Efficient portfolios in financial markets with proportional transaction costs

Abstract

In this article, we characterize efficient portfolios, i.e. portfolios which are optimal for at least one rational agent, in a very general multi-currency financial market model with proportional transaction costs. In our setting, transaction costs may be random, time-dependent, have jumps and the preferences of the agents are modeled by multivariate expected utility functions. We provide a complete characterization of efficient portfolios, generalizing earlier results of Dybvig (Rev Financ Stud 1:67–88, 1988) and Jouini and Kallal (J Econ Theory 66: 178–197, 1995). We basically show that a portfolio is efficient if and only if it is cyclically anticomonotonic with respect to at least one consistent price system that prices it. Finally, we introduce the notion of utility price of a given contingent claim as the minimal amount of a given initial portfolio allowing any agent to reach the claim by trading, and give a dual representation of it as the largest proportion of the market price necessary for all agents to reach the same expected utility level.

Appendix

This appendix collects the proofs of some technical results that have been used throughout the paper.

Proof of Proposition 4.1

It is straightforward that cyclic anticomonotonicity implies (4.3). Conversely, assume that $X$ and $Y$ satisfy property (4.3). To obtain that $X$ and $Y$ are cyclically anticomonotonic, it suffices to prove that there exist two random vectors $X^{\prime },Y^{\prime }$ with $X^{\prime }=X$ and $Y^{\prime }=Y$ a.s., such that $X^{\prime }$ and $Y^{\prime }$ are cyclically anticomonotonic. Denote $X=(X^1,\ldots ,X^d)$ and $Y=(Y^1,\ldots ,Y^d)$, and consider the product set for $n \in \mathbb N ^*$:

$$\begin{aligned} \mathbb N ^{d}_{n2^n} := \left\{ 1,\ldots ,n2^n\right\} \times \ldots \times \left\{ 1,\ldots ,n2^n\right\} . \end{aligned}$$

Then, for each vector of integers $I = (I^1,\ldots ,I^d)$ and $K=(K^1,\ldots ,K^d)$ in $\mathbb N ^{d}_{n2^n}$, define the sets $A_I$ and $B_K$ by

$$\begin{aligned} A_{I} :=\bigcap _{1 \le i \le d} \left\{ \frac{I^i}{2^{n}}\le X^i <\frac{ I^i +1}{2^{n}}\right\} , \quad B_{K} :=\bigcap _{1 \le j \le d} \left\{ \frac{I^j}{2^{n}}\le Y^j <\frac{I^j +1}{2^{n}}\right\} . \end{aligned}$$

in such a way that the family $\{ A_I \cap B_K : (I,K) \in ( N^{d}_{n2^n})^2 \}$ is a partition of the set $\{ (X ,Y ) \in [0,n]^d \times [0,n]^d \}$. The rest of the proof is structured in three main steps.

Step 1: construction of $X^{\prime }$. We define

$$\begin{aligned} X_{n}^{i}:=\sum _{\left( I,K\right) \in P_{n}}\frac{I^{i}}{2^{n}}\mathbf 1 _{A_{I}\bigcap B_{K}} \end{aligned}$$

with $P_{n}:=\left\{ \left( I,K\right) \in \left( \mathbb N _{n2^{n}}^{d}\right) ^{2}:\mathbb P \left( A_{I}\cap B_{K}\right) >0\right\}$. Defining

$$\begin{aligned} N_{n}:=\left( \bigcap _{\left( I,K\right) \in P_{n}}(A_{I}\cap B_{K})\right) ^{C} \end{aligned}$$

one can see that when $\omega \notin N_{n}$, we have $\Vert X^{n}(\omega )-X(\omega )\Vert \le 2^{-n}$ for every $n\in \mathbb N$. Thus, for every $\omega \in \Omega ,\,X^{n}$ converges and we can define its limit by setting

$$\begin{aligned} X^{\prime }(\omega ):=\lim _{n\rightarrow +\infty }X_{n}(\omega ). \end{aligned}$$

By construction, $X^{\prime }=X$ a.s., since $\mathbb P (N_{n})\rightarrow 0$. Indeed, one has $\mathbb{P }(N_{n})=\mathbb{P }(X\ge n\mathbf 1 ,Y\ge n\mathbf 1 )$ which goes to $0$ as $n\rightarrow +\infty$, being both $X,Y$ finite valued.

Step 2: construction of $Y^{\prime }$. In the same way, we define

$$\begin{aligned} Y_{n}^{j}=\sum _{\left( I,K\right) \in P_{n}}\frac{K^{j}}{2^{n}}\mathbf 1 _{A_{I}\bigcap B_{K}}. \end{aligned}$$

If $\omega \notin N_{n}$, we have $\Vert Y^{n}(\omega )-Y(\omega )\Vert \le 2^{-n}$ for every $n\in \mathbb N$. Therefore, for every $\omega ,\,Y^{n}$ converges and we can define its limit setting

$$\begin{aligned} Y^{\prime }(\omega ):=\lim _{n\rightarrow +\infty }Y_{n}(\omega ) \end{aligned}$$

yielding that $Y^{\prime }=Y$ almost surely.

Step 3: conclusion. The two sequences $Y_{n}$ and $X_{n}$ have been constructed in order to have cyclically anticomonotonic random vectors in the limit. Indeed, let $\Omega _{n}=\Omega \setminus N_{n}$ and $p\in \mathbb N ^{*}$. For $(\omega _{1},\omega _{2})\in \Omega _{n}^{2}$, we have

$$\begin{aligned}&\langle Y^{n}(\omega _{1}),X^{n}(\omega _{1})-X^{n}(\omega _{2})\rangle -\langle Y(\omega _{1}),X(\omega _{1})-X(\omega _{2})\rangle \\&\quad =\langle Y^{n}(\omega _{1})-Y(\omega _{1}),X^{n}(\omega _{1})-X^{n}(\omega _{2})\rangle \\&\qquad +\langle Y(\omega _{1}),X^{n}(\omega _{1})-X(\omega _{1})\rangle \\&\qquad -\langle Y(\omega _{1}),X^{n}(\omega _{2})-X(\omega _{2})\rangle \\&\quad \le \frac{n}{2^{n}}+\frac{n}{2^{n}}+\frac{n}{2^{n}}. \end{aligned}$$

Thus, for every $(\omega _{1},\ldots ,\omega _{p+1})\in \Omega _{n}^{p+1}$ with $\omega _{p+1}=\omega _{1}$, we have

$$\begin{aligned} \sum _{i=1}^{p}\left\langle Y^{n}(\omega _{i}),X^{n}(\omega _{i})-X^{n}(\omega _{i+1})\right\rangle -\sum _{i=1}^{p}\left\langle Y(\omega _{i}),X(\omega _{i})-X(\omega _{i+1})\right\rangle \le 3p\frac{n}{2^{n}}. \end{aligned}$$

(7.1)

Consider now the function

$$\begin{aligned} g_{n}(\omega _{1},\ldots ,\omega _{p}):=\left\langle Y_{n}(\omega _{1}),X_{n}(\omega _{1})-X_{n}(\omega _{2})\right\rangle +\cdots +\left\langle Y_{n}(\omega _{p}),X_{n}(\omega _{p})-X_{n}(\omega _{1})\right\rangle . \end{aligned}$$

It is constant on the set $\times _{i=1}^{p}A_{I^{i}}\cap B_{K^{i}}$, for $(I^{i},K^{i})\in P_{n},\,i=1,\ldots ,p$. Thus, we necessarily have

$$\begin{aligned} g_{n}(\omega _{1},\ldots ,\omega _{p})\le 3p\frac{n}{2^{n}}. \end{aligned}$$

Indeed, if it was not the case, Eq. (7.1) would yield that for some $(I^{i},K^{i})\in P_{n},\,i=1,\ldots ,p$, and for all $(\omega _{1},\ldots ,\omega _{p})\in \times _{i=1}^{p}A_{I^{i}}\cap B_{K^{i}}$ – a strictly positive probability set, by construction – the following inequality holds:

$$\begin{aligned} \left\langle Y(\omega _{1}),X(\omega _{1})-X(\omega _{2})\right\rangle +\cdots +\left\langle Y(\omega _{p}),X(\omega _{p})-X(\omega _{1})\right\rangle >0 \end{aligned}$$

which contradicts Eq. (4.3). We conclude that for every $(\omega _{1},\ldots ,\omega _{p})\in A^{p}$, where $A:=\Omega \setminus \cap _{n\ge 1}N_{n}$, we have

$$\begin{aligned} \left\langle Y^{\prime }(\omega _{1}),X^{\prime }(\omega _{1})-X^{\prime }(\omega _{2})\right\rangle +\cdots +\left\langle Y^{\prime }(\omega _{p}),X^{\prime }(\omega _{p})-X^{\prime }(\omega _{1})\right\rangle \le 0 \end{aligned}$$

with $\mathbb P (A)=1-\lim _{n}\mathbb P (N_{n})=1$.$\square$

Proof of Proposition 4.2

First, distinguish between two cases: If for all $X\in \mathcal L (X_{0})$ (implying in particular that $X\in L^{0}(\mathbb R _{+}^{d})$) we have $\mathbb E \left[ XY\right] =+\infty$, there is nothing to prove. Let us turn to the second case when there exists at least one $X\in \mathcal L (X_{0})$ such that $\mathbb E \left[ XY\right] <\infty$, so that the infimum in (4.5) is finite.

In the following, we denote $C_{X_{0}}$ a copula of $X_{0}$, and $C_{Y}$ a copula of $Y$ (we refer to Nelson’s book [24] for details on copulas). Consider now the set $\mathcal C (X_{0},Y)$ of all copulas on $\mathbb R ^{2d}$, such that for every $C\in \mathcal C (X_{0},Y)$, the marginal copula of the $d$ first variables is $C_{X_{0}}$, and the marginal copula of the $d$ last variables is $C_{Y}$, i.e.

$$\begin{aligned} C(u_{1},u_{2},\ldots ,u_{d},1,\ldots ,1)&= C_{X}(u_{1},u_{2},\ldots ,u_{d}),\\ C(1,\ldots ,1,v_{1},v_{2},\ldots ,v_{d})&= C_{Y}(u_{1},u_{2},\ldots ,u_{d}). \end{aligned}$$

It is straightforward to see that the set $\mathcal C (X_{0},Y)$ is closed with respect to the topology of pointwise convergence on $\mathcal C$, the set of all possible copulas on $\mathbb R ^{2d}$. Furthermore the set $\mathcal C$ is compact with respect to this topology (see Deheuvels [7], Theorem 2.3). Thus the set $\mathcal C (X_{0},Y)$ is itself compact with respect to the topology of pointwise convergence. Let $X_{n}$ be a sequence in ${\mathcal{L }X_{0}})$ such that

$$\begin{aligned} \lim _{n\rightarrow +\infty }\mathbb{E }\left( YX_{n}\right) =\inf \{\mathbb{E } (YX):X\in {\mathcal{L }(X_{0}})\}, \end{aligned}$$

and let $C_{n}$ denote the copula of $(X_{n},Y)$. Since $\mathcal C (X_{0},Y)$ is compact, we can assume without loss of generality. (up to extracting subsequences) that the sequence $C_{n}\in \mathcal C (X_{0},Y)$ converges pointwise to a copula $C\in \mathcal C (X_{0},Y)$. Consider the random vector $\tilde{X}_{0}$ such that the copula of $(\tilde{X}_{0},Y)$ is $C$. In particular, $\tilde{X}_{0}\in \mathcal L (X_{0})$. Notice that

$$\begin{aligned} \mathbb E \left[ \langle Y,X_{n}\rangle \right]&= \sum _{i=1}^{d}\mathbb E \left[ X_{n}^{i}Y^{i}\right] =\sum _{i=1}^{d}\int _{0}^{\infty }\int _{0}^{\infty }\mathbb P (X_{n}^{i}>t,Y^{i}>u)dtdu \\&= \sum _{i=1}^{d}\int _{0}^{\infty }\int _{0}^{\infty }\left[ 1\!-\!\mathbb P \left( X_{0}^{i}\le t\right) \!-\!\mathbb P \left( Y^{i}\le u\right) \right.\nonumber \\&\left.-C_{n}^{i}\left( \mathbb P \left( X_{0}^{i}\le t\right) ,\mathbb P \left( Y^{i}\le u\right) \right) \right]dtdu. \nonumber \end{aligned}$$

(7.2)

where $C_{n}^{i}$ denotes the marginal copula of the vector $(X_{n}^{i},Y^{i})$ and where we used the fact that $X_{n}^{i}$ has the same law as $X_{0}^{i}$ for all $i$. By the pointwise convergence of $C_{n}$ to the copula $C$ of $(\tilde{X}_{0},Y)$, we deduce that, for every $u\ge 0,\, t\ge 0$ and $i\in \left\{ 1,\ldots ,d\right\}$, we have

$$\begin{aligned} \lim _{n\rightarrow +\infty }C_{n}^{i}\left( \mathbb P \left( X_{0}^{i}\le t\right) ,\mathbb P \left( Y^{i}\le u\right) \right) =C^{i}\left( \mathbb P \left( X_{0}^{i}\le t\right) ,\mathbb P \left( Y^{i}\le u\right) \right) \end{aligned}$$

with $C^{i}$ is the marginal copula of the vector $(\tilde{X}_{0}^{i},Y^{i})$. Therefore by Fatou’s lemma and the Eq. (7.2), we have

$$\begin{aligned} \mathbb{E }\left[ \langle \tilde{X}_{0},Y\rangle \right]&= \sum _{i=1}^{d} \int _{0}^{\infty }\int _{0}^{\infty }1-\mathbb{P }\left( X_{0}^{i}\le t\right) -\mathbb{P }\left( Y^{i}\le u\right)\\&-C^{i}\left( \mathbb{P }\left( X_{0}^{i}\le t\right) ,\mathbb{P }\left( Y^{i}\le u\right) \right) dtdu \\&\le \liminf _{n\rightarrow +\infty }\mathbb{E }\left[ \langle X_{n},Y\rangle \right] \le \min \{\mathbb{E }\left[ XY\right] :X\in \mathcal L (X_{0})\} \end{aligned}$$

and thus, $\mathbb E \left[ \langle \tilde{X}_{0},Y\rangle \right] =\min \{ \mathbb E \left[ YX\right] :X\in \mathcal L (X_{0})\}$.

Now, let us prove that $\tilde{X}_{0}$ and $Y$ are cyclically anticomonotonic. Suppose that this is not the case. Thus, by Corollary there exist $\varepsilon >0,\,p\ge 1$ and some non negligible measurable sets $\Omega _{1},\ldots ,\Omega _{p}$ such that for all $(\omega _{1},\ldots ,\omega _{p})\in \Omega _{1}\times \ldots \times \Omega _{p}$, we have

$$\begin{aligned} \langle \tilde{X}_{0}(\omega _{1})-\tilde{X}_{0}(\omega _{2}),Y(\omega _{1})\rangle +\ldots +\langle \tilde{X}_{0}(\omega _{p})-\tilde{X} _{0}(\omega _{1}),Y(\omega _{p})\rangle \,\ge \varepsilon \end{aligned}$$

and, being the space $(\Omega ,\mathcal F ,\mathbb{P })$ atomless, we can choose the sets $\Omega _{1},\ldots ,\Omega _{p}$ in such a way that $\mathbb P (\Omega _{1})=\mathbb P (\Omega _{2})=\ldots =\mathbb P (\Omega _{p})$. Consider a random vector $X^{\prime }$, distributed as $\tilde{X}_{0}$ with

$$\begin{aligned} \left\{ \begin{array}{l} X_{\mid \Omega \setminus \cup _{i=1}^{p}\Omega _{i}}^{\prime }=(\tilde{X} _{0})_{\mid \Omega \setminus \cup _{i=1}^{p}\Omega _{i}} \\ X_{\mid \Omega _{i}}^{\prime }\sim (\tilde{X}_{0})_{\mid \Omega _{i+1}} \quad \text{ for} 1\le i\le p \end{array}\right. \end{aligned}$$

with the convention $\Omega _{p+1}=\Omega _{1}$.⁴ A consequence of such a construction is that $X^{\prime }(\Omega _{i})=\tilde{X}_{0}(\Omega _{i+1})$ a.s. for all $i=1,\ldots ,p$. Since $X^{\prime }$ and $\tilde{X} ^{0}$ coincide on $\Omega \setminus \cup _{i=1}^{p}\Omega _{i}$ we have

$$\begin{aligned} \mathbb E \left[ \langle X^{\prime },Y\rangle \right] -\mathbb E \left[ \langle \tilde{X}_{0},Y\rangle \right] =\sum _{i=1}^{p}\mathbb E \left[ \langle Y,(X^{\prime }-\tilde{X}_{0})_{\mid \Omega _{i}}\rangle \right]. \end{aligned}$$

Moreover, we have by construction that $\sum _{i=1}^{p}\langle Y,(X^{\prime }- \tilde{X}_{0})_{\mid \Omega _{i}}\rangle \le -\varepsilon$, which implies

$$\begin{aligned} \mathbb E _{{}}\left[ \langle X^{\prime },Y\rangle \right] -\mathbb E \left[ \langle \tilde{X}_{0},Y\rangle \right] \le -p\varepsilon . \end{aligned}$$

As a consequence, we have $\mathbb E \left[ X^{\prime }Y\right] <\mathbb E \left[ \tilde{X}_{0}Y\right]$, so that $\tilde{X}_{0}$ cannot be the minimizer. This is clearly a contradiction and ends the proof.$\square$

Proof of Lemma 5.1

Let $\lambda >0$ be a certain proportion of portfolio $x_{0}$. Using Theorem (3.1), we have that $X_{0}$ can be hedged by the initial holdings vector $\lambda x_{0}$, if and only if:

$$\begin{aligned} m (X_{0})\le \lambda \langle x_{0},m (\Omega )\rangle ,\quad \forall m \in \mathcal D . \end{aligned}$$

(7.3)

We claim that the latter condition is equivalent to the following

$$\begin{aligned} m (X_{0})\le \lambda \langle x_{0},m (\Omega )\rangle ,\quad \forall m\in \mathcal D ^\bot (x_0).\end{aligned}$$

(7.4)

Clearly (7.3) implies (7.4). Assume (7.4) and let $m$ be any measure in $\mathcal D$. Since $m (\Omega )\in K_{0}^{*}$ and $K_{0}^{*}= {\text{ cone}}(K_{0}^{*}(x_{0}))$, we have $m (\Omega )=\beta z_{0}$ where $\beta \ge 0$ and $z_{0}\in K_{0}^{*}(x_{0})$. If $\beta =0$ there is nothing to prove. Consider the case $\beta >0$ and define the measure $m_{\beta }:=m/\beta \in \mathcal D$. Moreover, $m_{\beta }\in \mathcal D ^{\bot }(x_{0})$ by construction. The claim is proved. To conclude the proof, notice that whenever $m\in \mathcal D ^{\bot }(x_{0})$ one has $m (\Omega )=x_{0}/\Vert x_{0}\Vert ^{2}+z^{\bot }$ with $\langle z^{\bot },x_{0}\rangle =0$, so that $\langle x_{0},m (\Omega )\rangle =1$. Thus, $X_{0}$ can be hedged by the initial portfolio $\lambda x_{0}$ if and only if

$$\begin{aligned} \sup _{m\in \mathcal D ^{\bot }(x_{0})}m (X_{0})\le \lambda \end{aligned}$$

which proves the first equality in 5.3. For the second one, it suffices to use the same arguments using the formulation of Theorem 3.1 in terms of consistent price systems together with the Remark 3.3 in [1] stating a one-to-one correspondence between consistent price systems and elements of $\mathcal D \cap {\text{ ca}}(\mathbb{R }_+ ^d)$.$\square$

Proof of Lemma 5.2

(i) Let $Y=\frac{dm}{d\mathbb P }$ for some $m\in \mathcal D ^\bot (x_0) \cap {\text{ ca}}(\mathbb{R }^d)$. First note that we obviously have $\mathcal B (X_{0})\subset \mathcal B ^{U}(X_{0})$ for all $U\in \mathcal U$, so that

$$\begin{aligned} \sup _{U\in \mathcal U }\inf _{X\in \mathcal B ^{U}(X_{0})}\mathbb E \left[ XY \right] \le \inf _{X\in \mathcal B (X_{0})}\mathbb E \left[ XY\right] \end{aligned}$$

We are now going to prove the converse inequality and that the infimum in the right hand side above is attained. By Proposition 4.2, we can choose $\tilde{X}_{0}\sim X_{0}$ such that $\tilde{X} _{0}$ and $Y$ are cyclically anticomonotonic and $\tilde{X}_{0}$ satisfies

$$\begin{aligned} \mathbb{E }[\tilde{X}_0Y]=\min \{\mathbb{E }\left[ YX\right] :X\in \mathcal{L } (X_{0})\}=:\lambda _{0}, \end{aligned}$$

so that the infimum in (5.8) is attained. Notice that $\lambda _{0}\le 1$ since $m\in \mathcal D ^{\bot }(x_{0})$ and $X_{0}\in \mathcal{A }_{T}^{x_{0}}$, so that $m (X_{0})=\mathbb{E }[YX_0]\le 1$. At this point, we would be tempted to follow the second part of the proof of Theorem to construct a utility function $U\in \mathcal{U }$ such that $\tilde{X}_{0}$ solves

$$\begin{aligned} \sup \{\mathbb{E }\left[ U(X)\right] :X\in \mathcal{A }_{T}^{x_{0}},\mathbb{E }\left[ XY \right] \le \lambda _{0}\}. \end{aligned}$$

However, it may happen that $\tilde{X}_{0}$ and $Y$ do not satisfy the condition (iv) of Theorem , i.e.

$$\begin{aligned} \mathrm{ess\,sup\,}\Vert Y\Vert =\infty \Rightarrow \mathrm{ess\,inf\,}\Vert \tilde{X}_{0}\Vert =0,\quad \mathrm{ess\,inf\,}\Vert Y\Vert =0\Rightarrow \mathrm{ess\,sup\,}\Vert \tilde{X}_{0}\Vert =\infty . \end{aligned}$$

(7.5)

Nevertheless, for any $\varepsilon >0$ and for all $i=1,\ldots ,d$, we can exhibit a random vector $\tilde{X}_{\varepsilon }$ satisfying the following properties

• 1. $\tilde{X}_{\varepsilon }$ and $Y$ are cyclically anticomonotonic and satisfy the properties (7.5) above;

• 2. $\mathbb{E }[Y\tilde{X}_0] - \varepsilon _0 \le \mathbb{E }[Y\tilde{X}_\varepsilon ] \le \mathbb{E }[Y\tilde{X}_0]$, for some $0<\varepsilon _0 \le \varepsilon$.

In order to obtain such random variables, one may proceed as follows: for any $\varepsilon >0$ and any $y\in \mathbb R ^d _+$, define

$$\begin{aligned} \tilde{X}_{\varepsilon ,y} ^i := e^{-\varepsilon Y^i} \tilde{X}_0 ^i \mathbf{1 } _{\{\Vert Y\Vert < y\}} + \left( \tilde{X}_0 + \frac{\varepsilon }{Y^i}\right) \mathbf{1 }_{\{\Vert Y\Vert \ge y\}}, \quad i=1,\ldots ,d. \end{aligned}$$

Consider now the function $y\mapsto \psi (y) := \mathbb{E }[Y (\tilde{X}_{\varepsilon ,y} - \tilde{X}_0)]$ and notice that $\psi (y) \rightarrow \varepsilon >0$ as $y\downarrow 0$, while $\psi (y)$ tends to a strictly negative value as $y \rightarrow +\infty$. Thus, one can choose $y=y(\varepsilon _0)$ such that property (b) above is satisfied. It is then easy to check that, by construction, property (a) above is fulfilled as well. Let us denote $\tilde{X}_{\varepsilon } := \tilde{X}_{\varepsilon ,y(\varepsilon _0)}$.

Thus, even though $\tilde{X}_{\varepsilon }$ might not be attainable, we can nonetheless reproduce step-by-step the second part of the proof of Theorem , and find a utility function $U_{\varepsilon }\in \mathcal{U }$ such that $\tilde{X}_{\varepsilon }$ solves

$$\begin{aligned} \sup \left\{ \mathbb{E }\left[ U_{\varepsilon }(X)\right] :X\in \mathcal{A }_{T}^{x_{0}},\mathbb{E }\left[ XY\right] \le \mathbb{E }[\tilde{X}_{\varepsilon }Y] \right\} . \end{aligned}$$

We first deduce that

$$\begin{aligned} \sup _{U\in \mathcal U }\inf _{X\in \mathcal B ^{U}(\tilde{X}_{\varepsilon })} \mathbb E \left[ XY\right] \ge \inf _{X\in \mathcal B ^{U_{\varepsilon }}( \tilde{X}_{\varepsilon })}\mathbb E \left[ XY\right] \ge \mathbb{E }[Y \tilde{X}_{\varepsilon }]. \end{aligned}$$

(7.6)

Moreover, $\tilde{X}_{0}$ is a contingent claim preferred by each agent to $\tilde{X}_{\varepsilon }$, i.e. $\mathbb{E }[U(\tilde{X}_0 )]\ge \mathbb{E }[U(\tilde{X}_{\varepsilon })]$ for all $U\in \mathcal U$. Indeed, let $U\in \mathcal U$. Since $Y\in \partial U(\tilde{X}_{0})$ (recall that $Y$ and $\tilde{X}_{0}$ are cyclically anticomonotonic), we have

$$\begin{aligned} \mathbb E \left[ U(\tilde{X}_{\varepsilon })\right] -\mathbb E \left[ U( \tilde{X}_{0})\right] \le \mathbb E \left[ \langle Y,\tilde{X}_{\varepsilon }-\tilde{X}_{0}\rangle \right] \le 0, \end{aligned}$$

by the right hand side in property (b) above.

Therefore $\mathcal B ^{U}(\tilde{X}_{0})\subset \mathcal B ^{U}(\tilde{X} _{\varepsilon })$ for all $U\in \mathcal U$, yielding

$$\begin{aligned} \sup _{U\in \mathcal U }\inf _{X\in \mathcal B ^{U}(\tilde{X}_{0})}\mathbb E \left[ XY\right] \ge \sup _{U\in \mathcal U }\inf _{X\in \mathcal B ^{U}( \tilde{X}_{\varepsilon })}\mathbb E \left[ XY\right] \ge \mathbb E \left[ Y \tilde{X}_{\varepsilon }\right] \ge \mathbb E \left[ Y\tilde{X}_{0}\right] -\varepsilon _{0}, \end{aligned}$$

where the second inequality is due to (7.6) while the last one comes from the left hand side in property (b) above. Since $\mathcal B ^{U}(\tilde{X}_{0})=\mathcal B ^{U}(X_{0})$, for all $U\in \mathcal U$, we finally obtain

$$\begin{aligned} \sup _{U\in \mathcal U }\inf _{X\in \mathcal B ^{U}(X_{0})}\mathbb E \left[ XY \right] =\sup _{U\in \mathcal U }\inf _{X\in \mathcal B ^{U}(\tilde{X}_{0})} \mathbb E \left[ XY\right] \ge \min _{X\in \mathcal B (X_{0})}\mathbb E \left[ YX\right] -\varepsilon _{0}. \end{aligned}$$

Letting $\varepsilon$ (and so $\varepsilon _{0}$) tend to zero ends the proof.

(ii) To prove equality (5.9), it suffices to apply Theorem 3.2 in Pratelli [25], which is a version of min-max theorem well suited to our case since it does not use any compactness assumption. Indeed, $m\mapsto m(X)$ is weak* continuous on the convex set $\mathcal D ^{\bot }(x_{0})$ and $X\mapsto m(X)$ is a linear and so convex function. Moreover $\mathcal B (X_{0})$ is a nonempty convex subset of $L^0 (\mathbb R ^d _+)$ and the its level sets $\{X\in \mathcal B (X_0) : m(X)\le b\}$ are closed and bounded with respect to the convergence in probability for any $b\ge 0$ and $m\in \mathcal D ^{\bot } (x_0) \cap {\text{ ca}}(\mathbb{R }^d _+)$. More precisely, Pratelli’s theorem is formulated for subsets of $L^0_+ = L^0(\mathbb{R }_+)$, the set of all positive real valued random variables. Nonetheless, a careful inspection of his proof reveals that the result still holds in the multidimensional case, i.e. for $L^0(\mathbb{R }_+ ^d)$.$\square$