Certainty equivalent and utility indifference pricing for incomplete preferences via convex vector optimization

Abstract

For incomplete preference relations that are represented by multiple priors and/or multiple—possibly multivariate—utility functions, we define a certainty equivalent as well as the utility indifference price bounds as set-valued functions of the claim. Furthermore, we motivate and introduce the notion of a weak and a strong certainty equivalent. We will show that our definitions contain as special cases some definitions found in the literature so far on complete or special incomplete preferences. We prove monotonicity and convexity properties of utility buy and sell prices that hold in total analogy to the properties of the scalar indifference prices for complete preferences. We show how the (weak and strong) set-valued certainty equivalent as well as the indifference price bounds can be computed or approximated by solving convex vector optimization problems. Numerical examples and their economic interpretations are given for the univariate as well as for the multivariate case.

Appendix: Proof of the results from Sect. 4

Proof of Proposition 4.7

• 1. We first show that $P^b(C_T)$ is a lower set, that is, $P^b(C_T) = P^b(C_T) - \mathbb {R}^d_+$. It is clear that $P^b(C_T) \subseteq P^b(C_T) - \mathbb {R}^d_+$. To show the reverse inclusion, let $p^b \in P^b(C_T)$ and $r \in \mathbb {R}^d_+$. By Assumption 4.2c., $\mathcal {A}(x_0 - p^b +r) \supseteq \mathcal {A}(x_0-p^b)$. Then, by definition of $V(\cdot , \cdot )$ and since $p^b \in P^b(C_T)$, we have $V(x_0-p^p+r, C_T) \supseteq V(x_0-p^b, C_T) \supseteq V(x_0, 0)$. Thus, $p^b - r \in P^b(C_T)$.

  Next we show that $P^b(C_T) \subseteq \mathbb {R}^d$ is a convex set. Let $p^1, p^2 \in P^b(C_T)$, $\lambda \in [0,1]$ and $p^{\lambda }:=\lambda p^1 + (1-\lambda )p^2$. First, note that $V(x_0-p^i, C_T) \supseteq V(x_0, 0)$ for $i=1,2$ implies

  $$\begin{aligned} \lambda V(x_0-p^1, C_T) + (1-\lambda )V(x_0-p^2, C_T) \supseteq V(x_0,0). \end{aligned}$$

  In order to show $p^{\lambda } \in P^b(C_T)$, it would be enough to prove

  $$\begin{aligned} V(x_0-p^{\lambda }, C_T) \supseteq \lambda V(x_0-p^1, C_T) + (1-\lambda )V(x_0-p^2, C_T). \end{aligned}$$

  (30)

  Consider $U(V_T^i)-r^i \in V(x_0-p^i, C_T)$, where $V_T^i \in \mathcal {A}(x_0-p^i)$ and $r^i \in \mathbb {R}^q_+$ for $i =1,2$. By concavity of $u \in \mathcal {U}$, we have

  $$\begin{aligned} \lambda U(V_T^1+C_T)+ (1-\lambda )U(V_T^2+C_T)&\le U(\lambda V_T^1 + (1-\lambda )V_T^2 + C_T). \end{aligned}$$

  Clearly, for some ${\tilde{r}}\in \mathbb {R}^q_+$, where $r^{\lambda }= \lambda r^1 + (1-\lambda )r^2$, it holds

  $$\begin{aligned} \lambda U(V_T^1+C_T)+ (1-\lambda )U(V_T^2+C_T) -r^{\lambda }&= U(\lambda V_T^1 + (1-\lambda )V_T^2 + C_T) -r^{\lambda }- {\tilde{r}}. \end{aligned}$$

  As $\lambda V_T^1 + (1-\lambda )V_T^2 \in \mathcal {A}(x_0 - p^{\lambda })$ by Assumption 4.2b., we have

  $$\begin{aligned} U(\lambda V_T^1 + (1-\lambda )V_T^2 + C_T) -r^{\lambda }- {\tilde{r}} \in V(x_0-p^{\lambda }, C_T), \end{aligned}$$

  which implies (30).

• 2. We show that $P^b(\cdot )$ is monotone with respect to $\le$ and $\curlyeqprec$. Let $C_T^1, C_T^2 \in L(\mathcal {F}, \mathbb {R}^d)$ with $C_T^1 \le C_T^2$ and $p^b \in P^b(C_T^1)$. Then, by Remark 4.3a., and by the definition of $P^b(\cdot )$,

  $$\begin{aligned} V(x_0-p^b, C_T^2) \supseteq V(x_0-p^b, C_T^1) \supseteq V(x_0,0). \end{aligned}$$

  This concludes that $P^b(C_T^1)\subseteq P^b(C_T^2)$, which implies $P^b(C_T^1) \curlyeqprec P^b(C_T^2)$.

• 3. We show $\lambda P^b(C_T^1) + (1-\lambda )P^b(C_T^2) \subseteq P^b(C_T^{\lambda })$, which implies (13). Let $p^i \in P^b(C_T^i)$, that is, $V(x_0, 0) \subseteq V(x_0-p^i, C^i_T)$ for $i = 1,2$. Clearly,

  $$\begin{aligned} V(x_0, 0) \subseteq \lambda V(x_0-p^1, C_T^1) + (1-\lambda )V(x_0-p^2, C_T^2). \end{aligned}$$

  Then, it would be enough to show that

  $$\begin{aligned} \lambda V(x_0-p^1, C_T^1) + (1-\lambda )V(x_0-p^2, C_T^2) \subseteq V(x_0 - p^{\lambda }, C_T^{\lambda }), \end{aligned}$$

  (31)

  where $p^{\lambda }:= \lambda p^1 + (1-\lambda )p^2$. Let $V_T^i \in \mathcal {A}(x_0-p^i)$ and $r^i \in \mathbb {R}^q_+$ for $i=1,2$. By the concavity of $U(\cdot )$, we have

  $$\begin{aligned} \lambda (U(V_T^1+C_T^1)-r^1)&+ (1-\lambda ) (U(V_T^2+C_T^2)-r^2) \\&= \lambda U(V_T^1+C_T^1) + (1-\lambda ) U(V_T^2+C_T^2) - r^{\lambda } \\&= U(\lambda V_T^1 + (1-\lambda )V_T^2 + C_T^{\lambda }) -r^{\lambda } -{\tilde{r}} \end{aligned}$$

  for some ${\tilde{r}} \in \mathbb {R}^q_+$, where $r^{\lambda } := \lambda r^1 + (1-\lambda )r^2$. Note that $\lambda V_T^1 + (1-\lambda )V_T^2 \in \mathcal {A}(x_0-p^{\lambda })$ by Assumption 4.2 b. Then, (31) is implied by

  $$\begin{aligned} \lambda (U(V_T^1+C_T^1)-r^1) + (1-\lambda ) (U(V_T^2+C_T^2)-r^2) \in V(x_0 - p^{\lambda }, C^T_{\lambda }). \end{aligned}$$

$\square$

Fig. 5

Set-valued buy price $P^b(C_T)$ (dark green) and sell price $P^s(C_T)$ (dark blue); set of all subhedging portfolios $\mathrm{SubHP \,}(C_T)$ (light green) and superhedging portfolios $\mathrm{SHP \,}(C_T)$ (light blue) for Example 6.7. (Color figure online)

Proof of Proposition 4.8

• 1. Let $p \in P^b(C_T)\cap P^s(C_T)$, that is, $V(x_0-p,C_T) \supseteq V(x_0,0)$ and $V(x_0+p,-C_T) \supseteq V(x_0,0)$. Let $v^b \in \bigcup _{V_T^b\in \mathcal {A}(x_0-p)}[U(V_T+C_T)-\mathbb {R}^q_+]$ and $v^s \in \bigcup _{V_T^b\in \mathcal {A}(x_0+p)}[U(V_T-C_T)-\mathbb {R}^q_+]$. Then, $v^b = U(V_T^b+C_T)-r^b$ and $v^s = U(V_T^s-C_T)-r^s$ for some $V_T^b\in \mathcal {A}(x_0-p)$, $V_T^s\in \mathcal {A}(x_0+p)$ and $r^b,r^s \in \mathbb {R}^q_+$. By the concavity of U, we have

  $$\begin{aligned} \frac{1}{2} v^b+\frac{1}{2} v^s \le U(\frac{1}{2} V_T^b+\frac{1}{2} V_T^s)-\frac{1}{2}(r^b+r^s). \end{aligned}$$

  Note that for any $V_T^b\in \mathcal {A}(x_0-p)$ and $V_T^s\in \mathcal {A}(x_0+p)$, we have $V_T^b + p, V_T^s - p \in \mathcal {A}(x_0)$ and $\frac{1}{2} V_T^b + \frac{1}{2} V_T^s \in \mathcal {A}(x_0)$ by Assumption 4.2d. Then, $U(\frac{1}{2} V_T^b+\frac{1}{2} V_T^s)-\frac{1}{2}(r^b+r^s) \in V(x_0,0)$ and hence, $\frac{1}{2} v^b+\frac{1}{2} v^s \in V(x_0,0)$. Now, for any $V^b \in V(x_0-p,C_T), V^s \in V(x_0+p,-C_T)$, there exists sequences $(v_n^b)\in \bigcup _{V_T^b\in \mathcal {A}(x_0-p)}[U(V_T+C_T)-\mathbb {R}^q_+], (v_m^s)\in \bigcup _{V_T^b\in \mathcal {A}(x_0+p)}[U(V_T-C_T)-\mathbb {R}^q_+]$ with $\lim _{n\rightarrow \infty } v^b_n = V^b$ and $\lim _{n\rightarrow \infty } v^s_n = V^s$. Since for each $n,m \ge 1$, $\frac{1}{2} v_n^b+\frac{1}{2} v_m^s \in V(x_0,0)$ and since $V(x_0,0)$ is closed, we have $\frac{1}{2} V^b+\frac{1}{2} V^s \in V(x_0,0)$, which proves that

  $$\begin{aligned} \frac{1}{2} V(x_0-p,C_T)+\frac{1}{2} V(x_0+p,-C_T) \subseteq V(x_0,0). \end{aligned}$$

  (32)

  As $V(x_0-p,C_T), V(x_0+p,-C_T), V(x_0,0)$ are lower closed sets and both $V(x_0-p,C_T), V(x_0+p,-C_T)$ are supersets of $V(x_0,0)$, (32) holds only if we have $V(x_0-p,C_T) = V(x_0,0) = V(x_0+p,-C_T)$.

• 2. Let $p\in P^b(C_T)\cap P^s(C_T)$. Then, by Proposition 4.8-1., $V(x_0-p,C_T)=V(x_0,0)$. Moreover, by Assumption 4.2d., and by Remark 4.3d., $V(x_0-p+\epsilon ,C_T)\supsetneq ~V(x_0,0)$ for any $\epsilon \in \mathrm{int \,}\mathbb {R}^d_+$. Thus, $p - \epsilon \notin P^b(C_T)\cap P^s(C_T)$ for any $\epsilon \in \mathrm{int \,}\mathbb {R}^d_+$. Then, $\mathrm{int \,}(P^b(C_T)\cap P^s(C_T)) = \emptyset$.

$\square$

Proof of Proposition 4.9

We will show that $P^b(C_T)$ is closed. By Proposition 4.7, it is enough to show that $\mathrm{bd \,}P^b(C_T) \subseteq P^b(C_T)$. Moreover, as $P^b(C_T)$ is a lower set for any $p \in \mathrm{bd \,}P^b(C_T)$, there exists a sequence $(p^n)_{n\ge 1} \in P^b(C_T)$ such that $p^{n+1} \ge p^{n}$ for all $n\ge 1$ and $\lim _{n \rightarrow \infty } p^n = p$. The proof will be complete if

$$\begin{aligned} V(x_0-p,C_T) \supseteq \bigcap _{n\ge 1}V(x_0-p^n,C_T) \end{aligned}$$

(33)

holds. Indeed, together with the fact that $V(x_0-p^n,C_T)\supseteq V(x_0,0)$ for all $n \ge 1$, (33) implies that $V(x_0-p,C_T) \supseteq \bigcap _{n\ge 1} V(x_0-p^n,C_T) \supseteq V(x_0,0)$; hence $p \in P^b(C_T)$.

In order to show (33), first note that $\mathcal {A}(x_0-p)= \bigcap _{n\ge 1} \mathcal {A}(x_0-p^n)$ by Assumption 4.2 e. Then, we have

$$\begin{aligned} V(x_0-p,C_T)&= \mathrm{cl \,}\bigcup _{V_T \in \mathcal {A}(x_0-p)} \left( U(V_T+C_T)-\mathbb {R}^q_+ \right) \\&=\mathrm{cl \,}\{U(V_T+C_T)-r |\;\forall n \ge 1: V_T \in \mathcal {A}(x_0-p^n), r \in \mathbb {R}^q_+\}\\&=\mathrm{cl \,}\bigcap _{n\ge 1}\bigcup _{V_T \in \mathcal {A}(x_0-p^n)}\left( U(V_T+C_T) - \mathbb {R}^q_+\right) . \end{aligned}$$

Let $y \in \bigcap _{n \ge 1}V(x_0-p^n,C_T)$. Since y is an element of the lower image $V(x_0-p^n,C_T)$, it is true that $\{y\} - \mathrm{int \,}\mathbb {R}^q_+ \subseteq \bigcup _{V_T \in \mathcal {A}(x_0-p^n)}\left( U(V_T+C_T) - \mathbb {R}^q_+\right)$ for all $n \ge 1$. Let $(y^k)_{k\ge 1} \in \{y\} - \mathrm{int \,}\mathbb {R}^q_+$ be a sequence with $\lim _{k \rightarrow \infty } y^k = y.$ Clearly, for each $k\ge 1$,

$y^k \in \bigcap _{n\ge 1}\bigcup _{V_T \in \mathcal {A}(x_0-p^n)}\left( U(V_T+C_T) - \mathbb {R}^q_+\right)$, hence

$$\begin{aligned} y \in \mathrm{cl \,}\bigcap _{n\ge 1}\bigcup _{V_T \in \mathcal {A}(x_0-p^n)}\left( U(V_T+C_T) - \mathbb {R}^q_+\right) . \end{aligned}$$

$\square$

Proof of Proposition 4.12

Assume for a proof by contradiction that $V(x_0,0)\subseteq \mathrm{int \,}V(x_0-p,C_T)$. Hence, there exists $\epsilon > 0$ such that $V(x_0-p,C_T)\supseteq V(x_0,0)+B(0,\epsilon )$. It is enough to show that there is ${\tilde{p}}>p$ such that

$$\begin{aligned} V(x_0-{\tilde{p}},C_T)+B(0,\epsilon )\supseteq V(x_0 - p,C_T) \end{aligned}$$

(34)

as this implies that $V(x_0-{\tilde{p}},C_T) \supseteq V(x_0,0)$, hence ${\tilde{p}} \in P^b(C_T)$, which contradicts that $p\in \mathrm{bd \,}P^b(C_T)$. To prove (34) note that as each $u\in \mathcal {U}$ is uniformly continuous, there exists $\delta > 0$ such that $\left\| x-y\right\| \le \delta$ implies that $\left| u(x)-u(y)\right| \le \frac{\epsilon }{\sqrt{q}}$. Let ${\tilde{\delta }} \in (0,\delta )$ and ${\tilde{p}}:= p+{\tilde{\delta }}\frac{e}{\left\| e\right\| }$, where $e\in \mathbb {R}^d$ is the vector of ones. Note that

$$\begin{aligned} V(x_0-{\tilde{p}},C_T) +B(0,\epsilon )&= \bigcup _{V_T\in \mathcal {A}(x_0-{\tilde{p}})}U(V_T+C_T)+B(0,\epsilon )-\mathbb {R}^q_+ \\&= \bigcup _{V_T\in \mathcal {A}(x_0-{p})}U(V_T+C_T-{\tilde{\delta }} \frac{e}{\left\| e\right\| }) + B(0,\epsilon )-\mathbb {R}^q_+, \end{aligned}$$

as $\mathcal {A}(x_0-{\tilde{p}}) = \mathcal {A}(x_0-p)-{\tilde{\delta }}\frac{e}{\left\| e\right\| }$. Let $U(V_T+C_T)-r \in V(x_0-p,C_T)$ for some $V_T \in \mathcal {A}(x_0-p)$ and $r\in \mathbb {R}^q_+$. Clearly, for each $u\in \mathcal {U}$ and $Q\in \mathcal {Q}$, we have

$$\begin{aligned} \biggl |{\mathbb {E}}_{Q}[u(V_T+C_T)]-{\mathbb {E}}_{Q}\left[ u\left( V_T+C_T-{\tilde{\delta }}\frac{e}{\left\| e\right\| }\right) \right] \biggr |\le \frac{\epsilon }{\sqrt{q}}. \end{aligned}$$

Then

$$\begin{aligned}&\left\| U(V_T+C_T)-U(V_T+C_T-{\tilde{\delta }}\frac{e}{\left\| e\right\| })\right\| \\&\quad = \bigg (\sum _{j=1}^s\sum _{i=1}^r \biggl |{\mathbb {E}}_{Q_j}(u^i(V_T+C_T)-u^i(V_T+C_T-{\tilde{\delta }}\frac{e}{\left\| e\right\| }))\biggr |^2\bigg )^{\frac{1}{2}}\le \epsilon . \end{aligned}$$

Hence, $U(V_T+C_T) -r \in U(V_T+C_T-{\tilde{\delta }}\frac{e}{\left\| e\right\| })+B(0,\epsilon )-\mathbb {R}^q_+$ and (34) holds. $\square$