Maxmin weighted expected utility: a simpler characterization

Abstract

Chateauneuf and Faro (J Math Econ 45:535–558, 2009) axiomatize a weighted version of maxmin expected utility over acts with nonnegative utilities, where weights are represented by a confidence function. We argue that their representation is only one of many possible, and we axiomatize a more natural form of maxmin weighted expected utility. We also provide stronger uniqueness results.

Appendices

Appendix 1: Proof of Theorem 4.2

Proof of Theorem 4.2

We assume that t is continuous and strictly decreasing, and that there exists some $\beta > 0$ such that $[\beta , \beta / \alpha _0^*(\phi )]\in \mathrm{range}(t)$. Recall that $\alpha _0^*(\phi ) = \max \{ \alpha _0, \min _{p \in \Delta (S)} \phi (p) \}$.

Let $\alpha '_0 = t^{-1}(\frac{\beta }{\alpha ^*_0})$ and for all $p\in \Delta (S)$, let $\phi '(p) = t^{-1}(\frac{\beta }{\phi (p)})$. It is easy to see that, for all acts f, g,

$$\begin{aligned}&\min _{p \in L_{\alpha _0} \phi } \frac{1}{\phi (p)} \int _S{ u(f) dp} \ge \min _{p \in L_{\alpha _0} \phi } \frac{1}{\phi (p)} \int _S{ u(g) dp} \\&\text {iff } \min _{\{ p : \phi (p) \ge \alpha '_0 \}} \frac{\beta }{\phi (p)} \int _S{ u(f) dp} \ge \min _{\{p : \phi (p) \ge \alpha '_0\}} \frac{\beta }{\phi (p)} \int _S{ u(g) dp} \\&\text {iff } \min _{p \in L_{\alpha '_0} \phi '} t({\phi '(p)}) \int _S{ u(f) dp} \ge \min _{p \in L_{\alpha '_0} \phi '} t({\phi '(p)}) \int _S{ u(g) dp}, \end{aligned}$$

since for all $p \in L_{\alpha _0} \phi$,

$$\begin{aligned} \frac{\beta }{\phi (p)} = t\left( t^{-1}\left( \frac{\beta }{\phi (p)}\right) \right) = t( \phi '(p) ). \end{aligned}$$

Now, we show that if $t(1) = \beta$, then $\phi '$ must be a regular* fuzzy set. Since $\phi$ is normal, there exists $p^*$ such that $\phi (p^*) = 1$. By definition of $\phi '$, $\phi '(p^*) = t^{-1}( \frac{\beta }{\phi (p^*)}) = t^{-1}( \beta ) = 1$, so $\phi '$ is normal.

To show that $\phi '$ is weakly* upper semi-continuous, we must show that the set $L_{\alpha } \phi ' = \{ p \in \Delta (S) : \phi '(p) \ge \alpha \}$ is weakly* closed for $\alpha \in [0,1]$. In other words, we have to show that the set $L_{\alpha } \phi '$ contains all of its limit points, for all $\alpha \in [0,1]$. Now, for $\alpha = 0$, $L_{\alpha } \phi ' = \Delta (S)$ and is closed. So consider the case $\alpha > 0$.

Recall from our definition of $\phi '$ that $\phi '(p) = t^{-1}( \frac{\beta }{\phi (p)} )$ for all p. Suppose $p_n \rightarrow p$. Observe that $\frac{\beta }{\phi (p_n)}$ is in the domain of $t^{-1}$ for all n, since $[\beta , \beta / \alpha _0^*(\phi )]\in \mathrm{range}(t)$. Note that for all p, $\phi '(p) \ge \alpha$ if and only if

$$\begin{aligned}&t^{-1}\left( \frac{\beta }{\phi (p)} \right) \ge \alpha \\ \text {iff }&\frac{\beta }{\phi (p)} \le t( \alpha ) \\ \text {iff }&\phi (p) \ge \frac{\beta }{t(\alpha )}, \end{aligned}$$

where $t(\alpha ) \ge \beta$ since $0 < \alpha \le 1$, t is monotonically decreasing, and $t(1) = \beta$. Since $\phi$ is assumed to be weakly* upper semi-continuous, and $\phi (p_n) \ge \frac{\beta }{t(\alpha )}$ for all n, we have $\phi (p) \ge \frac{\beta }{t(\alpha )}$. Therefore, $\phi '(p) \ge \alpha$, as required.

Finally, to show that $\phi '$ is quasi-concave, let $\gamma \in [0,1]$. Using the fact that t is strictly monotonically decreasing, we have that

$$\begin{aligned}&\phi ( \gamma p_1 + (1-\gamma ) p_2 ) \ge \min ( \phi ( p_1 ), \phi ( p_2 ) ) \\&\quad \Rightarrow \frac{\beta }{\phi ( \gamma p_1 + (1-\gamma ) p_2 ) } \le \max \left( \frac{\beta }{\phi ( p_1 ) }, \frac{\beta }{\phi ( p_2 )}\right) \\&\quad \Rightarrow t^{-1}\left( \frac{\beta }{\phi ( \gamma p_1 + (1-\gamma ) p_2 ) } \right) \ge \min \left( t^{-1}\left( \frac{\beta }{\phi ( p_1 ) }\right) , t^{-}\left( \frac{\beta }{\phi ( p_2 )}\right) \right) \\&\quad \Rightarrow \phi '( \gamma p_1 + (1-\gamma ) p_2 ) \ge \min ( \phi '( p_1 ), \phi '( p_2 ) ). \end{aligned}$$

For the other direction, suppose that $t(1) = \beta$ and that $\phi '$ is a regular* fuzzy confidence function. We want to show that $\phi$ defined by $\phi (p^*) = \frac{1}{t(\phi '(p^*))}$ is also regular* fuzzy. The arguments for this direction are analogous to those used to show the first direction. $\square$

Appendix 2: Details of Example 5.2

We now show that for all $n\ge 1$, $f_n \sim ^{+}_{\phi } \tilde{1}$.

Suppose $c \in [\frac{1}{2^{2m+1}}, \frac{1}{2^{2m-1}})$. The weighted expected utility of $f_n$ with respect to $p_c$ is

$$\begin{aligned} 2^m \left[ c 2^n + (1-c) \frac{ 2^n }{2^{2n+1} - 1}\right] ,\quad \text { for } m \in \{ 0,1,2, \ldots \}. \end{aligned}$$

If $m=n$, note that

$$\begin{aligned} 2^n \left[ \frac{1}{2^{2n+1}} 2^n + \frac{2^{2n+1} - 1}{2^{2n+1}} \frac{ 2^n }{2^{2n+1} - 1}\right] = 1. \end{aligned}$$

If $m < n$, then

$$\begin{aligned} 2^m \left[ c 2^n + (1-c) \frac{ 2^n }{2^{2n+1} - 1}\right]&\ge 2^m \left[ \frac{1}{2^{2m + 1}} 2^n + \frac{2^{2m-1} - 1}{2^{2m-1}} \frac{ 2^n }{2^{2n+1} - 1}\right] \\&= \frac{2^n}{2^{m + 1}} + \frac{2^{2m-1} - 1}{2^{m-1}} \frac{ 2^n }{2^{2n+1} - 1} \\&\ge \frac{2^n}{2^{m + 1}} \ge 1. \end{aligned}$$

If $m > n$, then

$$\begin{aligned} 2^m \left[ c 2^n + (1-c) \frac{ 2^n }{2^{2n+1} - 1}\right]&\ge 2^m \left[ \frac{1}{2^{2m + 1}} 2^n + \frac{2^{2m-1} - 1}{2^{2m-1}} \frac{ 2^n }{2^{2n+1} - 1}\right] \\&= \frac{2^n}{2^{m + 1}} + \frac{2^{2m-1} - 1}{2^{m-1}} \frac{ 2^n }{2^{2n+1} - 1} \\&\ge \frac{2^n}{2^{m + 1}} + \frac{2^{2m-1} - 1}{2^{m-1}} \frac{ 1 }{2^{n+1} } \\&\ge \frac{2^n}{2^{m + 1}} + \frac{2^{2m-1} }{2^{m+n}} - \frac{ 1 }{2^{m+n}} \\&\ge 2^{m-n-1} \ge 1. \end{aligned}$$

If $c \in [ \frac{1}{2}, 1]$, then the weighted expected utility of $f_n$ is

$$\begin{aligned} c 2^n + (1-c) \frac{ 2^n }{2^{2n+1} - 1}\ge c 2^n \ge 2^{n-{1}}. \end{aligned}$$

Therefore, for all n, the minimum weighted expected utility of $f_n$ is 1, so $f_n \sim ^{+}_{\phi } \tilde{1}$.

Now, let $\tilde{m}$ be a constant act with constant utility m. The act $\frac{1}{2} f_n + \frac{1}{2 \delta } \tilde{\delta }$ has utility $2^{n-1} + \frac{1}{2}$ in state $s_1$ and utility $\frac{ 2^{n-1} }{2^{2n+1} - 1} + \frac{1}{2}$ in state $s_2$. If $c \in [\frac{1}{2^{2m+1}}, \frac{1}{2^{2m-1}})$ for $m \ge 1$, then the weighted expected utility of $\frac{1}{2} f_n + \frac{1}{2 \delta } \tilde{\delta }$ with respect to $p_c$ is

$$\begin{aligned}&2^m \left[ c \left( 2^{n-1}+ \frac{1}{2}\right) + (1-c) \left( \frac{ 2^{n-1} }{2^{2n+1} - 1} + \frac{1}{2} \right) \right] \\&\quad \ge \frac{1}{2^{m+1}} (2^{n-1}) + \frac{2^{2m-1} - 1}{2^{m-1}} \left( \frac{1}{2} \right) \\&\quad \ge 2^{n-m-2} + 2^{m-2}. \end{aligned}$$

Suppose that $n \ge 4 + 2\log _2 \delta$ and $\delta \ge 1$. If $n \ge m+2 + \log _2 \delta$,

$$\begin{aligned} 2^{n-m-2} + 2^{m-2} > 2^{\log _2 \delta } = \delta . \end{aligned}$$

Otherwise, if $n < m+2+\log _2 \delta$, since $n \ge 4 + 2\log _2 \delta$, it follows that $m \ge \log _2 \delta + 2$, and

$$\begin{aligned} 2^{n-m-2} + 2^{m-2} > 2^{\log _2 \delta } = \delta . \end{aligned}$$

If $c \ge \frac{1}{2}$, then the weighted expected utility of $\frac{1}{2} f_n + \frac{1}{2 \delta } \tilde{\delta }$ with respect to $p_c$ is

$$\begin{aligned}&c \left( 2^{n-1} + \frac{1}{2} \right) + (1-c) \left( \frac{ 2^{n-1} }{2^{2n+1} - 1} + \frac{1}{2}\right) \\&\quad > \frac{1}{2} 2^{n-1} \ge \frac{1}{2} 2^{3 + 2\log _2 \delta } \ge 2^2 \delta ^2 > \delta , \end{aligned}$$

since $\delta \ge 1$. This means that if $n \ge 4 + 2\log _2 \delta$, the minimum weighted expected utility of $\frac{1}{2} f_n + \frac{1}{2 \delta } \tilde{\delta }$ is strictly greater than $\delta$.

Appendix 3: Proof of Theorem 5.5

We show here that if a family of preferences $\succeq$ satisfies Axioms 1–6, then $\succeq$ can be represented as maximizing weighted expected utility with respect to a regular confidence function and a utility function. We make use of many of the same techniques as used in Halpern and Leung (2012). The key differences are highlighted.

First, we establish a von-Neumann–Morgenstern expected utility function over constant acts. This part follows the CF proof, rather than the proof in Halpern and Leung (2012).

Lemma 8.1

If Axioms 1, 3 and 5 hold, then there exists a nonconstant function $U : X\rightarrow {\mathbb {R}}$, unique up to positive affine transformations, such that for all constant acts $l^*$ and $(l')^*$,

$$\begin{aligned} l^* \succeq (l')^* \Leftrightarrow \sum _{ \{y :\, l^*(y)>0 \}} l(y) U(y) \ge \sum _{ \{y :\, l'(y)>0 \}} l'(y) U(y). \end{aligned}$$

Proof

As noted by CF, it was shown by Herstein and Milnor (1953) that Axioms 1, 3 and 5 are sufficient to satisfy the premises of the von-Neumann–Morgenstern theorem. $\square$

Since U is nonconstant, we can choose a U such that the minimum value that it takes on is 0 (for some constant act), and the maximum value it takes on is at least 1. If c is the utility of some lottery $l_c$, let $l^*_c$ be a constant act such that $l^*(s) = l_c$, so that $u(l^*_c) = c$. The following lemma, whose proof is given in Halpern and Leung (2012) (Lemma 2), follows from Lemma 8.1.

Lemma 8.2

$u(l^*_c) \ge u(l^*_{c'})$ iff $l^*_c \succeq l^*_{c'}$; similarly, $u(l^*_c) =u(l^*_{c'})$ iff $l^*_c \sim l^*_{c'}$, and $u(l^*_c) > u(l^*_{c'})$ iff $l^*_c \succ l^*_{c'}$.

In Halpern and Leung (2012), a slightly different continuity axiom (Axiom 9) is used.

Axiom 9

(Mixture continuity). If $f\succ g\succ h$, then there exist $q,r\in (0,1)$ such that

$$\begin{aligned} qf+(1-q)h \succ g \succ rf+(1-r)h. \end{aligned}$$

It is not difficult to derive mixture continuity from completeness (Axiom 1) and Axiom 3. Therefore, from here on, we assume that the preference order satisfies mixture continuity.

We establish some useful notation for acts and utility acts (real-valued functions on S). Given a utility act b, let $f_b$, the act corresponding to b, be the act such that $f_b(s) = l_{b(s)}$, if such an act exists. Conversely, let $b_f$, the utility act corresponding to the act f, be defined by taking $b_f(s) = u(f(s))$. Note that monotonicity implies that if $f_b = g_b$, then $f \sim g$. That is, only utility acts matter. If c is real, we take $c^*$ to be the constant utility act such that $c^*(s) = c$ for all $s \in S$.

1.1 Defining a functional on utility acts

Our proof uses the same technique as that used in Halpern and Leung (2012). Specifically, like Gilboa and Schmeidler 1989, we define a functional I on utility acts such that the preference order on utility acts is determined by their value according to I (see Lemma 8.4). Using I, we can then determine the weight of each probability in $\Delta (S)$ and prove the desired representation theorem.

Recall that u represents $\succeq$ on constant acts and that only utility acts matter to $\succeq$. The space of all nonnegative utility acts is the set ${\mathcal {B}}^+$ of real-valued functions b on S where $b(s) \ge 0$ for all $s \in S$. We now define a functional I on utility acts in ${\mathcal {B}}^+,$ such that for all f, g with $b_f, b_g \in {\mathcal {B}}^+$, we have $I(b_f)\ge I(b_g)$ iff $f\succeq g$. Let

$$\begin{aligned} R_f = \{\alpha ': l_{\alpha '}^* \preceq f\}. \end{aligned}$$

If $0^* \le b \le 1^*$, then $f_b$ exists, and we define

$$\begin{aligned} I(b) = \sup (R_{f_b}). \end{aligned}$$

For the remaining utility acts $b\in {\mathcal {B}}^+$, we extend I by homogeneity. Let $||b|| = |\max _{s \in S}b(s)|$. Note that if $b \in {\mathcal {B}}^+$, then $0^* \le b/||b|| \le 1^*$, so we define

$$\begin{aligned} I(b) = ||b|| I(b/||b||). \end{aligned}$$

It is worth noting that while in Halpern and Leung (2012), I was extended from the nonpositive utility acts to the entire set of real-valued acts to invoke a separating theorem for Banach spaces, the extension is not performed here. Consequently, we will be using a different separating hyperplane theorem than in Halpern and Leung (2012).

Lemma 8.3

If $b_f \in {\mathcal {B}}^+$, then $f \sim l_{I(b_f)}^*$.

Proof

Suppose that $b_f \in {\mathcal {B}}^+$ and, by way of contradiction, that $l_{I(b_f)}^* \prec f$. If $f\sim l_0^*$, then it must be the case that $I(b_f)=0$, since $I(b_f)\ge 0$ by definition of $\sup$, and $f \sim l_0^* \prec l_{\epsilon }^*$ for all $\epsilon > 0$ by Lemma 8.2, so $I(b_f) < \epsilon$ for all $\epsilon < 0$. Therefore, $f \sim l_{I(b_f)}^*$. Otherwise, since $b_f \in {\mathcal {B}}^+$, by monotonicity, we must have $l_0^* \prec f$, and thus $l_0^* \prec f \prec l_{I(b_f)}^*$. By mixture continuity, there is some $q\in (0,1)$ such that $q\cdot l_0^* + (1-q) \cdot l_{I(b_f)}^* \sim l_{(1-q)I(b_f)} \succ f$, contradicting the fact that I(b) is the least upper bound of $R_f.$

If, on the other hand, $l^*_{I(b_f)} \succ f$, then $l^*_{I(b_f)} \succ f \succeq l^*_{\underline{c}}$, where the existence of $l^*_{\underline{c}}$ is guaranteed by Axiom 4. If $f \sim l^*_{\underline{c}},$ then it must be the case that $I(b_f)=\underline{c}$. This is because $I(b_f) \ge \underline{c},$ since $l^*_{\underline{c}}\succeq l^*_{\underline{c}}$, and $I(b_f) \le \underline{c}$ since for all $c' > \underline{c}$, $l^*_{c'} \succ f \sim l^*_{\underline{c}}$.

Otherwise, $l^*_{I(b_f)} \succ f \succ l^*_{\underline{c}}$, and by Axiom 3, there is some $q\in (0,1)$ such that $q\cdot l^*_{I(b_f)} + (1-q) l^*_{\underline{c}} \prec f$. Since $qI(b_f) + (1-q)\underline{c} > I(b_f)$, this contradicts the fact that $I(b_f)$ is an upper bound of $R_{f}$. Therefore, it must be the case that $l^*_{I(b_f)} \sim f$.

$\square$

We can now show that I has the required property.

Lemma 8.4

For all acts f, g such that $b_f, b_g \in {\mathcal {B}}^+$, $f \succeq g$ iff $I( b_f ) \ge I( b_g )$.

Proof

Suppose that $b_f, b_g \in {\mathcal {B}}^+$. By Lemma 8.3, $l^*_{I(b_f)} \sim f$ and $g \sim l^*_{I(b_g)}$. Thus, $f \succeq g$ iff $l^*_{I(b_f)} \succeq l^*_{I(b_g)}$, and by Lemma 8.2, $l^*_{I(b_f)} \succeq l^*_{I(b_g)}$ iff $I(b_f)\ge I(b_g)$. $\square$

We show that the axioms guarantee that I has a number of standard properties. The proof of each property is analogous to its counterpart in Halpern and Leung (2012), but here we deal with nonnegative utility acts, as opposed to nonpositive utility acts.

Lemma 8.5

• (a) If $c \ge 0$, then $I(c^*)=c$.

• (b) I satisfies positive homogeneity :  if $b \in {\mathcal {B}}^+$ and $c > 0$, then $I(cb) = cI(b)$.

• (c) I is monotonic: if $b, b' \in {\mathcal {B}}^+$ and $b \ge b'$, then $I(b) \ge I(b')$.

• (d) I is continuous: if $b, b_1, b_2, \ldots \in {\mathcal {B}}^+$, and $b_n \rightarrow b$, then $I(b_n) \rightarrow I(b)$.

• (e) I is superadditive: if $b, b' \in {\mathcal {B}}^+$, then $I(b+b') \ge I(b) + I(b')$.

Proof

For part (a), if c is in the range of u, then it is immediate from the definition of I and Lemma 8.2 that $I(c^*) = c$. If c is not in the range of u, then since [0, 1] is a subset of the range of u, we must have $c > 1$, and by definition of I, we have $I(c^*) = |c| I(c^*/|c|) = c$.

For part (b), first suppose that $||b|| \le 1$ and $b \in {\mathcal {B}}^+$ (i.e., $0^* \le b \le 1^*$). Then there exists an act f such that $b_f = b$. By Lemma 8.3, $f \sim l^*_{I(b)}$. We now consider the case that $c \le 1$ and $c > 1$ separately. If $c \le 1$, by Worst Independence, $c f_b + (1-c) l_0^* \sim c l^*_{I(b)} + (1-c) l_0^*$. By Lemma 8.4, $I(b_{c f_b + (1-c) l_0^*}) = I(b_{c l^*_{I(b)} + (1-c) l_0^*})$. It is easy to check that $b_{c f_b + (1-c) l_0^*} = cb$, and $b_{c l^*_{I(b)}} + (1-c) l_0^* = cI(b)^*$. Thus, $I(cb) = I(cI(b)^*)$. By part (a), $I(cI(b)^*) = cI(b)$. Thus, $I(cb) = cI(b)$, as desired.

If $c > 1$, there are two subcases. If $||cb|| \le 1$, since $1/c < 1$, by what we have just shown $I(b) = I(\frac{1}{c}(cb)) = \frac{1}{c}I(cb)$. Cross multiplying, we have that $I(cb) = cI(b)$, as desired. If $||cb||>1$, by definition, $I(cb) = ||cb|| I(bc/||cb||) = c||b||I(b/||b||)$ (since $bc/||cb|| = b/||b||$). Since $||b|| \le 1$, by the earlier argument, $I(b) = I(||b|| (b/||b||) = ||b||I(b/||b||)$, so $I(b/||b||) = \frac{1}{||b||} I(b)$. Again, it follows that $I(cb) = cI(b)$.

Now, suppose that $||b|| > 1$. Then, $I(b) = ||b|| I(b/||b||)$. Again, we have two subcases. If $||cb|| > 1$, then

$$\begin{aligned} I(cb) = ||cb|| I(cb/||cb||) = c||b|| I(b/||b||) = cI(b). \end{aligned}$$

If $||cb|| \le 1$, by what we have shown for the case $||b|| \le 1$,

$$\begin{aligned} I(b) = I\left( \frac{1}{c} (cb)\right) = \frac{1}{c}I(cb), \end{aligned}$$

so again $I(cb) = cI(b)$.

For part (c), first note that for $b, b' \in {\mathcal {B}}^+$, if $||b|| \le 1$ and $||b'|| \le 1$, then the acts $f_b$ and $f_{b'}$ exist. Moreover, since $b \ge b'$, we must have $(f_b(s))^* \succeq (f_{b'}(s))^*$ for all states $s \in S$. Thus, by monotonicity, $f_b \succeq f_{b'}$. If either $||b|| > 1$ or $||b'|| > 1$, let $n = \max (||b||,||b'||)$. Then, $||b/n|| \le 1$ and $||b'/n|| \le 1$. Thus, $I(b/n) \ge I(b'/n)$, by what we have just shown. By part (b), $I(b) \ge I(b')$.

For part (d), note that if $b_n \rightarrow b$, then for all k, there exists $n_k$ such that $b_n - (1/k)^* \le b_n \le b_n + (1/k)^*$ for all $n\ge n_k$. Moreover, by the monotonicity of I (part (c)), we have that $I(b - (1/k)^*) \le I(b_n) \le I(b + (1/k)^*)$. Thus, it suffices to show that $I(b - (1/k)^*) \rightarrow I(b)$ and that $I(b + (1/k)^*) \rightarrow I(b)$.

To show that $I(b - (1/k)^*) \rightarrow I(b)$, we must show that for all $\epsilon > 0$, there exists k such that $I(b- (1/k)^*) \ge I(b) - \epsilon$. By positive homogeneity (part (b)), we can assume without loss of generality that $||b - (1/2)^*|| \le 1$ and that $||b|| \le 1$. Fix $\epsilon > 0$. If $I(b - (1/2)^*) \ge I(b) - \epsilon$, then we are done. If not, then $I(b) > I(b)- \epsilon > I(b - (1/2)^*)$. Since $||b|| \le 1$ and $||b-(1/2)^*|| \le 1$, $f_b$ and $f_{b-(1/2)^*}$ exist. Moreover, by Lemma 8.4, $f_b \succ f_{(I(b) - \epsilon )^*} \succ f_{b-(1/2)^*}$. By mixture continuity, for some $p\in (0,1)$, we have $pf_b + (1-p) f_{(b-(1/2)^*} \succ f_{(I(b) - \epsilon )^*}$. It is easy to check that $b_{p f_b + (1-p) f_{b-(1/2)^*}} = b - ((1-p)/2)^*$. Thus, by Lemma 8.4, $f_{b-((1-p)/2)^*} \succeq f_{(I(b)-\epsilon )^*}$, and $I(b - ((1-p)/2)^*) > I(b) - \epsilon$. Choose k such that $1/k < (1-p)/2$. Then, by monotonicity [part (c)], $I(b-(1/k)^*) \ge I(b - ((1-p)/2)^*) > I(b) - \epsilon$, as desired.

The argument that $I(b + (1/k)^*) \rightarrow I(b)$ is similar and left to the reader.

For part (e), if $||b||, ||b'|| \le 1$, and $I(b), I(b') \ne 0$, consider $\frac{b}{I(b)}$ and $\frac{b'}{I(b')}$. Since $I( \frac{b}{I(b)} ) = I(\frac{b'}{I(b')}) = 1$, it follows from Lemma 8.3 that $f_{\frac{b}{I(b)}} \sim f_{\frac{b'}{I(b')}}$. By ambiguity aversion, for all $p\in (0,1]$, $p f_{\frac{b}{I(b)}} + (1-p) f_{\frac{b'}{I(b')}} \succeq f_{\frac{b}{I(b)}}$. Thus, taking $p = \frac{I(b)}{I(b) + I(b')}$, $I(\frac{b+b'}{I(b) + I(b')} ) = \frac{1}{I(b) + I(b')} I(b+b') = I( \frac{I(b)}{I(b) + I(b') } \frac{b}{I(b)} + \frac{I(b')}{I(b) + I(b') }\frac{b'}{I(b')} ) \ge I( \frac{b}{I(b)} ) = I( \frac{b'}{I(b')}) = 1$. Hence, $I( b + b' ) \ge I( b ) + I( b' )$.

If either $||b|| > 1$ or $||b'|| > 1$, and both $I(b) \ne 0$ and $I(b') \ne 0$, then the result easily follows by positive homogeneity [property (b)].

If either $I(b)=0$ or $I(b') = 0$, let $b_n = b+ \frac{1}{n}^*$ and $b'_n = b'+ \frac{1}{n}^*$. Clearly, $||b_n|| > 0$, $||b'_n|| > 0$, $b_n \rightarrow b$, and $b_n' \rightarrow b_n'$. By our argument above, $I(b_n + b_n') \ge I(b_n) + I(b_n')$ for all $n \ge 1$. The result now follows from continuity. $\square$

1.2 Defining the confidence function

In this section, we use I to define a confidence function $\phi$ that maps each $p \in \Delta (S)$ to a confidence value in [0, 1]. The heart of the proof involves showing that the resulting function $\phi$ so determined gives us the desired representation.

Given a confidence function $\phi$, for $b \in {\mathcal {B}}^+$, define

$$\begin{aligned} { WE }(b) = \inf _{p \in {\mathcal {P}}} \phi (p) \left( \sum _{s \in S} b(s) p(s)\right) . \end{aligned}$$

Define

$$\begin{aligned} \underline{ E }(b) = \inf _{p \in {\mathcal {P}}} \sum _{s \in S} b(s) p(s). \end{aligned}$$

and

$$\begin{aligned} E _{p}(b) = \sum _{s \in S} b(s) p(s). \end{aligned}$$

For each probability $p \in \Delta (S)$, define

$$\begin{aligned} \phi _{t}(p) = \inf \{\alpha \in {\mathbb {R}}: I(b) \le \alpha E _{p}(b)\quad \text { for all }b \in {\mathcal {B}}^+\}, \end{aligned}$$

(1)

and let $\phi _{t}(p) = \infty$ if the $\inf$ does not exist. Note that $\phi _t{(p)} \ge 1$, since $E _{p}((c)^*) = I((c)^*) = c$ for all distributions p and $c \in {\mathbb {R}}$. Moreover, it is immediate from the definition of $\phi _{t}({p})$ that $\phi _{t}({p}) E _{p}(b) \ge I(b)$ for all $b \in {\mathcal {B}}^+$. The next lemma shows that there exists a probability p where we have equality.

Lemma 8.6

• (a) For some distribution p, we have $\phi _{t}({p}) = 1$.

• (b) For all $b \in {\mathcal {B}}^+$, there exists p such that $\phi _{t}({p}) E _{p}(b) = I(b)$.

Proof

The proofs of both parts (a) and (b) use a separating hyperplane theorem. If U is a convex subset of ${\mathcal {B}}^+$, and $b \notin U$, then there is a linear functional $\lambda$ that separates U from b, that is, $\lambda (b') < \lambda (b)$ for all $b' \in U$. We proceed as follows.

For part (a), we must show that there exists a probability measure p such that for all $b\in {\mathcal {B}}^+$, we have $E _{p}(b) \ge I(b)$. This would show that $\phi _{t}(p) = 1$.

Let $U = \{b' \in {\mathcal {B}}^+: I(b') \ge 1 \}$. U is closed (by continuity of I) and convex (by positive homogeneity and superadditivity of I), and $(0)^* \notin U$. Thus, there exists a linear functional $\lambda$ such that $\lambda (b') > \lambda ((0)^*) = 0$ for $b' \in U$. We can assume without lost of generality that $\lambda ( 1^* ) = 1$.

We want to show that $\lambda$ is a positive linear functional, that is, that $\lambda (b) \ge 0$ if $b \ge 0^*$. Clearly, this holds for $b'$ such that $I(b') \ge 1$. If $b' \ge 0^*$, $I(b') < 1$, and $I(b') > 0$, note that $cI(b') = I(cb') \ge 1$ for some $c \ge 0$. Therefore, $I(b') \ge \frac{1}{c} \ge 0$. If $b' \ge 0^*$ and $I(b') = 0$, note that for all $c > 0$, $\lambda ( b' + c^* ) \ge 0$ by the previous case. Thus, $\lambda (b') \ge 0$. It follows that $\lambda$ is a positive functional.

Define the probability distribution p on S by taking $p(s) = \lambda (1_{s})$. To see that p is indeed a probability distribution, note that since $1_{s} \ge 0$ and $\lambda$ are positive, we must have $\lambda (1_{s}) \ge 0$. Moreover, $\sum _{s \in S} p(s) = \lambda (1^*) = 1$. In addition, for all $b' \in {\mathcal {B}}$, we have

$$\begin{aligned} \lambda (b') = \sum _{s \in S} \lambda (1_{s}) b'(s) = \sum _{s \in S} p(s) b'(s) = E_{p}(b'). \end{aligned}$$

Next, we claim that, for $b \in {\mathcal {B}}^+$,

$$\begin{aligned} \text {for all}\,\, c>0,\quad \text {if } I(b) > c,\,\, \text {then } \lambda (b) > c. \end{aligned}$$

(2)

To see why the claim is true, note that if $I(b) \ge c$, then $I(b/c) \ge 1$ by positive homogeneity, so $\lambda (b/c) \ge 1$ and $\lambda (b) \ge c$. Therefore, $\lambda (b) \ge I(b)$, as desired.

The proof of part (b) is similar to that of part (a). We want to show that, given $b \in {\mathcal {B}}^+$, there exists p such that $\phi _{t}(p) E_{p}(b) = I(b)$. First, consider the case where $||b|| \le 1$. If $I(b) = 0$, then there must exist some s such that $b(s) = 0$; otherwise, there exists $c > 0$ such that $b \ge c^*$, so $I(b) \ge c$. If $b(s) = 0$, let $p_s$ be such that $p_s(s) = 1$. Then $E_{p_s}(b) = 0$, so part (b) of the Lemma holds in this case.

If $||b|| \le 1$ and $I(b) > 0$, let $U = \{b': I(b') \ge I(b)\}$. Again, U is closed and convex, and $b \notin U$, so there exists a linear functional $\lambda$ such that $\lambda (b') > \lambda (b)$ for $b' \in U$. Since $1^* \in U$ and we can assume without loss of generality $\lambda (1^*) = 1$, we must have $\lambda (b) < 1$.

The same argument as that used in the proof of (a) shows that $\lambda$ is a positive functional.

Therefore, $\lambda$ determines a probability distribution p such that, for all $b' \in {\mathcal {B}}^+$, we have $\lambda (b') = E_{p}(b')$. p, of course, will turn out to be the desired distribution. To show this, we need to show that $\phi _{t}(p) = I(b)/E_{p}(b)$. By definition, $\phi _{t}(p) \ge I(b)/E_{p}(b)$. To show that $\phi _{t}(p) \le I(b)/E_{p}b$, we must show that $\frac{I(b) }{ E_{p}(b)} \ge \frac{I(b') }{E_{p} b'}$ for all $b' \in {\mathcal {B}}^+$. Equivalently, we must show that $I(b) \lambda (b')/\lambda (b) \ge I(b')$ for all $b' \in {\mathcal {B}}^+$.

Essentially the same argument used to prove (2) also shows that

$$\begin{aligned} \mathrm{for\,\, all }\,\, c>0,\mathrm{if }\frac{I(b')}{I(b)} \ge c,\quad \mathrm{then }\,\,\frac{\lambda (b') }{\lambda (b)} \ge c. \end{aligned}$$

In particular, if $\frac{I(b')}{I(b)} \ge c$, then by positive homogeneity, $\frac{I(b')}{c} \ge I(b)$, so $\frac{b'}{c}\in U$, and $\lambda (\frac{b'}{c}) > \lambda (b)$ and hence $\frac{\lambda (b') }{\lambda (b)} \ge c$.

It follows that $\lambda (b')/(\lambda (b)) \ge I(b')/(I(b))$ for all $b' \in {\mathcal {B}}^+$. Thus, $I(b) \lambda (b')/\lambda (b) \ge I(b')$ for all $b' \in {\mathcal {B}}^+$, as required.

Finally, if $||b|| > 1$, let $b' = b/||b||$. By the argument above, there exists a probability measure p such that $\phi _{t}(p)E_{p}(b/||b||) = I(b/||b||)$. Since $E_{p}(b/||b||) = E_{p}(b)/||b||$, and $I(b/||b||) = I(b)/||b||$, we must have that $\phi _{t}(p)E_{p}(b) = I(b)$. $\square$

We can now complete the proof of Theorem 5.5. By Lemma 8.6 and the definition of $\phi _{t}(p)$, for all $b\in {\mathcal {B}}^+$,

$$\begin{aligned} I(b) =&\inf _{p \in \Delta (S)} \phi _{t}(p) E_{p} (b). \end{aligned}$$

(3)

Recall that, by Lemma 8.4, for all acts f, g such that $b_f, b_g \in {\mathcal {B}}^+$, $f \succeq g$ iff $I( b_f ) \ge I( b_g )$. Thus, $f \succeq g$ iff

$$\begin{aligned} \inf _{p \in \Delta (S) }\left( \phi _{t}(p)\sum _{s\in S}u(f(s))p(s) \right) \ge \inf _{p \in \Delta (S) }\left( \phi _{t}(p)\sum _{s\in S}u(g(s))p(s) \right) . \end{aligned}$$

To get the confidence function $\phi$ from $\phi _t$, note that $\lim _{x \rightarrow 0^+} t(x) = \infty$ and $t(1) > 0$. We let $\phi (p) = t^{-1}( t(1) \phi _t( p ) )$, with the special case $\phi (p) = 0$ if $\phi _t(p) = \infty$. (Note that $t(1) \phi _t( p )$ is in the range of $t^{-1}$, since $\phi _t(p) \ge 1$, t is nonincreasing and $\lim _{x\rightarrow 0^+} t(x) = \infty$.)

1.3 Properties of the confidence function

In this section, we show that the confidence function $\phi$ that we constructed satisfies the properties claimed in Theorem 5.5.

We first show that $t\circ \phi = \phi _t$ has convex upper support. To that end, we show that if $c_1 \ge \phi _t(p_1)$ and $c_2 \ge \phi _t(p_2)$, then for all $\alpha \in (0,1)$,

$$\begin{aligned} \left( \alpha c_1p_1 + (1-\alpha ) c_2p_2\right) (S) \ge \phi _t\left( \frac{\alpha c_1p_1 + (1-\alpha ) c_2p_2}{\left( \alpha c_1p_1 + (1-\alpha ) c_2p_2\right) (S)}\right) . \end{aligned}$$

By the definition of $\phi _t$, it suffices to show that for all $b\in {\mathcal {B}}^+$,

$$\begin{aligned} I(b) \le \left( \alpha c_1p_1 + (1-\alpha ) c_2p_2\right) (S) E_{ \frac{\alpha c_1p_1 + (1-\alpha ) c_2p_2}{\left( \alpha c_1p_1 + (1-\alpha ) c_2p_2\right) (S)}} (b). \end{aligned}$$

(4)

It is easy to see that the inequality holds. Let $b\in {\mathcal {B}}^+$. The right-hand side of (4) is equal to

$$\begin{aligned} \sum _{s\in S} \left( (\alpha c_1p_1(s) + (1-\alpha ) c_2p_2(s) ) b(s) \right)&= \alpha c_1 E_{p_1} (b) + (1-\alpha ) c_2 E_{p_2} (b) \\&\ge \alpha \phi _t(p_1) E_{p_1} (b) + (1-\alpha ) \phi _t(p_2) E_{p_2} (b) \\&\ge \alpha I(b) + (1-\alpha ) I(b) \text { (by (3)) } \\&\ge I(b). \end{aligned}$$

We now show that $\phi$ is regular*. Since we have shown that, for some $p^*$, $\phi _t(p^*) =1$, we have $\phi (p^*) = t^{-1}( t(1) 1) = 1$. Therefore, $\phi$ is normal.

Secondly, we show that $\phi$ is weakly* upper semi-continuous. We show that if $\{ p_n \} \rightarrow p$ and $\phi (p_n) \ge \alpha$ for all n, then $\phi (p) \ge \alpha$. Suppose for the purpose of contradiction that $\phi (p) < \alpha$. Then, $\phi _t(p) = t(\phi (p)) > t(\alpha )$. By continuity of t, $\phi _t(p_n) = t( \phi (p_n) ) > t(\alpha )$ for all sufficiently large n, implying that $\phi (p_n) < \alpha$, contradicting the assumption that $\phi (p_n) \ge \alpha$. Therefore, $\phi (p) \ge \alpha$, as required.

We now show that $\phi$ is quasi-concave; that is, $\phi (\beta p_1 + (1-\beta ) p_2 ) \ge \min \{ \phi (p_1), \phi (p_2) \}$ for any $\beta \in [0,1]$. Since t is strictly decreasing, so is $t^{-1}$. Thus, $-t^{-1}$ is strictly increasing. Moreover, if $\phi _{t}$ is quasi-convex, then $-t^{-1} \circ \phi _{t}$ is also quasi-convex. Since the negative of a quasi-convex function is quasi-concave, $t^{-1} \circ \phi _{t}$ is quasi-concave. Therefore, if we show that $\phi _t$ is quasi-convex, this would show that $\phi = t^{-1} \circ \phi _{t}$ is quasi-concave.

Recall from (1) that

$$\begin{aligned} \phi _{t}(p) = \inf \{\alpha \in {\mathbb {R}}: I(b) \le \alpha E _{p}(b)\quad \text { for all } b \in {\mathcal {B}}^+\}. \end{aligned}$$

If $\max \{ \phi _t(p_1), \phi _t(p_2) \} \le c$ for $c\in {\mathbb {R}}$, then for all $b \in {\mathcal {B}}^+$, we have

$$\begin{aligned} I(b) \le c E _{p_1} (b), \end{aligned}$$

and

$$\begin{aligned} I(b) \le c E _{p_2} (b). \end{aligned}$$

Therefore, for all $b \in {\mathcal {B}}^+$ and all $\beta \in [0,1]$, by the linearity of $E_p(b)$ with respect to the parameter p,

$$\begin{aligned} I(b) \le c E _{\beta p_1 + (1-\beta )p_2}(b). \end{aligned}$$

This means that $\phi _t( \beta p_1 + (1-\beta )p_2 ) \le c$. Thus, $\phi _t( \beta p_1 + (1-\beta )p_2 ) \le \max \{ \phi _t(p_1), \phi _t(p_2) \}$. Therefore, $\phi _t$ is quasi-convex.

1.4 Uniqueness of the representation

In this section, we show that our constructed $\phi$ is the only regular* fuzzy confidence function, such that $t\circ \phi$ has convex upper support and such that $\succeq ^{+}_{t,\phi } = \succeq$. Our uniqueness result is similar in spirit to the uniqueness results of Gilboa and Schmeidler (1989), who show that the convex, closed, and nonempty set of probability measures in their representation theorem for MMEU is unique.

The proof of this result, like the proof of uniqueness in Gilboa and Schmeidler (1989), uses a separating hyperplane theorem to show the existence of acts on which two different representations must ‘disagree’. The proof presented here is essentially the same as that used in Halpern and Leung (2012), with only superficial changes to accommodate our definitions and notation.

Lemma 8.7

For all confidence functions $\phi '$, if $\succeq ^{+}_{t,\phi '} = \succeq$ and $t\circ \phi '$ has convex upper support, then $\phi = \phi '$.

Proof

Suppose for contradiction that there exists a regular* fuzzy confidence function $\phi ' \ne \phi ,$ such that $t\circ \phi '$ has convex upper support, and that $\succeq ^+_{t,\phi '} = \succeq ^+_{t,\phi }$. Consider the two upper supports $\overline{V}_{t\circ \phi }$ and $\overline{V}_{t\circ \phi '}$. $\overline{V}_{t\circ \phi }$ and $\overline{V}_{t\circ \phi '}$ are both closed. To see why, consider a sequence $\{p_n\}_{n\in {\mathbb {N}}}$ contained in $p_n \in \overline{V}_{t\circ \phi },$ such that $p_n \rightarrow p$. We show that $p \in \overline{V}_{t\circ \phi }$, by showing that for some $q\in \Delta (S)$, $\phi (q) > 0$ and $p \ge t(\phi ({q})) q$.

We first show that $p \ge t(\phi ({q})) q$ for some $q\in \Delta (S)$. Recall that for all n, there exists $q_n \in \Delta (S)$ such that $p_n \ge t(\phi ({q_n})) q_n$. Since $q_n \in \Delta (S)$, $q_{k_m} \rightarrow q$ for some subsequence $\{ q_{k_m} \}$ and $q \in \Delta (S)$. Therefore, we have

$$\begin{aligned} p&= \lim _{n\rightarrow \infty } {p_n}\\&\ge \limsup _{n\rightarrow \infty } t(\phi ({q_n})) q_n ,\quad \text {since } p_n \ge t(\phi (q_n)) q_n\\&= \lim _{n\rightarrow \infty } \sup _{m\ge n} t(\phi ({q_{k_m}})) q_{k_m} \\&= \lim _{n\rightarrow \infty } \sup _{m\ge n} t(\phi ({q_{k_m}})) \lim _{m\rightarrow \infty } q_{k_m} \\&= \lim _{n\rightarrow \infty } t(\inf _{m\ge n} \phi ({q_{k_m}})) \lim _{m\rightarrow \infty } q_{k_m},\quad \text {since } t \text { is nonincreasing and continuous} \\&= t(\liminf _{m\rightarrow \infty } \phi ({q_{k_m}})) \lim _{m\rightarrow \infty }q_{k_m},\quad \text {by continuity of } t \\&\ge t(\phi ({q})) q, \end{aligned}$$

since $\phi (q)\ge \limsup _{m\rightarrow \infty } \phi (q_{k_m}) \ge \liminf _{m\rightarrow \infty } \phi (q_{k_m})$ by upper semi-continuity of $\phi$, and t is nonincreasing.

It remains to show that $\phi (q) > 0$. To that end, suppose for the purpose of contradiction that $\phi (q) = 0$. Then it must be the case that $\lim _{m\rightarrow \infty } \phi (q_{k_m}) = 0$, since if there exists an $\epsilon > 0$ such that $\lim _{m\rightarrow \infty } \phi (q_{k_m}) \ge \epsilon$, then by upper semi-continuity of $\phi$ it must be the case that $\phi (q) \ge \epsilon$. Since $\lim _{x\rightarrow 0^+} t(x) = \infty$, we have that $\lim _{m\rightarrow \infty } t(\phi (q_{k_m})) = \infty$. However, recall that $p_n \ge t(\phi ({q_n})) q_n$ for all n. Since, $q_n \in \Delta (S)$ and hence does not vanish, $p_n$ cannot be a convergent sequence. Hence, it must be the case that $\phi (q) > 0$.

Therefore, $p \in \overline{V}_{t\circ \phi }$, as required, and that $\overline{V}_{t\circ \phi }$ is closed. The same argument shows that $\overline{V}_{t\circ \phi '}$ is closed.

Without loss of generality, let ${q} \in \overline{V}_{t\circ \phi '} \backslash \overline{V}_{t\circ \phi }$. Since $\overline{V}_{t\circ \phi }$ and $\{{q}\}$ are closed, convex, and disjoint, and $\{{q}\}$ is compact, the separating hyperplane theorem (Rockafellar 1970) says that there exists $\theta \in {\mathbb {R}}^{|S|}$ and $c\in {\mathbb {R}}$ such that

$$\begin{aligned} \theta \cdot {p}> c \quad \text { for all } {p}\in \overline{V}_{t\circ \phi } ,\text {and }\,\, \theta \cdot {q} < c. \end{aligned}$$

(5)

By scaling c appropriately, we can assume that $|\theta (s)| \le 1$ for all $s\in S$. Now, we argue that it must be the case that $\theta (s) \ge 0$ for all $s\in S$ (so that $\theta$ corresponds to the utility profile of some act with nonnegative utilities). Suppose that $\theta (s') < 0$ for some $s'\in S$. By (5), $\theta \cdot {p} > c \text { for all } {p}\in \overline{V}_{t\circ \phi }$. Let $p^* \in \overline{V}_{t\circ \phi }$ be any measure with $\phi (p^*)=1$, and let ${p^{**}} \in \overline{V}_{t\circ \phi }$ be defined by

$$\begin{aligned} {p^{**}}(s) = {\left\{ \begin{array}{ll} p^*(s) ,\quad \text {if } s\ne s' \\ \frac{ |S| \max \{|c|,\max _{s'' \in S }{|p^*(s'')|}\}}{|\theta (s')|} ,\quad \text {if } s = s'. \end{array}\right. } \end{aligned}$$

We have defined $p^{**}$ such that $p^{**} \ge p^*$, since for all $s\in S$, $p^{**}(s) \ge p^*(s)$. To see how, note that $p^{**}(s) = p^*(s)$ for $s\ne s'$, and $p^{**}(s) \ge \max _{s'' \in S}|p^*(s'')| \ge p^*(s)$ for $s=s'$. Therefore, $p^{**}$ is in $\overline{V}_{t\circ \phi }$.

Our definition of $p^{**}$ also ensures that $\theta \cdot {p^{**}} = \sum _{s\in S} p^{**}(s) \theta (s) \le c$, since

$$\begin{aligned} \sum _{s\in S} p^{**}(s) \theta (s) =&p^{**}(s') \theta (s') + \sum _{s\ne s'} p^{**}(s) \theta (s) \\ \le&p^{**}(s') \theta (s') + \sum _{s\ne s'} |p^{**}(s) | , \quad \text {since } |\theta (s)| \le 1\\ =&- |S| \max \{ |c|, \max _{s'' \in S}|p^*(s'')| \}+ \sum _{s\ne s'} |p^{**}(s)| \\ \le&{-|c|} \le c. \end{aligned}$$

This contradicts (5), which says that $\theta \cdot {p}> c \text { for all } {p}\in \overline{V}_{t\circ \phi }$. Thus, it must be the case that $\theta (s) \ge 0$ for all $s\in S$.

Consider the $\theta$ given by the separating hyperplane theorem, and let f be an act such that $u\circ f = \theta$. $f \sim l^*_d$ for some constant act $l^*_d$. Since $\overline{V}_{t\circ \phi }$ and $\overline{V}_{t\circ \phi '}$ as sets of generalized probabilities both represent $\succeq$, and $\overline{V}_{t\circ \phi }$ and $\overline{V}_{t\circ \phi '}$ both contain a normal probability measure,

$$\begin{aligned} \min _{{p} \in \overline{V}_{t\circ \phi }} {p}\cdot (u\circ f) = \min _{{p} \in \overline{V}_{t\circ \phi }} {p}\cdot (u\circ l^*_d) = d = \min _{{p} \in \overline{V}_{t\circ \phi '}} {p} \cdot (u\circ f). \end{aligned}$$

However, by (5),

$$\begin{aligned} \min _{{p}\in \overline{V}_{t\circ \phi }} {p}\cdot (u\circ f) > c > \min _{{p} \in \overline{V}_{t\circ \phi '}} {p} \cdot (u\circ f), \end{aligned}$$

which is a contradiction. $\square$

Appendix 4: Proof of Theorem 5.7

Proof

The proof is almost the same as the proof of Theorem 5.5. We point out the differences, which are mostly straightforward adaptations from ${\mathcal {B}}^+$ to ${\mathcal {B}}^-$. Lemma 8.1 and Lemma 8.2 hold without change. By Axiom 8, we can assume that the maximum value that u takes on is 0, and by Axiom 1 we can assume that the minimum is no greater than $-1$.

We now define a functional I on utility acts, as before. All occurrences of ${\mathcal {B}}^+$ in the proof of Theorem 5.5 needs to be replaced by ${\mathcal {B}}^-$, defined by the real-valued functions b on S where $b(s) \le 0$ for all $s\in S$.

More specifically, let

$$\begin{aligned} R_f = \{\alpha ': l_{\alpha '}^* \preceq f\}. \end{aligned}$$

If $0^* \ge b \ge (-1)^*$, then $f_b$ exists, and we define

$$\begin{aligned} I(b) = \sup (R_{f_b}). \end{aligned}$$

For the remaining utility acts $b\in {\mathcal {B}}^+$, we extend I by homogeneity, as before.

The analog of Lemma 8.3 for $b_f \in {\mathcal {B}}^-$ follows from analogous arguments used in the original proof. The case of $l^*_{I(b_f)} \prec f$, however, is a bit simpler than for the positive case.

Lemma 9.1

If $b_f \in {\mathcal {B}}^-$, then $f \sim l^*_{I(b_f)}$.

Proof

Suppose, by way of contradiction, that $l^*_{I(b_f)} \prec f$. If $f \sim l^*_0$, then $I(b_f) \ge 0$ by the definition of I. However, we also have $I(b_f) \le 0$ by Lemma 8.4, so $I(b_f)= 0$, and therefore $f \sim l^*_{I(b_f)}$, as required. Otherwise, $f\prec l^*_0$ by monotonicity, so $l^*_{I(b_f)} \prec f \prec l_0^*$, which, when taken together with mixture continuity, contradicts the definition of I. $\square$

The proof of Lemma 8.4 still holds. The analog of Lemma 8.5 also follows from similar arguments; we discuss some key differences below.

Lemma 9.2

• (a) If $c \le 0$, then $I(c^*)=c$.

• (b) I satisfies positive homogeneity: if $b \in {\mathcal {B}}^-$ and $c > 0$, then $I(cb) = cI(b)$.

• (c) I is monotonic: if $b, b' \in {\mathcal {B}}^-$ and $b \ge b'$, then $I(b) \ge I(b')$.

• (d) I is continuous: if $b, b_1, b_2, \ldots \in {\mathcal {B}}^-$, and $b_n \rightarrow b$, then $I(b_n) \rightarrow I(b)$.

• (e) I is superadditive: if $b, b' \in {\mathcal {B}}^-$, then $I(b+b') \ge I(b) + I(b')$.

Proof

For part (b), instead of making use of Axiom 4 (worst independence), we use Axiom 8 (best independence).

For part (e), note that since I(b) is nonpositive for $b \in {\mathcal {B}}^-$, $I(\frac{b}{I(b)})$ is not defined, unlike in the case of nonnegative utilities. We use the same proof as in Halpern and Leung (2012): Clearly, $I(\frac{b}{-I(b)}) = -1$. Therefore, $f_{\frac{b}{-I(b)}} \sim f_{\frac{b'}{-I(b')}} \sim l^*_{-1}$. From Axiom 6 (ambiguity aversion), taking $p=\frac{-I(b)}{-I(b) - I(b')}$, we have

$$\begin{aligned} I\left( \frac{-I(b)}{-I(b) - I(b')}\frac{b}{-I(b)}+ \frac{-I(b')}{-I(b) - I(b')} \frac{b'}{-I(b')}\right) \ge I\left( \frac{b}{-I(b)}\right) = -1, \end{aligned}$$

which implies that $I(b + b') \ge I(b) + I(b')$, as required. $\square$

We now use I to define a confidence function $\phi$. ${ WE }, \underline{ E },$ and $E$ are defined as before. For each probability $p \in \Delta (S)$, define

$$\begin{aligned} \phi _t(p) = \sup \{\alpha \in {\mathbb {R}}: I(b) \le \alpha E _{p}(b)\quad \text { for all }b \in {\mathcal {B}}^-\}. \end{aligned}$$

Note that $\phi _t{(p)} \le 1$, since $E _{p}((c)^*) = I((c)^*) = c$ for all distributions p and $c \in {\mathbb {R}}$. Moreover, $\phi _t{(p)} \ge 0$ for all $b \in {\mathcal {B}}^-$. The next lemma shows that there exists a probability p where we have equality. The proof of the lemma is similar to that of Lemma 8.6, and is left to the reader.

Lemma 9.3

• (a) For some distribution p, we have $\phi _t({p}) = 1$.

• (b) For all $b \in {\mathcal {B}}^-$, there exists p such that $\phi _t({p}) E _{p}(b) = I(b)$.

By Lemma 9.3 and the definition of $\phi _t(p)$, for all $b\in {\mathcal {B}}^-$,

$$\begin{aligned} I(b) =&\inf _{p \in \Delta (S)} \phi _t(p) E_{p} (b). \end{aligned}$$

We have $f \succeq g$

$$\begin{aligned} \text {iff }&\inf _{p \in \Delta (S) }\left( \phi _t(p)\sum _{s\in S}u(f(s))p(s) \right) \ge \inf _{p \in \Delta (S) }\left( \phi _t(p)\sum _{s\in S}u(g(s))p(s) \right) \\ \text {iff }&t(1) \inf _{p \in \Delta (S) }\left( \phi _t(p)\sum _{s\in S}u(f(s))p(s) \right) \ge t(1) \inf _{p \in \Delta (S) }\left( \phi _t(p)\sum _{s\in S}u(g(s))p(s) \right) . \end{aligned}$$

Since t is strictly increasing, $t(1) > t(0)$. Therefore, since $\phi _t(p) \in [0,1]$ and $t(0) \le 0$, $t(1) \phi _t(p)$ is in the range of t, and we can define

$$\begin{aligned} \phi (p) = t^{-1}( t(1) \phi _t(p)). \end{aligned}$$

We now have $f \succeq g$

$$\begin{aligned} \text {iff }\inf _{p \in \Delta (S) }\left( t(\phi (p))\sum _{s\in S}u(f(s))p(s) \right) \ge \inf _{p \in \Delta (S) }\left( t(\phi (p))\sum _{s\in S}u(g(s))p(s) \right) . \end{aligned}$$

Finally, the uniqueness of the representation follows from arguments analogous to those for nonnegative utilities. $\square$