\begin{aligned} \int _{H \times K}f dP= \int _{H} \int _{K} f(h,k) q(dk|h) p(dh). \end{aligned}$$

(36)

Proof

Using Definition 5.2 and applying Lemma 5.1 to $f^{+}$ and $f^{-}$ we obtain that

$$\begin{aligned} \int _{H \times K}f dP&= \int _{H \times K}f^{+} dP- \int _{H \times K}f^{-} dP \\&=\int _{H} \int _{K} f^{+}(h,k) q(dk|h) p(dh)-\int _{H} \int _{K} f^{-}(h,k) q(dk|h) p(dh). \end{aligned}$$

To establish (36), assume for a moment that the following linearity result have been proved: let $g_{i}: H\times K \rightarrow \mathbb {R} \cup \{\pm \infty \}$ be some $\mathcal {H}\otimes \mathcal {K}$-measurable functions such that $\int _{H\times K} g_{i}^{+}dP< \infty$ for $i=1,2$. Then

$$\begin{aligned} \int _{H} \left( g_{1} +g_{2}\right) dp= \int _{H}g_{1} dp+ \int _{H} g_{2} dp. \end{aligned}$$

(37)

We apply (37) with $g_{1}(h)=\int _{K}f^{+}(h,k)q(dh|k)$ and $g_{2}=-\int _{K}f^{-}(h,k)q(dh|k)$ since by Lemma 5.1,

$$\begin{aligned} \int _{H}g^{+}_{1}dp&= \int _{H} \left( \int _{K}f^{+}(h,k)q(dh|k)\right) p(dh) = \int _{H \times K}f^{+}(h,k)q(dh|k)p(dh)\\&=\int _{H \times K} f^{+}dP<\infty \end{aligned}$$

and clearly $\int _{H}g^{+}_{2}dp=0<\infty$. So we obtain that

$$\begin{aligned}&\int _{H} \int _{K} f^{+}(h,k) q(dk|h) p(dh) - \int _{H} \int _{K} f^{-}(h,k) q(dk|h) p(dh)\\&\quad = \int _{H}\left( \int _{K} f^{+}(h,k)q(dk|h) - \int _{K}f^{-}(h,k)q(dk|h)\right) p(dh)\\&\quad = \int _{H} \int _{K} f(h,k) q(dk|h) p(dh), \end{aligned}$$

where the second equality comes from the definition of the generalised integral of $f(h,\cdot )$ with respect to $q(\cdot |h)$ and (36) is proven. We prove now (37). If $\int _{H} g_{i}^{-} dp <\infty$ for $i=1,2$ this is trivial. From $\int _{H }g_{i}^{+} dp <\infty$ we get that $g^{+}_{i} <\infty$ p-almost surely for $i=1,2$, so the sum $g_{1}+g_{2}$ is p-almost surely well-defined, taking its value in $[-\infty ,\infty )$. As $(g_{1}+g_{2})^{+} \le g_{1}^{+}+g_{2}^{+}$, using the linearity property of the integral for non-negative functions we get that $\int _{H} \left( g_{1}+g_{2}\right) ^{+}(h) p(dh)\le \int _{H }g_{1}^{+} dp +\int _{H} g_{2}^{+} dp <\infty .$ Now from $g_{1}^{+}+g_{2}^{+}-g_{1}^{-}-g_{2}^{-}=g_{1}+g_{2} =\left( g_{1}+g_{2}\right) ^{+}-\left( g_{1}+g_{2}\right) ^{-},$ using again the linearity property of the integral for non-negative functions we get that $\int _{H} \left( g_{1}+g_{2}\right) ^{+} dp + \int _{H} g^{-}_{1} dp + \int _{H}g^{-}_{2}dp=\int _{H} \left( g_{1}+g_{2}\right) ^{-}dp + \int _{H} g^{+}_{1}dp + \int _{H} g^{+}_{2}dp.$ Checking the different cases, i.e. $\int _{H}g_{1}^{-} dp=\infty$ and $\int _{H}g_{2}^{-} dp<\infty$ (and the opposite case) as well as $\int _{H}g_{i}^{-} dp=\infty$ for $i=1,2$ we get that (37) is true. $\square$

1.2 Further measure theory issues

We present now specific applications or results that are used throughout the paper. We start with four extensions of the Fubini results presented previously. As noted in Remark 4.38, the introduction of the trace sigma-algebra is the price to pay in order to avoid using the convention $\infty -\infty =-\infty$.

Proposition 5.4

Fix some $t\in \{1,\ldots ,T\}$.

• (i) Let $f: \Omega ^t \rightarrow \mathbb {R}_{+}\cup \{+\infty \}$ be a non-negative ${\mathcal {F}}_t$-measurable function. Then $\omega ^{t-1} \rightarrow \int _{\Omega _t}f(\omega ^{t-1}, \cdot )dq_t(\cdot |\omega ^{t-1})$ is ${\mathcal {F}}_{t-1}$-measurable with values in $\mathbb {R}_{+}\cup \{+\infty \}$.

• (ii) Let $f:\Omega ^t \times \mathbb {R}^{d} \rightarrow \mathbb {R}_{+}\cup \{+ \infty \}$ be a non-negative ${\mathcal {F}}_t \otimes \mathcal {B}(\mathbb {R}^{d})$-measurable function. Then $(\omega ^{t-1},h) \rightarrow \int _{\Omega _t}f(\omega ^{t-1},\cdot ,h)dq_t(\cdot |\omega ^{t-1})$ is ${\mathcal {F}}_{t-1}\otimes \mathcal {B}(\mathbb {R}^{d})$-measurable with values in $\mathbb {R}_{+} \cup \{ +\infty \}$.

• (iii) Let $f: \Omega ^t \rightarrow \mathbb {R}_{+}\cup \{+\infty \}$ be a non-negative $\overline{{\mathcal {F}}}_{t-1}\otimes \mathcal {G}_{t}$-measurable function. Then $\omega ^{t-1} \rightarrow \int _{\Omega _t}f(\omega ^{t-1},\cdot )dq_t(\cdot |\omega ^{t-1})$ is $\overline{{\mathcal {F}}}_{t-1}$-measurable with values in $\mathbb {R}_{+}\cup \{+\infty \}$.

• (iv) Let $\mathcal {S} \in \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})$. Introduce $\left[ \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})\right] _{\mathcal {S}}:=\left\{ A \cap \mathcal {S},\; A \in \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})\right\}$ the trace sigma-algebra of $\mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})$ on $\mathcal {S}$. Let $f: \Omega ^{t-1} \times \mathbb {R}^{d} \times \Omega _{t} \rightarrow \mathbb {R}_{+}\cup \{+ \infty \}$ be a non-negative ${\mathcal {F}}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d}) \otimes \mathcal {G}_{t}$-measurable function. Then $(\omega ^{t-1},h) \in \mathcal {S} \rightarrow \int _{\Omega _t}f(\omega ^{t-1},h,\cdot )dq_t(\cdot |\omega ^{t-1})$ is $\left[ \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})\right] _{\mathcal {S}}$-measurable with values in $\mathbb {R}_{+} \cup \{ +\infty \}$.

Proof

Statement (i) is a direct application of Lemma 5.1 for $H=\Omega ^{t-1}$, $\mathcal {H}=\mathcal {F}_{t-1}$, $K=\Omega _{t}$, $\mathcal {K}=\mathcal {G}_{t}$ and $q(\cdot |\cdot )=q_{t}(\cdot |\cdot )$. Statement (ii) follows by an application of Lemma 5.1 for $H=\Omega ^{t-1}\times \mathbb {R}^{d}$, $\mathcal {H}={\mathcal {F}}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})$, $K=\Omega _{t}$, $\mathcal {K}=\mathcal {G}_{t}$ and $q(\cdot |\cdot )=\bar{q}_{t}(\cdot |\cdot )$, where for $(G,\omega ^{t-1},h) \in \mathcal {G}_{t}\times \Omega ^{t-1}\times \mathbb {R}^{d}$ $\bar{q}_{t}(G| \omega ^{t-1},h):=q_{t}(G|\omega ^{t-1})$ and $\bar{q}_{t}$ is clearly an ${\mathcal {F}}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})$ stochastic kernel on $\mathcal {G}_{t}$ given $\Omega ^{t-1}\times \mathbb {R}^{d}$. To prove statement (iii) note that since $\mathcal {F}_{t-1} \subset \overline{\mathcal {F}}_{t-1}$ it is clear that $q_{t}$ is a stochastic kernel on $(\Omega _{t},\mathcal {G}_{t})$ given $(\Omega ^{t-1},\overline{{\mathcal {F}}}_{t-1})$ (i.e. measurability is with respect to $\overline{{\mathcal {F}}}_{t-1}$). And statement (iii) follows immediately from an application of Lemma 5.1 for $H=\Omega ^{t-1}$, $\mathcal {H}=\overline{\mathcal {F}}_{t-1}$, $K=\Omega _{t}$, $\mathcal {K}=\mathcal {G}_{t}$ and $q(\cdot |\cdot )=q_{t}(\cdot |\cdot )$. We prove now the last statement. Let for $(G,\omega ^{t-1},h) \in \mathcal {G}_{t}\times S\times \mathbb {R}^{d}$, $\tilde{q}_{t}(G| \omega ^{t-1},h):=q_{t}(G|\omega ^{t-1})$, then $\tilde{q}_{t}$ is clearly a stochastic kernel on $(\Omega _{t},\mathcal {G}_{t})$ given $\left( \mathcal {S},\left[ \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})\right] _{\mathcal {S}}\right)$. Now let $f_{S}$ be the restriction of f to $\mathcal {S}\times \Omega _{t}$. Using similar arguments and the fact that

$$\begin{aligned} \left[ \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})\otimes \mathcal {G}_{t}\right] _{\mathcal {S}\times \Omega _{t}}=\left[ \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})\right] _{\mathcal {S}}\otimes \mathcal {G}_{t}, \end{aligned}$$

we obtain that $f_{S}$ is $\left[ \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})\right] _{\mathcal {S}}\otimes \mathcal {G}_{t}$-measurable. Finally, statement (iv) follows from another application of Lemma 5.1 for $H=S$, $\mathcal {H}=\left[ \mathcal {F}_{t-1} \otimes \mathcal {B}(\mathbb {R}^{d})\right] _{\mathcal {S}}$, $K=\Omega _{t}$, $\mathcal {K}=\mathcal {G}_{t}$ and $q(\cdot |\cdot )=\tilde{q}_{t}(\cdot |\cdot )$. $\square$

Lemma 5.5

Let $f : \Omega ^{t+1} \rightarrow \mathbb {R}_{+} \cup \{+\infty \}$ be $\mathcal {F}_{t+1}$-measurable, non-negative verifying $\int _{\Omega ^{t+1}} f dP_{t+1} <\infty$. Then $\omega ^{t} \rightarrow \int _{\Omega _{t+1}} f(\omega ^{t},\cdot )dq_{t+1}(\cdot |\omega ^{t})$ is $\mathcal {F}_{t}$-measurable. Furthermore, $N^{t}:=\{\int _{\Omega _{t+1}} f(\omega ^{t},\cdot )dq_{t+1}(\cdot |\omega ^{t})=\infty \} \in \mathcal {F}_{t}$ and $P_{t}(N^{t})=0.$

Proof

The first assertion of the lemma is a direct application of (i) of Proposition 5.4. So it is clear that $N^{t} \in \mathcal {F}_{t}$. Applying Lemma 5.1 we get that $\int _{\Omega ^{t}} \int _{\Omega _{t+1}} f(\omega ^{t},\omega _{t+1})q_{t+1}(d\omega _{t+1}|\omega ^{t}) P_{t}(d\omega ^{t})=\int _{\Omega ^{t+1}} f dP_{t+1} <\infty$ so necessarily $P(N^t)=0$. $\square$

The next lemma, loosely speaking, allows to obtain “nice” sections (i.e. set of full measure for a certain probability measure). We use it in the proofs of Theorem 4.16 and Lemma 5.7.

Lemma 5.6

Fix some $t\in \{1,\ldots ,T\}$. Let $\widetilde{\Omega }^t \in {\mathcal {F}}_t$ such that $P_t\left( \widetilde{\Omega }^t\right) =1$ and $\widetilde{\Omega }^{t-1} \in {\mathcal {F}}_{t-1}$ such that $P_{t-1}\left( \widetilde{\Omega }^{t-1}\right) =1$ and set $\overline{\Omega }^{t-1}:=\left\{ \omega ^{t-1} \in \widetilde{\Omega }^{t-1},\; q_t\left( \left( \widetilde{\Omega }^t\right) _{\omega ^{t-1}}|\omega ^{t-1}\right) =1\right\}$ where $\left( \widetilde{\Omega }^t\right) _{\omega ^{t-1}}$ has been defined in Lemma 5.1. Then $\overline{\Omega }^{t-1} \in \mathcal {F}_{t-1}$ and $P_{t-1}\left( \overline{\Omega }^{t-1}\right) =1$.

Proof

From Lemma 5.1, $\omega ^{t-1} \rightarrow q_t\left( \left( \widetilde{\Omega }^t\right) _{\omega ^{t-1}}|\omega ^{t-1}\right)$ is $\mathcal {F}_{t-1}$-measurable thus $\overline{\Omega }^{t-1} \in \mathcal {F}_{t-1}$. Assume that $P(\widetilde{\Omega }^{t-1}\backslash {\overline{\Omega }^{t-1}})>0$. Then, using the Fubini Theorem (see Lemma 5.1), $P(\widetilde{\Omega }^{t-1})=1$ and the definition of $\overline{\Omega }^{t-1}$ we have that

$$\begin{aligned} 1=P_{t}(\widetilde{\Omega }^{t})&=\int _{\overline{\Omega }^{t-1}} 1 \times P_{t-1}(d\omega ^{t-1})+ \int _{\widetilde{\Omega }^{t-1}\backslash {\overline{\Omega }^{t-1}}} q_t\left( \left( \widetilde{\Omega }^t\right) _{\omega ^{t-1}}|\omega ^{t-1} \right) P_{t-1}(d\omega ^{t-1})\\&< P_{t-1}\left( \widetilde{\Omega }^{t-1}\right) +P_{t-1} \left( \widetilde{\Omega }^{t-1}\backslash {\overline{\Omega }^{t-1}} \right) =1, \end{aligned}$$

which is absurd. $\square$

The following lemma is used throughout the paper. In particular, the last statement is used in the proof of the main theorem.

Lemma 5.7

Let $0\le t \le T-1$, $B \in \mathcal {B}(\mathbb {R})$, $H: \Omega ^{t} \rightarrow \mathbb {R}$ and $h_{t}: \Omega ^{t} \rightarrow \mathbb {R}^{d}$ be $\mathcal {F}_{t}$-measurable. Then

$$\begin{aligned} (\omega ^{t},h)&\rightarrow q_{t+1}(H(\omega ^{t})+h\Delta S_{t+1}(\omega ^t,\cdot ) \in B|\omega ^t) \text{ is } {\mathcal {F}}_{t} \otimes \mathcal {B}(\mathbb {R}^{d})\text{-measuable } \end{aligned}$$

(38)

$$\begin{aligned} \omega ^{t}&\rightarrow q_{t+1}(H(\omega ^{t})+h_{t}(\omega ^{t})\Delta S_{t+1}(\omega ^t,\cdot ) \in B|\omega ^t) \text{ is } \mathcal {F}_{t}\text{-measurable } \end{aligned}$$

(39)

Furthermore, if we also have that $P_{t+1}\left( H(\cdot )+h_{t}(\cdot )\Delta S_{t+1}(\cdot ) \in B\right) =1$, then there exists some $P_{t}$-full measure set $\overline{\Omega }^{t}$ such that $q_{t+1}(H(\omega ^{t})+h_{t}(\omega ^{t})\Delta S_{t+1}(\omega ^t,\cdot ) \in B|\omega ^t)=1$, for all $\omega ^{t} \in \overline{\Omega }^{t}$.

Proof

As $h \rightarrow h \Delta S_{t+1}(\omega ^{t},\omega _{t+1})$ is continuous for all $(\omega ^{t},\omega _{t+1}) \in \Omega ^{t}\times \Omega _{t+1}$ and $(\omega ^{t},\omega _{t+1})\rightarrow h \Delta S_{t+1}(\omega ^{t},\omega _{t+1})$ is $\mathcal {F}_{t+1}=\mathcal {F}_{t} \otimes \mathcal {G}_{t+1}$-measurable for all $h \in \mathbb {R}^{d}$ (recall that $S_{t}$ and $S_{t+1}$ are respectively ${\mathcal {F}}_t$ and ${\mathcal {F}}_{t+1}$ measurable by assumption), $(\omega ^{t},\omega _{t+1},h) \rightarrow h\Delta S_{t+1}(\omega ^t,\omega _{t+1})$ is $\mathcal {F}_{t}\otimes \mathcal {G}_{t+1} \otimes \mathcal {B}(\mathbb {R}^{d})$-measurable as a Carathéodory function. As H is $\mathcal {F}_{t}$-measurable we obtain for any $B \in \mathcal {B}(\mathbb {R})$ that $\psi : (\omega ^{t},\omega _{t+1},h) \rightarrow H(\omega ^{t}) + h\Delta S_{t+1}(\omega ^t,\omega _{t+1})$ and $f_{B}: (\omega ^{t},\omega _{t+1},h) \rightarrow 1_{\psi (\cdot ,\cdot ,\cdot ) \in B}(\omega ^{t},\omega _{t+1},h)$ are also $\mathcal {F}_{t}\otimes \mathcal {G}_{t+1} \otimes \mathcal {B}(\mathbb {R}^{d})$-measurable. Then (38) is proved using statement (i) of Proposition 5.4 applied to $f_{B}$. Now (39) is showed using similar arguments. For the last statement we apply Lemma 5.6 to $\widetilde{\Omega }^{t+1}:=\left\{ \omega ^{t+1}=(\omega ^{t},\omega _{t+1}) \in \Omega ^{t}\times \Omega _{t+1},\; H(\omega ^{t})+h_{t}(\omega ^{t})\Delta S_{t+1}(\omega ^{t},\omega _{t+1}) \in B\right\}$. $\square$

The next lemma is often used in conjunction with the Aumann Theorem.

Lemma 5.8

Let $f:\Omega ^t \rightarrow \mathbb {R}$ be $\overline{{\mathcal {F}}}_{t}$-measurable. Then there exists some ${{\mathcal {F}}}_{t}$-measurable $g: \Omega ^t \rightarrow \mathbb {R}$ such that $f=g$ $P_{t}$-almost surely, i.e. there exists $\Omega ^{t}_{fg} \in \mathcal {F}_{t}$ with $P_{t}\left( \Omega ^{t}_{fg}\right) =1$ and $\Omega ^{t}_{fg} \subset \{f=g\}$.

Proof

Let $f=1_B$ with $B \in \overline{{\mathcal {F}}}_{t}$ then $B=A\cup N$, with $A \in \mathcal {F}_{t}$ and $N \in {{\mathcal {N}}}_{P_{t}}$. Then $g=1_A$ is ${{\mathcal {F}}}_{t}$-measurable. Clearly, $\{f \ne g\}=N \in {{\mathcal {N}}}_{P_{t}}$, thus $f=g$ $P_{t}$ a.s. By taking linear combinations, the lemma is proven for step functions using the same argument for each indicator function. Then it is always possible to approximate some $\overline{{\mathcal {F}}}_t$-measurable function f by a sequence of step function $(f_n)_{n \ge 1}$. From the preceding step for all $n \ge 1$, we get some $\mathcal {F}_{t}$-measurable step functions $g_n$ such that $f_n=g_n$ $P_{t}$-almost surely. Let $g = \limsup g_n$, then g is $\mathcal {F}_{t}$-measurable and we conclude since $\{f \ne g\} \subset \cup _{n \ge 1} \{f_n \ne g_n\}$ which is again in ${{\mathcal {N}}}_{P_{t}}$. $\square$

Next we provide some simple but useful results on usc functions.

Lemma 5.9

Let C be a closed subset of $\mathbb {R}^{m}$ for some $m \ge 1$. Let $g: \mathbb {R}^{m} \rightarrow \mathbb {R} \cup \{\pm \infty \}$ be such that $g=-\infty$ on $\mathbb {R}^{m} \backslash {C}$. Then g is usc on $\mathbb {R}^{m}$ if and only if g is usc on C.

Proof

We prove that if g is usc on C then it is usc on $\mathbb {R}^{m}$ as the reverse implication is trivial. Let $\alpha \in \mathbb {R}$ be fixed and $(x_{n})_{n \ge 1}\subset S_{\alpha }:=\{x\in \mathbb {R}^m:g(x)\le \alpha \}$ be a sequence that converges to $x \in \mathbb {R}^{m}$. Then $x_{n} \in C$ for all $n \ge 1$ and as C is a closed set, $x \in C$. As g is usc on C, (i.e. the set $\{x \in C,\ g(x) \ge \alpha \}$ is closed for the induced topology of $\mathbb {R}^{m}$ on C) we get that $g(x) \ge \alpha$, i.e. $x \in S_{\alpha }$ and g is usc on $\mathbb {R}^{m}$. $\square$

Lemma 5.10

Let $S \subset \mathbb {R}$ be a closed subset of $\mathbb {R}$. Let $f: \mathbb {R} \rightarrow \mathbb {R} \cup \{\pm \infty \}$ be such that f is usc and non-decreasing on S. Then f is right-continuous on S.

Proof

Let $(x_{n})_{n \ge 1} \subset S$ be a sequence converging to some $x^{*}$ from above. Then $x^{*} \in S$ since S is closed. As f is non-decreasing on S, for all $n \ge 1$ we have that $f(x_{n}) \ge f(x^{*})$ and thus $\liminf _{n} f(x_{n}) \ge f(x^{*})$. Now as f is usc on S, we get that $\limsup _{n} f(x_{n}) \le f(x^{*})$. The right-continuity of f on S follows immediately. $\square$

We now establish a useful extension of Lemma 5.8.

Lemma 5.11

Let $f:\Omega ^t \times \mathbb {R} \rightarrow \mathbb {R} \cup \{\pm \infty \}$ be an $\overline{{\mathcal {F}}}_{t}\otimes \mathcal {B}(\mathbb {R})$-measurable function such that for all $\omega ^{t} \in \Omega ^{t}$, $f(\omega ^{t},\cdot )$ is usc and non-decreasing. There exists some ${{\mathcal {F}}}_{t}\otimes \mathcal {B}(\mathbb {R})$-measurable function g from $\Omega ^t \times \mathbb {R}$ to $\mathbb {R} \cup \{\pm \infty \}$ and some $\Omega ^{t}_{mes} \in \mathcal {F}_{t}$ such that $P_{t}(\Omega ^{t}_{mes})=1$ and $f(\omega ^{t},x)=g(\omega ^{t},x)$ for all $(\omega ^{t},x) \in \Omega ^{t}_{mes}\times \mathbb {R}$.

Proof

Let $n \ge 1$ and $k \in \mathbb {Z}$ be fixed. We apply Lemma 5.8 to $f\left( \cdot \right) =f\left( \cdot ,\frac{k}{2^{n}}\right)$ that is $\overline{\mathcal {F}}_{t}$-measurable by assumption and we get some $\mathcal {F}_{t}$-measurable $g_{n,k}: \Omega ^{t} \rightarrow \mathbb {R} \cup \{\pm \infty \}$ and some $\Omega ^{t}_{n,k} \in \mathcal {F}_{t}$ such that $P_{t}\left( \Omega ^{t}_{n,k}\right) =1$ and $\Omega ^{t}_{n,k} \subset \left\{ f\left( \cdot ,\frac{k}{2^{n}}\right) = g_{n,k}(\cdot )\right\}$. We set $\Omega ^{t}_{mes}:= \bigcap _{n \ge 1, k \in \mathbb {Z}} \Omega ^{t}_{n,k}.$ It is clear that $\Omega ^{t}_{mes} \in \mathcal {F}_{t}$ and that $P_{t}\left( \Omega ^{t}_{mes}\right) =1$. Now, we define for all $n \ge 1$, $g_{n}: \Omega ^{t} \times \mathbb {R} \rightarrow \mathbb {R} \cup \{\pm \infty \}$ by $g_{n}(\omega ^{t},x):= \sum _{k \in \mathbb {Z}} 1_{\big (\frac{k-1}{2^{n}},\frac{k}{2^{n}}\big ]}(x)g_{n,k}(\omega ^{t}).$ It is clear that $g_{n}$ is ${{\mathcal {F}}}_{t}\otimes \mathcal {B}(\mathbb {R})$-measurable for all $n \ge 1$. Finally, we define $g: \Omega ^{t} \times \mathbb {R} \rightarrow \mathbb {R} \cup \{\pm \infty \}$ by $g(\omega ^{t},x):= \lim _{n} g_{n}(\omega ^{t},x).$ Then g is again ${{\mathcal {F}}}_{t}\otimes \mathcal {B}(\mathbb {R})$-measurable and it remains to prove that $f=g$ on $\Omega ^{t}_{mes}\times \mathbb {R}$. Let $(\omega ^{t},x) \in \Omega ^{t}_{mes}\times \mathbb {R}$ be fixed. For all $n \ge 1$, there exists  $k_{n} \in \mathbb {Z}$ such that $\frac{k_n-1}{2^{n}} < x \le \frac{k_n}{2^{n}}$ and such that $g_{n}(\omega ^{t},x)=g_{n,k_n}(\omega ^{t})=f\left( \omega ^{t},\frac{k_n}{2^{n}}\right)$. Applying Lemma 5.10 to $f(\cdot )=f(\omega ^{t},\cdot )$ (and $S=\mathbb {R}$), we get that $f(\omega ^{t},\cdot )$ is right-continuous on $\mathbb {R}$. As $\left( \frac{k_n}{2^{n}}\right) _{n \ge 1}$ converges to x from above, it follows that $g(\omega ^{t},x)= \lim _{n} f\left( \omega ^{t},\frac{k_n}{2^{n}}\right) =f(\omega ^{t},x)$ and this concludes the proof. $\square$

Finally, we introduce the following definition and prove an extension of a well-known result on Carathéodory functions (see for example 4.10 in Aliprantis and Border (2006)).

Definition 5.12

Let S be a closed interval of $\mathbb {R}$. A function $f: \Omega ^{t} \times S \rightarrow \mathbb {R}$ is an extended Carathéodory function if for all $\omega ^{t} \in \Omega ^{t}$, $f(\omega ^{t},\cdot )$ is right-continuous and $f(\cdot ,x)$ is $\mathcal {F}_{t}$-measurable for all $x \in S$.

Lemma 5.13

Let $S \subset \mathbb {R}$ be a closed interval of $\mathbb {R}$ and $f: \Omega ^{t} \times S \rightarrow \mathbb {R}$ be an extended Carathéodory function. Then f is $\mathcal {F}_{t} \otimes \mathcal {B}(\mathbb {R})$-measurable.

Proof

For all $n \ge 1$, define $f_{n}:\Omega ^{t} \times \mathbb {R} \rightarrow \mathbb {R}$ by

$$\begin{aligned} f_{n}(\omega ^{t},x):= \sum _{k \in \mathbb {Z}} 1_{\big (\frac{k-1}{2^{n}},\frac{k}{2^{n}}\big ]}(x) 1_{S}\left( \frac{k}{2^{n}}\right) f\left( \omega ^{t},\frac{k}{2^{n}}\right) . \end{aligned}$$

It is clear that $f_{n}$ is $\mathcal {F}_{t} \otimes \mathcal {B}(\mathbb {R})$-measurable. From the right-continuity of f, we can show as in the proof of Lemma 5.11 that $f(\omega ^{t},x)=\lim _{n} f_{n}(\omega ^{t},x)$ for all $(\omega ^{t},x) \in \Omega ^{t} \times S$ and the proof is complete (recall that $\Omega \times S \in \mathcal {F}_{t}\otimes \mathcal {B}(\mathbb {R})$ as S is a closed subset of $\mathbb {R}$). $\square$

Remark 5.14

Note that we have the same result if we replace $\mathcal {F}_{t}$ with $\overline{\mathcal {F}_{t}}$.

1.3 Proof of technical results

Finally we provide the missing results and proofs of the paper.

Proof of Lemma 2.2

We refer to Section 6.1 of Carassus and Rásonyi (2016) for the definition and various properties of generalised conditional expectations. In particular since $E(h^{+})=\int _{\Omega ^{t}} h^{+}dP_{t}<\infty$, $E(h| \mathcal {F}_{s})$ is well-defined (in the generalised sense) for all $0 \le s \le t$ (see Lemma 6.2 of Carassus and Rásonyi (2016)). Similarly, from Proposition 5.3 we have that $\varphi : \Omega ^{s} \rightarrow \mathbb {R} \cup \{\pm \infty \}$ is well-defined (in the generalised sense) and ${\mathcal {F}}_s$-measurable. As $\varphi (X_1,\ldots ,X_s)$ is ${\mathcal {F}}_s$-measurable, it remains to prove that $E(gh)=E(g \varphi (X_1,\ldots ,X_s))$ for all $g: \Omega ^{s} \rightarrow \mathbb {R}_{+}$ non-negative, $\mathcal {F}_{s}$-measurable and such that E(gh) is well-defined in the generalised sense, i.e. such that $E\left( gh\right) ^{+}<\infty$ or $E\left( gh\right) ^{-}<\infty$. Recalling the notations of the beginning of Sect. 2 and using the Fubini Theorem for the third and fourth equality (see Proposition 5.3), we get that

$$\begin{aligned} E(gh)&= E(g(X_1,\ldots ,X_s)h(X_1,\ldots ,X_t))\\&=\int _{\Omega ^T} g(\omega _1,\ldots ,\omega _s)h(\omega _1,\ldots ,\omega _t)P(d \omega ^T)\\&= \int _{\Omega ^t} g(\omega _1,\ldots ,\omega _s)h(\omega _1,\ldots ,\omega _t)q_{t}(\omega _{t}|\omega ^{t-1}) \ldots q_{s+1}(\omega _{s+1}|\omega ^s)P_s(d \omega ^s)\\&= \int _{\Omega ^s} g(\omega _1,\ldots ,\omega _s)\times \left( \int _{\Omega _{s+1}\times \cdots \times \Omega _t} h(\omega _1,\ldots ,\omega _s,\omega _{s+1},\right. \\&\left. \quad \ldots ,\omega _t)q_{t}(\omega _{t}|\omega ^{t-1}) \ldots q_{s+1}(\omega _{s+1}|\omega ^s)\right) P_s(d \omega ^s)\\&= \int _{\Omega ^s} g(\omega _1,\ldots ,\omega _s)\varphi (\omega _1,\ldots ,\omega _s)P_s(d \omega ^s)\\&= E(g(X_1,\ldots ,X_s)\varphi (X_1,\ldots ,X_t)). \end{aligned}$$

$\square$

Proof of Lemma 3.2

Clearly, for all $\omega ^{t} \in \Omega ^{t}$, $\widetilde{D}^{t+1}(\omega ^{t})$ is a non-empty and closed subset of $\mathbb {R}^{d}$. We now show that $\widetilde{D}^{t+1}$ is measurable. Let O be a fixed open set in $\mathbb {R}^{d}$ and introduce $\mu _{O}$ defined for all $\omega ^{t} \in \Omega ^{t}$ by $\mu _{O}(\omega ^{t}) := q_{t+1}\left( \Delta S_{t+1}(\omega ^{t},.) \in O|\omega ^{t}\right) = \int _{\Omega _{t+1}}1_{\Delta S_{t+1}(\cdot ,\cdot ) \in O}(\omega ^{t},\cdot )dq_{t+1}(\cdot |\omega ^{t}).$ Using Lemma 5.7, $\mu _{O}$ is $\mathcal {F}_{t}$-measurable, since $\Delta S_{t+1}(\cdot ,\cdot )$ is $\mathcal {F}_{t}\otimes \mathcal {G}_{t+1}$-measurable, $O \in \mathcal {B}(\mathbb {R}^{d})$ and $1_{\Delta S_{t+1}(\cdot ,\cdot ) \in O}(\cdot ,\cdot )$ is $\mathcal {F}_{t}\otimes \mathcal {G}_{t+1}$-measurable. Then by definition of $\widetilde{D}^{t+1}(\omega ^{t})$ we get that $\{\omega ^{t} \in \Omega ^t,\; \widetilde{D}^{t+1}(\omega ^{t}) \cap O \ne \emptyset \} =\{\omega ^{t} \in \Omega ^t, \; \mu _{O}(\omega ^{t})>0\} \in \mathcal {F}_{t}$ and $\widetilde{D}^{t+1}$ is $\mathcal {F}_{t}$-measurable. Clearly, $D^{t+1}$ is a non-empty and closed-valued random set. We prove that $D^{t+1}$ is $\mathcal {F}_{t}$-measurable. As $\widetilde{D}^{t+1}$ is $\mathcal {F}_{t}$-measurable and closed valued, it admits a Castaing representation (see Theorem 14.5 of Rockafellar and Wets (1998)), i.e. a countable family of $\mathcal {F}_{t}$-measurable functions $(f_{n})_{n \ge 1}:\Omega ^t \rightarrow \mathbb {R}^{d}$ such that for all $\omega ^{t} \in \Omega ^{t}$, $\widetilde{D}^{t+1}(\omega ^t)= \overline{\left\{ f_{n}(\omega ^{t}),\; n\ge 1\right\} }$ . Let $\omega ^{t} \in \Omega ^{t}$ be fixed. It can be easily shown that

$$\begin{aligned} D^{t+1}(\omega ^{t}) =\overline{\left\{ f_{1}(\omega ^{t})+ \sum _{i=2}^{p} \lambda _{i} (f_{i}(\omega ^{t})-f_{1}(\omega ^{t})),\; (\lambda _{2},\dots , \lambda _{p}) \in \mathbb {Q}^{p-1},\; p\ge 2\right\} }. \end{aligned}$$

Since $D^{t+1}$ admits a Castaing representation it is $\mathcal {F}_{t}$-measurable. From Theorem 14.8 of Rockafellar and Wets (1998), $Graph(D^{t+1}) \in \mathcal {F}_{t} \otimes \mathcal {B}(\mathbb {R}^{d})$ (recall that $D^{t+1}$ is closed-valued). $\square$

Proof of Lemma 3.3

Introduce $C^{t+1}(\omega ^{t}):=\overline{\text{ Conv }} (\widetilde{D}^{t+1}(\omega ^{t}))$ the closed convex hull generated by $\widetilde{D}^{t+1}(\omega ^{t})$. We distinguish two cases. First assume that for all $h \in \mathbb {R}^{d}$, $h \ne 0$, $q_{t+1}(h\Delta S_{t+1}(\omega ^{t},.) \ge 0|\omega ^{t})<1$. Then the polar cone of $C^{t+1}(\omega ^{t})$, i.e. the set $\left( C^{t+1}(\omega ^{t})\right) ^{\circ }:=\{y \in \mathbb {R}^{d},\; yx\le 0, \, \forall \, x \in C^{t+1}(\omega ^{t})\}$ is reduced to $\{0\}$. Indeed if this is not the case there exists $y_{0} \in \mathbb {R}^{d}$ such that $-y_0x\ge 0$ for all $x \in C^{t+1}(\omega ^{t})$. As $A:=\{\Delta S_{t+1}(\omega ^{t},\cdot ) \in \widetilde{D}^{t+1}(\omega ^{t})\} \subset \{ -y_{0}\Delta S_{t+1}(\omega ^{t},\cdot )\ge 0\}$ and $q_{t+1}(A|\omega ^{t})=1$ we obtain that $q_{t+1}(-y_0 \Delta S_{t+1}(\omega ^{t},\cdot ) \ge 0|\omega ^{t})=1$ a contradiction. As $\left( \left( C^{t+1}(\omega ^{t})\right) ^{\circ }\right) ^{\circ }= \text{ cone }\left( C^{t+1}(\omega ^{t})\right)$ the cone generated by $C^{t+1}(\omega ^{t})$, we get that $\text{ cone }\left( C^{t+1}(\omega ^{t})\right) =\mathbb {R}^{d}$. Let $u \ne 0 \in \text{ cone }\left( C^{t+1}(\omega ^{t})\right)$ then $-u \in \text{ cone }\left( C^{t+1}(\omega ^{t})\right)$ and there exist $\lambda _{1}>0$, $\lambda _{2}>0$ and $v_{1},\; v_{2} \in C^{t+1}(\omega ^{t})$ such that $u=\lambda _{1} v_{1}$ and $-u=\lambda _{2} v_{2}$. Thus $0=\frac{\lambda _1}{\lambda _1+ \lambda _2} v_{1} + \frac{\lambda _2}{\lambda _1+ \lambda _2} v_{2} \in C^{t+1}(\omega ^{t})$ by convexity of $C^{t+1}(\omega ^{t})$.

Now we assume that there exists some $h _{0}\in \mathbb {R}^{d}$, $h_{0} \ne 0$ such that $q_{t+1}(h_0\Delta S_{t+1}(\omega ^{t},.) \ge 0|\omega ^{t})=1$. We exhibit some $h_0 \in C^{t+1}(\omega ^{t})$ such that $q_{t+1}(h_0\Delta S_{t+1}(\omega ^t,\cdot )\ge 0|\omega ^t)=1$. Then as $C^{t+1}(\omega ^{t}) \subset D^{t+1}(\omega ^{t})$, the assumptions of Lemma 3.4 will implie that $h_0=0$, hence $0 \in C^{t+1}(\omega ^{t})$. Introduce p the orthogonal projection on $C^{t+1}(\omega ^{t})$ (recall that $C^{t+1}(\omega ^{t})$ is a closed convex subset of $\mathbb {R}^{d}$). Then p is continuous and $\left( h-p(h)\right) \left( x-p(h)\right) \le 0$ for all $x \in C^{t+1}(\omega ^{t})$. Fix $\omega _{t+1} \in \{\Delta S_{t+1}(\omega ^{t},\cdot ) \in \widetilde{D}^{t+1}(\omega ^{t})\} \cap \{h_{0}\Delta S_{t+1}(\omega ^{t},\cdot )\ge 0\}$ and $\lambda \ge 0$. Choose $h=\lambda h_{0}$ and $x= \Delta S_{t+1}(\omega ^{t},\omega _{t+1}) \in C^{t+1}(\omega ^{t})$ in the previous ineqality (recall that $\widetilde{D}^{t+1}(\omega ^{t}) \subset C^{t+1}(\omega ^{t})$), we obtain that

$$\begin{aligned} 0 \le \lambda h_{0} \Delta S_{t+1}(\omega ^{t},\omega _{t+1})&= \left( \lambda h_0-p(\lambda h_0)\right) \Delta S_{t+1}(\omega ^{t},\omega _{t+1})\\&\quad + p(\lambda h_0) \Delta S_{t+1}(\omega ^{t},\omega _{t+1}) \\&\le \left( \lambda h_0-p(\lambda h_0)\right) p(\lambda h_0) +p(\lambda h_0) \Delta S_{t+1}(\omega ^{t},\omega _{t+1}). \end{aligned}$$

As this is true for all $\lambda \ge 0$ we may take the limit when $\lambda$ goes to zero and use the continuity of p to obtain that $p(0) \Delta S_{t+1}(\omega ^{t},\omega _{t+1}) \ge |p(0)|^{2} \ge 0.$ By definition of $\widetilde{D}^{t+1}(\omega ^{t})$, we have $q_{t+1}\left( \left\{ \Delta S_{t+1}(\omega ^{t},\cdot ) \in \widetilde{D}^{t+1}(\omega ^{t})\right\} |\omega ^t\right) =1$ and as $q_{t+1}(h_0\Delta S_{t+1}(\omega ^{t},.) \ge 0|\omega ^{t})=1$ as well we obtain that $q_{t+1}( p(0) \Delta S_{t+1}(\omega ^{t},\cdot ) \ge 0|\omega ^{t})=1$ and $h_0=p(0)$ is found. $\square$

The following lemma has been used in the proof of Lemma 3.4. It is the pendant of Lemma 2.5 of Nutz (2016).

Lemma 5.15

Let $\omega ^{t} \in \Omega ^{t}$ be fixed. Let $L^{t+1}(\omega ^{t}):=\left( D^{t+1}(\omega ^{t})\right) ^{\bot }$ be the orthogonal space of $D^{t+1}(\omega ^{t})$. Then for $h \in \mathbb {R}^{d}$ we have that $q_{t+1}(h \Delta S_{t+1}(\omega ^{t},\cdot )=0|\omega ^{t})=1$ if and only if $h \in L^{t+1}(\omega ^{t}).$

Proof

Assume that $h \in L^{t+1}(\omega ^{t})$. Then $\{ \Delta S_{t+1}(\omega ^{t},\cdot ) \in D^{t+1}(\omega ^t)\} \subset \{h \Delta S_{t+1}(\omega ^{t},\cdot )=0\}$. As $q_{t+1}(\Delta S_{t+1}(\omega ^{t},.) \in D^{t+1}(\omega ^{t})|\omega ^{t})=1$, we conclude that $q_{t+1}(h \Delta S_{t+1}(\omega ^{t},.)=0|\omega ^{t})=1.$ Conversely, assume that $h \notin L^{t+1}(\omega ^{t})$. We first show that there exists $v \in \widetilde{D}^{t+1}(\omega ^{t})$ such that $hv \ne 0$. If not, for all $v \in \widetilde{D}^{t+1}(\omega ^{t})$, $hv = 0$. Thus, for any $w \in {D}^{t+1}(\omega ^{t})$ with $w=\sum _{i=1}^m \lambda _i v_i$ where $\lambda _i \in \mathbb {R}$, $\sum _{i=1}^m \lambda _i=1$ and $v_i \in \widetilde{D}^{t+1}(\omega ^{t})$, we get that $hw = 0$, a contradiction. Furthermore there exists an open ball centered in v with radius $\varepsilon >0$, $B(v,\varepsilon )$, such that $hv' \ne 0$ for all $v' \in B(v,\varepsilon )$. Assume that $q_{t+1}( \Delta S_{t+1}(\omega ^{t},.) \in B(v,\varepsilon )|\omega ^{t})=0$ or equivalently that $q_{t+1}( \Delta S_{t+1}(\omega ^{t},.) \in \mathbb {R}^d {\setminus } B(v,\varepsilon )|\omega ^{t})=1$. By definition of the support, $\widetilde{D}^{t+1}(\omega ^{t}) \subset \mathbb {R}^d {\setminus } B(v,\varepsilon )$: this contradicts $v \in \widetilde{D}^{t+1}(\omega ^{t})$. Therefore $q_{t+1}( \Delta S_{t+1}(\omega ^{t},.) \in B(v,\varepsilon )|\omega ^{t})>0$. As $\{\Delta S_{t+1}(\omega ^{t},.) \in B(v,\varepsilon )\} \subset \{h\Delta S_{t+1}(\omega ^{t},\cdot ) \ne 0\}$ we have proved that $q_{t+1}(h \Delta S_{t+1}(\omega ^{t},.)=0|\omega ^{t}))<1$. $\square$

Proof of Proposition 4.26

We start with the proof of (13) when $h \in D_{x}$. Since D is a vectorial subspace of $\mathbb {R}^d$ and $0 \in \mathcal {H}_{x}$, the affine hull of $D_{x}$ is also a vector space that we denote by $\text{ Aff } (D_x)$. If $x \le 1$ we have by Assumption 4.20 that for all $\omega \in \overline{\Omega }$, $h \in D_{x}$,

$$\begin{aligned} V^{+}(\omega ,x+hY(\omega )) \le V^{+}\left( \omega ,1+ h Y(\omega )\right) . \end{aligned}$$

(40)

If $x>1$ using Assumption 4.22 [see (12) in Remark 4.23] we get that for all $\omega \in \overline{\Omega }$, $h \in D_{x}$

$$\begin{aligned} V^{+}(\omega ,x+hY(\omega ))&=V^{+}\left( 2x\left( \frac{1}{2} +\frac{h}{2x}Y(\omega )\right) \right) \nonumber \\&\le (2x)^{\overline{\gamma }}K\left( V^{+}\left( \omega ,1+ \frac{h}{2x} Y(\omega )\right) + C(\omega )\right) . \end{aligned}$$

(41)

When $Dim (\text{ Aff } (D_x) )=0$, using (40) and (41), we obtain

$$\begin{aligned} V^{+}(\omega ,x+hY(\omega ))&\le V^{+}(\omega ,1) + (2x)^{\overline{\gamma }}K\left( V^{+}\left( \omega ,1\right) + C(\omega )\right) \nonumber \\&\le ((2x)^{\overline{\gamma }}K+1)(V^{+}(\omega ,1)+C(\omega )). \end{aligned}$$

(42)

We assume now that $Dim(\text{ Aff } (D_x) )>0$. Then $x=0$ implies that $Y=0$ Q-a.s. Indeed else $D_0=\{0\}$: if there exists some $h \in D_0 \backslash {\{0\}}$ then $Q\left( \frac{h}{|h|}Y<0\right) >0$ by Assumption 4.18 which contradicts $h \in D_0$. So, by Assumption 4.20 we get that for all $\omega \in \overline{\Omega }$, $h \in D_{0}$, $V^{+}(\omega ,0+hY(\omega )) \le V^{+}(\omega ,1)$.

From now we assume that $x>0$. Then as $g \in D_{x}$ if and only if $\frac{g}{x} \in D_{1}$, we have that $\text{ Aff } (D_x) =\text{ Aff } (D_1)$. We set $d':=Dim(\text{ Aff } (D_1))$. Let $(e_{1},\dots ,e_{d'})$ be an orthonormal basis of $\text{ Aff } (D_1)$ (which is a sub-vector space of $\mathbb {R}^{d}$) and $\varphi$  :  $(\lambda _{1},\dots ,\lambda _{d'})\in \mathbb {R}^{d'} \rightarrow \Sigma _{i=1}^{d'} \lambda _{i} e_{i} \in \text{ Aff } (D_1)$. Then $\varphi$ is an isomorphism and it is also an homeomorphism between $\mathbb {R}^{d'}$ and $\text{ Aff } (D_1)$ (those spaces are of finite dimension and $\varphi$ is linear). Since $D_{1}$ is compact by Lemma 4.25, $\varphi ^{-1}(D_{1})$ is a compact subset of $\mathbb {R}^{d'}$ . So there exists some $c \ge 0$ such that for all $h=\Sigma _{i=1}^{d'} \lambda _{i} e_{i} \in D_1$, $|\lambda _{i}| \le c$ for all $i=1,\ldots ,d'$. We complete the family of vector $(e_{1},\dots ,e_{d'})$ in order to obtain an orthonormal basis of $\mathbb {R}^{d}$, denoted by $(e_{1},\dots ,e_{d'}, e_{d'+1},\dots e_{d})$. For all $\omega \in \Omega$, let $(y_{i}(\omega ))_{i=1,\dots ,d}$ be the coordinate of $Y(\omega )$ in this basis.

Now let $h\in D_x$ be fixed. Then $\frac{h}{2x} \in D_{\frac{1}{2}} \subset D_1$ and $\frac{h}{2x}=\Sigma _{i=1}^{d'} \lambda _{i} e_{i}$ for some $(\lambda _{1},\dots \lambda _{d'})\in \mathbb {R}^{d'}$ with $|\lambda _{i}| \le c$ for all $i=1,\ldots ,d'$. Note that as $\frac{h}{2x} \in D_1$, $\lambda _{i}=0$ for $i \ge d'+1$. Then as $(e_{1},\dots ,e_{d})$ is an orthonormal basis of $\mathbb {R}^d$, we obtain for all $\omega \in \overline{\Omega }$ that $1 +\frac{h}{2x} Y(\omega ) = 1 + \Sigma _{i=1}^{d'} \lambda _{i}y_{i}(\omega ) \le 1 + \Sigma _{i=1}^{d'} |\lambda _{i}| |y_{i}(\omega )| \le 1+ c \Sigma _{i=1}^{d'} |y_{i}(\omega )|.$ Thus from Assumption 4.20 for all $\omega \in \overline{\Omega }$ we get that $V^{+}\left( \omega ,1+\frac{h}{2x} Y(\omega )\right) \le V^{+}\left( \omega ,1+ c \Sigma _{i=1}^{d'} |y_{i}(\omega )|\right) .$ We set

$$\begin{aligned} L(\cdot ):= V^{+}\left( \omega ,1+ c \Sigma _{i=1}^{d'} |y_{i}(\omega )|\right) 1_{d'>0}+ V^{+}(\cdot ,1) + C(\cdot ). \end{aligned}$$

As $d'=Dim (\text{ Aff } (D_1))$ it is clear that L does not depend on x. It is also clear that L is ${\mathcal {H}}$-measurable. Then using (40), (41) and (42) we obtain that (13) holds true for all $\omega \in \overline{\Omega }$. As by Assumptions 4.22 and 4.24, $E(V^{+}(\cdot ,1)+C(\cdot )) < \infty$, to prove that $E L<\infty$, we have to show that $d'>0$ implies $E \left( V^{+}\left( \cdot ,1+ c \Sigma _{i=1}^{d'} |y_{i}(\cdot )|\right) \right) <\infty$.

Introduce W, the finite set of $\mathbb {R}^d$ whose coordinates on $(e_{1},\dots ,e_{d'})$ are 1 or $-1$ and 0 on $(e_{d'+1},\dots e_{d})$. Then $W \subset \text{ Aff } (D_1)$ and the vectors of W will be denoted by $\theta ^{j}$ for $j \in \{1,\dots , 2^{d'}\}$. Let ${\theta }^{\omega }$ be the vector whose coordinates on $(e_{1},\dots ,e_{d'})$ are $(sign(y_{i}(\omega )))_{i=1\dots d'}$ and 0 on $(e_{d'+1},\dots e_{d})$. Then it is clear that $\theta ^{\omega } \in W$ and we get that

$$\begin{aligned} V^{+}\left( \omega ,1+c \Sigma _{i=1}^{d'} |y_{i}(\omega )|\right) = V^{+}(\omega ,1+ c\theta ^{\omega } Y(\omega )) \le \sum _{j=1}^{2^{d'}} V^{+}(\omega ,1+ c \theta ^{j} Y(\omega )). \end{aligned}$$

So it is sufficient to prove that if $d'>0$ $E V^{+}(\cdot ,1+ c \theta ^{j}Y(\cdot ))<\infty$ for all $1\le j \le 2^{d'}$. Recall that $\theta ^{j} \in \text{ Aff } (D_1)$. Let $ri(D_{1})=\{y \in D_{1}, \, \exists \alpha >0 \, s.t \; \text{ Aff } (D_1) \cap B(y,\alpha ) \subset D_1\}$ denote the relative interior of $D_{1}$. As $D_{1}$ is convex and non-empty (recall $d'>0$), $ri(D_{1})$ is also non-empty and convex and we fix some $e^{*} \in ri(D_{1})$. We prove that $\frac{e^{*}}{2} \in ri(D_{1})$. Let $\alpha >0$ be such that $\text{ Aff } (D_1) \cap B(e^*,\alpha ) \subset D_1$ and $g \in \text{ Aff } (D_1) \cap B(\frac{e^*}{2},\frac{\alpha }{2})$. Then $2g \in \text{ Aff } (D_1) \cap B({e^*},{\alpha })$ and thus $2g \in D_1$. As $D_{1}$ is convex and $0 \in D_{1}$, we get that $g \in D_1$ and $\text{ Aff } (D_1) \cap B(\frac{e^*}{2},\frac{\alpha }{2}) \subset D_1$ which proves that $\frac{e^{*}}{2} \in ri(D_{1})$. Now let $\varepsilon _{j}$ be such that $\varepsilon _{j}(\frac{c}{2} \theta ^{j}-\frac{e^{*}}{2})\in B(0,\frac{\alpha }{2})$. It is easy to see that one can chose $\varepsilon _{j} \in (0,1)$. Then as $\bar{e}^{j}:=\frac{e^{*}}{2}+ \frac{\varepsilon _{j}}{2}(c \theta ^{j}-e^{*}) \in \text{ Aff } (D_1)\cap B(\frac{e^*}{2},\frac{\alpha }{2})$ we deduce that $\bar{e}^{j} \in D_{1}$. Using (12) we obtain that for Q-almost all $\omega$

$$\begin{aligned}&V^{+}(\omega ,1+c \theta ^{j}Y(\omega ))\nonumber \\&\quad =V^{+}(\omega ,1+e^{*}Y(\omega )+(c \theta ^{j}-e^{*})Y(\omega ))\\&\quad \le \left( \frac{2}{\varepsilon _{j}}\right) ^{\overline{\gamma }}K \left[ V^{+}\left( \omega ,\frac{\varepsilon _{j}}{2}(1+e^{*}Y(\omega )) +\frac{\varepsilon _{j}}{2}(c \theta ^{j}-e^{*})Y(\omega )+ \frac{1}{2}\right) +C(\omega )\right] \\&\quad \le \left( \frac{2}{\varepsilon _{j}}\right) ^{\overline{\gamma }} K\left[ V^{+}\left( \omega ,\frac{1}{2}+\frac{e^{*}}{2}Y(\omega ) +\frac{\varepsilon _{j}}{2}(c \theta ^{j}-e^{*})Y(\omega ) + \frac{1}{2}\right) +C(\omega )\right] \\&\quad \le \left( \frac{2}{\varepsilon _{j}}\right) ^{\overline{\gamma }} K\left[ V^{+}(\omega ,1+\bar{e}^{j}Y(\omega ))+C(\omega ))\right] , \end{aligned}$$

where the second inequality follows from the fact that $1+e^{*}Y(\cdot ) \ge 0$ Q-a.s and the monotonicity property of V. Note that the above inequalities are true even if $1+c \theta ^{j}Y(\omega )<0$ since (12) and the monotonicity property of V hold true for all $x \in \mathbb {R}$. From Assumption 4.24 and Assumption 4.22 we get that $E V^{+}(\cdot ,1+ c \theta ^{j}Y(\cdot )) <\infty$ and (13) is proven for $h \in D_{x}$. Now let $h \in \mathcal {H}_{x}$ and $h'$ its orthogonal projection on D, then $h Y(\cdot )=h'Y(\cdot )$ Q-a.s (see Remark 4.19). It is clear that $h' \in D_{x}$ thus $V^{+}(\cdot ,x+hY(\cdot ))=V^{+}(\cdot ,x+h'Y(\cdot ))$ Q-a.s and (13) is true also for $h \in \mathcal {H}_{x}$. $\square$

To conclude, the following lemma was used in the proof of Theorem 4.15.

Lemma 5.16

Assume that (NA) holds true. Let $\phi \in \Phi$ such that $V_T^{x,\phi } \ge 0$ P-a.s. then $V_t^{x,\phi } \ge 0$ $P_t$-a.s.

Proof

Assume that there is some t such that $P_t(V_{t}^{{x,\phi }} \ge 0)<1$ and let $n=\sup \{t | P_t(V_{t}^{{x,\phi }}<0)>0\}$. Then $P_n(V_{n}^{x,\phi }<0)>0$ and for all $s \ge n+1$, $P_s(V_{s}^{x,\phi } \ge 0)=1$. Let $\Psi _s(\omega )=0$ if $s\le n$ and $\Psi _s(\omega )=1_A \phi _s(\omega )$ if $s\ge n+1$ with $A=\{V_{n}^{\Phi }<0\}$. Then, we have that ${V}^{0,\Psi }_s = \sum _{k=1}^s \Psi _s \Delta S_s =\sum _{k=n+1}^s \Psi _s \Delta S_s = 1_A\left( {V}^{x,\phi }_s-{V}^{x,\phi }_n\right)$. If $s\ge n+1$ $P_s(V_{s}^{x,\phi }\ge 0)=1$ and $-{V}^{\Phi }_n>0$ on A, then $P_T(V_{T}^{0,\Psi } \ge 0)=1$ and $V_{T}^{0,\Psi } > 0$ on A. As by the (usual) Fubini Theorem $P_T(A)=P_n(V_{n}^{x,\phi }{<}0){>}0$, we get an arbitrage opportunity, a contradiction. $\square$