Weighted sets of probabilities and minimax weighted expected regret: a new approach for representing uncertainty and making decisions

Joseph Y. Halpern; Samantha Leung

doi:10.1007/s11238-014-9471-y

Theory and Decision

November 2015, Volume 79, Issue 3, pp 415–450 | Cite as

Weighted sets of probabilities and minimax weighted expected regret: a new approach for representing uncertainty and making decisions

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Joseph Y. Halpern
Samantha LeungEmail author

Article

First Online: 11 December 2014

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Abstract

We consider a setting where a decision maker’s uncertainty is represented by a set of probability measures, rather than a single measure. Measure-by-measure updating of such a set of measures upon acquiring new information is well known to suffer from problems. To deal with these problems, we propose using weighted sets of probabilities: a representation where each measure is associated with a weight, which denotes its significance. We describe a natural approach to updating in such a situation and a natural approach to determining the weights. We then show how this representation can be used in decision making, by modifying a standard approach to decision making—minimizing expected regret—to obtain minimax weighted expected regret (MWER). We provide an axiomatization that characterizes preferences induced by MWER both in the static and dynamic case.

Keywords

Decision theory Ambiguity aversion Minimax regret

JEL Classification

D010 D810

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Notes

Acknowledgments

The authors thank Joerg Stoye for useful comments. Work supported in part by NSF Grants IIS-0812045, IIS-0911036, and CCF-1214844, by AFOSR Grants FA9550-08-1-0438 and FA9550-09-1-0266, by the Multidisciplinary University Research Initiative (MURI) program administered by the AFOSR under Grant FA9550-12-1-0040, and by ARO Grants W911NF-09-1-0281 and W(INF-14-1-0017).

Appendix A: Proof of Lemma 1

A.1 Defining a functional on utility acts

Stoye (2011a) also started his proof of a representation theorem for MER by reducing to a single preference order $\succeq _{M^*}$. He then noted that, the expected regret of an act $f$ with respect to a probability $\Pr $ and menu $M^*$ is just the negative of the expected utility of $f$. Thus, the worst-case expected regret of $f$ with respect to a set $\mathcal {P}$ of probability measures is the negative of the worst-case expected utility of $f$ with respect to $\mathcal {P}$. Thus, it sufficed for Stoye to show that $\succeq _{M^*}$ had an MMEU representation, which he did by showing that $\succeq _{M^*}$ satisfied Gilboa and Schmeidler’s (1989) axioms for MMEU, and then appealing to their representation theorem.

This argument does not quite work for us, because now $\succeq $ does not satisfy the C-independence axiom. (This is because our preference order is based on weighted regret, not regret.) However, we can get a representation theorem for weighted regret using some of the techniques used by Gilboa and Schmeidler to get a representation theorem for MMEU, appropriately modified to deal with lack of C-independence. Specifically, like Gilboa and Schmeidler, 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 3). Using $I$, we can then determine the weight of each probability in $\Delta (S)$, and prove the desired representation theorem.

By standard results, $u$ represents $\succeq $ on constant acts, and $\succeq $ depends only on the utility achieved in each state (as opposed to the actual outcomes) of the acts. The space of all utility acts is the Banach space $\mathcal {B}$ of real-valued functions on $S$. Let $\mathcal {B}^-$ be the set of nonpositive functions in $\mathcal {B}$, where the function $b$ is nonpositive if $b(s) \le 0$ 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

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

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

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

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

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

Lemma 2

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 $I(b_f)\le 0$ by definition of $\inf $, and $f \sim l_0^* \succ l_{\epsilon }^*$ for all $\epsilon < 0$ by Lemma 9, 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 $l_0^* \succ f$, and thus $l_0^* \succ f \succ l_{I(b_f)}^*$. 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)} \prec f $, contradicting the fact that $I(b)$ is the greatest lower bound of $R_f.$

If, on the other hand, $l^*_{I(b_f)} \succ f$, then $l^*_{I(b_f)} \succ f \succeq l^*_{\underline{c}}$ for some $\underline{c}\in \mathbb {R}$. If $f \sim l^*_{\underline{c}}$ then it must be the case that $I(b_f)=\underline{c}$. $I(b_f) \le \underline{c}$ since $l^*_{\underline{c}}\succeq l^*_{\underline{c}}$, and $I(b_f) \ge \underline{c}$ since for all $c' < \underline{c}$, $l^*_{c'} \prec f \sim l^*_{\underline{c}}$.

Otherwise, $l^*_{I(b_f)} \succ f \succ l^*_{\underline{c}}$, and by mixture continuity, there is some $q\in (0,1)$ such that $q\cdot l^*_{I(b_f)} + (1-q) l^*_{\underline{c}} \succ f$. Since $qI(b_f) + (1-q)\underline{c} < I(b_f)$, this contradicts the fact that $I(b_f)$ is a lower 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 3

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 2, $l^*_{I(b_f)} \sim f$ and $g \sim l^*_{I(b_g)}$. Thus, $f \succeq g$ iff $l^*_{I(b_f)} \succeq l^*_{I(b_g)}$, and by Lemma 9, $l^*_{I(b_f)} \succeq l^*_{I(b_g)}$ iff $I(b_f)\ge I(b_g)$. $\square $

In order to invoke a standard separation result for Banach spaces, we extend the definition of $I$ to the Banach space $\mathcal {B}$. We extend $I$ to $\mathcal {B}$ by taking $I(b) = I(b^-)$ for $b \in \mathcal {B}-\mathcal {B}^-$, where for all $b \in \mathcal {B}$, $b^-$ is defined as

$$\begin{aligned} b^-(s) = {\left\{ \begin{array}{ll} b(s),&{}\text { if } b(s) \le 0, \\ 0, &{}\text { if } b(s) > 0. \end{array}\right. } \end{aligned}$$

Clearly $b^- \in \mathcal {B}^-$ and $b=b^-$ if $b \in \mathcal {B}^-$.

We show that the axioms guarantee that $I$ has a number of standard properties. Since we have artificially extended $I$ to $\mathcal {B}$, our arguments require more cases than those in Gilboa and Schmeidler (1989). (We remark that such an “artificial” extension seem unavoidable in our setting.) Moreover, we must work harder to get the result that we want. We need different arguments from that for MMEU (Gilboa and Schmeidler 1989), since the preference order induced by MMEU satisfies C-independence, while our preference order does not.

Lemma 4

(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 (a), If $c$ is in the range of $u$, then it is immediate from the definition of $I$ and Lemma 9 that $I(c^*) = c$. If $c$ is not in the range of $u$, then since $[-1,0]$ 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., $0^* \ge b \ge (-1)^*$). Then there exists an act $f$ such that $b_f = b$. By Lemma 2, $f \sim l^*_{I(b)}$. We now need to consider the case that $c \le 1$ and $c > 1$ separately. If $c \le 1$, by Independence, $c f_b + (1-c) l_0^* \sim c l^*_{I(b)} + (1-c) l_0^*$. By Lemma 3, $I(b_{c f_b + (1-c) l_0^*}) = I(b_{c l^*_{I(b)} + (1-c) l_0^*})$. 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 $I(b) = I(\frac{1}{c}(cb)) = \frac{1}{c}I(cb)$. Crossmultiplying, we have that $I(cb) = cI(b)$, as desired. And 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 what we have shows $I(b) = I(||b|| (b/||b||) = ||b||I(b/||b||)$, 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

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

And 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 if $b, b' \in \mathcal {B}^-$. 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), $I(b) \ge I(b')$. Finally, if either $b \in \mathcal {B}- \mathcal {B}^-$ or $b' \in \mathcal {B}- \mathcal {B}^-$, note that if $b \ge b'$, then $b^- \ge (b')^-$. By definition, $I(b) = I(b^-)$ and $I(b') = I(b')^-$; moreover, $b^-, (b')^- \in \mathcal {B}^-$. Thus, by the argument above, $I(b) \ge I(b^-)$.

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 3, $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 $b_{p f_b + (1-p) f_{b-(1/2)^*}} = b - (1-p)(1/2)^*$. Thus, by Lemma 3, $f_{b-(1-p)(1/2)^*} \succeq f_{(I(b)-\epsilon )^*}$, and $I(b - (1-p)1/2)^*) > I(b) - \epsilon $. Choose $k$ such that $1/k < (1-p)(1/2)$. Then $I(b-(1/k)^*) \ge I(b - (1-p)1/2)^*) > 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), first suppose that $b, b' \in \mathcal {B}^-$. 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 2 that $f_{\frac{b}{-I(b)}} \sim f_{\frac{b'}{-I(b')}}$. By ambiguity aversion, for all $p\in (0,1]$,

$$\begin{aligned} p f_{\frac{b}{-I(b)}} + (1-p) f_{\frac{b'}{-I(b')}} \succeq f_{\frac{b}{-I(b)}}. \end{aligned}$$

Taking $p = I(b)/( I(b) + I(b'))$, we have that

$$\begin{aligned} (I(b)/(I(b) + I(b') )f_{b/I(b)} + (I(b')/(I(b) + I(b') )f_{b'/I(b')} \succeq f_{b/I(b)}. \end{aligned}$$

Therefore, we have

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

Simplifying, we have

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

which, together with positive homogeneity of $I$ (part (b)), implies $I(b + b') \ge I(b) + I(b')$, as required.

If $b, b^- \in \mathcal {B}^-$ and 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 $b, b^- \in \mathcal {B}-$ and 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.

Finally, if either $b \in \mathcal {B}- \mathcal {B}^-$ or $b' \in \mathcal {B}- \mathcal {B}^-$, observe that

$$\begin{aligned} (b+b')^-(s) {\left\{ \begin{array}{ll} = b^-(s) + b'^-(s) ,&{}\text { if }~ b(s) \le 0,b'(s) \le 0 \\ = b^-(s) + b'^-(s) ,&{}\text { if }~ b(s) \ge 0 ,b'(s) \ge 0 \\ \ge b^-(s) + b'^-(s) ,&{}\text { if }~ b(s)>0 ,b'(s) \le 0 \\ \ge b^-(s) + b'^-(s) ,&{}\text { if }~ b(s)\le 0,b'(s) > 0. \end{array}\right. } \end{aligned}$$

Therefore, $(b+b')^- \ge b^- + b'^-$. Thus, $I( b + b' ) = I( (b+b')^- ) \ge I(b^- + b'^-) $ by the monotonicity of $I$, and $I(b^- + b'^-) \ge I(b^-) + I(b'^-)$ by superadditivity of $I$ on $\mathcal {B}^-$. Therefore, $I(b+b') \ge I(b) + I(b')$. $\square $

A.2 Defining the weights

In this section, we use $I$ to define a weight $\alpha _{\Pr }$ for each probability $\Pr \in \Delta (S)$. The heart of the proof involves showing that the resulting set $\mathcal {P}^+$ so determined gives us the desired representation.

Given a set $\mathcal {P}^+$ of weighted probability measures, for $b \in \mathcal {B}^-$, define

$$\begin{aligned} NWREG (b) = \inf _{\Pr \in \mathcal {P}} \alpha _{\Pr } \left( \sum _{s \in S} b(s) \Pr (s)\right) . \end{aligned}$$

Note that $ NWREG $ is the negative of the weighted regret when the menu is $\mathcal {B}^-$. Define

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

and

$$\begin{aligned} NREG _{\Pr }(b) = \sum _{s \in S} b(s) \Pr (s) = E_{\Pr }b. \end{aligned}$$

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

$$\begin{aligned} \alpha _{\Pr } = \sup \{\alpha \in \mathbb {R}: \alpha NREG _{\Pr }(b) \ge I(b) \text{ for } \text{ all } b \in \mathcal {B}^- \}. \end{aligned}$$

(1)

Note that $\alpha _{\Pr } \ge 0$ for all distributions $\Pr \in \Delta (S)$, since $0 \ge I(b)$ for $b \in \mathcal {B}^-$ (by monotonicity); and $\alpha _{\Pr } \le 1$, since $ NREG _{\Pr }((-1)^*) = I((-1)^*) = -1$ for all distributions $\Pr $. Thus, $\alpha _{\Pr } \in [0,1]$. Moreover, it is immediate from the definition of $\alpha _{\Pr }$ that $\alpha _{\Pr } NREG _{\Pr }(b) \ge I(b)$ for all $b \in \mathcal {B}^-$. The next lemma shows that there exists a probability $\Pr $ where we have equality.

Lemma 5

(a)

For some distribution $\Pr $, we have $\alpha _{\Pr } = 1$.
(b)

For all $b \in \mathcal {B}^-$, there exists $\Pr $ such that $\alpha _{\Pr } NREG _{\Pr }(b) = I(b)$.

Proof

The proofs of both part (a) and (b) use a standard separation result: If $U$ is an open 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 for some $\Pr $, for all $b\in \mathcal {B}^-$, $ NREG _{\Pr }(b) \ge I(b)$. Since $ NREG _{\Pr }(b) = E_{\Pr } b$, it suffices to show that $E_{\Pr }(b) \ge I(b)$ for all $b \in \mathcal {B}^-$.

Let $U = \{b' \in \mathcal {B}: I(b') > -1\}$. $U$ is open (by continuity of $I$), and convex (by positive homogeneity and superadditivity of $I$), and $(-1)^* \notin U$. Thus, there exists a linear functional $\lambda $ such that $\lambda (b') > \lambda ((-1)^*)$ for $b' \in U$. We want to show that $\lambda $ is a positive linear functional, that is, that $\lambda (b) \ge 0$ if $b \ge 0^*$. Since $0^* \in U$, and $\lambda (0^*) = 0$, it follows that $\lambda ((-1)^*) < 0$. Since $\lambda $ is linear, we can assume without loss of generality that $\lambda ((-1)^*) = -1$. Thus, for all $b'\in \mathcal {B}^-$, $I(b') > -1$ implies $\lambda (b') > -1$. Suppose that $c > 0$ and $b' \ge 0^*$. From the definition of $I$, it follows that $I(cb') = I(0^*) = 0 > -1$. So $c\lambda (b') = \lambda (cb') > -1$, so $\lambda (b') > -1/c$. (The fact that $I(cb') = I(0^*)$ follows from the definition of $I$ on elements in $\mathcal {B}- \mathcal {B}^-$.) Since this is true for all $c > 0$, it must be the case that $\lambda (b') \ge 0$. Thus, $\lambda $ is a positive functional.

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

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

Next note that, for $b \in \mathcal {B}^-$,

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

(2)

For if $I(b) > c$, then $I(b/|c|) > -1$ by positive homogeneity, so $\lambda (b/|c|) > -1$ and $\lambda (b) > c$. The result now follows. For if $b \in \mathcal {B}^-$, then $I(b) \le I(0^*) = 0$ by monotonicity. Thus, if $c < I(b)$, then $c < 0$, so, by (2), $\lambda (b) > c$. Since $\lambda (b) > c$ whenever $I(b) > c$, it follows that $ E_{\Pr }(b) = \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 $\Pr $ such that $\alpha _{\Pr } NREG _{\Pr }(b) = I(b)$. First suppose that $||b|| \le 1$. If $I(b) = 0$, then there must exist some $s$ such that $b(s) = 0$, for otherwise there exists $c < 0$ such that $b \le c^*$, so $I(b) \le c$. If $b(s) = 0$, let $\Pr _s$ be such that $\Pr _s(s) = 1$. Then $ NREG _{\Pr _s}(b) = 0$, so (b) holds in this case.

If $||b|| \le 1$ and $I(b) < 0$, let $U = \{b': I(b') > I(b)\}$. Again, $U$ is open and convex, and $b \notin U$, so there exists a linear functional $\lambda $ such that $\lambda (b') > \lambda (b)$ for $b' \in U$. Since $0^* \in U$ and $\lambda (0^*) = 0$, we must have $\lambda (b) < 0$. Since $(-1)^* \le b$, $(-1)^*$ is not in $U$, and therefore, we also have $\lambda ((-1)^*) < 0$. Thus, we can assume without loss of generality that $\lambda ((-1)^*) = -1$, and hence $\lambda ((1)^*) = 1$. The same argument as above shows that $\lambda $ is positive: for all $c>0$ and $b'\ge 0^*$, $I(cb') = 0$ as before. Since $I(b)<0$, it follows that $I(cb') > I(b)$, so $cb' \in U$ and $\lambda (cb') > \lambda (b) \ge \lambda ((-1)^*) = -1$. Thus, as before, for all $c>0$, $b' \ge 0^*$, $\lambda (b') > \frac{-1}{c}$, so $\lambda $ is a positive functional.

Therefore, $\lambda $ determines a probability distribution $\Pr $ such that, for all $b' \in \mathcal {B}^-$, we have $\lambda (b') = E_{\Pr }(b')$. This, of course, will turn out to be the desired distribution. To show this, we need to show that $\alpha _{\Pr } = I(b)/ NREG _{\Pr }(b)$. Clearly $\alpha _{\Pr } \le I(b)/ NREG _{\Pr }(b)$, since if $\alpha > I(b)/ NREG _{\Pr }(b)$, then $\alpha NREG _{\Pr }(b) < I(b)$ (since $ NREG _{\Pr }(b) = \lambda (b) < 0$). To show that $\alpha _{\Pr } \ge I(b)/ NREG _{\Pr }b$, we must show that $(I(b) / NREG _{\Pr }(b)) NREG _{\Pr }(b') \ge I(b')$ 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

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

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

Thus, if $I(b')/(-I(b)) > c$ and $c < 0$, then $I(b') > -c I(b)$, and hence $\lambda (b')/(-\lambda (b)) > c$. 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 $\Pr $ such that $\alpha _{\Pr } NREG _{\Pr }(b/||b||) = I(b/||b||)$. Since $ NREG _{\Pr }(b/||b||) = NREG _{\Pr }(b)/||b||$, and $I(b/||b||) = I(b)/||b||$, we must have that $\alpha _{\Pr } NREG _{\Pr }(b) = I(b)$. $\square $

We can now complete the proof of Lemma 1. By Lemma 5 and the definition of $\alpha _{\Pr }$, for all $b\in \mathcal {B}^-$,

$$\begin{aligned} I(b)&= \inf _{\Pr \in \Delta (S)} \alpha _{\Pr } NREG (b) \nonumber \\&= \inf _{\Pr \in \Delta (S) }\left( \alpha _{\Pr }\sum _{s\in S}b(s)\Pr (s) \right) \nonumber \\&= \sup _{\Pr \in \Delta (S) }\left( -\alpha _{\Pr }\sum _{s\in S}b(s)\Pr (s) \right) . \end{aligned}$$

(3)

Recall that, by Lemma 3, 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

$$\begin{aligned} \sup _{\Pr \in \Delta (S) }\left( -\alpha _{\Pr }\sum _{s\in S}u(f(s))\Pr (s) \right) \le \sup _{\Pr \in \Delta (S) }\left( -\alpha _{\Pr }\sum _{s\in S}u(g(s))\Pr (s) \right) . \end{aligned}$$

Note that, for $f \in M^* = \mathcal {B}^-$, we have $ reg _{M^*,\Pr }(f) = \sup (- u(f(s)) \Pr (s)$, since $0^*$ dominates all acts in $M^*$. Thus, $\succeq = \succeq _{M^*,\mathcal {P}^+}^{S,Y,U}$, where $\mathcal {P}^+ = \{(\Pr ,\alpha _{\Pr }: \Pr \in \Delta (S)\}$.

We have already observed that $U$ is unique up to affine transformations, so it remains to show that $\mathcal {P}^+$ is maximal. This follows from the definition of $\alpha _{\Pr }$. If $\succeq _M = \succeq _{M,(\mathcal {P}')^+}^{S,Y,U}$, and $(\alpha ',\Pr ) \in (\mathcal {P}')^+$, then we claim that $ \alpha ' \in \{\alpha \in \mathbb {R}: \alpha NREG _{\Pr }(b) \ge I(b) \text{ for } \text{ all } b \in \mathcal {B}^-\}$. If not, there would be some $b\in \mathcal {B}^-$ with $||b|| \le \frac{1}{2}$, such that $ \alpha ' NREG _{\Pr }(b) < I(b)$, which, by the definition of $\prec ^{S,Y,U}_{M^*,(\mathcal {P}')^*}$, means that $ l^*_{-1} \prec ^{S,Y,U}_{M^*,(\mathcal {P}')^+} f_b \prec ^{S,Y,U}_{M^*,(\mathcal {P}')^+} l^*_{I(b)}$. Recall that $I(b_f) = \inf \{ \gamma : l^*_{\gamma } \succeq _{M^*} f \}$. Moreover, since $\prec ^{S,Y,U}_{M^*,(\mathcal {P}')^+}$ satisfies mixture continuity, there exists some $p \in (0,1)$ such that $f_b \prec ^{S,Y,U}_{M^*,(\mathcal {P}')^+} p l^*_{-1} + (1-p) l^*_{I(b)} \prec ^{S,Y,U}_{M^*,(\mathcal {P}')^+} \prec ^{S,Y,U}_{M^*,(\mathcal {P}')^+} l^*_{I(b)}$. This contradicts the definition of $I(b)$. Therefore, $ \alpha ' \in \{\alpha \in \mathbb {R}: \alpha NREG _{\Pr }(b) \ge I(b) \text{ for } \text{ all } b \in \mathcal {B}^-\}$, and hence $\alpha ' \le \alpha _{\Pr }$.

A.3 Uniqueness of representation

In this section, we show that the canonical set of weighted probabilities we constructed, when viewed as a set of subnormal probability measures, is regular and includes at least one proper probability measure. Moreover, this set of sub-probability measures is the only regular set that induces the preference order $\succeq $ on nonpositive acts. Our uniqueness result is analogous to the uniqueness results of Gilboa and Schmeidler (1989), who show that the convex, closed, and non-empty set of probability measures in their representation theorem for MMEU is unique. The argument is based on two lemmas: Lemma 6 says that the canonical set of sub-probability measures is regular; and Lemma 7 says that a set of sub-probability measures representing $\succeq $ over nonpositive acts that is regular and contains at least one proper probability measure is unique. The proof of this second lemma, 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’. However, a slightly different argument is required in our case, since our acts must have utilities corresponding to nonpositive vectors in $\mathbb {R}^{|S|}$.

Lemma 6

Let $\mathcal {P}^+$ be the canonical set of weighted probability measures representing $\succeq $. The set $C(\mathcal {P}^+)$ of sub-probability measures is regular.

Proof

It is useful to note that, by definition, $\mathbf {p} \in C(\mathcal {P}^+)$ if and only if

$$\begin{aligned} E_{ \mathbf {p}}(b) \ge I(b)\quad \text{ for } \text{ all } b \in \mathcal {B}^- \end{aligned}$$

(where expectation with respect to a subnormal probability measure is defined in the obvious way).

Recall that a set is regular if it is convex, closed, and downward-closed. We first show that $C(\mathcal {P}^+)$ is downward-closed. Suppose that $\mathbf {p} \in C(\mathcal {P}^+)$ and $\mathbf {q} \le \mathbf {p}$ (i.e., $\mathbf {q}(s) \le \alpha \Pr (s)$ for all $s\in S$. Since $\mathbf {p} \in C(\mathcal {P}^+)$, $E_{ \mathbf {p}}(b) \ge I(b)$ for all $b \in \mathcal {B}^-$. Since $\mathbf {q} \le \mathbf {p}$ and, if $b \in \mathcal {B}^-$, we have $b \le 0^*$, it follows that $E_{ \mathbf {q}}(b) \ge E_{ \mathbf {p}}(b) \ge I(b)$ for all $b \in \mathcal {B}^-$, and thus $\mathbf {q}\in C(\mathcal {P}^+)$.

To see that $C(\mathcal {P}^+)$ is closed, let $\mathbf {p} = \lim _{n\rightarrow \infty } \mathbf {p}_n$, where each $\mathbf {p}_n \in C(\mathcal {P}^+)$. Since $\mathbf {p}_n \in C(\mathcal {P}^+)$ it must be the case that $E_{\mathbf {p}_n}(b) \ge I(b)$ for all $b \in \mathcal {B}^-$. By the continuity of expectation, it follows that $E_{\mathbf {p}}(b) \ge I(b)$ for all $b \in \mathcal {B}^-$. Thus, $\mathbf {p} \in C(\mathcal {P}^+)$.

To show that $C(\mathcal {P}^+)$ is convex, suppose that $\mathbf {p}, \mathbf {q} \in C(\mathcal {P}^+)$. Then $E_{ \mathbf {p}}(b) \ge I(b)$ and $E_{ \mathbf {q}}(b) \ge I(b)$ for all $b \in \mathcal {B}^-$. It easily follows that for all $a \in (0,1)$, $E_{ a\mathbf {p} + (1-a) \mathbf {q}}(b) \ge I(b)$ for all $b \in \mathcal {B}^-$. Thus, $a\mathbf {p} + (1-a) \mathbf {q} \in C(\mathcal {P}^+)$. $\square $

Lemma 7

A set of sub-probability measures representing $\succeq _{}$ over nonpositive acts that is regular, and has at least one proper probability measure is unique.

Proof

Suppose for contradiction that there exists two regular sets of subnormal probability distributions, $C_1$ and $C_2$, that represent $\succeq _{}$ and have at least one proper probability measure.

First, without loss of generality, let $\mathbf {q} \in C_2 \backslash C_1$. We actually look at an extension of $C_1$ that is downward-closed in each component to $-\infty $. Let $\overline{C}_1 = \{ \mathbf {p} \in \mathbb {R}^{|S|} : \mathbf {p} \le \mathbf {p'} \}$. Note an element $\mathbf {p}$ of $\overline{C}_1$ may not be a subnormal probability measure; we do not require that $\mathbf {p}(s) \ge 0$ for all $s \in S$. Since $\overline{C}_1$ and $\{\mathbf {q}\}$ are closed, convex, and disjoint, and $\{\mathbf {q}\}$ is compact, the separating hyperplane theorem (Rockafellar 1970) says that there exists $\theta \in \mathbb {R}^{|S|}$ and $c\in \mathbb {R}$ such that

$$\begin{aligned} \theta \cdot \mathbf {p}> c\quad \text { for all } \mathbf {p}\in \overline{C}_1 \text {, \quad and \quad } \theta \cdot \mathbf {q} < c. \end{aligned}$$

(4)

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 $\theta (s) \le 0$ for all $s\in S$ (so that $\theta $ corresponds to the utility profile of some act with nonpositive utilities). Suppose that $\theta (s') > 0 $ for some $s'\in S$. By (4), $\theta \cdot \mathbf {p} > c \text { for all } \mathbf {p}\in \overline{C}_1$. However, consider $\mathbf {p^*} \in \overline{C}_1$ defined by

$$\begin{aligned} \mathbf {p^*}(s) = {\left\{ \begin{array}{ll} 0, &{}\hbox { if }~s\ne s' \\ \frac{-|c|}{\theta (s)}, &{}\hbox { if }~s = s'. \end{array}\right. } \end{aligned}$$

Clearly, $\theta \cdot \mathbf {p^*} \le c $, contradicting (4). Thus it must be the case that $\theta (s) \le 0$ 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 $. By continuity, $f \sim _{} l^*_d$ for some constant act $l^*_d$. Since $C_1$ and $C_2$ both represent $\succeq _{}$, and $C_1$ and $C_2$ both contain a proper probability measure,

$$\begin{aligned} \min _{\mathbf {p} \in C_1} \mathbf {p}\cdot (u\circ f) = \min _{\mathbf {p} \in C_1} \mathbf {p}\cdot (u\circ l^*_d) = d = \min _{\mathbf {p} \in C_2} \mathbf {p} \cdot (u\circ f) . \end{aligned}$$

However, by (4),

$$\begin{aligned} \min _{\mathbf {p}\in C_1} \mathbf {p}\cdot (u\circ f) > c > \min _{\mathbf {p} \in C_2} \mathbf {p} \cdot (u\circ f), \end{aligned}$$

which is a contradiction. $\square $

Appendix B: An axiomatic characterization of MWER with preference relations

We consider an axiomatization based on primitive preference orders $\succeq _M$ indexed by menus. (Because regret is menu-dependent, we cannot consider a single preference order $\succeq $.) As Stoye (2011b) points out, one disadvantage of considering such preference orders is that they are not observable. For example, suppose that $f_1 \succ _M f_2 \succ _M f_3$. By presenting the menu $M = \{f_1,f_2,f_3\}$ to the decision maker, we can observe that he prefers $f_1$. But there is no way to observe that $f_2 \succ _M f_3$. The traditional approach (seeing which of $f_2$ and $f_3$ the decision maker prefers when presented with the menu $M' = \{f_2,f_3\}$) will not work, because the decision maker’s preferences with menu $M'$ may be different from those with menu $M$. (Bleichrodt (2009) studies a similar problem).

In Section 4, we provided a characterization of MWER with choice functions as the primitives. Despite the fact that a regret-based preference order is not observable, an axiomatization using menu-dependent preference orders allows us to compare the axioms for weighted regret to those for other decision rules.

We state the axioms in a way that lets us clearly distinguish the axioms for SEU, MMEU, MER, and MWER. The axioms are universally quantified over acts $f$, $g$, and $h$, menus $M$ and $M'$, and $p \in (0,1)$. We assume that $f,g \in M$ when we write $f \succeq _M g$.⁷ We use $l^*$ to denote a constant act that maps all states to $l$.

Axiom 10

(Transitivity) $f\succeq _M g\succeq _M h\Rightarrow f\succeq _M h$.

Axiom 11

(Completeness) $f\succeq _M g \text { or } g\succeq _M f$.

Let $\mathcal {M}_B$ denote the set of all menus that are bounded above; that is, $\mathcal {M}_B = \{M: \sup _{g\in M}u(g(s)) \text{ is } \text{ finite } \}$.

Axiom 12

(Nontriviality) $f\succ _M g$ for some acts $f$ and $g$ and menu $M \in \mathcal {M}_B$.

Up to now, we have taken the set of menus to be $\mathcal {M}_B$. This assumption is necessary (and sufficient) for regret to be well defined. Later, we use Axiom 12 in the context of different classes of menus. In particular, we are interested in the set of finite menus and the set of finitely generated convex menus, that is, the menus $M$ such such that there is a finite set $A_M$ of acts such that $M$ consists of all the convex combinations of acts in $A_M$. We denote these sets $\mathcal {M}_F$ and $\mathcal {M}_C$, respectively. When we use Axiom 12 in such contexts, $\mathcal {M}_B$ in Axiom 12 is understood to be replaced by $\mathcal {M}_C$ and $\mathcal {M}_F$, respectively. Observe that $\mathcal {M}_{C} \subseteq \mathcal {M}_B$ and $\mathcal {M}_F \subseteq \mathcal {M}_B$.

Axiom 13

(Monotonicity) If $(f(s))^* \succeq _{\{(f(s))^*,(g(s))^*\}} (g(s))^*$ for all $s\in S$, then $f\succeq _M g$.

Axiom 14

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

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

Menu-independent versions of Axioms 10–14 are standard (for example, (menu-independent versions of) these axioms are in Gilboa and Schmeidler (1989)). Clearly (menu-independent versions of) Axioms 10, 11, 13, and 14 hold for MMEU, and SEU; Axiom 12 is assumed in all the standard axiomatizations, and is used to get a unique representation.

Axiom 15

(Ambiguity Aversion)

$$\begin{aligned} f\sim _M g\Rightarrow pf+(1-p)g \succeq _{M\cup \{pf+(1-p)g\}} g. \end{aligned}$$

Ambiguity aversion says that the decision maker weakly prefers to hedge her bets. It also holds for MMEU, MER, and SEU, and is assumed in the axiomatizations for MMEU and MER. It is not assumed for the axiomatization of SEU, since it follows from the Independence axiom, discussed next. Independence also holds for MWER, provided that we are careful about the menus involved. Given a menu M and an act $h$, let $pM + (1-p)h$ be the menu $\{pf + (1-p)h: p \in M\}$.

Axiom 16

(Independence)

$$\begin{aligned} f\succeq _M g \text{ iff } pf + (1-p)h \succeq _{pM+(1-p)h} pg+(1-p)h. \end{aligned}$$

Independence holds in a strong sense for SEU, since we can ignore the menus. The menu-independent version of Independence is easily seen to imply ambiguity aversion. Independence does not hold for MMEU.

Although we have menu independence for SEU and MMEU, we do not have it for MER or MWER. The following two axioms are weakened versions of menu independence that do hold for MER and MWER.

Axiom 17

(Menu independence for constant acts) If $l^*$ and $(l')^*$ are constant acts, then $l^* \succeq _M (l')^*$ iff $l^* \succeq _{M'} (l')^*$.

In light of this axiom, when comparing constant acts, we omit the menu.

An act $h$ is never strictly optimal relative to M if, for all states $s \in S$, there is some $f \in M$ such that $(f(s))^* \succeq (h(s))^*$.

Axiom 18

(Independence of Never Strictly Optimal Alternatives (INA)) If every act in $M'$ is never strictly optimal relative to $M$, then $f\succeq _M g$ iff $f\succeq _{M\cup M'} g.$

Theorem 3

For all $Y$, $U$, $S$, and $\mathcal {P}^+$, the family of preference orders $\succeq _{M,\mathcal {P}^+}^{S,Y,U}$ for $M \in \mathcal {M}_B$ (resp., $\mathcal {M}_F$, $\mathcal {M}_C$) satisfies Axioms 10–18. Conversely, if a family of preference orders $\succeq _{M}$ on the acts in $\Delta (Y)^S$ for $M \in \mathcal {M}_B$ (resp., $\mathcal {M}_F$, $\mathcal {M}_C$) satisfies Axioms 10–18, then there exist a utility $U$ on $Y$ and a weighted set $\mathcal {P}^+$ of probabilities on $S$ such that $C(\mathcal {P}^+)$ is regular and $\succeq _M = \succeq _{M,\mathcal {P}^+}^{S,Y,U}$ for all $M \in \mathcal {M}_B$ (resp., $\mathcal {M}_C$, $\mathcal {M}_F$). Moreover, $U$ is unique up to affine transformations, and $C(\mathcal {P}^+)$ is unique in the sense that if $\fancyscript{Q}^+$ represents $\succeq _M$, and $C(\fancyscript{Q}^+)$ is regular, then $C(\fancyscript{Q}^+) = C(\mathcal {P}^+)$.

Proof

Showing that $\succeq _{M,\mathcal {P}^+}^{S,Y,U}$ satisfies Axioms 10–18 is fairly straightforward; we leave details to the reader. Essentially the same proof works for $\mathcal {M}_B$, $\mathcal {M}_C$, and $\mathcal {M}_F$. The proof of the converse is quite nontrivial, although it follows the lines of the proof of other representation theorems. We start by considering $\mathcal {M}_B$.

Using standard techniques, we can show that the axioms guarantee the existence of a utility function $U$ on prizes that can be extended to lotteries in the obvious way, so that $l^* \succeq (l')^*$ iff $U(l) \ge U(l')$. We then use techniques of Stoye (2011b) to show that it suffices to get a representation theorem for a single menu, rather than all menus: the menu consisting of all acts $f$ such that $U(f(s)) \le 0$ for all states $s \in S$. This allows us to use techniques in the spirit of those used by Gilboa and Schmeidler (1989) to represent (unweighted) MMEU.

We show here that if a family of menu-dependent preferences $\succeq _M$ satisfies Axioms 10–18, then $\succeq _M$ can be represented as minimizing expected regret with respect to a set of weighted probabilities and a utility function. Since the proof is somewhat lengthy and complicated, we split it into several steps, each in a separate subsection.

Simplifying the problem: Our proof starts in much the same way as the proof by Stoye (2011b) of a representation theorem for regret. Lemma 8 guarantees the existence of a utility function $U$ on prizes that can be extended to lotteries in the obvious way, so that $l^* \succeq (l')^*$ iff $U(l) \ge U(l')$. In other words, preferences over all constant acts are represented by the maximization of $U$ on the corresponding lotteries that the constant acts map to. Lemma 8 is a consequence of standard results. Our menus are arbitrary sets of acts, as opposed to convex hulls of a finite number of acts in Stoye (2011b); Lemma 10 shows that Stoye’s technique can be adapted to work when menus are arbitrary sets of acts. Finally, following Stoye (2011b), we reduce the proof of existence of a minimax weighted regret representation for the family $\succeq _M$ to the proof of existence of a minimax weighted regret representation for a single menu-independent preference order $\succeq $ (Lemma 11). $\square $

Lemma 8

If Axioms 1–3, 5, 7, and 8 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 $(l')^*$ and menus $M$,

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

Proof

By menu independence for constant acts, the family of preferences $\succeq _M$ all agree when restricted to constant acts. The lemma then follows from standard results (see, e.g., Kreps (1988)), since menu independence for constant acts, combined with independence, gives the standard independence (substitution) axiom from expected utility theory. $\square $

As is commonly done, given $U$, we define $u(l) = \sum _{\{y:\, l(y) > 0\}} l(y) U(y)$. Thus, $u(l)$ is the expected utility of lottery $l$. We extend $u$ to constant acts by taking $u(l^*) = u(l)$. Thus, Lemma 8 says that, for all menus $M$, $l^* \succeq (l')^*$ iff $u(l^*) \ge u((l')^*)$. If $c$ is the utility of some lottery, let $l^*_c$ be a constant lottery that $u(l^*_c) = c$. The following is now immediate. We state it as a lemma so that we can refer to it later.

Lemma 9

$u(l^*_c) \ge u(l^*_{c'})$ 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 $l^*_c \succ l^*_{c'}$.

The key step in showing that we can reduce to a single menu is to show that, roughly speaking, for each menu, there exists a menu-dependent function $g_M$ such that $u(g_M(s)) = -\sup _{f \in M} u(f(s))$. Stoye (2011b) proved a similar result, but he assumed that all menus were obtained by taking the convex hull of a finite set of acts. Because we allow arbitrary bounded menus, this result is not quite true for us. For example, suppose that the range of $u$ is $(-1,\infty ]$. Then there may be a menu $M$ such that $\sup _{f \in M} u(f(s)) = 5$, so $-\sup _{f \in M} u(f(s)) = -5$. But there is no act $g$ such that $u(g(s)) = -5$, since $u$ is bounded below by $-1$. The following weakening of this result suffices for our purpose.

Lemma 10

There exists a utility function $U$ such that for every menu $M$, there exists $\epsilon \in (0,1]$ and constant act $l^*$ such that for all $f,g \in M$, $f \succeq _M g \Leftrightarrow t(f) \succeq _{t(M)} t(g)$, where $t$ has the form $t(f) = \epsilon f + (1-\epsilon )l^*$ and $t(M)=\{t(f) : f \in M\}$. Moreover, there exists an act $g_{t(M)}$ such that $u(g_{t(M)}(s) ) = -\sup _{f\in t(M)}u(f(s))$ for all $s \in S$.

Proof

The nontriviality and monotonicity axioms imply there must exist prizes $x$ and $y$ such that $U(x) > U(y)$. We consider four cases.

Case 1: The range of $U$ is bounded above and below. Then we can rescale so that the range of $U$ is $[-1,1]$. Thus, there must be prizes $x$ and $y$ such that $U(x)=1$ and $U(y)=-1$. For all $c \in [-1,1]$, there must be a prize $x'$ that is a convex combination of $x$ and $y$ such that $u(x') = c$, so we can clearly define a function $g_M$ such that, for all $s \in S$, we have $u(g_{M}(s) ) = -\sup _{f\in M}u(f(s))$. Furthermore, we know that such a $g_M$ exists because it can be formed as an act which maps each state to an appropriate lottery over the prizes $x$ and $y$. More generally, we know that an act with a certain utility profile exists if its utility for each state is within the range of $U$. This fact will be used in the other cases as well. Thus, in this case we can take $t$ to be the identity (i.e., $\epsilon = 1$).

Case 2: The range of $U$ is $(-\infty ,\infty )$. Again, for all $c \in (\infty ,\infty )$, there must exist a prize $x$ such that $u(x) = c$. Since menus are assumed to be bounded above, we can again define the required function $g$ and take $\epsilon = 1$.

Case 3: The range of $U$ is bounded above and unbounded below. Then we can assume without loss of generality that the range is $(-\infty ,1]$, and for all $c$ in the range, there is a prize $x$ such that $u(x) = c$. For all menus $M$, $\epsilon >0$, and acts $f, g \in M$, by Independence, we have that

$$\begin{aligned} f \succeq _M g \Leftrightarrow \epsilon f + (1-\epsilon ) l_1^* \succeq _{\epsilon M + (1-\epsilon ) l_1^* } \epsilon g + (1-\epsilon ) l_1^*. \end{aligned}$$

There exists an $\epsilon > 0$ such that for all $s\in S$,

$$\begin{aligned} 1 \ge \sup _{f \in M} \epsilon u(f(s)) + (1-\epsilon ) \ge -1. \end{aligned}$$

Let $t(f) = \epsilon f + (1-\epsilon ) l_1^*$. Clearly there exists an act $g_{t(M)}$ such that $ u(g_{t(M)}(s) ) = -\sup _{f\in t(M)}u(f(s))$ for all $s \in S$.

Case 4: The range of $U$ is bounded below and unbounded above. By the upper-boundedness axiom, every menu has an upper bound on its utility range. Therefore, for every menu $M$, $\epsilon >0$, and all acts $f$ and $g$ in $M$, by Independence,

$$\begin{aligned} f\succeq _M g \Leftrightarrow \epsilon f + (1-\epsilon ) l_{-1}^* \succeq _{\epsilon M + (1-\epsilon ) l_{-1}^*} \epsilon g + (1-\epsilon ) l_{-1}^*. \end{aligned}$$

There exists $\epsilon >0$ such that for all $s\in S$,

$$\begin{aligned} \sup _{f \in M} \epsilon u(f(s)) + (1-\epsilon ) u(l_{-1}^*(s) ) \le 1. \end{aligned}$$

Let $t(f) = \epsilon f + (1-\epsilon ) l_{-1}^*$. Again, it is easy to see that $g_{t(M)}$ exists. $\square $

In light of Lemma 10, we henceforth assume that the utility function $u$ derived from $U$ is such that its range is either $(-\infty ,\infty )$, $[-,1,1]$, $(-\infty ,1]$, or $[-1,\infty )$. In any case, its range always includes $[-1,1]$.

Before proving the key lemma, 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 $f \sim _M g$ for all menus $M$. That is, only utility acts matter. If $c$ is a real, we take $c^*$ to be the constant utility act such that $c^*(s) = c$ for all $s \in S$.

Lemma 11

Let $M^*$ be the menu consisting of all acts $f$ such that $(-1)^* \le b_f \le 0^*$. Then $(U,\mathcal {P}^+)$ represents $\succeq _{M^*}$ (i.e., $\succeq _{M^*} = \succeq _{M^*,\mathcal {P}^+}^{S,X,U})$ iff $(U,\mathcal {P}^+)$ represents $\succeq _M$ for all menus $M$.

Proof

Our arguments are similar in spirit to those of Stoye (2011b).

By Lemma 10, there exists $t$ such that $t(f) = \epsilon f + (1-\epsilon )h$ for a constant function $h$ such that

$$\begin{aligned} f\succeq _M g \text{ iff } t(f) \succeq _{t(M)} t(g); \end{aligned}$$

moreover, for this choice of $t$, the act $g_{t(M)}$ defined in Lemma 10 exists.

By Independence,

$$\begin{aligned} t(f) \succeq _{t(M)} t(g) \text{ iff } \frac{1}{2}t(f)+\frac{1}{2}g_{t(M)} \succeq _{\frac{1}{2}t(M) +\frac{1}{2}g_{t(M)}} \frac{1}{2}t(g)+\frac{1}{2}g_{t(M)}. \end{aligned}$$

Let $M^*$ be the menu that contains all acts with utilities in $[-1,0]$. By INA, we know that for all acts $f$ and $g$, and menus $M$ for which $g_M$ is defined, we have

$$\begin{aligned} f\succeq _M g \text{ iff } \frac{1}{2}f+\frac{1}{2}g_M \succeq _{M^*} \frac{1}{2}g+\frac{1}{2}g_M. \end{aligned}$$

This is because acts of the form $\frac{1}{2}f+\frac{1}{2}g_M$ are never strictly optimal with respect to the menu $\frac{1}{2}M + \frac{1}{2}g_M$. At every state $s$ there must be some act in $\frac{1}{2}M+\frac{1}{2}g_M$ that has utility 0 at $s$ (namely, the mixture that involves an act $f \in M$ whose utility at $s$ is maximal; that is, $u(f(s))\ge \hbox {max}_{f^{\prime }\in M}u(f^{\prime }(s))$.

Thus,$ f\succeq _M g \text{ iff } \frac{1}{2}t(f)+\frac{1}{2}g_{t(M)} \succeq _{M^*} \frac{1}{2}t(g)+\frac{1}{2}g_{t(M)}.$

Since the MWER representation also satisfies Independence and INA, we know that for all menus $M$, and acts $f$ and $g$ in $M$, $f \succeq _{M,\mathcal {P}^+}^{S,X,U} g \Leftrightarrow t(f) \succeq _{t(M),\mathcal {P}^+}^{S,X,U} t(g) \Leftrightarrow \frac{1}{2}t(f) + \frac{1}{2}g_{t(M)} \succeq _{M^*,\mathcal {P}^+}^{S,X,U} \frac{1}{2}t(g) + \frac{1}{2}g_{t(M)}.$ Therefore, to show that $\succeq _M$ has a MWER representation with respect to $(U,\mathcal {P}^+)$, it suffices to show that $\succeq _{M^*}$ has a MWER representation with respect to $(U,\mathcal {P}^+)$. $\square $

In the sequel, we drop the menu subscript when we refer to the family of preferences, and just write $\succeq $ (to denote $\succeq _{M^*}$); by Lemma 11, it suffices to consider $\succeq _{M^*}$.

It is straightforward to check that $\succeq _{M^*}$ satisfies completeness, transitivity, nontriviality, monotonicity, mixture continuity, independence, INA, and ambiguity aversion. Therefore, by Lemma 1, there exists some $(U,\mathcal {P}^+)$ representing $\succeq _{M^*}$. By Lemma 11, $(U,\mathcal {P}^+)$ represents $\succeq _M$ for all menus $M$, as required.

Since the axioms hold for all menus in $\mathcal {M}_B$, they clearly continue to hold if we restrict to $\mathcal {M}_F$ and $\mathcal {M}_C$. To prove the converse in the case of $\mathcal {M}_F$, we first argue that if the preference orders $\succeq _M$ for $M \in \mathcal {M}_F$ satisfy the axioms, then they uniquely determine preference orders $\succeq _M$ for menus $M \in \mathcal {M}_B$ that also satisfy the axioms. Clearly, it also then follows that the set of preference orders $\succeq _M$ for $M \in \mathcal {M}_C$ determines $\succeq _M$ for $M \in \mathcal {M}_B$. The proof immediately follows from this observation and the proof in the case of $\mathcal {M}_B$.

Consider a bounded menu $M$. The utility frontier of menu $M$ is a function mapping each state to the maximum utility achieved in that state by any act in $M$. Since $S$ is assumed to be finite, there exists a finite subset $M' \subseteq M$ such that the utility frontier of $M'$ is the same as the utility frontier of $M$. Therefore, for all acts $f,g \in M$,

$$\begin{aligned} f \succeq _M g \Leftrightarrow f \succeq _{M' \cup \{f,g\}} g, \end{aligned}$$

by Axiom 5. Since $M'$ is finite, we have shown what we need.

Finally, for $\mathcal {M}_C$, suppose that $M$ is a convex set of acts generated by the finite set $A_M$. Then, for all $f,g \in A_M$,

$$\begin{aligned} f \succeq _{A_M} g \Leftrightarrow f \succeq _M g, \end{aligned}$$

by Axiom 5, since no interior points in $M$ can be strictly optimal; hence, interior points can be removed from $M$ without changing preferences. Thus, the result for $\mathcal {M}_C$ follows from the result for $\mathcal {M}_F$. $\square $

It is instructive to compare Theorem 3 to other representation results in the literature. Anscombe and Aumann (1963) showed that the menu-independent versions of axioms 10–14 and 16 characterize SEU. The presence of Axiom 16 (menu-independent Independence) greatly simplifies things. Gilboa and Schmeidler (1989) showed that axioms 10–15 together with one more axiom that they call certainty-independence characterizes MMEU. Certainty-independence, or C-independence for short, is a weakening of independence (which, as we observed, does not hold for MMEU), where the act $h$ is required to be a constant act. Since MMEU is menu-independent, we state it in a menu-independent way.

Axiom 19

(C-Independence) If $h$ is a constant act, then $f\succeq g$ iff $pf+(1-p)h \succeq pg + (1-p)h$.

Table 6 describes the relationship between the axioms characterizing the decision rules.

Table 6

Characterizing axioms for several decision rules.

	SEU	REG	MER	MWER	MMEU
Ax. 1-6,8-10	$\checkmark $	$\checkmark $	$\checkmark $	$\checkmark $	$\checkmark $
Ind	$\checkmark $	$\checkmark $	$\checkmark $	$\checkmark $
C-Ind	$\checkmark $				$\checkmark $
Ax. 12	$\checkmark $	$\checkmark $	$\checkmark $
Symmetry	$\checkmark $	$\checkmark $

Appendix C: Characterizing MWER with likelihood updating

We can in fact directly translate the MDC axiom into a setting with preference relations instead of choice functions.

Definition 2

(Null event) An event $E$ is null if, for all $f, g \in \Delta (Y)^S$ and menus $M$ with $fEg, g \in M$, we have $fEg \sim _M g$.

MDC. For all non-null events $E$, $f \succeq _{E,M} g$ iff $fEh \succeq _{MEh} gEh$ for some $h \in M$.⁸

The key feature of MDC is that it allows us to reduce all the conditional preference orders $\succeq _{E,M}$ to the unconditional order $\succeq _M$, to which we can apply Theorem 3.

Theorem 4

For all $Y$, $U$, $S$, $\mathcal {P}^+$, and $M \in \mathcal {M}_B$, the family of preference orders $\succeq _{\mathcal {P}^+{\mid }E,M}^{S,Y,U}$ for events $E$ such that $\overline{\mathcal {P}}^+(E) > 0$ satisfies Axioms 10–18 and MDC. Conversely, if a family of preference orders $\succeq _{E,M}$ on the acts in $\Delta (Y)^S$ satisfies Axioms 10–18 and MDC for $M \in \mathcal {M}_B$ (resp., $\mathcal {M}_F$, $\mathcal {M}_C$), then there exists a utility $U$ on $Y$ and a weighted set $\mathcal {P}^+$ of probabilities on $S$ such that $C(\mathcal {P}^+)$ is regular, and for all non-null $E$, $\succeq _{E,M} = \succeq _{\mathcal {P}^+{\mid }E,M}^{S,Y,U}$. Moreover, $U$ is unique up to affine transformations, and $C(\mathcal {P}^+)$ is unique in the sense that if $\fancyscript{Q}^+$ represents $\succeq _{E,M}$, and $C(\fancyscript{Q}^+)$ is regular, then $C(\fancyscript{Q}^+) = C(\mathcal {P}^+)$.

Proof

Since $\succeq _M = \succeq _{S,M}$ satisfies Axioms 10–18, there must exist a weighted set $\mathcal {P}^+$ of probabilities on $S$ and a utility function $U$ such that $f\succeq _M g$ iff $f \succeq _{M,\mathcal {P}^+}^{S,Y,U} g$. The rest of the proof is identical to that of Theorem 2; we do not repeat it here. $\square $

Analogs of MDC have appeared in the literature before in the context of updating preference orders. In particular, Epstein and Schneider (1993) discuss a menu-independent version of MDC, although they do not characterize updating in their framework. Ghirardato (2002) characterizes update for a menu-independent version of DC. Sinischalchi (2011) also uses an analog of MDC in his axiomatization of measure-by-measure updating of MMEU. Like us, he starts with an axiomatization for unconditional preferences, and adds an axiom called constant act dynamic consistency (CDC), somewhat analogous to MDC, to extend the axiomatization of MMEU to deal with conditional preferences. CDC in the form in Siniscalchi (2011) was first proposed by Pires (2002), based on an observation of Jaffray (1992).

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Authors and Affiliations

Joseph Y. Halpern
- 1
Samantha Leung
- 1
Email author

1.Computer Science DepartmentCornell UniversityIthacaUSA

	SEU	REG	MER	MWER	MMEU
Ax. 1-6,8-10	\(\checkmark \)	\(\checkmark \)	\(\checkmark \)	\(\checkmark \)	\(\checkmark \)
Ind	\(\checkmark \)	\(\checkmark \)	\(\checkmark \)	\(\checkmark \)
C-Ind	\(\checkmark \)				\(\checkmark \)
Ax. 12	\(\checkmark \)	\(\checkmark \)	\(\checkmark \)
Symmetry	\(\checkmark \)	\(\checkmark \)

Weighted sets of probabilities and minimax weighted expected regret: a new approach for representing uncertainty and making decisions

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