Preferences with changing ambiguity aversion

Abstract

We provide two extensions of Gilboa and Schmeidler (J Math Econ 18:141–153, 1989)’s maxmin expected utility decision rule to accommodate a decision maker’s changing ambiguity attitudes. The two rules are, respectively, a weighted maxmin rule and a variant constraint rule. The former evaluates an act by a weighted average of its worst and best possible expected utilities over a set of priors, with the weights depending on the act. The latter evaluates an act by its worst expected utility over a neighborhood of a set of approximating priors, with the size of the neighborhood depending on the act. Canonical representations of the two rules are provided for classes of preference relations that exhibit, respectively, ambiguity aversion à la Schmeidler (Econometrica 57:571–587, 1989) and ambiguity aversion à la Ghirardato and Marinacci (J Econ Theory 102:251–289, 2002). In the second part of this paper, we study wealth effect under ambiguity. We propose axioms on absolute and relative ambiguity aversion and derive three representations for the ambiguity averse preference relations exhibiting decreasing (increasing) absolute ambiguity aversion. In particular, decreasing absolute ambiguity aversion implies that as the baseline utility of an act increases, a weighted maxmin decision maker puts less weight on the worst case, and a variant constraint decision maker considers a smaller neighborhood of approximating priors.

Appendix: Proofs

We denote by ${{\mathbb {N}}}$ the set of positive integers. For each $\varphi \in {{\mathbb {R}}}^S$, let $\varphi ^*:=\max \limits _{s\in S}\varphi (s)$ and $\varphi _*:=\min \limits _{s\in S}\varphi (s)$. A functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ is normalized if for each $t\in {{\mathbb {R}}}$, $I(t\mathbf 1 )=t$. It is constant additive if for each $\varphi \in {{\mathbb {R}}}^S$ and each $t\in {{\mathbb {R}}}$, $I(\varphi +t\mathbf 1 )=I(\varphi )+t$. It is constant superadditive if for each $\varphi \in {{\mathbb {R}}}^S$ and each $t\in {{\mathbb {R}}}_+$, $I(\varphi +t\mathbf 1 )\ge I(\varphi )+t$. Lastly, it is superadditive if for each pair $\varphi ,\varphi '\in {{\mathbb {R}}}^S$, $I(\varphi +\varphi ')\ge I(\varphi )+I(\varphi ')$.

1.1 Proofs in Section 3

Lemma 1

A preference relation ${\ \succsim \ }$ satisfies A.1–A.4 and A.6 if and only if there exist an affine onto function $u:X\rightarrow {{\mathbb {R}}}$ and a normalized, non-decreasing, and continuous functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ such that for each pair $f,g\in {\mathcal {F}}$,

$$\begin{aligned} f{\ \succsim \ }g\ \Longleftrightarrow \ I(u(f))\ge I(u(g)). \end{aligned}$$

(22)

Moreover, u is unique up to a positive affine transformation, and given u, there is a unique normalized functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ satisfying (22).

Proof

To prove the sufficiency of the axioms, let ${\ \succsim \ }$ satisfy A.1–A.4 and A.6. Note that A.6 implies the usual non-degeneracy axiom: There are $f,g\in {\mathcal {F}}$ such that $f\succ g$. Then by applying the same techniques of Lemma 57 of Cerreia-Vioglio et al. (2011b), one can show that there exist a non-constant affine function $u:X\rightarrow {{\mathbb {R}}}$ and a normalized, non-decreasing, and continuous functional $I:u(X)^S\rightarrow {{\mathbb {R}}}$ satisfying (22). Although A.5.1 is assumed in their Lemma 57, the axiom is only used to show that I is quasi-concave, and dropping it does not affect the existence of u and I satisfying the other properties. Moreover, since ${\ \succsim \ }$ satisfies A.6, by Lemma 29 of Maccheroni et al. (2006), $u(X)={{\mathbb {R}}}$.

The necessity of the axioms can be readily seen, and the uniqueness follows from routine arguments. Thus, we omit the proofs. $\square$

Proof of Proposition 1

To prove the sufficiency of the axioms, let ${\ \succsim \ }$ satisfy A.1–A.4 and A.6. By Lemma 1, there exist an affine onto function $u:X\rightarrow {{\mathbb {R}}}$ and a normalized, non-decreasing, and continuous functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ satisfying (22). Then by Propositions 1 and 5 of Cerreia-Vioglio et al. (2011a) and their proof of Proposition 5,²³ there exist a non-empty, closed, and convex set $D\subseteq \Delta$ and a function $\lambda :{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ such that (i) $\langle u,D\rangle$ is a Bewley representation of the unambiguous preference relation ${\ \succsim \ }^*$ induced from ${\ \succsim \ }$, and (ii) for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} I(\varphi )=\lambda (\varphi )\min \limits _{p\in D}E_p\varphi +(1-\lambda (\varphi ))\max \limits _{p\in D}E_p\varphi . \end{aligned}$$

(23)

In particular, for each $\varphi \in {{\mathbb {R}}}^S$ such that $\min \limits _{p\in D}E_p\varphi \ne \max \limits _{p\in D}E_p\varphi$,

$$\begin{aligned} \lambda (\varphi )=\dfrac{\max \limits _{p\in D}E_p\varphi -I(\varphi )}{\max \limits _{p\in D}E_p\varphi -\min \limits _{p\in D}E_p\varphi }. \end{aligned}$$

(24)

Since I is continuous, $\lambda$ is continuous on $\{\varphi \in {{\mathbb {R}}}^S:\min \limits _{p\in D}E_p\varphi \ne \max \limits _{p\in D}E_p\varphi \}.$ Thus, it can be readily seen that $\langle u,D,\lambda \rangle$ is a Bewley WM representation of ${\ \succsim \ }$.

To prove the necessity of the axioms, let ${\ \succsim \ }$ admit a Bewley WM representation $\langle u,D,\lambda \rangle$. Clearly, ${\ \succsim \ }$ satisfies A.1. Since u is affine, it satisfies A.2. Since u is onto, by Lemma 29 of Maccheroni et al. (2006), it satisfies A.6.

To show that ${\ \succsim \ }$ satisfies A.3, define $J:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} J(\varphi ):=\lambda (\varphi )\min \limits _{p\in D}E_p\varphi +(1-\lambda (\varphi ))\max \limits _{p\in D}E_p\varphi , \end{aligned}$$

and it suffices to show that J is continuous. Let $\varphi \in {{\mathbb {R}}}^S$ and $\{\varphi ^n\}_{n=1}^{\infty }$ be a sequence of elements in ${{\mathbb {R}}}^S$ such that $\lim \nolimits _{n\rightarrow \infty }\varphi ^n=\varphi$. If $\min \limits _{p\in D}E_p\varphi <\max \limits _{p\in D}E_p\varphi$, then $\lambda$ is continuous at $\varphi$, and thus, $\lim \nolimits _{n\rightarrow \infty }J(\varphi ^n)=J(\varphi )$. If $\min \limits _{p\in D}E_p\varphi =\max \limits _{p\in D}E_p\varphi$, then $\lim \nolimits _{n\rightarrow \infty }\min \limits _{p\in D}E_p\varphi ^n=\lim \nolimits _{n\rightarrow \infty }\max \limits _{p\in D}E_p\varphi ^n=J(\varphi )$, and thus, $\lim \nolimits _{n\rightarrow \infty }J(\varphi ^n)=J(\varphi )$.

To show that ${\ \succsim \ }$ satisfies A.4, let $f,g\in {\mathcal {F}}$ be such that for each $s\in S$, $f(s){\ \succsim \ }g(s)$, and we want to show that $f{\ \succsim \ }g$. Since for each $s\in S$, $f(s){\ \succsim \ }g(s)$, and since $\langle u,D,\lambda \rangle$ is a Bewley WM representation of ${\ \succsim \ }$, $u(f)\ge u(g)$. Then for each $p\in D$, $E_pu(f)\ge E_pu(g)$. Besides, by the definition of a Bewley WM representation, $\langle u,D\rangle$ is a Bewley representation of the unambiguous preference relation ${\ \succsim \ }^*$ induced from ${\ \succsim \ }$. Hence, $f{\ \succsim \ }^*g$. Thus, by the definition of ${\ \succsim \ }^*$, $f{\ \succsim \ }g$.

Lastly, to prove the uniqueness of the representation,²⁴ let $\langle u,D,\lambda \rangle$ and $\langle u',D',\lambda '\rangle$ be two Bewley WM representations of some ${\ \succsim \ }$. Since both u and $u'$ are affine functions representing ${\ \succsim \ }$ restricted to X, by routine arguments, $u'$ is a positive affine transformation of u. Moreover, by the definition of a Bewley WM representation, both $\langle u,D\rangle$ and $\langle u',D'\rangle$ are Bewley representations of the unambiguous preference relation ${\ \succsim \ }^*$ induced from ${\ \succsim \ }$. By Proposition 5 of Ghirardato et al. (2004), $D=D'$. Finally, suppose that $u=u'$, let $\varphi \in {{\mathbb {R}}}^S$ be such that $\min \limits _{p\in D}E_p\varphi \ne \max \limits _{p\in D}E_p\varphi$, and we want to show that $\lambda (\varphi )=\lambda '(\varphi )$. Let $f\in {\mathcal {F}}$ be such that $u(f)=\varphi$. Since $x_f\sim f$ and both $\langle u,D,\lambda \rangle$ and $\langle u',D',\lambda '\rangle$ are WM representations of ${\ \succsim \ }$,

$$\begin{aligned} u(x_f)=\lambda (u(f))\min \limits _{p\in D}E_pu(f)+(1-\lambda (u(f)))\max \limits _{p\in D}E_pu(f), \end{aligned}$$

(25)

and

$$\begin{aligned} u'(x_f)=\lambda (u'(f))\min \limits _{p\in D'}E_pu'(f)+(1-\lambda (u'(f)))\max \limits _{p\in D'}E_pu'(f). \end{aligned}$$

(26)

Since $u=u'$, $u(f)=\varphi$, and $D=D'$, by (25) and (26),

$$\begin{aligned} \lambda (\varphi )\min \limits _{p\in D}E_p\varphi +(1-\lambda (\varphi ))\max \limits _{p\in D}E_p\varphi =\lambda '(\varphi )\min \limits _{p\in D}E_p\varphi +(1-\lambda '(\varphi ))\max \limits _{p\in D}E_p\varphi . \end{aligned}$$

(27)

Since $\max \limits _{p\in D}E_p\varphi \ne \min \limits _{p\in D}E_p\varphi$, by (27), $\lambda (\varphi )=\lambda '(\varphi )$. $\square$

Proof of Theorem 1

To show the sufficiency of the axioms, let ${\ \succsim \ }$ satisfy A.1–A.4, A.5.1, and A.6. By the proof of Proposition 1, it admits a Bewley WM representation $\langle u,D,\lambda \rangle$, with $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ given by (23) being non-decreasing and continuous. We first check that $\lambda \in \Lambda (D)$. Let $\varphi ,\varphi '\in {{\mathbb {R}}}^S$. If $\varphi '\ge \varphi$, then by the monotonicity of I, $I(\varphi ')\ge I(\varphi )$, which implies (3). To check (4), suppose that $\varphi ,\varphi '$ satisfy (3), and let $\varphi '':=\frac{\varphi +\varphi '}{2}$. Since $\varphi ,\varphi '$ satisfy (3), $I(\varphi ')\ge I(\varphi )$. Thus, by the monotonicity and continuity of I, there is $t\in {{\mathbb {R}}}_+$ such that $I(\varphi '-t\mathbf 1 )=I(\varphi )$. Let $f,g\in {\mathcal {F}}$ be such that $u(f)=\varphi$ and $u(g)=\varphi '-t\mathbf 1$. Then $I(u(f))=I(u(g))$, so $f\sim g$. Since $f\sim g$, by A.5.1, $\frac{1}{2}f+\frac{1}{2}g{\ \succsim \ }f$. Thus, $I(\frac{1}{2}\varphi +\frac{1}{2}(\varphi '-t\mathbf 1 ))=I(u(\frac{1}{2}f+\frac{1}{2}g))\ge I(u(f))=I(\varphi )$. Since I is non-decreasing, $I(\frac{1}{2}\varphi +\frac{1}{2}\varphi ')\ge I(\frac{1}{2}\varphi +\frac{1}{2}(\varphi '-t\mathbf 1 ))$. Thus, $I(\varphi '')\ge I(\varphi )$, which implies (4).

We then prove that $\langle u,D,\lambda \rangle$ is also a canonical WM representation of ${\ \succsim \ }$. That is, let $\langle u',D',\lambda '\rangle$ be another WM representation of ${\ \succsim \ }$ and we shall prove that $D\subseteq D'$. Since ${\ \succsim \ }$ satisfies A.1–A.6, by Theorems 3 and 5 of Cerreia-Vioglio et al. (2011b), it admits an ambiguity averse representation $\langle v,G\rangle$, where $G:{{\mathbb {R}}}\times \Delta \rightarrow (-\infty ,\infty ]$ is given by, for each $(t,p)\in {{\mathbb {R}}}\times \Delta$,

$$\begin{aligned} G(t,p)=\sup \{v(x_f):f\in {\mathcal {F}},E_pv(f)\le t\}. \end{aligned}$$

(28)

Recall the set $D^*$ defined in (5). By Proposition 9 and Theorem 10 of Cerreia-Vioglio et al. (2011b), $\langle v,D^*\rangle$ is a Bewley representation of the unambiguous preference relation ${\ \succsim \ }^*$ induced from ${\ \succsim \ }$. On the other hand, by the definition of a Bewley WM representation, $\langle u,D\rangle$ is also a Bewley representation of ${\ \succsim \ }^*$. Thus, by Proposition 5 of Ghirardato et al. (2004), $D=D^*$. Hence, to show that $D\subseteq D'$, it is equivalent to show that $D^*\subseteq D'$.

Suppose to the contrary that $D^*\nsubseteq D'$. Since $D^*=cl(\{p\in \Delta :G(t,p)<\infty \text { for some }t\in {{\mathbb {R}}}\})$ and $D'$ is closed, there exist $q\in \Delta {\setminus } D'$ and $t\in {{\mathbb {R}}}$ such that $G(t,q)<\infty$. Then by a standard separation theorem, there is $\varphi \in {{\mathbb {R}}}^S$ such that $E_q\varphi<0<\min \limits _{p\in D'}E_p\varphi$. Let $x\in X$ be such that $v(x)=t$. Since $E_qv(x)=t$, by (28), $G(t,q)\ge v(x)=t$. Let $n\in {{\mathbb {N}}}$ be such that $E_qn\varphi<t\le G(t,q)<\min \limits _{p\in D'}E_pn\varphi$. Let $g\in {\mathcal {F}}$ be such that $v(g)=n\varphi$. Since $E_qv(g)=E_qn\varphi <t$, by (28), $v(x_g)\le G(t,q)$. Observe that $\min \limits _{p\in D'}E_pu'(g)\le \lambda '(u'(g))\min \limits _{p\in D'}E_pu'(g)+(1-\lambda '(u'(g)))\max \limits _{p\in D'}E_pu'(g)=u'(x_g)$. Since both v and $u'$ are affine functions representing ${\ \succsim \ }$ restricted to X, by routine arguments, v is a positive affine transformation of $u'$. Then $\min \limits _{p\in D'}E_pv(g)\le v(x_g)$. Recall that $G(t,q)<\min \limits _{p\in D'}E_pn\varphi =\min \limits _{p\in D'}E_pv(g)$. Thus, $G(t,q)<v(x_g)$, which contradicts $v(x_g)\le G(t,q)$, as desired.

To show the necessity of the axioms, let ${\ \succsim \ }$ admit a canonical WM representation $\langle u,D,\lambda \rangle$ with $\lambda \in \Lambda (D)$. Clearly, it satisfies A.1. Since $\lambda \in \Lambda (D)$, one can readily verify that it satisfies A.4 and A.5.1. Recall that Proposition 1 shows the necessity of A.2, A.3, and A.6 for a preference relation to admit a Bewley WM representation. Since the arguments there rely on the same properties of u, D, and $\lambda$ as we have here, they can be used to show the necessity of the axioms for a preference relation to admit a canonical WM representation.

Lastly, we show the uniqueness of a canonical WM representation $\langle u,D,\lambda \rangle$ of some preference relation ${\ \succsim \ }$. By the definition of a canonical WM representation, D is the smallest admissible set and thus unique. As argued in the proof of the sufficiency of the axioms, D coincides with the Bewley set. Moreover, the uniqueness of u and $\lambda$ follows from the same arguments as used for a Bewley WM representation in the proof of Proposition 1. $\square$

Proof of Corollary 1

Let $\succsim$ admit an ambiguity averse representation $\langle u,G\rangle$. Define $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi ):=\min \limits _{p\in \Delta }G(E_p\varphi ,p)$. By the proof of Theorem 3 of Cerreia-Vioglio et al. (2011b), I is normalized, non-decreasing, and continuous. Moreover, by their Theorem 3, $\succsim$ satisfies A.5.1. Thus, by the proofs of Proposition 1 and Theorem 1, $\succsim$ admits a canonical WM representation $\langle u,D^*,\lambda \rangle$ in which $\lambda$ belongs to $\Lambda (D^*)$ and is given by (24) on $\{\varphi \in {{\mathbb {R}}}^S:\min \limits _{p\in D^*}E_p\varphi \ne \max \limits _{p\in D^*}E_p\varphi \}$. $\square$

Proposition 8

The preference relation $\succsim$ in Example 2 admits a Bewley weighted maxmin representation but not a canonical weighted maxmin representation.

Proof

We first prove that $\succsim$ admits a Bewley WM representation. By Proposition 1, it is equivalent to prove that $\succsim$ satisfies A.1–A.4 and A.6. Clearly, it satisfies A.1. Since for each $t\in {{\mathbb {R}}}$, $V(t\mathbf 1 )=t$, it satisfies A.2 and A.6.

To show that ${\ \succsim \ }$ satisfies A.3, it suffices to show the continuity of V. Let $f\in {{\mathbb {R}}}^S$ and $\{f^n\}_{n=1}^{\infty }$ be a sequence of elements in ${{\mathbb {R}}}^S$ that converges to f. We shall show that $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$ in each of the following two cases.

Case 1: $\max \{f(1),f(2)\}\ne f(3)$. $\max \{f(1),f(2)\}<f(3)$. Then $V(f)=\min \limits _{p\in D_1}E_pf$. Moreover, for sufficiently large n, $\max \{f^n(1),f^n(2)\}<f^n(3)$, so that $V(f^n)=\min \limits _{p\in D_1}E_pf^n$. Hence, $\lim \nolimits _{n\rightarrow \infty }V(f^n)=\lim \nolimits _{n\rightarrow \infty }\min \limits _{p\in D_1}E_pf^n=\min \limits _{p\in D_1}E_pf=V(f)$. Similarly, one can show that when $\max \{f(1),f(2)\}>f(3)$, $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$.

Case 2: $\max \{f(1),f(2)\}=f(3)$. Then $V(f)=\min \limits _{p\in D_2}E_pf$. Suppose that $f(1)<f(2)$. Thus, $f(1)<f(2)=f(3)$ and for sufficiently large n, $f^n(1)<f^n(2)$. Recall $p''=(\frac{2}{3},\frac{1}{6},\frac{1}{6})$. Then $p''\in D_2$. For each $p\in D_2$, since

$$\begin{aligned}&\frac{1}{6}\ge \left( p_1-\frac{1}{3}\right) ^2+\left( p_2-\frac{1}{3}\right) ^2+\left( p_3-\frac{1}{3}\right) ^2 \ge \left( p_1-\frac{1}{3}\right) ^2+2\left( \frac{1-p_1}{2}-\frac{1}{3}\right) ^2\nonumber \\&\quad =\frac{3}{2}\left( p_1-\frac{1}{3}\right) ^2, \end{aligned}$$

(29)

$p_1\le \frac{2}{3}=p''_1$. Since $f(1)<f(2)=f(3)$, and since $p''\in D_2$ and for each $p\in D_2$, $p_1\le p''_1$, $\min \limits _{p\in D_2}E_pf=E_{p''}f=E_{p'}f$. Then $V(f)=\min \limits _{p\in D_2}E_pf=E_{p'}f$, and thus,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }E_{p'}f^n=\lim \limits _{n\rightarrow \infty }\min \limits _{p\in D_2}E_pf^n=V(f). \end{aligned}$$

(30)

For sufficiently large n, since $f^n(1)<f^n(2)$, and since $p_1'>q_1'$ and $p_3'=q_3'$, $\min \limits _{p\in D_1}E_pf^n=E_{p'}f^n$, and thus,

$$\begin{aligned} \text {either }V(f^n)=\min \limits _{p\in D_1}E_pf^n=E_{p'}f^n \text { or }V(f^n)=\min \limits _{p\in D_2}E_pf^n. \end{aligned}$$

(31)

By (30) and (31), $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$. Similarly, one can show that when $f(1)>f(2)$, $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$. Suppose that $f(1)=f(2)$. Then $f(1)=f(2)=f(3)$, and thus, $\lim \nolimits _{n\rightarrow \infty }\min \limits _{p\in D_1}E_{p}f^n=\lim \nolimits _{n\rightarrow \infty }\min \limits _{p\in D_2}E_p f^n=\min \limits _{p\in D_2}E_pf=V(f)$. For each $n\in {{\mathbb {N}}}$, either $V(f^n)=\min \limits _{p\in D_1}E_{p}f^n$ or $V(f^n)=\min \limits _{p\in D_2}E_{p}f^n$. Thus, $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$.

Lastly, to show that ${\ \succsim \ }$ satisfies A.4, let $f,g\in {{\mathbb {R}}}^S$ be such that $f\ge g$. We shall show that $V(f)\ge V(g)$ in each of the following three cases.

Case 1: Either $\max \{f(1),f(2)\}<f(3)$ and $\max \{g(1),g(2)\}<g(3)$, or $\max \{f(1),f(2)\}\ge f(3)$ and $\max \{g(1),g(2)\}\ge g(3)$. Then either $V(f)=\min \limits _{p\in D_1}E_pf$ and $V(g)=\min \limits _{p\in D_1}E_pg$, or $V(f)=\min \limits _{p\in D_2}E_pf$ and $V(g)=\min \limits _{p\in D_2}E_pg$. In either scenario, since $f\ge g$, $V(f)\ge V(g)$.

Case 2: $\max \{f(1),f(2)\}< f(3)$ and $\max \{g(1),g(2)\}\ge g(3)$. Then $V(f)=\min \limits _{p\in D_1}E_pf$ and $V(g)=\min \limits _{p\in D_2}E_pg$. Let $f'\in {{\mathbb {R}}}^S$ be such that

$$\begin{aligned} f'(1)=f(1), \ \ \ \ f'(2)=f(2), \ \ \ \ f'(3)=\max \{f(1),f(2)\}. \end{aligned}$$

Since $\max \{f(1),f(2)\}<f(3)$, $f'(3)<f(3)$. Thus, $f'\le f$. Since $f\ge g$, $f'(3)=\max \{f(1),f(2)\}\ge \max \{g(1),g(2)\}\ge g(3)$. Thus, $f'\ge g$. Consider the sequence $\{\frac{1}{n}f+\frac{n-1}{n}f'\}_{n=1}^{\infty }$ of elements of ${{\mathbb {R}}}^S$. For each $n\in {{\mathbb {N}}}$, since $\max \{\frac{1}{n}f(1)+\frac{n-1}{n}f'(1),\frac{1}{n}f(2)+\frac{n-1}{n}f'(2)\}=\max \{f(1),f(2)\}<\frac{1}{n}f(3)+\frac{n-1}{n}f'(3)$, $V(\frac{1}{n}f+\frac{n-1}{n}f')=\min \limits _{p\in D_1}E_p(\frac{1}{n}f+\frac{n-1}{n}f')$. Since V is continuous (as shown when proving that A.3 holds), $V(f')=\lim \nolimits _{n\rightarrow \infty }V(\frac{1}{n}f+\frac{n-1}{n}f')=\lim \nolimits _{n\rightarrow \infty }\min \limits _{p\in D_1}E_p(\frac{1}{n}f+\frac{n-1}{n}f')=\min \limits _{p\in D_1}E_pf'$. Since $f'\le f$, $\min \limits _{p\in D_1}E_pf'\le \min \limits _{p\in D_1}E_pf$. Thus, $V(f')\le V(f)$. Since $\max \{f'(1),f'(2)\}=f'(3)$ and $f'\ge g$, $V(f')=\min \limits _{p\in D_2}E_pf'\ge \min \limits _{p\in D_2}E_pg=V(g)$. Hence, $V(f)\ge V(f')\ge V(g)$.

Case 3: $\max \{f(1),f(2)\}\ge f(3)$ and $\max \{g(1),g(2)\}<g(3)$. Then $V(f)=\min \limits _{p\in D_2}E_pf$ and $V(g)=\min \limits _{p\in D_1}E_pg$. Let $f'\in {{\mathbb {R}}}^S$ be such that

Since $\max \{f(1),f(2)\}\ge f(3)$, $f'(1)\le f(1)$ and $f'(2)\le f(2)$. Thus, $f'\le f$. Since no matter $f(1)\ge f(2)$ or $f(1)< f(2)$, $\max \{f'(1),f'(2)\}\ge f'(3)$, $V(f')=\min \limits _{p\in D_2}E_pf'$. Since $f'\le f$, $\min \limits _{p\in D_2}E_pf'\le \min \limits _{p\in D_2}E_pf$. Hence, $V(f')\le V(f)$. Let $g'\in {{\mathbb {R}}}^S$ be such that

Since $\max \{g(1),g(2)\}<g(3)$, $g'\ge g$ and $\max \{g'(1),g'(2)\}=g'(3)$. Then $V(g')=\min \limits _{p\in D_2}E_pg'$. Since $f\ge g$, $f'\ge g'$. Thus, $\min \limits _{p\in D_2}E_pf'\ge \min \limits _{p\in D_2}E_pg'$. Hence, $V(f')\ge V(g')$. Consider the sequence $\{\frac{1}{n}g+\frac{n-1}{n}g'\}_{n=1}^{\infty }$ of elements of ${{\mathbb {R}}}^S$. Since $\max \{g(1),g(2)\}<g(3)$, for each $n\in {{\mathbb {N}}}$, $\max \{\frac{1}{n}g(1)+\frac{n-1}{n}g'(1),\frac{1}{n}g(2)+\frac{n-1}{n}g'(2)\}<\frac{1}{n}g(3)+\frac{n-1}{n}g'(3)$, and thus, $V(\frac{1}{n}g+\frac{n-1}{n}g')=\min \limits _{p\in D_1}E_p(\frac{1}{n}g+\frac{n-1}{n}g')$. Since V is continuous, $V(g')=\lim \nolimits _{n\rightarrow \infty }V(\frac{1}{n}g+\frac{n-1}{n}g')=\lim \nolimits _{n\rightarrow \infty }\min \limits _{p\in D_1}E_p(\frac{1}{n}g+\frac{n-1}{n}g')=\min \limits _{p\in D_1}E_pg'$. Since $g'\ge g$, $\min \limits _{p\in D_1}E_pg'\ge \min \limits _{p\in D_1}E_pg$. Thus, $V(g')\ge V(g)$. Hence, $V(f)\ge V(f')\ge V(g')\ge V(g)$.

Now we prove that ${\ \succsim \ }$ does not admit a canonical WM representation. Suppose to the contrary that it admits a canonical WM representation $\langle u,D,\lambda \rangle$. Let $D_3:=\{p\in \Delta :p_3\ge \frac{1}{4}\}$.

We first show that for each $f\in {{\mathbb {R}}}^S$, $V(f)\in [\min \limits _{p\in D_2}E_pf,\max \limits _{p\in D_2}E_pf]$. Let $f\in {{\mathbb {R}}}^S$. Suppose that $f(1)\le f(2)<f(3)$. Then $V(f)=\min \limits _{p\in D_1}E_pf=E_{p'}f$. Since $f(1)\le f(2)<f(3)$, and since $p'_2<p^*_2$ and $p'_3<p^*_3$, $E_{p'}f\le E_{p^*}f$. Recall $p''=(\frac{2}{3},\frac{1}{6},\frac{1}{6})$. Since $f(1)\le f(2)<f(3)$, and since $p'_1=p''_1$ and $p'_3>p''_3$, $E_{p'}f\ge E_{p''}f$. Thus, $V(f)\in [E_{p''}f,E_{p^*}f]$. Since $p'',p^*\in D_2$, $[E_{p''}f,E_{p^*}f]\subseteq [\min \limits _{p\in D_2}E_pf,\max \limits _{p\in D_2}E_pf]$. Hence, $V(f)\in [\min \limits _{p\in D_2}E_pf,\max \limits _{p\in D_2}E_pf]$. If $f(2)<f(1)<f(3)$, by analogous arguments, $V(f)\in [\min \limits _{p\in D_2}E_pf,\max \limits _{p\in D_2}E_pf]$. Lastly, if $\max \{f(1),f(2)\}\ge f(3)$, then $V(f)=\min \limits _{p\in D_2}E_pf$, and thus, $V(f)\in [\min \limits _{p\in D_2}E_pf,\max \limits _{p\in D_2}E_pf]$.

We then show that for each $f\in {{\mathbb {R}}}^S$, $V(f)\in [\min \limits _{p\in D_3}E_pf,\max \limits _{p\in D_3}E_pf]$. Let $f\in {{\mathbb {R}}}^S$. Suppose that $\max \{f(1),f(2)\}<f(3)$. Then $V(f)=\min \limits _{p\in D_1}E_pf$. Since $D_1\subseteq D_3$, $\min \limits _{p\in D_1}E_pf\in [\min \limits _{p\in D_3}E_pf,\max \limits _{p\in D_3}E_pf]$. Thus, $V(f)\in [\min \limits _{p\in D_3}E_pf,\max \limits _{p\in D_3}E_pf]$. Suppose that $\max \{f(1),f(2)\}\ge f(3)$. Then $V(f)=\min \limits _{p\in D_2}E_pf$. Since $p^*\in D_2\cap D_3$, $\min \limits _{p\in D_2}E_pf\le E_{p^*}f\le \max \limits _{p\in D_3}E_pf$. Thus, $V(f)\le \max \limits _{p\in D_3}E_pf$. To show that $V(f)\ge \min \limits _{p\in D_3}E_pf$, we further consider the following three cases. First, suppose that $f(1)\le f(3)\le f(2)$. Recall that for each $p\in D_2$, by (29), $p_1\le \frac{2}{3}$. Since $f(1)\le f(3)\le f(2)$, and since $(\frac{3}{4},0,\frac{1}{4})\in D_3$ and for each $p\in D_2$, $p_1\le \frac{2}{3}<\frac{3}{4}$, $\min \limits _{p\in D_2}E_pf\ge \frac{3}{4}f(1)+\frac{1}{4}f(3)\ge \min \limits _{p\in D_3}E_pf$. Thus, $V(f)\ge \min \limits _{p\in D_3}E_pf$. Second, if $f(2)\le f(3)\le f(1)$, by analogous arguments, $V(f)\ge \min \limits _{p\in D_3}E_pf$. Lastly, if $f(3)<\min \{f(1),f(2)\}$, then $\min \limits _{p\in D_2}E_pf\ge f(3)=\min \limits _{p\in D_3}E_pf$, and thus, $V(f)\ge \min \limits _{p\in D_3}E_pf$.

We claim that $D\subseteq D_2\cap D_3$. To see this, define $\lambda _2:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $f\in {{\mathbb {R}}}^S$,

$$\begin{aligned} \lambda _2(f):=\left\{ \begin{array}{ll} \dfrac{\max \limits _{p\in D_2}E_pf-V(f)}{\max \limits _{p\in D_2}E_pf-\min \limits _{p\in D_2}E_pf} &{} \hbox { if } \min \limits _{p\in D_2}E_pf\ne \max \limits _{p\in D_2}E_pf, \\ 1 &{} \hbox {if }\min \limits _{p\in D_2}E_pf=\max \limits _{p\in D_2}E_pf. \end{array} \right. \end{aligned}$$

Thus, for each $f\in {{\mathbb {R}}}^S$, $V(f)=\lambda _2(f)\min \limits _{p\in D_2}E_pf+(1-\lambda _2(f))\max \limits _{p\in D_2}E_pf$, and since $V(f)\in [\min \limits _{p\in D_2}E_pf,\max \limits _{p\in D_2}E_pf]$, $\lambda _2(f)\in [0,1]$. Since V is continuous, $\lambda _2$ is continuous on $\{f\in {{\mathbb {R}}}^S:\min \limits _{p\in D_2}E_pf\ne \max \limits _{p\in D_2}E_pf\}$. Let v be the identity mapping on ${{\mathbb {R}}}$. It can be readily seen that $\langle v,D_2,\lambda _2\rangle$ is a weighed maxmin representation of ${\ \succsim \ }$. Analogously, since $V(f)\in [\min \limits _{p\in D_3}E_pf,\max \limits _{p\in D_3}E_pf]$, one can define $\lambda _3:{{\mathbb {R}}}^S\rightarrow [0,1]$ so that $\langle v,D_3,\lambda _3\rangle$ is also a WM representation of ${\ \succsim \ }$. Since $\langle u,D,\lambda \rangle$ is a canonical WM representation of ${\ \succsim \ }$, $D\subseteq D_2\cap D_3$.

Finally, to derive a contradiction, fix $g\in {{\mathbb {R}}}^S$ such that $g(1)<g(2)=g(3)$. Then $V(g)=\min \limits _{p\in D_2}E_pg$. Recall that for each $p\in D_2$, by (29), $p_1\le \frac{2}{3}$, and it can also be readily seen that $p_1=\frac{2}{3}$ if only if $p_2=p_3=\frac{1}{6}$. Thus, $V(g)=E_{p''}g$, and for each $p\in D_2{\setminus }\{p''\}$, $E_pg>V(g)$. Since $p''\notin D_3$ and $D\subseteq D_2\cap D_3$, $D\subseteq D_2{\setminus }\{p''\}$. Thus, $\min \limits _{p\in D}E_pg>V(g)=v(x_g)$. Since both v and u are affine functions representing ${\ \succsim \ }$ restricted to X, by routine arguments, u is a positive affine transformation of v. Then $\min \limits _{p\in D}E_pu(g)>u(x_g)$, and thus, $\lambda (u(g))\min \limits _{p\in D}E_pu(g)+(1-\lambda (u(g)))\max \limits _{p\in D}E_pu(g)>u(x_g)$. Since $\langle u,D,\lambda \rangle$ is a WM representation of ${\ \succsim \ }$, $g\succ x_g$, which is not possible. Hence, there is no canonical WM representation of ${\ \succsim \ }$.$\square$

Proposition 9

The preference relation $\succsim$ in Example 3 admits both a Bewley weighted maxmin representation and a canonical weighted maxmin representation, whereas the Bewley set of priors for ${\ \succsim \ }$ is not the smallest admissible set.

Proof

We first prove that $\succsim$ admits a Bewley WM representation. By Proposition 1, it is equivalent to prove that $\succsim$ satisfies A.1–A.4 and A.6. Clearly, it satisfies A.1. Since for each $t\in {{\mathbb {R}}}$, $V(t\mathbf 1 )=t$, it satisfies A.2 and A.6.

To show that ${\ \succsim \ }$ satisfies A.3, it suffices to show the continuity of V. Let $f\in {{\mathbb {R}}}^S$ and $\{f^n\}_{n=1}^{\infty }$ be a sequence of elements in ${{\mathbb {R}}}^S$ that converges to f. Before showing that $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$, we assume first that $\max \{f(1),f(2)\}<f(3)$ and for each $n\in {{\mathbb {N}}}$, $\max \{f^n(1),f^n(2)\}<f^n(3)$, and check that $\lim \nolimits _{n\rightarrow \infty }\alpha (f^n)=\alpha (f)$. Suppose that $f(1)<f(2)<f(3)$. Then $\alpha (f)=\text {med}\{0,\frac{1-E_{p''}f}{1-E_{p''}f+E_{p'}f},1\}$. Moreover, for sufficiently large n, $f^n(1)<f^n(2)<f^n(3)$, so that $\alpha (f^n)=\text {med}\{0,\frac{1-E_{p''}f^n}{1-E_{p''}f^n+E_{p'}f^n},1\}$. Since the median operator is continuous, $\lim \nolimits _{n\rightarrow \infty }\alpha (f^n)=\alpha (f)$. Suppose that $f(2)<f(1)<f(3)$. By analogous arguments, $\lim \nolimits _{n\rightarrow \infty }\alpha (f^n)=\alpha (f)$. Suppose that $f(1)=f(2)<f(3)$. Since $f(1)=f(2)<f(3)$, and since $p'_3=q'_3$ and $p''_3=q''_3$, $\frac{1-E_{p''}f}{1-E_{p''}f+E_{p'}f}=\frac{1-E_{q''}f}{1-E_{q''}f+E_{q'}f}$. Then $\alpha (f)=\lim \nolimits _{n\rightarrow \infty }\text {med}\{0,\frac{1-E_{p''}f^n}{1-E_{p''}f^n+E_{p'}f^n},1\}=\lim \nolimits _{n\rightarrow \infty }\text {med}\{0,\frac{1-E_{q''}f^n}{1-E_{q''}f^n+E_{q'}f^n},1\}$. For each $n\in {{\mathbb {N}}}$, either $\alpha (f^n)=\text {med}\{0,\frac{1-E_{p''}f^n}{1-E_{p''}f^n+E_{p'}f^n},1\}$ or $\alpha (f^n)=\text {med}\{0,\frac{1-E_{q''}f^n}{1-E_{q''}f^n+E_{q'}f^n},1\}$. Thus, $\lim \nolimits _{n\rightarrow \infty }\alpha (f^n)=\alpha (f)$. We now show that $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$ in each of the following two cases.

Case 1: $\max \{f(1),f(2)\}\ne f(3)$. Suppose that $\max \{f(1),f(2)\}<f(3)$. Then $V(f)=\min \limits _{p\in D_{1}(f)}E_{p}f$. Moreover, for sufficiently large n, $\max \{f^n(1),f^n(2)\}<f^n(3)$, so that $V(f^n)=\min \limits _{p\in D_{1}(f^n)}E_{p}f^n$. Since $\alpha$ is continuous at f, $D_1$ is continuous (i.e., both upper and lower hemicontinuous) at f. Thus, by the maximum theorem, $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$. Suppose that $\max \{f(1),f(2)\}>f(3)$. Then $V(f)=\min \limits _{p\in D_{2}}E_{p}f$. Moreover, for sufficiently large n, $\max \{f^n(1),f^n(2)\}>f^n(3)$, so that $V(f^n)=\min \limits _{p\in D_{2}}E_{p}f^n$. Thus, $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$.

Case 2: $\max \{f(1),f(2)\}=f(3)$. Then $V(f)=\min \limits _{p\in D_2}E_pf$. Suppose that $f(1)<f(2)$. Thus, $f(1)<f(2)=f(3)$ and for sufficiently large n, $f^n(1)<f^n(2)$. Recall that as shown in the proof of Proposition 8, for each $p\in D_2$, by (29), $p_1\le \frac{2}{3}=p''_1$, and thus, $\min \limits _{p\in D_2}E_pf=E_{p''}f=E_{p'}f$. Then

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }E_{p'}f^n=\lim \limits _{n\rightarrow \infty }E_{p''}f^n=\lim \limits _{n\rightarrow \infty }\min \limits _{p\in D_2}E_pf^n=V(f). \end{aligned}$$

(32)

For sufficiently large n, since $f^n(1)<f^n(2)$, and since $p_1'>q_1'$, $p_1''>q_1''$, $p_3'=q_3'$, and $p_3''=q_3''$, for each $\beta \in [0,1]$, $E_{\beta p'+(1-\beta )p''}f^n<E_{\beta q'+(1-\beta )q''}f^n$, and thus,

$$\begin{aligned}&\text {either }V(f^n)=\min \limits _{p\in D_1(f^n)}E_pf^n=\alpha (f^n)E_{p'}f^n+(1-\alpha (f^n))E_{p''}f^n\nonumber \\&\quad \text { or }V(f^n)=\min \limits _{p\in D_2}E_pf^n. \end{aligned}$$

(33)

By (32) and (33), $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$. Similarly, one can show that when $f(1)>f(2)$, $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$. Suppose that $f(1)=f(2)$. Then $f(1)=f(2)=f(3)$, and thus, for each $p\in \Delta$, $\lim \nolimits _{n\rightarrow \infty }E_{p}f^n=\lim \nolimits _{n\rightarrow \infty }\min \limits _{p\in D_2}E_pf^n=\min \limits _{p\in D_2}E_pf=V(f)$. For each $n\in {{\mathbb {N}}}$, either $V(f^n)=E_{\alpha (f^n)p'+(1-\alpha (f^n))p''}f^n$, or $V(f^n)=E_{\alpha (f^n)q'+(1-\alpha (f^n))q''}f^n$, or $V(f^n)=\min \limits _{p\in D_2}E_{p}f^n$. Thus, $\lim \nolimits _{n\rightarrow \infty }V(f^n)=V(f)$.

Lastly, we show that ${\ \succsim \ }$ satisfies A.4. To facilitate our proof, we first derive an equivalent expression of V. Let $f\in {{\mathbb {R}}}^S$. Suppose that $f(1)\le f(2)<f(3)$. Since $f(1)\le f(2)<f(3)$, and since $p'_1=p''_1$ and $p_3'>p''_3$, $E_{p'}f>E_{p''}f$. Thus,

Moreover, since $f(1)\le f(2)<f(3)$, and since $p_1'>q_1'$, $p_1''>q_1''$, $p_3'=q_3'$, and $p_3''=q_3''$, for each $\beta \in [0,1]$, $E_{\beta p'+(1-\beta )p''}f<E_{\beta q'+(1-\beta )q''}f$. Thus,

$$\begin{aligned} V(f)=\min \limits _{p\in D_1(f)}E_pf=E_{\alpha (f)p'+(1-\alpha (f))p''}f=\alpha (f)E_{p'}f+(1-\alpha (f))E_{p''}f. \end{aligned}$$

(34)

Similarly, if $f(2)<f(1)<f(3)$, then

and $V(f)=\alpha (f)E_{q'}f+(1-\alpha (f))E_{q''}f$.

Hence, for each $f\in {{\mathbb {R}}}^S$,

Now let $f,g\in {{\mathbb {R}}}^S$ be such that $f\ge g$. We shall show that $V(f)\ge V(g)$ in each of the following four cases.

Case 1: Either $f(1)\le f(2)<f(3)$ and $g(1)\le g(2)<g(3)$, or $f(2)<f(1)<f(3)$ and $g(2)<g(1)<g(3)$. Consider the former scenario first. Suppose that $E_{p'}f\le 0$. Then $V(f)=E_{p'}f$ and $E_{p'}g\le E_{p'}f\le 0$. Thus, $V(g)=E_{p'}g\le V(f)$. Suppose that $E_{p'}f>0$ and $E_{p''}f<1$. Then $V(f)=\frac{E_{p'}f}{1-E_{p''}f+E_{p'}f}>0$ and $E_{p''}g\le E_{p''}f<1$. If $E_{p'}g\le 0$, then $V(g)=E_{p'}g\le 0<V(f)$. If $E_{p'}g>0$, then $V(g)=\frac{E_{p'}g}{1-E_{p''}g+E_{p'}g}\le \frac{E_{p'}f}{1-E_{p''}f+E_{p'}f}=V(f)$, where the inequality holds since $0<E_{p'}g\le E_{p'}f$ and $0<1-E_{p''}f\le 1-E_{p''}g$. Suppose that $E_{p''}f\ge 1$. Then $V(f)=E_{p''}f\ge 1$. If $E_{p'}g\le 0$, then $V(g)=E_{p'}g\le 0<1\le V(f)$. If $E_{p'}g>0$ and $E_{p''}g<1$, then $V(g)=\frac{E_{p'}g}{1-E_{p''}g+E_{p'}g}<1\le V(f)$. If $E_{p''}g\ge 1$, then $V(g)=E_{p''}g\le E_{p''}f=V(f)$. In the latter scenario, by analogous arguments, $V(f)\ge V(g)$.

Case 2: $\max \{f(1),f(2)\}\ge f(3)$ and $\max \{g(1),g(2)\}\ge g(3)$. Then $V(f)=\min \limits _{p\in D_2}E_pf\ge \min \limits _{p\in D_2}E_pg=V(g)$.

Case 3: Either $f(1)\le f(2)<f(3)$ and $g(2)<g(1)<g(3)$, or $f(2)<f(1)<f(3)$ and $g(1)\le g(2)<g(3)$. Consider the former scenario first. Let $f'\in {{\mathbb {R}}}^S$ be such that

$$\begin{aligned} f'(1)=f(1), \ \ \ \ f'(2)=f(1), \ \ \ \ f'(3)=f(3). \end{aligned}$$

Then $f\ge f'\ge g$. Since $f\ge f'$, and since $f(1)\le f(2)<f(3)$ and $f'(1)=f'(2)<f'(3)$, by the result of Case 1, $V(f)\ge V(f')$. Consider the sequence $\{\frac{n-1}{n}f'+\frac{1}{n}g\}_{n=1}^{\infty }$ of elements in ${{\mathbb {R}}}^S$. For each $n\in {{\mathbb {N}}}$, since $f'\ge g$, $\frac{n-1}{n}f'+\frac{1}{n}g\ge g$; since $f'(2)=f'(1)<f'(3)$ and $g(2)<g(1)<g(3)$, $\frac{n-1}{n}f'(2)+\frac{1}{n}g(2)<\frac{n-1}{n}f'(1)+\frac{1}{n}g(1)<\frac{n-1}{n}f'(3)+\frac{1}{n}g(3)$. For each $n\in {{\mathbb {N}}}$, since $\frac{n-1}{n}f'+\frac{1}{n}g\ge g$, and since $\frac{n-1}{n}f'(2)+\frac{1}{n}g(2)<\frac{n-1}{n}f'(1)+\frac{1}{n}g(1)<\frac{n-1}{n}f'(3)+\frac{1}{n}g(3)$ and $g(2)<g(1)<g(3)$, by the result of Case 1, $V(\frac{n-1}{n}f'+\frac{1}{n}g)\ge V(g)$. Since V is continuous (as shown when proving that A.3 holds), $V(f')=\lim \nolimits _{n\rightarrow \infty }V(\frac{n-1}{n}f'+\frac{1}{n}g)$. Thus, $V(f')\ge V(g)$. Hence, $V(f)\ge V(f')\ge V(g)$. In the latter scenario, by analogous arguments, $V(f)\ge V(g)$.

Case 4: Either $\max \{f(1),f(2)\}<f(3)$ and $\max \{g(1),g(2)\}\ge g(3)$, or $\max \{f(1),f(2)\}\ge f(3)$ and $\max \{g(1),g(2)\}<g(3)$. Consider the former scenario first. Let $f'\in {{\mathbb {R}}}^S$ be such that

$$\begin{aligned} f'(1)=f(1), \ \ \ \ f'(2)=f(2), \ \ \ \ f'(3)=\max \{f(1),f(2)\}. \end{aligned}$$

Then $f\ge f'\ge g$ and $\max \{f'(1),f'(2)\}=f'(3)$. Since $f'\ge g$, and since $\max \{f'(1),f'(2)\}=f'(3)$ and $\max \{g(1),g(2)\}\ge g(3)$, by the result of Case 2, $V(f')\ge V(g)$. Consider the sequence $\{\frac{n-1}{n}f'+\frac{1}{n}f\}_{n=1}^{\infty }$ of elements in ${{\mathbb {R}}}^S$. For each $n\in {{\mathbb {N}}}$, since $f\ge f'$, $f\ge \frac{n-1}{n}f'+\frac{1}{n}f$; since $f'(3)=\max \{f(1),f(2)\}<f(3)$, $\max \{\frac{n-1}{n}f'(1)+\frac{1}{n}f(1),\frac{n-1}{n}f'(2)+\frac{1}{n}f(2)\}=\max \{f(1),f(2)\}<\frac{n-1}{n}f'(3)+\frac{1}{n}f(3)$. For each $n\in {{\mathbb {N}}}$, since $f\ge \frac{n-1}{n}f'+\frac{1}{n}f$, and since $\max \{f(1),f(2)\}<f(3)$ and $\max \{\frac{n-1}{n}f'(1)+\frac{1}{n}f(1),\frac{n-1}{n}f'(2)+\frac{1}{n}f(2)\}<\frac{n-1}{n}f'(3)+\frac{1}{n}f(3)$, by the results of Cases 1 and 3, $V(f)\ge V(\frac{n-1}{n}f'+\frac{1}{n}f)$. Since V is continuous, $V(f')=\lim \nolimits _{n\rightarrow \infty }V(\frac{n-1}{n}f'+\frac{1}{n}f)$. Thus, $V(f)\ge V(f')$. Hence, $V(f)\ge V(f')\ge V(g)$. In the latter scenario, by similar arguments, $V(f)\ge V(g)$.

Now we prove that ${\ \succsim \ }$ admits a canonical WM representation with $D:=D_2\cap \{p\in \Delta :p_3\ge \frac{1}{6}\}$ being the smallest admissible set. Let $f\in {{\mathbb {R}}}^S$. We first show that $V(f)\in [\min \limits _{p\in D} E_pf,\max \limits _{p\in D}E_pf]$ in each of the following two cases.

Case 1: Either $f(1)\le f(2)<f(3)$ or $f(2)<f(1)<f(3)$. Consider the former scenario first. Then $E_{p''}f<E_{p'}f<E_{p^*}f$, and as shown in (34), $V(f)=\alpha (f)E_{p'}f+(1-\alpha (f))E_{p''}f$. Thus, $V(f)\in [E_{p''}f,E_{p^*}f]$. Since $V(f)\in [E_{p''}f,E_{p^*}f]$ and $p'',p^*\in D$, $V(f)\in [\min \limits _{p\in D} E_pf,\max \limits _{p\in D}E_pf]$. In the latter scenario, by analogous arguments, $V(f)\in [\min \limits _{p\in D} E_pf,\max \limits _{p\in D}E_pf]$.

Case 2: Either $f(1)\le f(2)$ and $f(3)\le f(2)$, or $f(1)>f(2)$ and $f(3)\le f(1)$. Consider the former scenario first. Then $V(f)=\min \limits _{p\in D_2}E_pf$. Since $p^*\in D_2$ and $p^*\in D$, $V(f)\le E_{p^*}f\le \max \limits _{p\in D}E_pf$. To show that $V(f)\ge \min \limits _{p\in D}E_pf$, suppose to the contrary that $V(f)<\min \limits _{p\in D}E_pf$. Then f is not a constant act. Let $\bar{p}\in \arg \min \limits _{p\in D_2} E_pf$. Then $E_{\bar{p}}f=V(f)<\min \limits _{p\in D}E_pf$. Thus, $\bar{p}\notin D$, i.e., $\bar{p}_3<\frac{1}{6}$. Recall that as argued in the proof of Proposition 8, for each $p\in D_2$, by (29), $p_1\le \frac{2}{3}$, and $p_1=\frac{2}{3}$ if and only if $p_2=p_3=\frac{1}{6}$. Since $\bar{p}\in D_2$ and $\bar{p}_3<\frac{1}{6}$, $\bar{p}_1<\frac{2}{3}$. Since $f(1),f(3)\le f(2)$ and f is not a constant act, and since $\bar{p}_1<\frac{2}{3}=p''_1$ and $\bar{p}_3<\frac{1}{6}=p''_3$, $E_{\bar{p}}f>E_{p''}f$. Since $p''\in D_2$, $E_{p''}f\ge \min \limits _{p\in D_2}E_pf$. Thus, $E_{\bar{p}}f>\min \limits _{p\in D_2}E_pf$, which contradicts $\bar{p}\in \arg \min \limits _{p\in D_2} E_pf$. In the latter scenario, by analogous arguments, $V(f)\in [\min \limits _{p\in D} E_pf,\max \limits _{p\in D}E_pf]$.

Next, we show that ${\ \succsim \ }$ admits a WM representation with D being an admissible set. Since for each $f\in {\mathbb {R}}^S$, $V(f)\in [\min \limits _{p\in D} E_pf,\max \limits _{p\in D}E_pf]$, there is $\lambda :{\mathbb {R}}^S\rightarrow [0,1]$ such that for each $f\in {\mathbb {R}}^S$, $V(f)=\lambda (f)\min \limits _{p\in D}E_pf+(1-\lambda (f))\max \limits _{p\in D}E_pf$. Since V is continuous, $\lambda$ is continuous on $\{f\in {{\mathbb {R}}}^S:\min \limits _{p\in D}E_pf\ne \max \limits _{p\in D}E_pf\}$. Let u be the identity mapping on ${{\mathbb {R}}}$. Then $\langle u,D,\lambda \rangle$ is a WM representation of $\succsim$.

We then show that D is actually the smallest admissible set, so that $\langle u,D,\lambda \rangle$ is a canonical WM representation of $\succsim$. Let $\langle u',D',\lambda '\rangle$ be another WM representation of $\succsim$. To show that $D\subseteq D'$, suppose to the contrary that there is $\bar{p}\in D{\setminus } D'$. By a standard separation theorem, there is a non-constant act $\bar{f}\in {\mathbb {R}}^S$ such that $E_{\bar{p}}\bar{f}<\min \limits _{p\in D'}E_p\bar{f}$. Let $t\in {\mathbb {R}}$ be such that $E_{p''}(\bar{f}+t\mathbf 1 )\ge 1$ and $E_{q''}(\bar{f}+t\mathbf 1 )\ge 1$. We claim that $V(\bar{f}+t\mathbf 1 )\le E_{\bar{p}}(\bar{f}+t\mathbf 1 )$. To see this, suppose first $\max \{\bar{f}(1)+t,\bar{f}(2)+t\}\ge \bar{f}(3)+t$. Then $V(\bar{f}+t\mathbf 1 )=\min \limits _{p\in D_2}E_p(\bar{f}+t\mathbf 1 )$. Since $\bar{p}\in D$ and $D\subseteq D_2$, $\min \limits _{p\in D_2}E_p(\bar{f}+t\mathbf 1 )\le E_{\bar{p}}(\bar{f}+t\mathbf 1 )$. Thus, $V(\bar{f}+t\mathbf 1 )\le E_{\bar{p}}(\bar{f}+t\mathbf 1 )$. Suppose next that $\bar{f}(1)+t\le \bar{f}(2)+t<\bar{f}(3)+t$. Since $\bar{f}(1)+t\le \bar{f}(2)+t<\bar{f}(3)+t$ and $E_{p''}(\bar{f}+t\mathbf 1 )\ge 1$, $V(\bar{f}+t\mathbf 1 )=E_{p''}(\bar{f}+t\mathbf 1 )$. Recall that as argued in the proof of Proposition 8, for each $p\in D_2$, $p_1\le \frac{2}{3}$. Since $\bar{p}\in D_2$, $\bar{p}_1\le \frac{2}{3}=p''_1$. Since $\bar{p}\in D$, $\bar{p}_3\ge \frac{1}{6}=p''_3$. Since $\bar{f}(1)+t\le \bar{f}(2)+t<\bar{f}(3)+t$, and since $\bar{p}_1\le p''_1$ and $\bar{p}_3\ge p''_3$, $E_{p''}(\bar{f}+t\mathbf 1 )\le E_{\bar{p}}(\bar{f}+t\mathbf 1 )$. Thus, $V(\bar{f}+t\mathbf 1 )\le E_{\bar{p}}(\bar{f}+t\mathbf 1 )$. Lastly, if $\bar{f}(2)+t<\bar{f}(1)+t<\bar{f}(3)+t$, by analogous arguments as in the case $\bar{f}(1)+t\le \bar{f}(2)+t<\bar{f}(3)+t$, $V(\bar{f}+t\mathbf 1 )\le E_{\bar{p}}(\bar{f}+t\mathbf 1 )$.

Since $V(\bar{f}+t\mathbf 1 )\le E_{\bar{p}}(\bar{f}+t\mathbf 1 )$ and $E_{\bar{p}}\bar{f}<\min \limits _{p\in D'}E_p\bar{f}$, $V(\bar{f}+t\mathbf 1 )<\min \limits _{p\in D'}E_p(\bar{f}+t\mathbf 1 )$. Recall that u is the identity mapping on ${{\mathbb {R}}}$. Thus, $u(x_{\bar{f}+t\mathbf 1 })<\min \limits _{p\in D'}E_p(\bar{f}+t\mathbf 1 )$. Since both u and $u'$ are affine functions representing $\succsim$ restricted to X, by routine arguments, $u'$ is a positive affine transformation of u. Then $u'(x_{\bar{f}+t\mathbf 1 })<\min \limits _{p\in D'}E_pu'(\bar{f}+t\mathbf 1 )\le \lambda '(u'(\bar{f}+t\mathbf 1 ))\min \limits _{p\in D'}E_pu'(\bar{f}+t\mathbf 1 )+(1-\lambda '(u'(\bar{f}+t\mathbf 1 )))\max \limits _{p\in D'}E_pu'(\bar{f}+t\mathbf 1 )$. Since $\langle u',D',\lambda '\rangle$ is a WM representation of $\succsim$, $\bar{f}+t\mathbf 1 \succ x_{\bar{f}+t\mathbf 1 }$, which is not possible. Hence, $D\subseteq D'$, as desired.

Finally, we show that D is not the Bewley set. Suppose to the contrary that D is the Bewley set. That is, there is a WM representation $\langle u'',D,\lambda ''\rangle$ of ${\ \succsim \ }$ such that $\langle u'',D\rangle$ is a Bewley representation of the unambiguous preference relation ${\ \succsim \ }^*$ induced from ${\ \succsim \ }$. Then $u''$ is a positive affine transformation of the identity mapping u. Observe that $p'\notin D_2$. Thus, $p'\notin D$. By a standard separation theorem, there is a non-constant act $f'\in {\mathbb {R}}^S$ such that $E_{p'}f'<\min \limits _{p\in D}E_pf'$. Let $x:=\min \limits _{p\in D}E_pf'$. Thus, $E_{p'}f'<x$ and $\min \limits _{p\in D}E_pu''(f')=u''(x)$. Let $g\in {\mathbb {R}}^S$ be such that $g(1)<g(2)<g(3)$, $f'(1)+g(1)<f'(2)+g(2)<f'(3)+g(3)$, and $E_{p'}(x\mathbf 1 +g)\le 0$. Since $E_{p'}f'<x$, $E_{p'}(\frac{1}{2}f'+\frac{1}{2}g)<E_{p'}(\frac{1}{2}x\mathbf 1 +\frac{1}{2}g)\le 0$. Then $V(\frac{1}{2}f'+\frac{1}{2}g)=E_{p'}(\frac{1}{2}f'+\frac{1}{2}g)<E_{p'}(\frac{1}{2}x\mathbf 1 +\frac{1}{2}g)=V(\frac{1}{2}x+\frac{1}{2}g)$. Thus, $\frac{1}{2}x+\frac{1}{2}g\succ \frac{1}{2}f'+\frac{1}{2}g$. Since $\langle u'',D\rangle$ is a Bewley representation of ${\ \succsim \ }^*$, and since for each $p\in D$, $E_p u''(f')\ge \min \limits _{p\in D}E_pu''(f')=u''(x)$, $f'\succsim ^*x$. Thus, $\frac{1}{2}f'+\frac{1}{2}g\succsim \frac{1}{2}x+\frac{1}{2}g$, which contradicts $\frac{1}{2}x+\frac{1}{2}g\succ \frac{1}{2}f'+\frac{1}{2}g$. Hence, D is not the Bewley set.$\square$

Lemma 2

Let $\varphi \in {{\mathbb {R}}}^S$, $t\in [\varphi _*,\varphi ^*]$, and $B:=\{p\in \Delta :E_p\varphi =t\}$. Let K be a non-empty closed subset of $\Delta$ such that for each $p\in K$, $E_p\varphi \ge t$. Let $c:=\min \limits _{p\in B}d(p,K)$. Then $\min \limits _{p\in \Delta :d(p,K)\le c}E_p\varphi =t$.

Proof

Let $\bar{p}\in B$ be such that $d(\bar{p},K)=c$. Since $\bar{p}\in B$, $E_{\bar{p}}\varphi =t$. Since $d(\bar{p},K)=c$, $E_{\bar{p}}\varphi \ge \min \limits _{p\in \Delta :d(p,K)\le c}E_p\varphi$. Thus, $t\ge \min \limits _{p\in \Delta :d(p,K)\le c}E_p\varphi$. Suppose that $t>\min \limits _{p\in \Delta :d(p,K)\le c}E_p\varphi$. Let $p'\in \Delta$ and $q'\in K$ be such that $d(p',q')=d(p',K)\le c$ and $E_{p'}\varphi =\min \limits _{p\in \Delta :d(p,K)\le c}E_p\varphi$. Then $t>E_{p'}\varphi$. Since $q'\in K$ and for each $p\in K$, $E_p\varphi \ge t$, $E_{q'}\varphi \ge t$. Since $E_{q'}\varphi \ge t>E_{p'}\varphi$, $p'\ne q'$ and there is $\alpha \in [0,1)$ such that $t=E_{\alpha p'+(1-\alpha )q'}\varphi$. Thus, $\alpha p'+(1-\alpha )q'\in B$ and $d(\alpha p'+(1-\alpha )q',q')<d(p',q')$. Hence,

$$\begin{aligned} c=\min \limits _{p\in B}d(p,K)\le d(\alpha p'+(1-\alpha )q',q')<d(p',q')=d(p',K)\le c, \end{aligned}$$

which is not possible. Hence, $t=\min \limits _{p\in \Delta :d(p,K)\le c}E_p\varphi$. $\square$

Proof of Theorem 2

To prove the sufficiency of the axioms, let ${\ \succsim \ }$ satisfy A.1–A.4, A.5.2, and A.6. By Lemma 1, there exist an affine onto function $u:X\rightarrow {{\mathbb {R}}}$ and a normalized, non-decreasing, and continuous functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ satisfying (22).

Let $K:=\{p\in \Delta :\text {for each }\varphi \in {{\mathbb {R}}}^S, I(\varphi )\le E_p\varphi \}$. Equivalently, K is the benchmark set. The equivalence has been shown by Ghirardato and Marinacci (2002) for biseparable preference relations (see their Theorem 12). It can be readily seen that their proof applies here. Since K is the benchmark set, by A.5.2, K is non-empty. Since I is continuous, K is closed. By the definition of K, it is convex.

Define $B:{{\mathbb {R}}}^S\rightrightarrows \Delta$ and $\sigma :{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}_+$ by setting for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} B(\varphi ):=\{p\in \Delta :I(\varphi )=E_p\varphi \}, \end{aligned}$$

and

$$\begin{aligned} \sigma (\varphi ):=\min \limits _{p\in B(\varphi )}d(p,K). \end{aligned}$$

(35)

For each $\varphi \in {{\mathbb {R}}}^S$, it can be readily seen that $B(\varphi )$ is closed, and since I is normalized and non-decreasing, $I(\varphi )\in [\varphi _*,\varphi ^*]$, so that $B(\varphi )$ is non-empty. Thus, $\sigma$ is well-defined. Since I is non-decreasing, $\sigma \in \Sigma (K)$. Moreover, by Lemma 2, for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi )=\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi )}E_p\varphi$. Thus, by (22), for each pair $f,g\in {\mathcal {F}}$,

$$\begin{aligned} f{\ \succsim \ }g\ \Longleftrightarrow \ \min \limits _{p\in \Delta :d(p,K)\le \sigma (u(f))}E_pu(f)\ge \min \limits _{p\in \Delta :d(p,K)\le \sigma (u(g))}E_pu(g), \end{aligned}$$

Therefore, if we can show that $\sigma :{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}_+$ is continuous on $\{u(f)\in {{\mathbb {R}}}^S:f\in {\mathcal {F}},f\not \sim x_{*f}\}$ and lower semicontinuous on $\{u(f)\in {{\mathbb {R}}}^S:f\in {\mathcal {F}},f\sim x_{*f}\}$, then $\langle u,K,\sigma \rangle$ is a VC representation of ${\ \succsim \ }$.

To show the continuity property of $\sigma$, let $\varphi \in {{\mathbb {R}}}^S$ and $f\in {\mathcal {F}}$ be such that $u(f)=\varphi$, and let $\{\varphi ^n\}_{n=1}^{\infty }$ be a sequence of elements in ${{\mathbb {R}}}^S$ such that $\lim \nolimits _{n\rightarrow \infty }\varphi ^n=\varphi$. Suppose first that $f\not \sim x_{*f}$. Then by (22), $I(u(f))\ne I(u(x_{*f}))$, and thus, $I(\varphi )\ne I(\varphi _*\mathbf 1 )$. Since $I(\varphi )\ne I(\varphi _*\mathbf 1 )$, and since I is normalized and non-decreasing, $\varphi _*<I(\varphi )\le \varphi ^*$. We shall show that $\lim \nolimits _{n\rightarrow \infty }\sigma (\varphi ^n)=\sigma (\varphi )$ in each of the following two cases.

Case 1: $I(\varphi )=\varphi ^*$. Let $q^*\in K$. By the definition of K, $I(\varphi )\le E_{q^*}\varphi$. Since $I(\varphi )\le E_{q^*}\varphi \le \varphi ^*=I(\varphi )$, $E_{q^*}\varphi =I(\varphi )$. Thus, $q^*\in B(\varphi )$. Since $q^*\in K$ and $q^*\in B(\varphi )$, $\sigma (\varphi )=\min \limits _{p\in B(\varphi )}d(p,K^*)=d(q^*,q^*)=0$. To show that $\lim \nolimits _{n\rightarrow \infty }\sigma (\varphi ^n)=0$, suppose to the contrary that there exist $\epsilon >0$ and a subsequence $\{\varphi ^{n_m}\}_{m=1}^{\infty }$ of $\{\varphi ^{n}\}_{n=1}^{\infty }$ such that for each $m\in {{\mathbb {N}}}$, $\sigma (\varphi ^{n_m})>\epsilon$. Let $q_*\in \Delta$ be such that $E_{q_*}\varphi =\varphi _*$. Since $I(\varphi )>\varphi _*$, $I(\varphi )>E_{q_*}\varphi$. Thus, $q_*\notin K$. Let $\alpha \in (0,1)$ be such that $d(\alpha q_*+(1-\alpha )q^*,q^*)<\epsilon$. Since $q^*\in K$, $d(\alpha q_*+(1-\alpha )q^*,K)\le d(\alpha q_*+(1-\alpha )q^*,q^*)<\epsilon$. Then for each $m\in {{\mathbb {N}}}$, $d(\alpha q_*+(1-\alpha )q^*,K)<\sigma (\varphi ^{n_m})$, and thus, $I(\varphi ^{n_m})=\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi ^{n_m})}E_p\varphi ^{n_m}\le E_{\alpha q_*+(1-\alpha )q^*}\varphi ^{n_m}$. Since I is continuous, $I(\varphi )=\lim \nolimits _{m\rightarrow \infty }I(\varphi ^{n_m})$. Thus, $I(\varphi )=\lim \nolimits _{m\rightarrow \infty }I(\varphi ^{n_m})\le \lim \nolimits _{m\rightarrow \infty }E_{\alpha q_*+(1-\alpha )q^*}\varphi ^{n_m}=\alpha E_{q_*}\varphi +(1-\alpha )E_{q^*}\varphi =\alpha \varphi _*+(1-\alpha )I(\varphi )<I(\varphi )$, which contradicts $I(\varphi )=\varphi ^*$, as desired.

Case 2: $\varphi _*<I(\varphi )<\varphi ^*$. To show the continuity of $\sigma$ at $\varphi$, by the maximum theorem, it suffices to show the continuity of the correspondence B at $\varphi$. To show the upper hemicontinuity of B at $\varphi$, let $\bar{p}\in \Delta$ and let $\{p^n\}_{n=1}^{\infty }$ be a sequence of elements in $\Delta$ such that $\lim \nolimits _{n\rightarrow \infty }p^n=\bar{p}$ and for each $n\in {{\mathbb {N}}}$, $p^n\in B(\varphi ^n)$. We shall show that ${\bar{p}}\in B(\varphi )$. For each $n\in {{\mathbb {N}}}$, since $p^n\in B(\varphi ^n)$, $E_{p^n}\varphi ^n=I(\varphi ^n)$. Thus, $E_{\bar{p}}\varphi =\lim \nolimits _{n\rightarrow \infty }E_{p^n}\varphi ^n=\lim \nolimits _{n\rightarrow \infty }I(\varphi ^n)$, and since I is continuous, $E_{\bar{p}}\varphi =I(\varphi )$. Hence, $\bar{p}\in B(\varphi )$. To show the lower hemicontinuity of B at $\varphi$, let $\bar{p}\in B(\varphi )$. We shall show that there exist a sequence $\{p^m\}_{m=1}^{\infty }$ and a subsequence $\{\varphi ^{n_m}\}_{m=1}^{\infty }$ of $\{\varphi ^n\}_{n=1}^{\infty }$ such that $\lim \nolimits _{m\rightarrow \infty }p^m=\bar{p}$ and for each $m\in {{\mathbb {N}}}$, $p^m\in B(\varphi ^{n_m})$. Since $\bar{p}\in B(\varphi )$, $E_{\bar{p}}\varphi =I(\varphi )$. For each $\epsilon >0$, let $A(\epsilon ):=\{p\in \Delta :d(p,\bar{p})\le \epsilon \}$. For each $\epsilon >0$, since $\varphi _*<E_{\bar{p}}\varphi =I(\varphi )<\varphi ^*$, $\min \limits _{p\in A(\epsilon )}E_{p}\varphi<I(\varphi )<\max \limits _{p\in A(\epsilon )}E_{p}\varphi$, and since I is continuous, for sufficiently large n, $\min \limits _{p\in A(\epsilon )}E_{p}\varphi ^n<I(\varphi ^n)<\max \limits _{p\in A(\epsilon )}E_{p}\varphi ^n$. Thus, for each $\epsilon >0$, when n is sufficiently large, there is $p\in A(\epsilon )$ such that $E_{p}\varphi ^n=I(\varphi ^n)$, so that $p\in A(\epsilon )\cap B(\varphi ^n)$. Thus, there exist a sequence $\{p^m\}_{m=1}^{\infty }$ and a subsequence $\{\varphi ^{n_m}\}_{m=1}^{\infty }$ of $\{\varphi ^n\}_{n=1}^{\infty }$ such that for each $m\in {{\mathbb {N}}}$, $p^m\in A(\frac{1}{m})\cap B(\varphi ^{n_m})$. Since for each $m\in {{\mathbb {N}}}$, $p^m\in A(\frac{1}{m})$, $\lim \nolimits _{m\rightarrow \infty }p^m=\bar{p}$.

Now suppose that $f\sim x_{*f}$. We want to show that $\liminf \limits _{n\rightarrow \infty }\sigma (\varphi ^n)\ge \sigma (\varphi )$. Since for each $n\in {{\mathbb {N}}}$, $\sigma (\varphi ^n)\ge 0$, $\liminf \limits _{n\rightarrow \infty }\sigma (\varphi ^n)\ge 0$. If $\varphi$ is constant, then $B(\varphi )=\Delta$, and thus, $\sigma (\varphi )=\min \limits _{p\in B(\varphi )}d(p,K^*)=0\le \liminf \limits _{n\rightarrow \infty }\sigma (\varphi ^n)$. Suppose that $\varphi$ is not constant. Suppose to the contrary that $\liminf \limits _{n\rightarrow \infty }\sigma (\varphi ^n)<\sigma (\varphi )$. Then there is a convergent subsequence $\{\varphi ^{n_m}\}_{m=1}^{\infty }$ of $\{\varphi ^n\}_{n=1}^{\infty }$ such that $\lim \nolimits _{m\rightarrow \infty }\sigma (\varphi ^{n_m})<\sigma (\varphi )$. Let $t\in {{\mathbb {R}}}$ be such that $\lim \nolimits _{m\rightarrow \infty }\sigma (\varphi ^{n_m})<t<\sigma (\varphi )$. For each $p\in \Delta$ such that $d(p,K)\le t$, since $\sigma (\varphi )=\min \limits _{q\in B(\varphi )}d(q,K)>t$, $p\notin B(\varphi )$, and thus, $E_p\varphi \ne I(\varphi )$. Since $f\sim x_{*f}$, by (22), $I(\varphi )=I(u(f))=I(u(x_{*f}))=I(\varphi _*\mathbf 1 )$, and since I is normalized, $I(\varphi )=\varphi _*$. For each $p\in \Delta$ such that $d(p,K)\le t$, since $E_p\varphi \ne I(\varphi )$ and $E_p\varphi \ge \varphi _*=I(\varphi )$, $E_p\varphi >I(\varphi )$. Thus,

$$\begin{aligned} \min \limits _{p\in \Delta :d(p,K)\le t}E_p\varphi >I(\varphi ). \end{aligned}$$

(36)

Since $\lim \nolimits _{m\rightarrow \infty }\sigma (\varphi ^{n_m})<t$, for sufficiently large m, $\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi ^{n_m})}E_p\varphi ^{n_m} \ge \min \limits _{p\in \Delta :d(p,K)\le t}E_p\varphi ^{n_m}$. Since I is continuous, and since for each $m\in {{\mathbb {N}}}$, $I(\varphi ^{n_m})=\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi ^{n_m})}E_p\varphi ^{n_m}$, $I(\varphi )=\lim \nolimits _{m\rightarrow \infty }I(\varphi ^{n_m})=\lim \nolimits _{m\rightarrow \infty }\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi ^{n_m})}E_p\varphi ^{n_m}$. Thus,

$$\begin{aligned}&\quad I(\varphi )=\lim \limits _{m\rightarrow \infty }\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi ^{n_m})}E_p\varphi ^{n_m}\ge \lim \limits _{m\rightarrow \infty }\min \limits _{p\in \Delta :d(p,K)\le t}E_p\varphi ^{n_m}\\&\quad =\min \limits _{p\in \Delta :d(p,K)\le t}E_p\varphi , \end{aligned}$$

which contradicts (36), as desired.

We complete the proof of the sufficiency of the axioms by showing that K is actually the largest essential set, so that $\langle u,K,\sigma \rangle$ is a canonical VC representation of ${\ \succsim \ }$. Let $\langle u',K',\sigma '\rangle$ be another VC representation of ${\ \succsim \ }$. To show that $K'\subseteq K$, let $p'\in K'$ and $\varphi \in {{\mathbb {R}}}^S$. By the definition of K, it is equivalent for us to show that $I(\varphi )\le E_{p'}\varphi$. Let $f\in {\mathcal {F}}$ such that $u(f)=\varphi$. Since $\langle u',K',\sigma '\rangle$ is a VC representation of ${\ \succsim \ }$, $u'(x_f)=\min \limits _{p\in \Delta :d(p,K')\le \sigma '(u'(f))}E_{p}u'(f)$. Since $p'\in K'$, $\min \limits _{p\in \Delta :d(p,K')\le \sigma '(u'(f))}E_{p}u'(f)\le E_{p'}u'(f)$. Thus, $u'(x_f)\le E_{p'}u'(f)$. Since both u and $u'$ are affine functions representing ${\ \succsim \ }$ restricted to X, by routine arguments, $u'$ is a positive affine transformation of u. Thus, $u(x_f)\le E_{p'}u(f)$. Since I is normalized, by (22), $I(\varphi )=I(u(f))=I(u(x_f))=u(x_f)$. Hence, $I(\varphi )\le E_{p'}\varphi$.

To prove the necessity of the axioms, let ${\ \succsim \ }$ admit a canonical VC representation $\langle u,K,\sigma \rangle$ with $\sigma \in \Sigma (K)$. Clearly, it satisfies A.1. Since u is affine, it satisfies A.2. Since $\sigma \in \Sigma (K)$, one can readily verify that it satisfies A.4. Since u is onto, by Lemma 29 of Maccheroni et al. (2006), it satisfies A.6.

To show that ${\ \succsim \ }$ satisfies A.3, define $C:{{\mathbb {R}}}^S\rightrightarrows \Delta$ and $J:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} C(\varphi ):=\{p\in \Delta :d(p,K)\le \sigma (\varphi )\}, \end{aligned}$$

and

$$\begin{aligned} J(\varphi ):=\min \limits _{p\in C(\varphi )}E_p\varphi , \end{aligned}$$

and it suffices to show that J is continuous. Let $\varphi \in {{\mathbb {R}}}^S$ and $\{\varphi ^n\}_{n=1}^{\infty }$ be a sequence of elements in ${{\mathbb {R}}}^S$ such that $\lim \nolimits _{n\rightarrow \infty }\varphi ^n=\varphi$. If $\varphi$ is constant, then for each $n\in {{\mathbb {N}}}$, $|J(\varphi ^n)-J(\varphi )|\le \sup \limits _{s\in S}|\varphi ^n(s)-\varphi (s)|$, and thus, $0\le \lim \nolimits _{n\rightarrow \infty }|J(\varphi ^n)-J(\varphi )|\le \lim \nolimits _{n\rightarrow \infty }\sup \limits _{s\in S}|\varphi ^n(s)-\varphi (s)|=0$, so that $\lim \nolimits _{n\rightarrow \infty }J(\varphi ^n)=J(\varphi )$. Suppose that $\varphi$ is not constant. Let $f\in {\mathcal {F}}$ be such that $u(f)=\varphi$. We shall show that $\lim \nolimits _{n\rightarrow \infty }J(\varphi ^n)=J(\varphi )$ in each of the following two cases.

Case 1: $f\not \sim x_{*f}$. Then $\sigma$ is continuous at $\varphi$. To show the continuity of J at $\varphi$, by the maximum theorem, it suffices to show the continuity of the correspondence C at $\varphi$.

To show the upper hemicontinuity of C at $\varphi$, let ${\bar{p}}\in \Delta$ and let $\{p^n\}_{n=1}^{\infty }$ be a sequence of elements in $\Delta$ such that $\lim \nolimits _{n\rightarrow \infty }p^n={\bar{p}}$ and for each $n\in {{\mathbb {N}}}$, $p^n\in C(\varphi ^n)$. We shall show that ${\bar{p}}\in C(\varphi )$. For each $n\in {{\mathbb {N}}}$, since $p^n\in C(\varphi ^n)$, $d(p^n,K)\le \sigma (\varphi ^n)$. Since $d(\cdot ,K)$ is continuous on $\Delta$ and $\sigma$ is continuous at $\varphi$, $d({\bar{p}},K)=\lim \nolimits _{n\rightarrow \infty }d(p^n,K)$ and $\sigma (\varphi )=\lim \nolimits _{n\rightarrow \infty }\sigma (\varphi ^n)$. Thus, $d({\bar{p}},K)\le \sigma (\varphi )$. Hence, $p\in C(\varphi )$.

To show the lower hemicontinuity of C at $\varphi$, let ${\bar{p}}\in C(\varphi )$. We shall show that there is a sequence $\{p^n\}_{n=1}^{\infty }$ such that $\lim \nolimits _{n\rightarrow \infty }p^n={\bar{p}}$ and for each $n\in {{\mathbb {N}}}$, $p^n\in C(\varphi ^n)$. Since ${\bar{p}}\in C(\varphi )$, $d({\bar{p}},K)\le \sigma (\varphi )$. Let $q\in K$ be such that $d({\bar{p}},q)=d({\bar{p}},K)$. Then $d({\bar{p}},q)\le \sigma (\varphi )$. For each $n\in {{\mathbb {N}}}$, let

where $\epsilon ^n\in (0,1]$ is such that $d(\epsilon ^n q+(1-\epsilon ^n){\bar{p}},q)=\sigma (\varphi ^n)$. Then for each $n\in {{\mathbb {N}}}$, $d(p^n,K)\le d(p^n,q)\le \sigma (\varphi ^n)$, and thus, $p^n\in C(\varphi ^n)$. Moreover, for each $n\in {{\mathbb {N}}}$, if $p^n={\bar{p}}$, then $d({\bar{p}},p^n)=0$; if $p^n=\epsilon ^n q+(1-\epsilon ^n){\bar{p}}$, then $d(p^n,q)=d(\epsilon ^n q+(1-\epsilon ^n){\bar{p}},q)=\sigma (\varphi ^n)<d({\bar{p}},q)$, and thus $d({\bar{p}},p^n)=d({\bar{p}},q)-d(p^n,q)=d({\bar{p}},q)-\sigma (\varphi ^n)\le \sigma (\varphi )-\sigma (\varphi ^n)$. Then by the continuity of $\sigma$ at $\varphi$, $\lim \nolimits _{n\rightarrow \infty }p^n={\bar{p}}$.

Case 2: $f\sim x_{*f}$. Then $\sigma$ is lower semicontinuous at $\varphi$. To show the continuity of J at $\varphi$, suppose to the contrary that there exist $\epsilon >0$ and a subsequence $\{\varphi ^{n_m}\}_{m=1}^{\infty }$ of $\{\varphi ^n\}_{n=1}^{\infty }$ such that for each $m\in {{\mathbb {N}}}$, $|J(\varphi ^{n_m})-J(\varphi )|>\epsilon$. Since $f\sim x_{*f}$, and since $\langle u,K,\sigma \rangle$ is a VC representation of ${\ \succsim \ }$, $J(u(f))=J(u(x_{*f}))=\varphi _*$. Thus, $J(\varphi )=\varphi _*$, and for each $m\in {{\mathbb {N}}}$, $|J(\varphi ^{n_m})-\varphi _*|>\epsilon$. Then for sufficiently large m, $|J(\varphi ^{n_m})-\varphi ^{n_m}_*|>\frac{\epsilon }{2}$, and since J is non-decreasing, $J(\varphi ^{n_m})-\varphi ^{n_m}_*=|J(\varphi ^{n_m})-\varphi ^{n_m}_*|>\frac{\epsilon }{2}$. Let $\alpha \in (0,1)$ be such that $\alpha (\varphi ^*-\varphi _*)<\frac{\epsilon }{2}$. Then for sufficiently large m, $\alpha (\varphi ^{n_m*}-\varphi ^{n_m}_*)<\frac{\epsilon }{2}<J(\varphi ^{n_m})-\varphi ^{n_m}_*$, and thus, $(1-\alpha )\varphi ^{n_m}_*+\alpha \varphi ^{n_m*}<J(\varphi ^{n_m})$. Let $B^{\infty }:=\{p\in \Delta :E_p\varphi =(1-\alpha )\varphi _*+\alpha \varphi ^*\}$, and for each $m\in {{\mathbb {N}}}$, $B^m:=\{p\in \Delta :E_p\varphi ^{n_m}=(1-\alpha )\varphi ^{n_m}_*+\alpha \varphi ^{n_m*}\}$. Observe that $B^{\infty }$ is non-empty and closed, and so is $B^m$ for each $m\in {{\mathbb {N}}}$. Moreover, for sufficiently large m, if $p\in B^m$, then $E_p\varphi ^{n_m}=(1-\alpha )\varphi ^{n_m}_*+\alpha \varphi ^{n_m*}<J(\varphi ^{n_m})$, so by the definition of J, $p\notin C(\varphi ^{n_m})$, and thus, by the definition of C, $d(p,K)>\sigma (\varphi ^{n_m})$. Hence, for sufficiently large m, $\min \limits _{p\in B^m}d(p,K)>\sigma (\varphi ^{n_m})$. Thus, $\liminf \limits _{m\rightarrow \infty }\min \limits _{p\in B^m}d(p,K)\ge \liminf \limits _{m\rightarrow \infty }\sigma (\varphi ^{n_m})\ge \liminf \limits _{n\rightarrow \infty }\sigma (\varphi ^n)$. Since $\varphi$ is not constant, $\min \limits _{p\in B^{\infty }}d(p,K)=\lim \nolimits _{m\rightarrow \infty }\min \limits _{p\in B^m}d(p,K)$, using similar arguments as in showing the continuity of $\sigma$ defined by (35) in the case of $\varphi _*<I(\varphi )<\varphi ^*$.²⁵ Since $\sigma$ is lower semicontinuous at $\varphi$, $\liminf \limits _{n\rightarrow \infty }\sigma (\varphi ^n)\ge \sigma (\varphi )$. Thus, $\min \limits _{p\in B^{\infty }}d(p,K)\ge \sigma (\varphi )$. Let $c:=\min \limits _{p\in B^{\infty }}d(p,K)$. Then $c\ge \sigma (\varphi )$. Hence,

$$\begin{aligned} \min \limits _{p\in \Delta :d(p,K)\le c}E_p\varphi \le \min \limits _{p\in C(\varphi )}E_p\varphi =J(\varphi )=\varphi _*<(1-\alpha )\varphi _*+\alpha \varphi ^*, \end{aligned}$$

(37)

where the last inequality holds since $\varphi$ is not constant and $\alpha \in (0,1)$. Observe that for each $p\in K$ and each $m\in {{\mathbb {N}}}$, $p\in C(\varphi ^{n_m})$, so that $J(\varphi ^{n_m})\le E_p\varphi ^{n_m}$. Thus, for each $p\in K$, when m is sufficiently large, $(1-\alpha )\varphi ^{n_m}_*+\alpha \varphi ^{n_m*}<J(\varphi ^{n_m})\le E_p\varphi ^{n_m}$, and hence $E_p\varphi =\lim \nolimits _{m\rightarrow \infty }E_p\varphi ^{n_m}\ge \lim \nolimits _{m\rightarrow \infty }(1-\alpha )\varphi ^{n_m}_*+\alpha \varphi ^{n_m*}=(1-\alpha )\varphi _{*}+\alpha \varphi ^*$. By lemma 2, $\min \limits _{p\in \Delta :d(p,K)\le c}E_p\varphi =(1-\alpha )\varphi _*+\alpha \varphi ^*$, which contradicts (37), as desired.

To show that ${\ \succsim \ }$ satisfies A.5.2, let $q\in K$. Let ${\ \succsim \ }_q$ be the SEU preference relation defined by setting for each pair $f,g\in {\mathcal {F}}$, $f{\ \succsim \ }_q g\Longleftrightarrow \sum \limits _{s\in S}q_sf(s){\ \succsim \ }\sum \limits _{s\in S}q_sg(s)$. We shall show that ${\ \succsim \ }$ is more ambiguity averse than the SEU preference relation ${\ \succsim \ }_q$. Let $f\in {\mathcal {F}}$ and $x\in X$ be such that $x{\ \succsim \ }_q f$. Then $x{\ \succsim \ }\sum \limits _{s\in S}q_sf(s)$. Since $\langle u,K,\sigma \rangle$ is a varaint constraint representation of ${\ \succsim \ }$ and since u is affine, $u(x)\ge u(\sum \limits _{s\in S}q_sf(s))=E_qu(f)$. Since $q\in K$, $E_qu(f)\ge \min \limits _{p\in \Delta :d(p,K)\le \sigma (u(f))}E_pu(f)$. Then $u(x)\ge \min \limits _{p\in \Delta :d(p,K)\le \sigma (u(f))}E_pu(f)$, and thus, $x{\ \succsim \ }f$.

Lastly, to prove the uniqueness of the representation, let $\langle u,K,\sigma \rangle$ and $\langle u',K',\sigma '\rangle$ be two canonical VC representations of some preference relation ${\ \succsim \ }$. By routine arguments, $u'$ is a positive affine transformation of u. By the definition of a canonical VC representation, K is the largest essential set, and so is $K'$. Thus, $K=K'$. As argued in the proof of the sufficiency of the axioms, K coincides with the benchmark set. Finally, suppose that $u=u'$, let $f\in {\mathcal {F}}$ be such that $f\not \sim x_{*f}$, and we want to show that $\sigma (u(f))=\sigma '(u'(f))$. Let $\varphi :=u(f)$. Suppose that $\sigma (\varphi )<\sigma '(\varphi )$. Let $p_*,q\in \Delta$ be such that $E_{p_*}\varphi =\varphi _*$, $E_q\varphi =\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi )}E_p\varphi$, and $d(q,K)\le \sigma (\varphi )$. Since $\langle u,K,\sigma \rangle$ is a VC representation of ${\ \succsim \ }$, and since $f\not \sim x_{*f}$, $u(x_f)=\min \limits _{p\in \Delta :d(p,K)\le \sigma (u(f))}E_pu(f)\ne u(x_{*f})=\varphi _*$. Thus, $E_q\varphi =u(x_f)\ne \varphi _*$. Since $d(q,K)\le \sigma (\varphi )<\sigma '(\varphi )$ and $d(\cdot ,K)$ is continuous on $\Delta$, there is $\epsilon \in (0,1)$ such that $d(\epsilon p_*+(1-\epsilon )q,K)<\sigma '(\varphi )$. Thus, $\min \limits _{p\in \Delta :d(p,K)\le \sigma '(\varphi )}E_p\varphi \le E_{\epsilon p_*+(1-\epsilon )q}\varphi =\epsilon \varphi _*+(1-\epsilon )E_q\varphi \ne u(x_f)$. Since $\langle u',K',\sigma '\rangle$ is a VC representation of ${\ \succsim \ }$, and since $u=u'$, $K=K'$, and $\min \limits _{p\in \Delta :d(p,K)\le \sigma '(\varphi )}E_p\varphi \ne u(x_f)$, $f\not \sim x_f$, which is a contradiction. By analogous arguments, $\sigma (\varphi )>\sigma '(\varphi )$ is not possible. Hence, $\sigma (\varphi )=\sigma '(\varphi )$, or equivalently, $\sigma (u(f))=\sigma (u'(f))$. $\square$

Proof of Proposition 2

Suppose that ${\ \succsim \ }$ admits both a canonical WM representation $\langle u,D,\lambda \rangle$ and a canonical VC representation $\langle u',K,\sigma \rangle$. To show that $K\subseteq D$, suppose to the contrary that there is $q\in K{\setminus } D$. Since $q\notin D$, by a standard separation theorem, there is $\varphi \in {{\mathbb {R}}}^S$ such that $E_{q}\varphi <\min \limits _{p\in D}E_p\varphi$. Let $f\in {\mathcal {F}}$ and $x\in X$ be such that $u(f)=\varphi$ and $u(x)=\min \limits _{p\in D}E_p\varphi$. Since both u and $u'$ are affine functions representing ${\ \succsim \ }$ restricted to X, by routine arguments, $u'$ is a positive affine transformation of u. Thus, $u'(x)=\min \limits _{p\in D}E_pu'(f)>E_{q}u'(f)$. Since $q\in K$, $E_{q}u'(f)\ge \min \limits _{p\in \Delta :d(p,K)\le \sigma (u'(f))}E_pu'(f)$. Hence, $u'(x)>\min \limits _{p\in \Delta :d(p,K)\le \sigma (u'(f))}E_pu'(f)$, and since $\langle u',K,\sigma \rangle$ is a VC representation of ${\ \succsim \ }$, $x\succ f$. Since $\langle u,D,\lambda \rangle$ is a WM representation of ${\ \succsim \ }$, and since

$$\begin{aligned} u(x)=&\min \limits _{p\in D}E_p\varphi =\min \limits _{p\in D}E_pu(f)\le \lambda (u(f))\min \limits _{p\in D}E_pu(f)\\&+(1-\lambda (u(f)))\max \limits _{p\in D}E_pu(f), \end{aligned}$$

$f{\ \succsim \ }x$, which contradicts $x\succ f$, as desired. $\square$

Proof of Proposition 3

To prove the “only if” direction, let ${\ \succsim \ }$ be a MEU preference relation. Thus, there exist an affine onto function $u:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ and a non-empty, closed, and convex set $D\subseteq \Delta$ such that for each pair $f,g\in {\mathcal {F}}$, $f{\ \succsim \ }g\ \Longleftrightarrow \ \min \limits _{p\in D}E_pu(f)\ge \min \limits _{p\in D}E_pu(g)$. It is known that a MEU preference relation satisfies A.1–A.4, A.5.1, A.5.2, and A.6. By Ghirardato et al. (2004), ${\ \succsim \ }$ admits a Bewley WM representation with D being the Bewley set.²⁶ By Theorem 1, it admits a canonical WM representation with the smallest admissible set being the Bewley set. Thus, D is the smallest admissible set, and when $\lambda :{{\mathbb {R}}}^S\rightarrow [0,1]$ is a constant function equal to 1, $\langle u,D,\lambda \rangle$ is a canonical WM representation of ${\ \succsim \ }$. Moreover, by Ghirardato and Marinacci (2002), D is the benchmark set.²⁷ By Theorem 2, ${\ \succsim \ }$ admits a canonical VC representation with the largest essential set being the benchmark set. Thus, D is the largest essential set, and when $\sigma :{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}_+$ is a constant function equal to 0, $\langle u,D,\sigma \rangle$ is a canonical VC representation of ${\ \succsim \ }$.

To prove the “if” direction, let ${\ \succsim \ }$ admit both a canonical WM representation $\langle u,D,\lambda \rangle$ and a canonical VC representation $\langle u',K,\sigma \rangle$ with $K=D$. To show that ${\ \succsim \ }$ is a MEU preference relation, it suffices to show that for each $f\in {\mathcal {F}}$, $\lambda (u(f))=1$. Let $f\in {\mathcal {F}}$. Since $\langle u',K,\sigma \rangle$ is a VC representation of ${\ \succsim \ }$, $u'(x_f)=\min \limits _{p\in \Delta :d(p,K)\le \sigma (u'(f))}E_pu'(f)$. Since both u and $u'$ are affine functions representing ${\ \succsim \ }$ restricted to X, by routine arguments, u is a positive affine transformation of $u'$. Thus, $u(x_f)=\min \limits _{p\in \Delta :d(p,K)\le \sigma (u'(f))}E_pu(f)\le \min \limits _{p\in K}E_pu(f)$. Since $\langle u,D,\lambda \rangle$ is a WM representation of ${\ \succsim \ }$, and since $K=D$, $u(x_f)=\lambda (u(f))\min \limits _{p\in K}E_pu(f)+(1-\lambda (u(f)))\max \limits _{p\in K}E_pu(f)\ge \min \limits _{p\in K}E_pu(f)$. Thus, $u(x_f)=\min \limits _{p\in K}E_pu(f)$, or equivalently, $\lambda (u(f))=1$. $\square$

Proof of Corollary 2

Let $\succsim$ admit an ambiguity averse representation $\langle u,G\rangle$. Define $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi ):=\min \limits _{p\in \Delta }G(E_p\varphi ,p)$. By the proof of Theorem 3 of Cerreia-Vioglio et al. (2011b), I is normalized, non-decreasing, and continuous. Moreover, by their Proposition 11, $K^*=\bigcap \limits _{x\in X}\pi (x)$ where $\pi (\cdot )$ is defined in (17). Thus, $p\in K^*$ if and only if for each $x\in X$ and each $f\in {\mathcal {F}}$, $u(x)\ge E_pu(f)\Longrightarrow x{\ \succsim \ }f$. Then by definition, $K^*$ is the benchmark set. Since $K^*$ is the benchmark set and $K^*\ne \emptyset$, $\succsim$ satisfies A.5.2. Thus, by the proof of Theorem 2, $\succsim$ admits a canonical VC representation $\langle u,K^*,\sigma \rangle$ in which $\sigma$ belongs to $\Sigma (K^*)$ and is given by (35). $\square$

Proof of Proposition 4

Let $\succsim$ admit a multiplier representation $\langle u,q,\theta \rangle$. Equivalently, $\succsim$ admits an ambiguity averse representation $\langle u,G\rangle$ such that for each $(t,p)\in {{\mathbb {R}}}\times \Delta$, $G(t,p)=t+\theta R(p||q)$.²⁸ It can be readily seen that $K^*=\{q\}$, and

Define $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi ):=\min \limits _{p\in \Delta }[E_p\varphi +\theta R(p||q)]$. When $\theta <\infty$, by the variational formula (see, e.g., Proposition 1.4.2 of Dupuis and Ellis (1997)), for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} I(\varphi )=-\theta \log E_qe^{-\frac{\varphi }{\theta }}. \end{aligned}$$

(38)

By the proof of Corollary 1, $\succsim$ admits a canonical WM representation $\langle u,D^*,\lambda \rangle$ in which $\lambda$ is given by (24) on $\{\varphi \in {{\mathbb {R}}}^S:\min \limits _{p\in D^*}E_p\varphi \ne \max \limits _{p\in D^*}E_p\varphi \}$. Suppose that $\theta <\infty$. Then $D^*=\{p\in \Delta :p\ll q\}$ and for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} \min \limits _{p\in D^*}E_p\varphi =\min \limits _{s\in S:q_s>0}\varphi (s)\text { and }\max \limits _{p\in D^*}E_p\varphi =\max \limits _{s\in S:q_s>0}\varphi (s). \end{aligned}$$

(39)

For each $\varphi \in {{\mathbb {R}}}^S$, if $\min \limits _{p\in D^*}E_p\varphi \ne \max \limits _{p\in D^*}E_p\varphi$, then by substituting (38) and (39) into (24), we know that the value of $\lambda (\varphi )$ is given by (10), and if $\min \limits _{s\in S:q_s>0}\varphi (s)=\max \limits _{s\in S:q_s>0}\varphi (s)$, it can be readily seen that $\lambda (\varphi )$ can take any value including that given by (10). Suppose that $\theta =\infty$. Then $D^*=\{q\}$ and for each $\varphi \in {{\mathbb {R}}}^S$, $\min \limits _{p\in D^*}E_p\varphi =\max \limits _{p\in D^*}E_p\varphi =E_q\varphi =I(\varphi )$. It can be readily seen that $\lambda$ can take any value and we simply let $\lambda$ be a constant function equal to 1.

Since $K^*=\{q\}\ne \emptyset$, by the proof of Corollary 2, $\succsim$ admits a canonical VC representation $\langle u,K^*,\sigma \rangle$ in which $\sigma$ is given by (35). For each $\varphi \in {{\mathbb {R}}}^S$, if $\theta <\infty$, then by substituting (38) into (35), we know that the value of $\sigma (\varphi )$ is given by (11), and if $\theta =\infty$, then by substituting $I(\varphi )=E_q\varphi$ into (35), $\sigma (\varphi )=0$. $\square$

1.2 Proofs in Section 4

Lemma 3

A preference relation ${\ \succsim \ }$ satisfies A.1, A.2.1, A.3, A.4, A.5.1, and A.6 if and only if there exist an affine onto function $u:X\rightarrow {{\mathbb {R}}}$ and a normalized, non-decreasing, quasi-concave, continuous, and constant superadditive functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ satisfying (22). Moreover, u is unique up to a positive affine transformation, and given u, there is a unique normalized functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ satisfying (22).

Proof

To prove the sufficiency of the axioms, let $\succsim$ satisfy A.1, A.2.1, A.3, A.4, A.5.1, and A.6. Note that A.2.1 implies A.2. Then by applying the same techniques of Lemma 57 of Cerreia-Vioglio et al. (2011b), one can show that there exist a non-constant affine function $u:X\rightarrow {{\mathbb {R}}}$ and a normalized, non-decreasing, quasi-concave, and continuous functional $I:u(X)^S\rightarrow {{\mathbb {R}}}$ satisfying (22). Moreover, since ${\ \succsim \ }$ satisfies A.6, by Lemma 29 of Maccheroni et al. (2006), $u(X)={{\mathbb {R}}}$. Thus, we only need to check that I is constant superadditive.

Let $\varphi \in {{\mathbb {R}}}^S$ and $t\in {{\mathbb {R}}}_+$. Since $u(X)={{\mathbb {R}}}$, there are $x,y\in X$ and $f\in {\mathcal {F}}$ such that $u(x)=0$, $u(y)=2t$, and $u(f)=2\varphi$. Then $u(\frac{1}{2}f+\frac{1}{2}y)=\varphi +t\mathbf 1$ and $u(\frac{1}{2}f+\frac{1}{2}x)=\varphi$. Since u is affine and onto, there is $z\in X$ satisfying $u(\frac{1}{2}f+\frac{1}{2}x)=u(\frac{1}{2}z+\frac{1}{2}x)$, so that $\frac{1}{2}f+\frac{1}{2}x\sim \frac{1}{2}z+\frac{1}{2}x$. Since $t\ge 0$, $y{\ \succsim \ }x$. Then by A.2.1, $\frac{1}{2}f+\frac{1}{2}y{\ \succsim \ }\frac{1}{2}z+\frac{1}{2}y$. Thus,

$$\begin{aligned}&I(\varphi +t\mathbf 1 )=I\left( u\left( \frac{1}{2}f+\frac{1}{2}y\right) \right) \ge I\left( u\left( \frac{1}{2}z+\frac{1}{2}y\right) \right) \\&\quad =\frac{1}{2}u(z)+\frac{1}{2}u(y)=\frac{1}{2}u(z)+\frac{1}{2}u(x)+\frac{1}{2}u(y)=u\left( \frac{1}{2}z+\frac{1}{2}x\right) +t\\&\quad =I\left( u\left( \frac{1}{2}f+\frac{1}{2}x\right) \right) +t=I(\varphi )+t. \end{aligned}$$

Hence, I is constant superadditive.

The necessity of the axioms can be readily seen, and the uniqueness follows from routine arguments. Thus, we omit the proofs. $\square$

Proof of Theorem 3

Assume first that ${\ \succsim \ }$ satisfies A.1, A.2.1, A.3, A.4, A.5.1, and A.6, i.e., statement 1 holds. We will prove three representation results, respectively, in statements 2, 3, and 4. When proving each representation result, we will also prove the additional properties of the representation stated in the end of Theorem 3.

Since ${\ \succsim \ }$ satisfies A.1, A.2.1, A.3, A.4, A.5.1, and A.6, by Lemma 3, there exist an affine onto function $u:X\rightarrow {{\mathbb {R}}}$ and a normalized, non-decreasing, quasi-concave, continuous, and constant superadditive functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ satisfying (22).

We first prove that statement 2 holds. The following observations will be useful. Let $\varphi \in {{\mathbb {R}}}^S$. First, for each $k>0$ and each $t\in {{\mathbb {R}}}$, $\frac{1}{k}[I(k\varphi -t\mathbf 1 )+t]\in [\varphi _*,\varphi ^*]$. This can be readily shown by the fact that I is normalized and non-decreasing. Second, for each $k>0$, $\frac{1}{k}[I(k\varphi -t\mathbf 1 )+t]$ is non-increasing in t on ${{\mathbb {R}}}$ since I is constant superadditive. Third, $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi -t\mathbf 1 )+t]$ is non-increasing in k on $(0,\infty )$. To see this, suppose to the contrary that there are $k,k'\in (0,\infty )$ such that $k<k'$ and $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi -t\mathbf 1 )+t]<\lim \nolimits _{t\rightarrow \infty }\frac{1}{k'}[I(k'\varphi -t\mathbf 1 )+t]$. Hence, there is $\bar{t}\in {{\mathbb {R}}}$ such that for each pair $t,t'\in [\bar{t},\infty )$, $\frac{1}{k}[I(k\varphi -t\mathbf 1 )+t]<\frac{1}{k'}[I(k'\varphi -t'{} \mathbf 1 )+t']$. That is,

$$\begin{aligned} I(k\varphi -t\mathbf 1 )<\frac{k}{k'}I(k'\varphi -t'{} \mathbf 1 )-t+\frac{k}{k'}t'. \end{aligned}$$

(40)

Pick $t,t'\ge \bar{t}$ such that $\frac{k}{k'}(-t')+(1-\frac{k}{k'})I(k'\varphi -t'{} \mathbf 1 )=-t.$ Thus, $k\varphi -t\mathbf 1 =\frac{k}{k'}(k'\varphi -t'{} \mathbf 1 )+(1-\frac{k}{k'})I(k'\varphi -t'{} \mathbf 1 )\mathbf 1$. Since I is normalized and quasi-concave, and by the choice of $t,t'$, we have

$$\begin{aligned} I(k\varphi -t\mathbf 1 )\ge I(k'\varphi -t'{} \mathbf 1 )=\frac{k}{k'}I(k'\varphi -t'{} \mathbf 1 )-t+\frac{k}{k'}t', \end{aligned}$$

which contradicts (40), as desired.

Define $J:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} J(\varphi ):=\lim \limits _{k\rightarrow \infty }\lim \limits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi -t\mathbf 1 )+t]. \end{aligned}$$

(41)

Because of the three observations above, J is well-defined, and for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi )\ge J(\varphi )$. Since I is normalized, so is J. Since I is non-decreasing, so is J. By definition, J is constant additive and positive homogeneous of degree 1. We now check that J is superadditive. Suppose to the contrary that there are $\varphi ,\varphi '\in {{\mathbb {R}}}^S$ such that $J(\varphi +\varphi ')<J(\varphi )+J(\varphi ')$. Since J is positive homogeneous of degree 1, $J(\frac{1}{2}\varphi +\frac{1}{2}\varphi ')<\frac{1}{2}J(\varphi )+\frac{1}{2}J(\varphi ')$. Thus, there is $k>0$ and $\bar{t}\in {{\mathbb {R}}}$ such that whenever $t,t',t''\ge \bar{t}$,

$$\begin{aligned} \frac{1}{k}\left[ I\left( k\left( \frac{1}{2}\varphi +\frac{1}{2}\varphi '\right) -t''{} \mathbf 1 \right) + t''\right] \!<\!\frac{1}{2k}[I(k\varphi \!-\!t\mathbf 1 )+t]+\frac{1}{2k}[I(k\varphi '-t'{} \mathbf 1 )+t']. \end{aligned}$$

(42)

Let $t,t'\ge \bar{t}$ be such that $I(k\varphi -t\mathbf 1 )=I(k\varphi '-t'{} \mathbf 1 )$. Let $t''=\frac{t+t'}{2}$. Then (42) becomes

$$\begin{aligned} I\left( \frac{1}{2}(k\varphi -t\mathbf 1 )+\frac{1}{2}(k\varphi '-t'{} \mathbf 1 )\right) <\frac{1}{2}I(k\varphi -t\mathbf 1 )+\frac{1}{2}I(k\varphi '-t'{} \mathbf 1 ). \end{aligned}$$

(43)

Since $I(k\varphi -t\mathbf 1 )=I(k\varphi '-t'{} \mathbf 1 )$ and I is quasi-concave,

$$\begin{aligned} I\left( \frac{1}{2}(k\varphi -t\mathbf 1 )+\frac{1}{2}(k\varphi '-t'{} \mathbf 1 )\right) \ge \frac{1}{2}I(k\varphi -t\mathbf 1 )+\frac{1}{2}I(k\varphi '-t'{} \mathbf 1 ), \end{aligned}$$

which contradicts (43), as desired.

Since J is normalized, non-decreasing, constant additive, positive homogeneous of degree 1, and superadditive, by Lemma 3.5 of Gilboa and Schmeidler (1989), there is a unique non-empty, closed, and convex set $D\subseteq \Delta$ such that for each $\varphi \in {{\mathbb {R}}}^S$, $J(\varphi )=\min \limits _{p\in D}E_p\varphi$.

Fix $\varphi \in {{\mathbb {R}}}^S$. We claim that $I(\varphi )\in [\min \limits _{p\in D}E_p\varphi ,\max \limits _{p\in D}E_p\varphi ]$. Since $I(\varphi )\ge J(\varphi )$ and $J(\varphi )=\min \limits _{p\in D}E_p\varphi$, $I(\varphi )\ge \min \limits _{p\in D}E_p\varphi$. To see that $I(\varphi )\le \max \limits _{p\in D}E_p\varphi$, let $t\in {{\mathbb {R}}}$ be such that $I(\varphi )=I(-\varphi +t\mathbf 1 )$. Since $I(\varphi )=I(-\varphi +t\mathbf 1 )$ and I is quasi-concave,

$$\begin{aligned} I\left( \frac{1}{2}\varphi +\frac{1}{2}(-\varphi +t\mathbf 1 )\right) \ge \frac{1}{2}I(\varphi )+\frac{1}{2}I(-\varphi +t\mathbf 1 ). \end{aligned}$$

(44)

Since I is normalized, $I(\frac{1}{2}\varphi +\frac{1}{2}(-\varphi +t\mathbf 1 ))=\frac{1}{2}t$. Since $I(-\varphi +t\mathbf 1 )\ge \min \limits _{p\in D}E_p(-\varphi +t\mathbf 1 )$, $\frac{1}{2}I(\varphi )+\frac{1}{2}I(-\varphi +t\mathbf 1 )\ge \frac{1}{2}I(\varphi )+\frac{1}{2}\min \limits _{p\in D}E_p(-\varphi )+\frac{1}{2}t$. Thus, by (44),

$$\begin{aligned} \frac{1}{2}t\ge \frac{1}{2}I(\varphi )+\frac{1}{2}\min \limits _{p\in D}E_p(-\varphi )+\frac{1}{2}t, \end{aligned}$$

which implies that $I(\varphi )\le -\min \limits _{p\in D}E_p(-\varphi )=\max \limits _{p\in D}E_p\varphi$.

Define $\lambda :{{\mathbb {R}}}^S\rightarrow [0,1]$ by setting for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} \lambda (\varphi ):=\left\{ \begin{array}{ll} \dfrac{\max \limits _{p\in D}E_p\varphi -I(\varphi )}{\max \limits _{p\in D}E_p\varphi -\min \limits _{p\in D}E_p\varphi } &{} \text{ if } \min \limits _{p\in D}E_p\varphi \ne \max \limits _{p\in D}E_p\varphi \text{, }\\ 1 &{} \hbox {if}\ \min \limits _{p\in D}E_p\varphi =\max \limits _{p\in D}E_p\varphi . \end{array} \right. \end{aligned}$$

By the definition of $\lambda$, for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi )=\lambda (\varphi )\min \limits _{p\in D}E_p\varphi +(1-\lambda (\varphi ))\max \limits _{p\in D}E_p\varphi$. Since I is continuous, $\lambda$ is continuous on $\{\varphi \in {{\mathbb {R}}}^S:\min \limits _{p\in D}E_p\varphi \ne \max \limits _{p\in D}E_p\varphi \}$. Thus, $\langle u,D,\lambda \rangle$ is a WM representation of ${\ \succsim \ }$. By the same arguments as in the first paragraph of the proof of Theorem 1, $\lambda \in \Lambda (D)$. Since I is constant superadditive, for each $\varphi \in {{\mathbb {R}}}^S$, $\lambda (\varphi +t\mathbf 1 )$ is non-increasing in t on ${{\mathbb {R}}}$. By the third observation in the second paragraph, for each $\varphi \in {{\mathbb {R}}}^S$, $\lim \nolimits _{t\rightarrow \infty }\lambda (k\varphi -t\mathbf 1 )$ is non-decreasing in k on $(0,\infty )$. Moreover, since for each $\varphi \in {{\mathbb {R}}}^S$, $\lim \nolimits _{k\rightarrow \infty }\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi -t\mathbf 1 )+t]=J(\varphi )=\min \limits _{p\in D}E_p\varphi$, it can be readily shown that $\lim \nolimits _{k\rightarrow \infty }\lim \nolimits _{t\rightarrow \infty }\lambda (k\varphi -t\mathbf 1 )=1$.

Now we show that D is the smallest admissible set. Since A.2.1 implies A.2, ${\ \succsim \ }$ satisfies A.1–A.4, A.5.1, and A.6. Thus, by Theorem 1, there is a canonical WM representation $\langle u',D',\lambda '\rangle$ of ${\ \succsim \ }$, where $D'$ coincides with the Bewley set, and moreover, the uniqueness property of the representation implies that we can assume $u'=u$. To show that D is the smallest admissible set, it suffices to show that $D\subseteq D'$. Since D and $D'$ are closed and convex, by Proposition A.1 of Ghirardato et al. (2004), it is sufficient to show that for each pair $\varphi ,\varphi '\in {{\mathbb {R}}}^S$, if for each $p\in D'$, $E_p\varphi \ge E_{p}\varphi '$, then for each $p\in D$, $E_p\varphi \ge E_{p}\varphi '$.

Let $\varphi ,\varphi '\in {{\mathbb {R}}}^S$ and $g,g'\in \mathcal {F}$ be such that $\varphi =u(g)$, $\varphi '=u(g')$, and for each $p\in D'$, $E_p\varphi \ge E_p\varphi '$. Let $\alpha \in (0,1]$ and $h\in \mathcal {F}$. For each $k>0$ and $t\in {{\mathbb {R}}}$, let $g_{k,t},g_{k,t}'\in \mathcal {F}$ be such that $u(g_{k,t})=ku(\alpha g+(1-\alpha )h)-t\mathbf 1$ and $u(g_{k,t}')=ku(\alpha g'+(1-\alpha )h)-t\mathbf 1$. For each $k>0$, each $t\in {{\mathbb {R}}}$, and each $p\in D'$, since $E_pu(g)=E_p\varphi \ge E_p\varphi '=E_pu(g')$, $E_pu(g_{k,t})\ge E_pu(g_{k,t}')$. Since $\langle u,D',\lambda '\rangle$ is a WM representation of ${\ \succsim \ }$ with $D'$ being the Bewley set, $\langle u,D'\rangle$ is a Bewley representation of the unambiguous preference relation ${\ \succsim \ }^*$ induced from ${\ \succsim \ }$. Thus, for each $k>0$ and each $t\in {{\mathbb {R}}}$, $g_{k,t}{\ \succsim \ }^*g_{k,t}'$, and hence, $g_{k,t}{\ \succsim \ }g_{k,t}'$. For each $k>0$ and each $t\in {{\mathbb {R}}}$, since $g_{k,t}{\ \succsim \ }g_{k,t}'$, $I(u(g_{k,t}))\ge I(u(g_{k,t}'))$, and thus, $I(ku(\alpha g+(1-\alpha )h)-t\mathbf 1 )\ge I(ku(\alpha g'+(1-\alpha )h)-t\mathbf 1 )$. Hence,

$$\begin{aligned}&J(u(\alpha g+(1-\alpha )h))\\&\quad =\lim \limits _{k\rightarrow \infty }\lim \limits _{t\rightarrow \infty }\frac{1}{k}[I(ku(\alpha g+(1-\alpha )h)-t\mathbf 1 )+t]\\&\quad \ge \lim \limits _{k\rightarrow \infty }\lim \limits _{t\rightarrow \infty }\frac{1}{k}[I(ku(\alpha g'+(1-\alpha )h)-t\mathbf 1 )+t]\\&\quad =J(u(\alpha g'+(1-\alpha )h)). \end{aligned}$$

Define ${\ \succsim \ }'$ on ${\mathcal {F}}$ by setting for each pair $f,f'\in {\mathcal {F}}$,

$$\begin{aligned} f{\ \succsim \ }' f'\ \Longleftrightarrow \ \min \limits _{p\in D}E_pu(f)\ge \min \limits _{p\in D}E_pu(f'). \end{aligned}$$

Let ${\ \succsim \ }'^*$ be the unambiguous preference relation induced from ${\ \succsim \ }'$. Since $J(u(\alpha g+(1-\alpha )h))\ge J(u(\alpha g'+(1-\alpha )h))$, and since for each $f\in {\mathcal {F}}$, we have shown that $J(u(f))=\min \limits _{p\in D}E_pu(p)$, $\alpha g+(1-\alpha )h{\ \succsim \ }'\alpha g'+(1-\alpha )h$. Since $\alpha g+(1-\alpha )h{\ \succsim \ }' \alpha g'+(1-\alpha )h$, and since $\alpha \in (0,1]$ and $h\in {\mathcal {F}}$ are arbitrarily chosen, $g{\ \succsim \ }'^* g'$. By Ghirardato et al. (2004),²⁹ for each $p\in D$, $E_pu(g)\ge E_pu(g')$, and thus, $E_p\varphi \ge E_p\varphi '$, as desired.

Next, we prove that statement 3 holds. By the similar arguments as in the second paragraph, we obtain the following observations. Let $\varphi \in {{\mathbb {R}}}^S$. First, for each $k>0$ and each $t\in {{\mathbb {R}}}$, $\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]\in [\varphi _*,\varphi ^*]$. Second, for each $k>0$, $\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]$ is non-decreasing in t on ${{\mathbb {R}}}$. Third, $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]$ is non-increasing in k on $(0,\infty )$.

Define $J':{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} J'(\varphi ):=\lim \limits _{k\searrow 0}\lim \limits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]. \end{aligned}$$

(45)

Because of the three observations above, $J'$ is well-defined, and for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi )\le J'(\varphi )$. Moreover, by the similar arguments as in the third paragraph, $J'$ is normalized, non-decreasing, constant additive, positive homogeneous of degree 1, and superadditive. Thus, by Lemma 3.5 of Gilboa and Schmeidler (1989), there is a unique non-empty, closed, and convex set $K\subseteq \Delta$ such that for each $\varphi \in {{\mathbb {R}}}^S$, $J'(\varphi )=\min \limits _{p\in K}E_p\varphi$.

We show that $K=\{p\in \Delta :$ for each $\varphi \in {{\mathbb {R}}}^S,I(\varphi )\le E_p\varphi \}$. Since for each $p'\in K$ and each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi )\le J'(\varphi )=\min \limits _{p\in K}E_p\varphi \le E_{p'}\varphi$, $K\subseteq \{p\in \Delta :$ for each $\varphi \in {{\mathbb {R}}}^S,I(\varphi )\le E_p\varphi \}$. To show that $\{p\in \Delta :$ for each $\varphi \in {{\mathbb {R}}}^S,I(\varphi )\le E_p\varphi \}\subseteq K$, suppose to the contrary that there is $p'\in \Delta {\setminus } K$ such that for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi )\le E_{p'}\varphi$. Then for each $\varphi \in {{\mathbb {R}}}^S$, each $k>0$, and each $t\in {{\mathbb {R}}}$, $I(k\varphi +t\mathbf 1 )\le E_{p'}(k\varphi +t\mathbf 1 )=kE_{p'}\varphi +t$, and thus, $\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]\le E_{p'}\varphi$. Hence, for each $\varphi \in {{\mathbb {R}}}^S$, $\min \limits _{p\in K}E_p\varphi =J'(\varphi )=\lim \nolimits _{k\searrow 0}\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]\le E_{p'}\varphi$. However, since $p'\notin K$, by a standard separation theorem, there is $\varphi '\in {{\mathbb {R}}}^S$ such that $E_{p'}\varphi '<\min \limits _{p\in K}E_p\varphi '$, which contradicts that for each $\varphi \in {{\mathbb {R}}}^S$, $\min \limits _{p\in K}E_p\varphi \le E_{p'}\varphi$, as desired.

Define $B:{{\mathbb {R}}}^S\rightrightarrows \Delta$ and $\sigma :{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}_+$ by setting for each $\varphi \in {{\mathbb {R}}}^S$,

$$\begin{aligned} B(\varphi ):=\{p\in \Delta :I(\varphi )=E_p\varphi \}, \end{aligned}$$

and

$$\begin{aligned} \sigma (\varphi ):=\min \limits _{p\in B(\varphi )}d(p,K). \end{aligned}$$

Since K is non-empty and $K=\{p\in \Delta :$ for each $\varphi \in {{\mathbb {R}}}^S,I(\varphi )\le E_p\varphi \}$, the latter set is also non-empty. Then by the same arguments as in the proof of the sufficiency of the axioms in Theorem 2, $\langle u,K,\sigma \rangle$ is a canonical VC representation of ${\ \succsim \ }$, and for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi )=\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi )}E_p\varphi$. Moreover, by the analogous arguments as in the first paragraph of the proof of Theorem 1, $\sigma \in \bar{\Sigma }(K)$.

To complete the proof that statement 1 implies statement 3, we fix an arbitrary $\varphi \in {{\mathbb {R}}}^S$ and show the following three properties of $\sigma$. First, $\sigma (\varphi +t\mathbf 1 )$ is non-increasing in t on ${{\mathbb {R}}}$. Let $t,t'\in {{\mathbb {R}}}$ be such that $t\le t'$. We want to show that $\sigma (\varphi +t\mathbf 1 )\ge \sigma (\varphi +t'{} \mathbf 1 )$. Let $p^*\in B(\varphi +t\mathbf 1 )$ be such that $d(p^*,K)=\min \limits _{p\in B(\varphi +t\mathbf 1 )}d(p,K)$. By the definition of $\sigma$, $d(p^*,K)=\sigma (\varphi +t\mathbf 1 )$. Since $p^*\in B(\varphi +t\mathbf 1 )$, by the definition of B, $I(\varphi +t\mathbf 1 )=E_{p^*}(\varphi +t\mathbf 1 )$. Since $I(\varphi +t\mathbf 1 )=E_{p^*}(\varphi +t\mathbf 1 )$ and $t\le t'$, and since I is constant superadditive, $I(\varphi +t'{} \mathbf 1 )\ge E_{p^*}(\varphi +t'{} \mathbf 1 )$. Assume first that $I(\varphi +t'{} \mathbf 1 )=E_{p^*}(\varphi +t'{} \mathbf 1 )$. Then $p^*\in B(\varphi +t'{} \mathbf 1 )$, and thus, $\sigma (\varphi +t\mathbf 1 )=d(p^*,K)\ge \min \limits _{p\in B(\varphi +t'{} \mathbf 1 )}d(p,K)=\sigma (\varphi +t'{} \mathbf 1 )$, as desired. Assume now that $I(\varphi +t'{} \mathbf 1 )>E_{p^*}(\varphi +t'{} \mathbf 1 )$. Suppose to the contrary that $\sigma (\varphi +t\mathbf 1 )<\sigma (\varphi +t'{} \mathbf 1 )$. Then $d(p^*,K)=\sigma (\varphi +t\mathbf 1 )<\sigma (\varphi +t'{} \mathbf 1 )$. Thus, $I(\varphi +t'{} \mathbf 1 )=\min \limits _{p\in \Delta :d(p,K)\le \sigma (\varphi +t'{} \mathbf 1 )}E_{p}(\varphi +t'{} \mathbf 1 )\le E_{p^*}(\varphi +t'{} \mathbf 1 )<I(\varphi +t'{} \mathbf 1 )$, which is not possible. Hence, $\sigma (\varphi +t\mathbf 1 )\ge \sigma (\varphi +t'{} \mathbf 1 )$.

Second, $\lim \nolimits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )$ is non-decreasing in k on $(0,\infty )$. For each $k>0$, define $B_k:[0,\infty ]\rightrightarrows \Delta$ by setting for each $t\in [0,\infty ]$,

$$\begin{aligned} B_k(t):=\left\{ \begin{array}{ll} B(k\varphi +t\mathbf 1 ) &{} \hbox {if }t<\infty ,\\ \{p\in \Delta :\lim \limits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t] =E_p\varphi \}&{} \hbox {if}\ t=\infty . \end{array} \right. \end{aligned}$$

Thus, for each $k>0$ and each $t\in {{\mathbb {R}}}$, $\sigma (k\varphi +t\mathbf 1 )=\min \limits _{p\in B_k(t)}d(p,K)$. Before proving the second property of $\sigma$, we first fix $k>0$ and show that $\lim \nolimits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )=\min \limits _{p\in B_k(\infty )}d(p,K)$, or equivalently, $\lim \nolimits _{t\rightarrow \infty }\min \limits _{p\in B_k(t)}d(p,K)=\min \limits _{p\in B_k(\infty )}d(p,K)$.

By the maximum theorem, it suffices to prove that $B_k$ is continuous at $\infty$. The upper hemicontinuity can be readily seen, and thus, we omit the proof. For the lower hemicontinuity, let $\{t^n\}_{n=1}^{\infty }$ be a sequence of elements in ${{\mathbb {R}}}_+$ such that $\lim \nolimits _{n\rightarrow \infty }t^n=\infty$. Fix $p^{\infty }\in B_k(\infty )$. We will show that there is a sequence $\{p^n\}_{n=1}^{\infty }$ of elements in $\Delta$ such that for each $n\in {\mathbb {N}}$, $p^n\in B_k(t^n)$, and $\lim \nolimits _{n\rightarrow \infty }p^n=p^{\infty }$. Since $p^{\infty }\in B_k(\infty )$, $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t] =E_{p^{\infty }}\varphi$. For each $n\in {\mathbb {N}}$, by the first observation in the first paragraph of our proof for statement 1 implying statement 3, $\frac{1}{k}[I(k\varphi +t^n\mathbf 1 )-t^n]\ge \varphi _*$, and by the second observation, $\frac{1}{k}[I(k\varphi +t^n\mathbf 1 )-t^n]\le \lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]=E_{p^{\infty }}\varphi$. Let $p_*\in \Delta$ be such that $E_{p_*}\varphi =\varphi _*$. For each $n\in {{\mathbb {N}}}$, since $E_{p_*}\varphi =\varphi _*\le \frac{1}{k}[I(k\varphi +t^n\mathbf 1 )-t^n]\le E_{p^{\infty }}\varphi$, if $\varphi _*<E_{p^{\infty }}\varphi$, then there is a unique $\alpha ^n\in [0,1]$ satisfying that

$$\begin{aligned} \frac{1}{k}[I(k\varphi +t^n\mathbf 1 )-t^n]=E_{\alpha ^np_*+(1-\alpha ^n)p^{\infty }}\varphi , \end{aligned}$$

(46)

and if $\varphi _*=E_{p^{\infty }}\varphi$, each $\alpha ^n\in [0,1]$ satisfies (46), and we choose $\alpha ^n:=0$ in this case. Let for each $n\in {\mathbb {N}}$, $p^n:=\alpha ^np_*+(1-\alpha ^n)p^{\infty }$. Thus, for each $n\in {\mathbb {N}}$, $p^n\in \Delta$ and $\frac{1}{k}[I(k\varphi +t^n\mathbf 1 )-t^n]=E_{p^n}\varphi$, the latter of which implies that $I(k\varphi +t^n\mathbf 1 )=E_{p^n}(k\varphi +t^n\mathbf 1 )$, i.e., $p^n\in B_k(t_n)$. When $\varphi _*<E_{p^{\infty }}\varphi$, since $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t] =E_{p^{\infty }}\varphi$, $\lim \nolimits _{n\rightarrow \infty }p^n=p^{\infty }$. When $\varphi _*=E_{p^{\infty }}\varphi$, by definition, for each $n\in {\mathbb {N}}$, $p^n=p^{\infty }$, so that $\lim \nolimits _{n\rightarrow \infty }p^n=p^{\infty }$.

To show that second property of $\sigma$, let $k,k'>0$ be such that $k\le k'$. We want to show that $\lim \nolimits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )\le \lim \nolimits _{t\rightarrow \infty }\sigma (k'\varphi +t\mathbf 1 )$. By the above result, it is equivalent to show that $\min \limits _{p\in B_k(\infty )}d(p,K)\le \min \limits _{p\in B_{k'}(\infty )}d(p,K)$. Let $p^*\in B_k(\infty )$, ${p^*}'\in B_{k'}(\infty )$ and $q^*,{q^*}'\in K$ be such that $d(p^*,q^*)=\min \limits _{p\in B_k(\infty )}d(p,K)$ and $d({p^*}',{q^*}')=\min \limits _{p\in B_{k'}(\infty )}d(p,K)$. We want to show that $d(p^*,q^*)\le d({p^*}',{q^*}')$. Since $p^*\in B_k(\infty )$ and ${p^*}'\in B_{k'}(\infty )$, $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t] =E_{p^*}\varphi$ and $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k'}[I(k'\varphi +t\mathbf 1 )-t] =E_{{p^*}'}\varphi$. By the third observation, $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k'}[I(k'\varphi +t\mathbf 1 )-t]\le \lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]$, and thus, $E_{{p^*}'}\varphi \le E_{{p^*}}\varphi$. By the definition of $J'$ in (45) and the third observation again, $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]\le J'(\varphi )$, and thus, $E_{p^*}\varphi \le J'(\varphi )$. Since $J'(\varphi )=\min \limits _{p\in K}E_p\varphi$ and ${q^*}'\in K$, $J'(\varphi )\le E_{{q^*}'}\varphi$. Then $E_{{p^*}'}\varphi \le E_{p^*}\varphi \le J'(\varphi )\le E_{{q^*}'}\varphi$. Hence, there is $\alpha \in [0,1]$ such that $E_{p^*}\varphi =E_{\alpha {p^*}'+(1-\alpha ){q^*}'}\varphi$, and thus, $\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]=E_{\alpha {p^*}'+(1-\alpha ){q^*}'}\varphi$, i.e., $\alpha {p^*}'+(1-\alpha ){q^*}'\in B_k(\infty )$. Since $\alpha {p^*}'+(1-\alpha ){q^*}'\in B_k(\infty )$ and ${q^*}'\in K$, $\min \limits _{p\in B_k(\infty )}d(p,K)\le d(\alpha {p^*}'+(1-\alpha ){q^*}',{q^*}')$. Thus, $d(p^*,q^*)=\min \limits _{p\in B_k(\infty )}d(p,K)\le d(\alpha {p^*}'+(1-\alpha ){q^*}',{q^*}')=\alpha d({p^*}',{q^*}')\le d({p^*}',{q^*}')$, as desired.

The last property of $\sigma$ is that $\lim \nolimits _{k\searrow 0}\lim \nolimits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )=0$. Let $p_*\in \Delta$ be such that $E_{p_*}\varphi =\varphi _*$. Let $q^*\in K$ be such that $E_{q^*}\varphi =\min \limits _{p\in K}E_p\varphi$. For each $k>0$ and each $t\in {{\mathbb {R}}}$, since $J'(k\varphi +t\mathbf 1 )=\min \limits _{p\in K}E_p(k\varphi +t\mathbf 1 )=kE_{q^*}\varphi +t$ and $J'(k\varphi +t\mathbf 1 )\ge I(k\varphi +t\mathbf 1 )$, $kE_{q^*}\varphi +t\ge I(k\varphi +t\mathbf 1 )$; since I is normalized and non-decreasing, $I(k\varphi +t\mathbf 1 )\ge k\varphi _*+t$. Hence, for each $k>0$ and each $t\in {{\mathbb {R}}}$, $kE_{q^*}\varphi +t\ge I(k\varphi +t\mathbf 1 )\ge k\varphi _*+t=kE_{p_*}\varphi +t$.

Assume that $E_{q^*}\varphi =E_{p_*}\varphi$. Then for each $k>0$ and each $t\in {{\mathbb {R}}}$, $kE_{q^*}\varphi +t=kE_{p_*}\varphi +t$, and since $kE_{q^*}\varphi +t\ge I(k\varphi +t\mathbf 1 )\ge kE_{p_*}\varphi +t$, $I(k\varphi +t\mathbf 1 )=kE_{q^*}\varphi +t=E_{q^*}(k\varphi +t\mathbf 1 )$, i.e., $q^*\in B(k\varphi +t\mathbf 1 )$. Since for each $k>0$ and each $t\in {{\mathbb {R}}}$, $q^*\in B(k\varphi +t\mathbf 1 )$ and $q\in K$, $\sigma (k\varphi +t\mathbf 1 )=\min \limits _{p\in B(k\varphi +t\mathbf 1 )}d(p,K)=0$. Thus, $\lim \nolimits _{k\searrow 0}\lim \nolimits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )=0$.

Assume that $E_{q^*}\varphi >E_{p_*}\varphi$. For each $k>0$ and each $t\in {{\mathbb {R}}}$, since $kE_{q^*}\varphi +t\ge I(k\varphi +t\mathbf 1 )\ge kE_{p_*}\varphi +t$, $E_{q^*}\varphi \ge \frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]\ge E_{p_*}\varphi$. Hence, for each $k>0$, $E_{q^*}\varphi \ge \lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]\ge E_{p_*}\varphi$. Since $E_{q^*}\varphi > E_{p_*}\varphi$, for each $k>0$,

$$\begin{aligned} \alpha _k:=\dfrac{1}{E_{q^*}\varphi -E_{p_*}\varphi }\left[ E_{q^*}\varphi -\lim \limits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]\right] \end{aligned}$$

is well-defined. It can be readily seen that for each $k>0$, $\alpha _k\in [0,1]$ and $E_{\alpha _kp_*+(1-\alpha _k)q^*}\varphi =\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]$. For each $k>0$, recall the definition of $B_k(\infty )$ in the proof of the second property of $\sigma$, and recall our previous result that $\lim \nolimits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )=\min \limits _{p\in B_k(\infty )}d(p,K)$. For each $k>0$, since $E_{\alpha _kp_*+(1-\alpha _k)q^*}\varphi =\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]$, $\alpha _kp_*+(1-\alpha _k)q^*\in B_k(\infty )$, and moreover, since $q^*\in K$, $\min \limits _{p\in B_k(\infty )}d(p,K)\le d(\alpha _kp_*+(1-\alpha _k)q^*,q^*)$. Thus, for each $k>0$,

$$\begin{aligned} 0\le \lim \limits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )=\min \limits _{p\in B_k(\infty )}d(p,K)\le d(\alpha _kp_*+(1-\alpha _k)q^*,q^*)=\alpha _k d(p_*,q^*). \end{aligned}$$

(47)

Since $J'(\varphi )=\min \limits _{p\in K}E_p\varphi =E_{q^*}\varphi$ and by the definition of $J'$ in (45), $\lim \nolimits _{k\searrow 0}\lim \nolimits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]=E_{q^*}\varphi$. Thus,

$$\begin{aligned}&\lim \limits _{k\searrow 0}\alpha _k=\dfrac{1}{E_{q^*}\varphi -E_{p_*}\varphi }\left[ E_{q^*}\varphi -\lim \limits _{k\searrow 0}\lim \limits _{t\rightarrow \infty }\frac{1}{k}[I(k\varphi +t\mathbf 1 )-t]\right] \\&\quad =\dfrac{1}{E_{q^*}\varphi -E_{p_*}\varphi }[E_{q^*}\varphi -E_{q^*}\varphi ]=0. \end{aligned}$$

Thus, by taking the limits with respect to k on both sides of (47),

$$\begin{aligned} 0\le \lim \limits _{k\searrow 0}\lim \limits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )\le 0, \end{aligned}$$

which implies that $\lim \nolimits _{k\searrow 0}\lim \nolimits _{t\rightarrow \infty }\sigma (k\varphi +t\mathbf 1 )=0$, as desired.

Lastly, we prove that statement 4 holds. Since A.2.1 implies A.2, ${\ \succsim \ }$ satisfies A.1–A.4, A.5.1, and A.6. Then by Theorems 3 and 5 of Cerreia-Vioglio et al. (2011b), ${\ \succsim \ }$ admits an ambiguity averse representation $\langle u',G\rangle$, and the uniqueness property of their representation implies that we can assume $u'=u$. Moreover, they show that $G:{{\mathbb {R}}}\times \Delta \rightarrow (-\infty ,\infty ]$ is given by, for each $(t,p)\in {{\mathbb {R}}}\times \Delta$,

$$\begin{aligned} G(t,p)=\sup \{u(x_f):f\in {\mathcal {F}},E_pu(f)\le t\}. \end{aligned}$$

(48)

Fix $p\in \Delta$. Let $t,t'\in {{\mathbb {R}}}$ be such that $t\le t'$. We want to show that $G(t,p)-t\le G(t',p)-t'$. Let $f\in {\mathcal {F}}$ be such that $E_pu(f)\le t$. Since u is onto, there is $f'\in {\mathcal {F}}$ satisfying that $u(f')=u(f)+(t'-t)\mathbf 1$, so that $E_pu(f')\le t'$. Since u and I satisfy (22), and since I is normalized and constant superadditive, $u(x_{f'})=I(u(f'))=I(u(f)+(t'-t)\mathbf 1 )\ge I(u(f))+t'-t=u(x_f)+t'-t$. Thus, $u(x_{f'})-t'\ge u(x_f)-t$. Since for each $f\in {\mathcal {F}}$ such that $E_pu(f)\le t$, there is $f'\in {\mathcal {F}}$ such that $E_pu(f')\le t'$ and $u(x_{f'})-t'\ge u(x_f)-t$, by (48), $G(t',p)-t' \ge G(t,p)-t$, as desired.

We now prove that statements 2, 3, and 4 all imply statement 1. Assume that ${\ \succsim \ }$ admits a WM representation $\langle u,D,\lambda \rangle$ with the properties in statement 2. To show that ${\ \succsim \ }$ satisfies A.1, A.3, A.4, A.5.1, and A.6, the same arguments for the necessity of the axioms in the proof of Theorem 1 can be applied since they rely on the same properties of u, D, and $\lambda$ as we have here. Using the fact that for each $\varphi \in {{\mathbb {R}}}^S$, $\lambda (\varphi +t\mathbf 1 )$ is non-increasing in t, it can be readily seen that ${\ \succsim \ }$ satisfies A.2.1.

Assume that ${\ \succsim \ }$ admits a VC representation $\langle u,K,\sigma \rangle$ with the properties in statement 3. To show that ${\ \succsim \ }$ satisfies A.1, A.3, A.4, and A.6, the same arguments for the necessity of the axioms in the proof of Theorem 2 can be applied since they rely on the same properties of u, K, and $\sigma$ as we have here. Using the fact that for each $\varphi \in {{\mathbb {R}}}^S$, $\sigma (\varphi +t\mathbf 1 )$ is non-increasing in t, it can be readily seen that ${\ \succsim \ }$ satisfies A.2.1. Using the fact that $\sigma \in {\bar{\Sigma }}(K)$, one can readily show that ${\ \succsim \ }$ satisfies A.5.1.

Assume that ${\ \succsim \ }$ admits an ambiguity averse representation $\langle u,G\rangle$ with the property in statement 4. By Theorems 3 and 5 of Cerreia-Vioglio et al. (2011b), ${\ \succsim \ }$ satisfies A.1, A.3, A.4, A.5.1, and A.6. Using the fact that for each $p\in \Delta$, $G(t,p)-t$ is non-decreasing in t, one can readily show that ${\ \succsim \ }$ satisfies A.2.1. $\square$

Proof of Proposition 5

Let ${\ \succsim \ }$ satisfies A.1, A.2.1, A.3, A.4, A.5.1, and A.6. Let $\langle u,D,\lambda \rangle$ be a canonical WM representation of ${\ \succsim \ }$. It can be readily seen that A.2.1 implies A.2. Thus, by Theorem 1, D is the Bewley set, i.e., $\langle u,D\rangle$ is a Bewley representation of the unambiguous preference relation ${\ \succsim \ }^*$ induced from ${\ \succsim \ }$. Then by Proposition 11 of Cerreia-Vioglio et al. (2011b), $D=cl(co(\bigcup \limits _{f\in \mathcal {F}}\pi (f)))$. To prove that for each $x\in X$, $D=cl(co(\bigcup \limits _{f\sim x}\pi (f)))$, it suffices to prove that for each pair $y,z\in X$, $cl(co(\bigcup \limits _{f\sim y}\pi (f)))=cl(co(\bigcup \limits _{f\sim z}\pi (f)))$. Fix $y,z\in X$. Assume without loss of generality that $y\succ z$.

Define $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}^S$, $I(\varphi ):=\lambda (\varphi )\min \limits _{p\in D}E_p\varphi +(1-\lambda (\varphi ))\max \limits _{p\in D}E_p\varphi$. By the similar arguments as in the proof of the necessity of the axioms in Proposition 1 and in the first paragraph of the proof of Theorem 1, I is non-decreasing, continuous, and quasi-concave. By the same arguments as in the second paragraph of the proof of Lemma 3, I is constant superadditive.

For each $x\in X$ and each $\varphi \in {{\mathbb {R}}}^S$, let $t_x(\varphi )\in {{\mathbb {R}}}$ be such that $I(\varphi -t_x(\varphi )\mathbf 1 )=u(x)$. Since I is constant superadditive, $t_x(\varphi )$ is uniquely determined. For each $x\in X$, define ${\ \succsim \ }_x$ on ${\mathcal {F}}$ by setting for each pair $f,g\in {\mathcal {F}}$,

$$\begin{aligned} f{\ \succsim \ }_x g\ \Longleftrightarrow \ t_x(u(f))\ge t_x(u(g)). \end{aligned}$$

Let $\sim _x$ denote its indifference component.

Fix $x\in X$. The following four observations about $\sim _x$ will be useful. First, for each $x'\in X$, $t_x(u(x'))=u(x')-u(x)$, and in particular, $t_x(u(x))=0$. This follows immediately from the definitions of I and $t_x(u(x'))$. Second, for each $f\in {\mathcal {F}}$,

$$\begin{aligned} (i)\ x{\ \succsim \ }f\ \Longleftrightarrow \ x{\ \succsim \ }_x f,\text { and }(ii)\ x\sim f\ \Longleftrightarrow \ x\sim _x f. \end{aligned}$$

(49)

To see (i) of (49), note that

$$\begin{aligned}&x{\ \succsim \ }f\ \Longleftrightarrow \ u(x)\ge I(u(f))\ \Longleftrightarrow \ 0\ge t_x(u(f))\ \Longleftrightarrow \ t_x(u(x))\\&\quad \ge t_x(u(f))\ \Longleftrightarrow \ x{\ \succsim \ }_{x}f, \end{aligned}$$

and by similar arguments, (ii) of (49) is true. Third, for each $\varphi \in {\mathcal {F}}$ and each $c\in {\mathbb {R}}$, $t_x(\varphi +c\mathbf 1 )=t_x(\varphi )+c$. This is because I is constant superadditive and $I(\varphi +c\mathbf 1 -t_x(\varphi +c\mathbf 1 )\mathbf 1 )=u(x)=I(\varphi -t_x(\varphi )\mathbf 1 )=I(\varphi +c\mathbf 1 -(t_x(\varphi )+c)\mathbf 1 )$. Fourth, ${\ \succsim \ }_x$ satisfies A.1–A.4, A.5.1, and A.6. By construction, ${\ \succsim \ }_x$ satisfies A.1. Since u is affine and onto, by the first observation, so is $t_x$ when restricted to constant vectors, and thus, ${\ \succsim \ }_x$ satisfies A.2 and A.6. Since I is continuous and quasi-concave, by routine arguments, so is $t_x$, and thus ${\ \succsim \ }_x$ satisfies A.3 and A.5.1. Since I is constant superadditive, it can also be readily seen that ${\ \succsim \ }_x$ satisfies A.4.

For each $x\in X$, define $\pi _x:\mathcal {F}\Rightarrow \Delta$ by setting for each $f\in {\mathcal {F}}$,

$$\begin{aligned} \pi _x(f):=\left\{ p\in \Delta :\sum \limits _{s\in S}p_s f(s){\ \succsim \ }_x\sum \limits _{s\in S}p_s g(s)\text { where }g\in {\mathcal {F}} \text { implies }f{\ \succsim \ }_x g\right\} . \end{aligned}$$

We claim that $\bigcup \limits _{f\in {\mathcal {F}}}\pi _x(f)=\bigcup \limits _{f\sim x}\pi (f)$. We first show that $\bigcup \limits _{f\sim x}\pi (f)\subseteq \bigcup \limits _{f\in {\mathcal {F}}}\pi _x(f)$. Let $x\in X$, $f\in {\mathcal {F}}$ with $f\sim x$, and $p\in \pi (f)$. We want to show that $p\in \pi _x(f)$. Let $g\in {\mathcal {F}}$ be such that $\sum \limits _{s\in S}p_s f(s){\ \succsim \ }_x\sum \limits _{s\in S}p_s g(s)$. Then by the first observation in the previous paragraph, $\sum \limits _{s\in S}p_s f(s){\ \succsim \ }\sum \limits _{s\in S}p_s g(s)$. Since $p\in \pi (f)$, $\sum \limits _{s\in S}p_s f(s){\ \succsim \ }\sum \limits _{s\in S}p_s g(s)$ implies $f{\ \succsim \ }g$. Since $f\sim x$ and $f{\ \succsim \ }g$, $x{\ \succsim \ }g$. Since $f\sim x$ and $x{\ \succsim \ }g$, by applying the second observation in the previous paragraph twice, $f{\ \succsim \ }_x g$, as desired.

We now show that $\bigcup \limits _{f\in {\mathcal {F}}}\pi _x(f)\subseteq \bigcup \limits _{f\sim x}\pi (f)$. Let $x\in X$, $f\in {\mathcal {F}}$, and $p\in \pi _x(f)$. We need to show that there is $f'\in {\mathcal {F}}$ such that $f'\sim x$ and $p\in \pi (f')$. Since u is onto, we can pick $f'\in {\mathcal {F}}$ satisfying that

$$\begin{aligned} u(f')=u(f)-t_x(u(f))\mathbf 1 . \end{aligned}$$

(50)

Since $I(u(f'))=I(u(f)-t_x(u(f))\mathbf 1 )=u(x)$, $f'\sim x$. To show that $p\in \pi (f')$, let $g'\in {\mathcal {F}}$ be such that $\sum \limits _{s\in S}p_s f'(s){\ \succsim \ }\sum \limits _{s\in S}p_s g'(s)$, and we need to show that $f'{\ \succsim \ }g'$, or equivalently, $x{\ \succsim \ }g'$. Since u is onto, we can pick $g\in {\mathcal {F}}$ satisfying that

$$\begin{aligned} u(g)=u(g')+t_x(u(f))\mathbf 1 . \end{aligned}$$

(51)

Since $\sum \limits _{s\in S}p_s f'(s){\ \succsim \ }\sum \limits _{s\in S}p_s g'(s)$ and u is affine, $\sum \limits _{s\in S}p_s u(f'(s))\ge \sum \limits _{s\in S}p_s u(g'(s))$. Thus, by (50) and (51), $\sum \limits _{s\in S}\!p_s u(f(s))\!\ge \! \sum \limits _{s\in S}p_s u(g(s))$. Hence, $\sum \limits _{s\in S}p_s f(s)\!{\ \succsim \ }\!\sum \limits _{s\in S}\!p_s g(s)$. Then by the first observation, $\sum \limits _{s\in S}p_s f(s){\ \succsim \ }_x\sum \limits _{s\in S}p_s g(s)$. Since $p\in \pi _x(f)$, $\sum \limits _{s\in S}p_s f(s){\ \succsim \ }_x\sum \limits _{s\in S}p_s g(s)$ implies $f{\ \succsim \ }_x g$, i.e., $t_x(u(f))\!\ge \! t_x(u(g))$. Since $t_x(u(f))\ge t_x(u(g))$, by the third observation and by (50) and (51), $t_x(u(f'))\ge t_x(u(g'))$, i.e., $f'{\ \succsim \ }_xg'$. Since $f'\sim x$, by the second observation, $f'\sim _x x$. Thus, $x{\ \succsim \ }_x g'$, and again by the second observation, $x{\ \succsim \ }g'$, as desired.

Next, we claim that ${\ \succsim \ }_z$ is more ambiguity averse than ${\ \succsim \ }_y$. To show this, let $x\in X$ and $f\in {\mathcal {F}}$ be such that $x{\ \succsim \ }_y f$. We need to show that $x{\ \succsim \ }_z f$. Since $x{\ \succsim \ }_y f$, by the first observation, $u(x)-u(y)=t_y(u(x))\ge t_y(u(f))$. Since $I(u(f)-t_y(u(f))\mathbf 1 )=u(y)$ and $u(x)-u(y)\ge t_y(u(f))$, and since I is non-decreasing, $I(u(f)-(u(x)-u(y))\mathbf 1 )\le u(y)$. Thus, since I is constant superadditive and $y\succ z$,

$$\begin{aligned}&I(u(f)-(u(x)-u(z))\mathbf 1 )\\&\quad =I(u(f)-(u(x)-u(y))\mathbf 1 -(u(y)-u(z))\mathbf 1 )\\&\quad \le I(u(f)-(u(x)-u(y))\mathbf 1 )-(u(y)-u(z))\\&\quad \le u(y)-(u(y)-u(z))=u(z). \end{aligned}$$

Since $I(u(f)-(u(x)-u(z))\mathbf 1 )\le u(z)=I(u(f)-t_z(u(f))\mathbf 1 )$, and since I is constant superadditive, $u(x)-u(z)\ge t_z(u(f))$. Thus, by the first observation, $t_z(u(x))=u(x)-u(z)\ge t_z(u(f))$, i.e., $x{\ \succsim \ }_z f$, as desired.

Since ${\ \succsim \ }_z$ is more ambiguity averse than ${\ \succsim \ }_y$, by Propositions 6 and 11 and Theorem 10 of Cerreia-Vioglio et al. (2011b), and by the first observation, $cl(co(\bigcup \limits _{f\in \mathcal {F}}\pi _y(f)))\subseteq cl(co(\bigcup \limits _{f\in \mathcal {F}}\pi _z(f)))$.³⁰ We have shown that for each $x\in X$, $\bigcup \limits _{f\in \mathcal {F}}\pi _x(f)=\bigcup \limits _{f\sim x}\pi (f)$. Thus, $cl(co(\bigcup \limits _{f\sim y}\pi (f)))\subseteq cl(co(\bigcup \limits _{f\sim z}\pi (f)))$.

Conversely, we show that $cl(co(\bigcup \limits _{f\sim z}\pi (f)))\subseteq cl(co(\bigcup \limits _{f\sim y}\pi (f)))$. It is sufficient to show that $\bigcup \limits _{f\sim z}\pi (f)\subseteq cl(co(\bigcup \limits _{f\sim y}\pi (f)))$. Suppose to the contrary that there exist $g\in {\mathcal {F}}$ with $g\sim z$ and $q\in \pi (g){\setminus } cl(co(\bigcup \limits _{f\sim y}\pi (f)))$. Then by a standard separation theorem, there is $\varphi \in {{\mathbb {R}}}^S$ such that for each $p\in cl(co(\bigcup \limits _{f\sim y}\pi (f)))$, $E_q\varphi<0<E_p\varphi$. Pick $n\in N$ such that $E_qn\varphi <E_qu(g)$ and for each $p\in cl(co(\bigcup \limits _{f\sim y}\pi (f)))$, $E_pn\varphi >u(y)$. Let $h\in \mathcal {F}$ be such that $u(h)=n\varphi$. Since $E_qu(g)>E_qn\varphi =E_qu(h)$, and since $q\in \pi (g)$, $g{\ \succsim \ }h$. Since $g\sim z$ and $g{\ \succsim \ }h$, $z{\ \succsim \ }h$. By the fourth observation, ${\ \succsim \ }_y$ satisfies A.1–A.4, A.5.1, and A.6. Thus, by the first observation and by applying Proposition 11 of Cerreia-Vioglio et al. (2011b) for ${\ \succsim \ }_y$, if for each $p\in cl(co(\bigcup \limits _{f\in {\mathcal {F}}}\pi _y(f)))$, $E_pu(h)\ge u(y)$, then $h{\ \succsim \ }_y^* y$, where ${\ \succsim \ }^*$ is the unambiguous preference relation induced from ${\ \succsim \ }_y$. Since $cl(co(\bigcup \limits _{f\in {\mathcal {F}}}\pi _y(f)))=cl(co(\bigcup \limits _{f\sim y}\pi (f)))$, and since for each $p\in cl(co(\bigcup \limits _{f\sim y}\pi (f)))$, $E_pu(h)=E_pn\varphi >u(y)$, $h{\ \succsim \ }_y^*y$, which in particular implies that $h{\ \succsim \ }y$. Since $h{\ \succsim \ }y$ and $y\succ z$, $h\succ z$, which contradicts that $z{\ \succsim \ }h$, as desired.$\square$

Proof of Proposition 6

Let ${\ \succsim \ }$ satisfy A.1–A.4, and A.6. We first prove the sufficiency of A.2.3. Assume that ${\ \succsim \ }$ also satisfies A.2.3. By Lemma 1, there exist an affine onto utility function $u:X\rightarrow {{\mathbb {R}}}$ and a normalized, monotone, and continuous functional $I:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ satisfying (22).

We claim that I is constant additive. Since ${\ \succsim \ }$ satisfies A.2.3, by the similar arguments as in the second paragraph of the proof of Lemma 3, for each $\varphi \in {{\mathbb {R}}}^S$ and each $t\in {{\mathbb {R}}}$, $I(\varphi +t\mathbf 1 )\ge I(\varphi )+t$, and hence $I((\varphi +t\mathbf 1 )-t\mathbf 1 )\ge I(\varphi +t\mathbf 1 )-t$. Then for each $\varphi \in {{\mathbb {R}}}^S$ and each $t\in {{\mathbb {R}}}$,

$$\begin{aligned} I(\varphi +t\mathbf 1 )\ge I(\varphi )+t=I(\varphi +t\mathbf 1 -t\mathbf 1 )+t\ge I(\varphi +t\mathbf 1 ), \end{aligned}$$

which implies that $I(\varphi +t\mathbf 1 )=I(\varphi )+t$, as desired.

Let $\alpha \in (0,1)$, $x,y\in X$, and $f,g\in {\mathcal {F}}$ be such that $f\unrhd g$ and $\alpha f+(1-\alpha )x{\ \succsim \ }\alpha g+(1-\alpha )x$. Thus, since u is affine and I satisfies 22, $I(\alpha u(f)+(1-\alpha )u(x))\ge I(\alpha u(g)+(1-\alpha )u(x))$. Then since I is constant additive,

$$\begin{aligned}&I(\alpha u(f)+(1-\alpha )u(y))\\&\quad =I(\alpha u(f)+(1-\alpha )u(x))+(1-\alpha )(u(y)-u(x))\\&\quad \ge I(\alpha u(g)+(1-\alpha )u(x))+(1-\alpha )(u(y)-u(x))\\&\quad =I(\alpha u(g)+(1-\alpha )u(y)). \end{aligned}$$

Hence, $\alpha f+(1-\alpha )y{\ \succsim \ }\alpha g+(1-\alpha )y$, as desired.

The necessity of A.2.3 follows from the fact that for each $f\in {\mathcal {F}}$ and $z\in X$, $f\unrhd z$, which can be readily seen by definition.$\square$

Proof of Proposition 7

Let $u:{{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ be an affine onto function and $\phi :{{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ an increasing, concave, and twice differentiable function. For each countably additive Borel probability measure $\mu$ on $\Delta$, define $I_{\mu }:{{\mathbb {R}}}^S\rightarrow {{\mathbb {R}}}$ by setting for each $\varphi \in {{\mathbb {R}}}$,

$$\begin{aligned} I_{\mu }(\varphi ):=\phi ^{-1}\left( \int \limits _{p\in \Delta }\phi (E_p\varphi )d\mu (p)\right) . \end{aligned}$$

By definition, I is normalized.

To show the “if” direction, assume that $-\frac{\phi ''}{\phi '}$ is non-increasing. Let ${\ \succsim \ }$ admit a smooth ambiguity representation $\langle u,\phi ,\mu \rangle$ for some countably additive Borel probability measure $\mu$ on $\Delta$. We want to show that ${\ \succsim \ }$ satisfies A.2.1. Since $\langle u,\phi ,\mu \rangle$ represents ${\ \succsim \ }$, I satisfies (22). By the analogous version of Lemma 52 in Cerreia-Vioglio et al. (2011b), and by the analogous arguments as in the proof of their Proposition 53, I is constant superadditive. It can be readily seen that since I satisfies (22), and since I is normalized and constant superadditive, ${\ \succsim \ }$ satisfies A.2.1.

Conversely, assume that each preference relation that admits a smooth ambiguity representation $\langle u,\phi ,\mu \rangle$ for some countably additive Borel probability measure $\mu$ on $\Delta$ satisfies A.2.1. We want to show that $-\frac{\phi ''}{\phi '}$ is non-increasing. By the analogous version of Lemma 52 in Cerreia-Vioglio et al. (2011b), it is equivalent to show that for each $t\in {{\mathbb {R}}}_+$, $J_t:\phi ({{\mathbb {R}}})\rightarrow {{\mathbb {R}}}$, defined by setting for each $c\in \phi ({{\mathbb {R}}})$,

$$\begin{aligned} J_t(c)=\phi [\phi ^{-1}(c)+t], \end{aligned}$$

is convex on $\phi ({{\mathbb {R}}})$. Note that since $\phi$ is increasing, $\phi ^{-1}$ is well-defined. Let $c,c'\in \phi ({{\mathbb {R}}})$ and $\alpha \in [0,1]$. Let $x,x'\in X$ be such that $\phi (u(x))=c$ and $\phi (u(x'))=c'$. Let $s,s'\in S$ and $f\in {\mathcal {F}}$ be such that $f(s)=x$ and $f(s')=x'$. Let $\delta _s$ and $\delta _{s'}$ be the degenerate probability measures which assign probability one, respectively, to s and $s'$. Let $\mu$ be a probability measure on $\Delta$ which assigns probability $\alpha$ to $\delta _s$ and $1-\alpha$ to $\delta _{s'}$. Let ${\ \succsim \ }$ admit the smooth ambiguity representation $\langle u,\phi ,\mu \rangle$. Then ${\ \succsim \ }$ satisfies A.2.1. By the similar arguments as in the proof of Lemma 3, $I_{\mu }$ is constant superadditive. Then for each $t\in {{\mathbb {R}}}_+$ and each $\alpha \in [0,1]$,

$$\begin{aligned}&J_t(\alpha c+(1-\alpha )c')=J_t(\alpha \phi (u(f(s)))+(1-\alpha )\phi (u(f(s'))))\\&\quad =J_t\left( \int \limits _{p\in \Delta }\phi (E_pu(f))d\mu (p)\right) =\phi [I_{\mu }(u(f))+t] \le \phi [I_{\mu }(u(f)+t\mathbf 1 )]\\&\quad =\int \limits \phi (E_p(u(f)+t\mathbf 1 ))d\mu (p)=\alpha \phi (u(x)+t)+(1-\alpha )\phi (u(x')+t)\\&\quad =\alpha \phi [\phi ^{-1}(c)+t]+(1-\alpha )\phi [\phi ^{-1}(c')+t]=\alpha J_t(c)+(1-\alpha )J_t(c'), \end{aligned}$$

as desired.

Similarly, it can be shown that each preference relation that admits a smooth ambiguity representation $\langle u,\phi ,\mu \rangle$ for some countably additive Borel probability measure $\mu$ on $\Delta$ satisfies A.2.2 if and only if $-\frac{\phi ''}{\phi '}$ is non-decreasing.$\square$