Minimizing regret in dynamic decision problems

Abstract

The menu-dependent nature of regret minimization creates subtleties when it is applied to dynamic decision problems. It is not clear whether forgone opportunities should be included in the menu. We explain commonly observed behavioral patterns as minimizing regret when forgone opportunities are present. If forgone opportunities are included, we can characterize when a form of dynamic consistency is guaranteed.

Appendices

Proof of Theorem 4.1

We restate the theorem (and elsewhere in the appendix) for the reader’s convenience.

Theorem 4.1. For a dynamic decision problem D, if and $\mu (h)=M$ for some fixed menu M, then there will be no preference reversals in D.

Proof

Before proving the result, we need some definitions. Say that an information set I refines an information set $I'$ if, for all $h \in I$, some prefix $h'$ of h is in $I'$. Suppose that there is a history h such that $f, g \in M_h$ and $I(h) = I$. Let fIg denote the plan that agrees with f at all histories $h'$ such that $I(h')$ refines I and agrees with g otherwise. As we now show, fIg gives the same outcome as f on states in $E = E(h)$ and the same outcome as g on states in $E^c$; moreover, $fIg \in M_h$.

Suppose that $s(h) = s$ and that $s \in E$. Since $E(h) = E$, there exists a history $h' \in I(h)$ such that $s(h') = s'$ and $R(h') = R(h)$. Since $f, g \in M_h$, there must exist some k such that $f^k(\langle s\rangle ) = g^k(\langle s\rangle ) = h$ [where, as usual, $f^0(\langle s\rangle ) = \langle s\rangle$ and for $k' \ge 1$, $f^{k'}(\langle s\rangle ) = f(f^{k'-1}(\langle s\rangle ))$]. We claim that for all $k' \le k$, $f^{k'}(\langle s'\rangle ) = g^{k'}(\langle s'\rangle )$, and $f^{k'}(\langle s'\rangle )$ is in the same information set as $f^{k'}(\langle s\rangle )$. The proof is by induction on $k'$. If $k' = 0$, the result follows from the observation that since $\langle s\rangle$ is a prefix of h, there must be some prefix of $h'$ in $I(\langle s\rangle )$. For the inductive step, suppose that $k' \ge 1$. We must have $f^{k'}(\langle s\rangle ) = g^{k'}(\langle s\rangle )$ (otherwise g would not be in $M_h$). Since $g^{k'-1}(\langle s\rangle ) = f^{k'-1}(\langle s\rangle )$ and $f^{k'-1}(\langle s'\rangle ) = g^{k'-1}(\langle s'\rangle )$ are in the same information set, by the inductive hypothesis, g must perform the same action at $g^{k'-1}(\langle s\rangle )$ and $g^{k'-1}(\langle s'\rangle )$, and must perform the same action at $f^{k'-1}(\langle s\rangle )$ and $f^{k'-1}(\langle s'\rangle )$. Since $g^{k'}(\langle s\rangle )$ and $f^{k'}(\langle s\rangle )$ are both prefixes of h, g and f perform the same action at $f^{k'-1}(\langle s\rangle ) = g^{k'-1}(\langle s\rangle )$. It follows that f and g perform the same action at $f^{k'-1}(\langle s'\rangle ) = g^{k'-1}(\langle s'\rangle )$, and so $f^{k'}(\langle s'\rangle ) = g^{k'}(\langle s'\rangle )$. Thus, $g^{k'}(\langle s'\rangle )$ must be a prefix of $h'$, and so must be in the same information set as $f^{k'}(\langle s\rangle )$. This completes the inductive proof.

Since $f^k(\langle s'\rangle ) = g^k(\langle s'\rangle ) =h'$, it follows that $f^k(\langle s'\rangle ) = (fIg)^k(\langle s'\rangle )$. Below I, all the information sets are refinements of I, so by definition, for $k' \le k$, we must $f^{k'}(\langle s'\rangle ) = (fIg)^{k'}(\langle s'\rangle )$. Thus, f and fIg give the same outcome for $s'$, and hence all states in E. Note it follows that $(fIg)^k(\langle s\rangle ) = h$, so $fIg \in M_h$.

For $s' \notin E$ and all $k'$, it cannot be the case that $I((fIg)^{k'}(\langle s'\rangle ))$ is a refinement of I, since the first state in $(fIg)^{k'}(\langle s'\rangle ))$ is $s'$, and no history in a refinement of I has a first state of $s'$. Thus, $fIg^{k'}(\langle s'\rangle ) = g^{k'}(\langle s'\rangle )$ for all $k'$, so f and fIg give the same outcome for $s'$, and hence all states in $E^c$.

Returning to the proof of the proposition, suppose that $f \in C_{\mu ,h}(M_h)$, $h'$ is a history extending h, and $f \in M_{h'}$. We want to show that $f \in C_{\mu ,h'}(M_{h'})$. By perfect recall, $E(h') \subseteq E(h)$. Suppose, by way of contradiction, that $f \notin C_{\mu ,h'}(M_{h'})$. Since $f \in C_{\mu ,h'}(M_{h'})$, we cannot have $E(h') = E(h)$, so $E(h') \subset E(h)$. Choose $f' \in C_{\mu ,E(h')}(M_{h'})$ and $g \in C_{\mu ,E(h')^c \cap E(h)}(M_{h'})$ (note that $C_{\mu ,E(h')}(M_{h'}) \ne \emptyset$ and $C_{\mu ,E(h')^c \cap E(h)}(M_{h'}) \ne \emptyset$ by Axiom 3). Since $f', g \in M_{h'}$ (by Axiom 3), $f'I(h')g$ is in $M_{h'}$. Since $f'I(h')g$ and $f'$, when viewed as acts, agree on states in $E(h')$, we must have $f'I(h')g \in C_{\mu ,E(h')}(M_{h'})$ by Axiom 2. Similarly, since $f'I(h')g$ and g, when viewed as acts, agree on states in $E(h')^c \cap E(h)$, we must have $f'I(h')g \in C_{\mu ,E(h')^c \cap E(h)}(M_{h'})$. Therefore, by Axiom 1, $f'I(h')g \in C_{\mu ,h}(M_{h'})$. Also by Axiom 1, since $f \notin C_{\mu ,h'}(M_{h'})$, we must have $f \notin C_{\mu ,h}(M_{h'})$. By Axiom 4, this implies that $f \notin C_{\mu ,h}(M_{h})$ (since $M_{h'} \subseteq M_h$), giving us the desired contradiction. $\square$

Proof of Theorem 4.3

Theorem 4.3. If ${\mathcal {P}}^+$ is a set of weighted distributions on $(S,\Sigma )$ such that $C({\mathcal {P}}^+)$ is closed, then the following are equivalent:

• (a) For all decision problems D based on $(S,\Sigma )$ and all menus M in D, Axioms 1–4 hold for choice functions represented by ${\mathcal {P}}^+|^l E$ (resp., ${\mathcal {P}}^+|^p E)$.

• (b) For all decision problems D based on $(S,\Sigma )$, states $s \in S$, and acts $f\in M_{\langle s \rangle }$, the weighted regret of f with respect to $M_{\langle s \rangle }$ and ${\mathcal {P}}^+$ is separable.

We actually prove the following stronger result.

Theorem 7.1

If ${\mathcal {P}}^+$ is a set of weighted distributions on $(S,\Sigma )$ such that $C({\mathcal {P}}^+)$ is closed, then the following are equivalent:

• (a) For all decision problems D based on $(S,\Sigma )$, Axioms 1–4 hold for menus of the form $M_{\langle s \rangle }$ for choice functions represented by ${\mathcal {P}}^+|^l E$ (resp., ${\mathcal {P}}^+|^p E)$.

• (b) For all decision problems D based on $(S,\Sigma )$ and all menus M in D, Axioms 1–4 hold for choice functions represented by ${\mathcal {P}}^+|^l E$ (resp., ${\mathcal {P}}^+|^p E)$.

• (c) For all decision problems D based on $(S,\Sigma )$, states $s \in S$, and acts $f\in M_{\langle s \rangle }$, the weighted regret of f with respect to $M_{\langle s \rangle }$ and ${\mathcal {P}}^+$ is separable.

• (d) For all decision problems D based on $(S,\Sigma )$, menus M in D, and acts $f\in M$, the weighted regret of f with respect to M and ${\mathcal {P}}^+$ is separable.

Proof

Fix an arbitrary state space S, measurable events $E, F \subseteq S$, and a set ${\mathcal {P}}^+$ of weighted distributions on $(S,\Sigma )$. The fact that (b) implies (a) and (d) implies (c) follows immediately. Therefore, it remains to show that (a) implies (d) and that (c) implies (b).

Since the proof is identical for prior-by-prior updating ($|^p$) and for likelihood updating ($|^l$), we use | to denote the updating operator. That is, the proof can be read with | denoting $|^p$, or with | denoting $|^l$.

To show that (a) implies (d), we first show that Axiom 1 implies that for all decision problems D based on $(S,\Sigma )$, menu M in D, sets ${\mathcal {P}}^+$ of weighted probabilities, and acts $f \in M$,

$$\begin{aligned} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|F} (f)\ge & {} \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f)\right. \nonumber \\&\left. + \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f)\right) . \end{aligned}$$

(1)

Suppose, by way of contradiction, that (1) does not hold. Then for some decision problem D based on $(S,\Sigma )$, measurable events $E,F \subseteq S$, menu M in D, and act $f\in M$, we have that

$$\begin{aligned} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|F} (f)< & {} \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f) \right. \\&\left. + \Pr (E^c \cap F) {\mathrm {reg}}_{M,F}^{{\mathcal {P}}^+|(E^c \cap F)}(f)\right) . \end{aligned}$$

We define a new decision problem $D'$ based on $(S,\Sigma )$. The idea is that in $D'$, we will have a plan $a_{f'}$ such that $a_{f'} \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',E\cap F}}(M'')$ and $a_{f'}\in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',E^c \cap F}}(M'')$ and $a_{f'}\notin C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',F}}(M'')$ for some $M''\subseteq M'$, where $M'$ is the menu at the initial decision node for the DM.

We construct $D'$ as follows. $D'$ is a depth-two tree; that is, nature makes a single move, and then the DM makes a single move. At the first step, nature choose a state $s \in F$. At the second step, the DM chooses from the set $\{ a_g : g \in M \} \cup \{a_{f'}\}$ of actions. With a slight abuse of notation, we let $a_g$ also denote the plan in $T'$ that chooses the action $a_g$ at the initial history $\langle s\rangle$. Therefore, the initial menu in decision problem $D'$ is $M'=\{ a_g : g \in M \} \cup \{a_{f'}\}$.

The utilities for the actions/plans in $D'$ are defined as follows. For actions $\{a_g : g\in M\}$, the utility of $a_g$ in state s is just the utility of the outcome resulting from applying plan g in state s in decision problem D. The action $a_{f'}$ has utilities

$$\begin{aligned} u(a_{f'}(s)) = {\left\{ \begin{array}{ll} {\sup }_{g\in M} u(g(s)) - {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E \cap F)}(f) \text { if } s\in E\cap F \\ {\sup }_{g\in M} u(g(s)) - {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f) \text { if } s\in E^c \cap F. \end{array}\right. } \end{aligned}$$

For all states $s\in F$, we have that $u(a_{f'}(s)) \le \sup _{g\in M} u(g(s))$. As a result, for all states $s\in F$, we have that

$$\begin{aligned} \sup _{g\in M} u(g(s)) = \sup _{a_g \in M'} u(a_g(s)). \end{aligned}$$

Since the regret of a plan in state s depends only on its payoff in s and the best payoff in s, it is not hard to see that the regrets of $a_g$ with respect to $M'$ is the same as the regret of g with respect to M. More precisely, for all $g \in M$,

$$\begin{aligned} {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_g)&= {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(g), \\ {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E^c \cap F)}(a_g)&= {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(g), \text { and} \\ {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|F}(a_g)&= {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|F}(g). \end{aligned}$$

By definition of $a_{f'}$, for each state $s\in E\cap F$, we have ${\mathrm {reg}}_{M'}(a_{f'},s) = {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f)$, and for each state $s\in E^c \cap F$, we have ${\mathrm {reg}}_{M'}(a_{f'},s) = {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f)$. Thus, for all $\Pr \in {\mathcal {P}}$, if $\Pr (E\cap F) \ne 0$, then ${\mathrm {reg}}_M^{\Pr |(E\cap F)}(f) = {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f)$, and if $\Pr (E^c\cap F) \ne 0$, then ${\mathrm {reg}}_M^{\Pr |(E^c\cap F)}(f) = {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c\cap F)}(f)$. If for all $(\Pr ,\alpha ) \in {\mathcal {P}}^+|(E\cap F)$, $\alpha \Pr ( E\cap F) = 0$, then ${\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_{f'}) = {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_f) = 0$. Otherwise, since there is some measure in ${\mathcal {P}}^+|(E\cap F)$ that has weight 1, we must have ${\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_{f'}) = {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_f)$. Similarly, ${\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E^c \cap F)}(a_{f'}) = {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E^c \cap F)}(a_f)$. Thus,

$$\begin{aligned} {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|F}(a_{f'})= & {} {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f) \right. \\&\left. + \Pr (E^c \cap F){\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f) \right) \\> & {} {\mathrm {reg}}_M^{{\mathcal {P}}^+|F}(f) \,\,\,\,\,\,\hbox {[by assumption]}\\= & {} {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|F}(a_f) \,\,\,\,\hbox {[by construction]}. \end{aligned}$$

Therefore, we have $a_{f'} \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',E\cap F}}(\{a_{f'}, a_f\})$, $a_{f'} \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',E^c \cap F}}(\{a_{f'}, a_f\})$, and $a_{f'}\notin C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',F}}(\{a_{f'}, a_f\})$, violating Axiom 1.

By an analogous argument, we show that the opposite weak inequality,

$$\begin{aligned} {\mathrm {reg}}_M^{{\mathcal {P}}^+|F} (f)\le & {} \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f)\right. \nonumber \\&\left. + \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f)\right) , \end{aligned}$$

(2)

is also implied by Axiom 1. Suppose, by way of contradiction, that (2) does not hold. Then for some decision problem D based on $(S,\Sigma )$, measurable events $E,F \subseteq S$, menu M in D, and act $f\in M$, we have that

$$\begin{aligned} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|F} (f)> & {} \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f) \right. \\&\left. + \Pr (E^c \cap F) {\mathrm {reg}}_{M,F}^{{\mathcal {P}}^+|(E^c \cap F)}(f)\right) . \end{aligned}$$

We define a decision problem $D'$ based on $(S,\Sigma )$ just as in the previous case. Specifically, we have that ${\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_{f'}) = {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_f)$, and that ${\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E^c \cap F)}(a_{f'}) = {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E^c \cap F)}(a_f)$. The one difference from the previous case is that we now have

$$\begin{aligned} {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|F}(a_{f'})= & {} {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f) \right. \\&\left. + \Pr (E^c \cap F){\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f) \right) \\< & {} {\mathrm {reg}}_M^{{\mathcal {P}}^+|F}(f) \,\,\,\,\,\,\hbox {[by assumption]}\\= & {} {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|F}(a_f) \,\,\,\,\hbox {[by construction].} \end{aligned}$$

Therefore, we have $a_{f} \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',E\cap F}}(\{a_{f'}, a_f\})$, $a_{f} \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',E^c \cap F}}(\{a_{f'}, a_f\})$, and $a_{f}\notin C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',F}}(\{a_{f'}, a_f\})$, violating Axiom 1.

To complete the proof that (a) implies (d), we show that Axiom 1 also implies that for all decision problems D based on $(S,\Sigma )$, menus M in D, sets ${\mathcal {P}}^+$ of weighted probabilities, and acts $f \in M$, if ${\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f) > 0$, then

$$\begin{aligned} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|F} (f) > \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f). \end{aligned}$$

(3)

Suppose, by way of contradiction, that (3) does not hold. Then for some decision problem D based on $(S,\Sigma )$, events $E,F\subseteq S$, menu M in D, and act $f\in M$ such that ${\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(f) > 0$ and

$$\begin{aligned} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|F} (f)\le \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f). \end{aligned}$$

We now define a new decision problem $D'$ based on $(S,\Sigma )$. The idea is that in $D'$, we have a plan $a_{f}$ such that $a_{f} \notin C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M,E\cap F}}(M')$ but $a_{f} \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M,F}}(M')$ for some $M'\subseteq M$.

Construct $D'$ exactly as before. That is, in the first step, nature chooses a state $s\in S$, and in the second step, the DM chooses from the set of actions/plans $M' = \{ a_g : g \in M \} \cup \{ a_{g'}\}$. For each $g\in M$, define the actions $a_g$ as before. We define a new action $a_{g'}$ with utilities

$$\begin{aligned} u(a_{g'}(s)) = {\left\{ \begin{array}{ll} {\sup }_{g\in M} u(g(s)), \text { if } s\in E\cap F \\ {\sup }_{g\in M} u(g(s)) - {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f), \text { if } s\in E^c \cap F. \end{array}\right. } \end{aligned}$$

It is almost immediate from the definition of $a_{g'}$ that we have

$$\begin{aligned} {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|F}(a_{g'})&= \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E^c \cap F){\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(f) \right) \ge {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|F}(a_f). \end{aligned}$$

However, we also have

$$\begin{aligned} {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_g) = 0 < {\mathrm {reg}}_{M'}^{{\mathcal {P}}^+|(E\cap F)}(a_f). \end{aligned}$$

Therefore, we have $a_{f} \notin C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',E\cap F}}(\{a_{g'}, a_f\})$ but $a_{f}\in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{{M',F}}(\{a_{g'}, a_f\})$, violating Axiom 1.

We next show that (c) implies (b). Specifically, we show that SEP for the initial menus of all decision problems D is sufficient to guarantee that Axioms 1–4 hold for menu M and all choice sets $M' \subseteq M$. It is easy to check that Axioms 2–4 hold for MWER, so we need to check only Axiom 1.

Consider an arbitrary decision problem D, menu M in D, $M'\subseteq M$, and a plan f in $M'$. We construct a new decision problem $D'$ such that the initial menu of $D'$ is “equivalent” to M. Just as before, let $D'$ be a two-stage decision problem where in the first stage, nature chooses $s\in S$, and in the second stage, the DM chooses from the set $M_0 = \{ a_g : g \in M\}$, where $a_g$ is defined as before. Again, we associate each action $a_g$ with the plan that chooses $a_g$ in $D'$. $M_0$ is then “equivalent” to M in the sense that

$$\begin{aligned} {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E\cap F)}(a_g)&= {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E\cap F)}(g), \\ {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_g)&= {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|(E^c \cap F)}(g), \text { and} \\ {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|F}(a_g)&= {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|F}(g). \end{aligned}$$

Suppose that $f \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{M,E\cap F}(M')$ and $f \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{M,E^c \cap F}(M')$. This means that for all $g \in M'$, we have ${\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E\cap F)}(a_f) \le {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E\cap F)}(a_g)$ and ${\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_f) \le {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_g)$. Therefore, we have

$$\begin{aligned}&{\mathrm {reg}}_{M}^{{\mathcal {P}}^+|F}(f) = {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|F}(a_f) \\&\quad = \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E\cap F)}(a_f) + \Pr (E^c \cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_f) \right) \\&\quad \le \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F){\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_g) + \Pr (E^c \cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_g) \right) \\&\quad = {\mathrm {reg}}_M^{{\mathcal {P}}^+|F}(g), \end{aligned}$$

which means that $f \in C^{{\mathrm {reg}},{\mathcal {P}}^+}_{M,F}(M')$, as required.

Next, consider an act $g \in M'$ such that $g \notin C^{{\mathrm {reg}},{\mathcal {P}}^+}_{M,E\cap F}(M')$. This means that ${\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_f) < {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_g)$ and ${\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_f) \le {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_g)$. Let $(\alpha _{\Pr ^*},\Pr ^*) \in C({\mathcal {P}}^+)$ be such that

$$\begin{aligned}&\alpha _{{\Pr }^*} ({\Pr }^*(E\cap F){\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_g) + {\Pr }^*(E^c \cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_g)\\&\quad = {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F){\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_g) + \Pr (E^c \cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_g) \right) . \end{aligned}$$

Such a pair $(\alpha _{\Pr ^*},\Pr ^*)$ exists, since we have assumed that $C({\mathcal {P}}^+)$ is closed. If $\alpha _{\Pr ^*}{\Pr }^*(E\cap F) = 0$, then ${\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|F}(a_g) = \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E^c \cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}\right. \left. (a_g) \right)$. By separability, it must be the case that ${\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_g) = 0$, contradicting our assumption that $0 \le {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_f) < {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_g)$. Therefore, it must be that $\alpha _{\Pr ^*}\Pr ^*(E\cap F) > 0$, and

$$\begin{aligned}&{\mathrm {reg}}_M^{{\mathcal {P}}^+|F}(f) = {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|F}(a_f) \\&\quad = \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_f) + \Pr (E^c \cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_f) \right) \\&\quad < \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E\cap F){\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E \cap F)}(a_g) + \Pr (E^c \cap F) {\mathrm {reg}}_{M_0}^{{\mathcal {P}}^+|(E^c \cap F)}(a_g) \right) \\&\quad = {\mathrm {reg}}_M^{{\mathcal {P}}^+|F}(g), \end{aligned}$$

which means that $g \notin C^{{\mathrm {reg}},{\mathcal {P}}^+}_{M,F}(M')$. $\square$

Proof of Theorem 4.5

To prove Theorem 4.5, we need the following lemma.

Lemma 8.1

For all utility functions u, sets ${\mathcal {P}}^+$ of weighted probabilities, acts f, and menus M containing f, ${\mathrm {reg}}_{M}^{{\mathcal {P}}^+}(f) = {\mathrm {reg}}_{M}^{C({\mathcal {P}}^+)}(f)$.

Proof

Simply observe that

$$\begin{aligned} {\mathrm {reg}}_M^{{\mathcal {P}}^+}(f)&= \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+}\left( \alpha \sum _{s\in S}\Pr (s){\mathrm {reg}}_M(f,s) \right) \\&= \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+}\left( \sum _{s\in S}\alpha \Pr (s){\mathrm {reg}}_M(f,s) \right) \\&= \sup _{\{ p :\, p \le \alpha \Pr , (\Pr ,\alpha ) \in {\mathcal {P}}^+\}}\left( \sum _{s\in S} p(s) {\mathrm {reg}}_M(f,s) \right) \\&= {\mathrm {reg}}_{M}^{C({\mathcal {P}}^+)}(f), \end{aligned}$$

by definition. $\square$

The next lemma uses an argument almost identical to one used in Lemma 7 of Halpern and Leung (2012).

Lemma 8.2

If $C({\mathcal {P}}^+|^{\chi } F)$ is convex and q is a subprobability on F not in $\overline{C({\mathcal {P}}^+|^{\chi }F)}$, then there exists a non-negative vector $\theta$ such that for all $(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F$, we have

$$\begin{aligned} \sum _{s \in F} \alpha \Pr (s) \theta (s) < \sum _{s\in F} q(s) \theta (s). \end{aligned}$$

Proof

Given a set ${\mathcal {P}}^+$ of weighted probabilities, let $C'({\mathcal {P}}^+) = \{ p : p \in {\mathbb {R}}^{|S|} \text { and } p \le \alpha \Pr \text { for some } (\Pr ,\alpha ) \in {\mathcal {P}}^+ \}$. Note that an element $q \in C'({\mathcal {P}}^+)$ may not be a subprobability measure, since we do not require that $q(s) \ge 0$. Since $\overline{C'({\mathcal {P}}^+|^{\chi } F)}$ and $\{{q}\}$ are closed, convex, and disjoint, and $\{{q}\}$ is compact, the separating hyperplane theorem Rockafellar (1970) says that there exist $\theta \in {\mathbb {R}}^{|S|}$ and $c\in {\mathbb {R}}$ such that

$$\begin{aligned} \theta \cdot {p}< c \text { for all } {p}\in \overline{C'({\mathcal {P}}^+|^{\chi } F)}, \text { and } \theta \cdot {q} > c. \end{aligned}$$

(4)

Since $\{ \alpha \Pr : (\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi } F \} \subseteq \overline{C'({\mathcal {P}}^+|^{\chi } F)}$, we have that for all $(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F$,

$$\begin{aligned} \sum _{s \in F} \alpha \Pr (s) \theta (s) < \sum _{s\in F} q(s) \theta (s). \end{aligned}$$

Now we argue that it must be the case that $\theta (s) \ge 0$ for all $s\in F$. Suppose that $\theta (s') < 0$ for some $s'\in F$. Define $p^*$ by setting

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

Note that $p^{*} \le \mathbf {0}$, since for all $s\in S$, $p^{*}(s) \le 0$. Therefore, $p^{*} \in C'({\mathcal {P}}^+|^\chi F)$.

Our definition of $p^{*}$ also ensures that $\theta \cdot {p^{*}} = \sum _{s\in S} p^{*}(s) \theta (s) = p^{*}(s') \theta (s') = |c| \ge c$. This contradicts (4), which says that $\theta \cdot {p} < c \text { for all } {p}\in C'({\mathcal {P}}^+|^{\chi } F)$. Thus, it must be the case that $\theta (s) \ge 0$ for all $s\in S$. $\square$

We are now ready to prove Theorem 4.5, which we restate here.

Theorem 4.5 If $C({\mathcal {P}}^+)$ is closed and convex, then Axiom 1 holds for the family of choices $C_{M}^{{\mathrm {reg}},{\mathcal {P}}^+ |^{\chi } E}$ if and only if ${\mathcal {P}}^+$ is $\chi$-rectangular.

We prove the two directions of implication in the theorem separately. Note that the proof that $\chi$-rectangularity implies Axiom 1 does not require $C({\mathcal {P}}^+)$ to be convex.

Claim 8.3

If ${\mathcal {P}}^+$ is $\chi$-rectangular, then Axiom 1 holds for the family of choices $C_{M}^{{\mathrm {reg}},{\mathcal {P}}^+ |^{\chi } E}$.

Proof

By Theorem 4.3, it suffices to show that SEP holds. For the first part of SEP, we must show that

$$\begin{aligned}&{\mathrm {reg}}_M^{{\mathcal {P}}^+|^{\chi }F} (f) \nonumber \\&\quad = \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F)}(f) + \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f)\right) . \end{aligned}$$

(5)

Unwinding the definitions, (5) is equivalent to

$$\begin{aligned}&{\mathrm {reg}}_M^{{\mathcal {P}}^+|^{\chi }F} (f) = {\sup }_{({\Pr }_3,\alpha _3) \in {\mathcal {P}}^+|^{\chi }F} \alpha _{{\Pr }_3} \left( {\Pr }_3(E\cap F) {\sup }_{({\Pr }_1,\alpha _1) \in {\mathcal {P}}^+|^{\chi }F}{\alpha _{1,E\cap F}^{\chi }} \right. \\&\qquad \left. {\sum }_{s \in E\cap F} {\Pr }_1(s|(E \cap F)) {\mathrm {reg}}_{M}(f,s)) \right. \\&\qquad \left. + {\Pr }_3(E^c \cap F) {\sup }_{({\Pr }_2,\alpha _2) \in {\mathcal {P}}^+|^{\chi }F} \alpha _{2,E^c\cap F}^{\chi } {\sum }_{s\in E^c \cap F} {\Pr }_2(s|(E^c \cap F)) {\mathrm {reg}}_{M}(f,s)) \right) . \end{aligned}$$

The $\sup$s in this expression are taken on by some $(\Pr _1^*,\alpha _1^*),(\Pr _2^*,\alpha _2^*),(\Pr _3^*,\alpha _3^*) \in \overline{{\mathcal {P}}^+|^{\chi }F}$. By $\chi$-rectangularity, we have that for all $(\Pr _1,\alpha _1),(\Pr _2,\alpha _2),(\Pr _3,\alpha _3) \in {\mathcal {P}}^+|^{\chi }F$,

$$\begin{aligned}&\alpha _{{\Pr }_3} {{\Pr }_3(E \cap F)}{\alpha _{1,E\cap F}^{\chi }}{\Pr }_1|(E \cap F) \nonumber \\&\qquad + \alpha _{{\Pr }_3}{{\Pr }_3(E^c \cap F)} {\alpha _{2,E^c \cap F}^{\chi }}{\Pr }_2|(E^c \cap F) \in \overline{C({\mathcal {P}}^+|^{\chi }F)}. \end{aligned}$$

(6)

Thus, for all $\epsilon > 0$,

$$\begin{aligned}&{\mathrm {reg}}_M^{{\mathcal {P}}^+|^{\chi }F}(f)\\&\quad = {\mathrm {reg}}_{M}^{C({\mathcal {P}}^+|^{\chi }F)}(f) \,\,\,\,\hbox {[by Lemma C.1]}\\&\quad \ge \alpha _{3}^* \left( {\Pr }_3^*(E\cap F) {(\alpha ^*_{1,E\cap F})^{\chi }} {\sum }_{s \in E\cap F} {\Pr }_1^*(s|(E \cap F)) {\mathrm {reg}}_{M}(f,s)) \right. \\&\qquad \left. + {\Pr }_3^*(E^c \cap F) {(\alpha ^*_{2,E^c\cap F})^{\chi }} {\sum }_{s\in E^c \cap F} {\Pr }_2^*(s|(E^c \cap F)) {\mathrm {reg}}_{M}(f,s)) \right) - \epsilon \ \hbox {[by (6)]}. \end{aligned}$$

Therefore,

$$\begin{aligned}&{\mathrm {reg}}_M^{{\mathcal {P}}^+|^{\chi }F}(f)\\&\quad \ge \alpha _{3}^* \left( {\Pr }_3^*(E\cap F) {(\alpha ^*_{1,E\cap F})^{\chi }} {\sum }_{s \in E\cap F} {\Pr }_1^*(s|(E \cap F)) {\mathrm {reg}}_{M}(f,s)) \right. \\&\qquad \left. +\, {\Pr }_3^*(E^c \cap F) {(\alpha ^*_{2,E^c\cap F})^{\chi }} {\sum }_{s\in E^c \cap F} {\Pr }_2^*(s|(E^c \cap F)) {\mathrm {reg}}_{M}(f,s)) \right) \\&\quad = {\sup }_{({\Pr }_3,\alpha _3) \in {\mathcal {P}}^+|^{\chi }F} \alpha _{3} \left( {\Pr }_3(E\cap F) {\sup }_{({\Pr }_1,\alpha _1) \in {\mathcal {P}}^+|^{\chi }F}{\alpha _{1,E\cap F}^{\chi } } {\sum }_{s \in E\cap F} {\Pr }_1(s|(E \cap F)) {\mathrm {reg}}_{M}(f,s)) \right. \\&\qquad \left. +\, {\Pr }_3(E^c \cap F) {\sup }_{({\Pr }_2,\alpha _2) \in {\mathcal {P}}^+|^{\chi }F}{\alpha _{2,E^c \cap F}^{\chi } } {\sum }_{s\in E^c \cap F} {\Pr }_2(s|(E^c \cap F)) {\mathrm {reg}}_{M}(f,s)) \right) \\&\qquad \quad [\hbox {by the choice of }({\Pr }^*_i,\alpha ^*_i), i = 1,2,3] \\&\quad = {\sup }_{({\Pr },\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha \left( {\Pr }(E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F)}(f) + {\Pr }(E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f)\right) , \end{aligned}$$

as required.

It remains to show the opposite inequality in (5), namely, that

$$\begin{aligned}&{\mathrm {reg}}_M^{{\mathcal {P}}^+|^{\chi }F} (f) \\&\quad \le \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F)}(f) + \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f)\right) . \end{aligned}$$

It suffices to note that the right-hand side is equal to

$$\begin{aligned}&{\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \left( {\alpha }\Pr (E \cap F) {\sup }_{({\Pr }_1,\alpha _1) \in {\mathcal {P}}^+|^{\chi } F}{\alpha _{1,E\cap F}^{\chi } }\right. \\&\qquad \left. {\sum }_{s \in E\cap F} {\Pr }_1(s|E \cap F) {\mathrm {reg}}_{M}(f,s))+ {\alpha } \Pr (E^c \cap F ) {\sup }_{({\Pr }_2,\alpha _2) \in {\mathcal {P}}^+|^{\chi } F} \alpha _{2,E^c \cap F}^{\chi } \right. \\&\qquad \left. {\sum }_{s\in E^c \cap F} {\Pr }_2(s|E^c \cap F) {\mathrm {reg}}_{M}(f,s)) \right) \\&\quad \ge \overline{E}_{{\mathcal {P}}^+|^{\chi }F}({\mathrm {reg}}_M(f))\ \ \ \hbox {[by rectangularity]} \\&\quad = {\mathrm {reg}}_M^{{\mathcal {P}}^+|^{\chi }F}(f) . \end{aligned}$$

This completes the proof that (5) holds.

For the second part of SEP, suppose that $\overline{{\mathcal {P}}}^+(E\cap F) > 0$ and ${\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F)}(f) \ne 0$. If ${\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f) = 0$ then, since $\overline{{\mathcal {P}}}^+(E\cap F) > 0$, we have that ${\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi } F}(f) > 0 = \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha Pr(E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f)$, as desired. Otherwise, by part (b) of $\chi$-rectangularity, for all $\delta > 0$, there exists $(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi } F$ such that $\alpha ( \delta \Pr (E \cap F) + \Pr (E^c \cap F)) > \sup _{(\Pr ',\alpha ') \in {\mathcal {P}}^+} \alpha ' \Pr '(E^c \cap F)$. Therefore, using the first part of SEP, we have

$$\begin{aligned}&{\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F)}(f)\\&\quad = {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F) }(f) \right. \\&\qquad \left. + \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f)\right) \\&\quad = {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f) {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha \\&\qquad \left( \Pr (E\cap F) \frac{{\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F) }(f)}{{\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f)} + \Pr (E^c \cap F) \right) \\&\quad > {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f) {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha Pr(E^c \cap F) \\&\qquad [\hbox {by part (b) of } \chi -\hbox {rectangularity}]\\&\quad = {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha Pr(E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f), \end{aligned}$$

as required. $\square$

Claim 8.4

If $C({\mathcal {P}}^+)$ is convex and Axiom 1 holds for the family of choices $C_{M}^{{\mathrm {reg}},{\mathcal {P}}^+ |^{\chi } E}$, then ${\mathcal {P}}^+$ is $\chi$-rectangular.

Proof

Suppose that $\chi$-rectangularity does not hold. Then one of the three conditions of rectangularity must fail.

First suppose that it is (a); that is, for some $(\Pr _1,\alpha _1), (\Pr _2,\alpha _2), (\Pr _3,\alpha _3) \in {\mathcal {P}}^+$, we have $\Pr _1( E\cap F) > 0$ and $\Pr _2(E^c \cap F) > 0$ and

$$\begin{aligned}&\alpha _3 {\Pr }_3( E \cap F) \alpha ^{\chi }_{1,E\cap F} {\Pr }_1|(E \cap F) + \alpha _3 {\Pr }_3(E^c\cap F) \alpha ^{\chi }_{2, E^c \cap F}\\&\quad {\Pr }_2|(E^c\cap F) \notin \overline{C({\mathcal {P}}^+)|^{\chi } F}. \end{aligned}$$

Let $p^* = \alpha _3 {\Pr }_3( E \cap F) \alpha _{1,E\cap F}^{\chi } {\Pr }_1| (E \cap F) + \alpha _3 {\Pr }_3(E^c\cap F) \alpha _{2,E^c \cap F}^{\chi } {\Pr }_2|(E^c\cap F)$. Since we have assumed that $C({\mathcal {P}}^+)$ is convex, we have that $C({\mathcal {P}}^+|^{\chi } F)$ is also convex. By Lemma 8.2, there exists a non-negative vector $\theta$ such that for all $\alpha \Pr \in \overline{C({\mathcal {P}}^+|^{\chi }F)}$, we have

$$\begin{aligned} \sum _{s \in F} \alpha \Pr (s) \theta (s) < \sum _{s\in F} p^*(s) \theta (s). \end{aligned}$$

We construct a decision problem D based on $(S,\Sigma )$. D has two stages: in the first stage, nature chooses a state $s \in S$, but only states in $F\subseteq S$ are chosen with positive probability, so when the DM plays, his beliefs are characterized by ${\mathcal {P}}^+|^{\chi }F$. In the second stage, the DM chooses an action from the set $M = \{f,g\}$, with utilities defined as follows:

$$\begin{aligned} u(f,s)= & {} -\theta ({s}), \text { and } \\ u(g,s)= & {} 0 \text { for all s}. \end{aligned}$$

The act f will have regret precisely $\theta (s)$ in state $s\in S$. By Lemma 8.2,

$$\begin{aligned}&{\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+} \alpha \left( \Pr (E \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F)}(f) + \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(f)\right) \\&\quad \ge \alpha _{{\Pr }_3} \left( {\Pr }_3(E \cap F) {\mathrm {reg}}_{M}^{\alpha _{1,E \cap F}^{\chi } {\Pr }_1|(E \cap F)}(f) + \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{\alpha _{2,E^c \cap F}^{\chi }{\Pr }_2|(E^c \cap F)}(f)\right) \\&\quad = {\sum }_{s\in F} p^*(s) \theta (s) \\&\quad > {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }F}(f), \end{aligned}$$

violating SEP. By Theorem 4.3, Axiom 1 cannot hold.

Now suppose that condition (b) in rectangularity does not hold. That is, for some $\delta > 0$, for all $(\alpha ,\Pr ) \in {\mathcal {P}}^+$, $\alpha ( \delta \Pr (E \cap F) + \Pr (E^c \cap F) ) \le \sup _{(\Pr ',\alpha ') \in {\mathcal {P}}^+} \alpha ' \Pr '(E^c \cap F)$. We construct a decision problem D based on $(S,\Sigma )$. D has two stages: in the first stage, nature chooses a state $s \in S$. In the second stage, the DM chooses an action from the set $M = \{f,g\}$, with utilities defined as follows:

$$\begin{aligned} u(f,s)= & {} 0 \hbox { for all }s \in S, \\ u(g,s)= & {} -\delta \hbox { if } s \in E\cap F\\ u(g,s)= & {} -1 \hbox { if }s \notin E \cap F. \end{aligned}$$

Then we have that ${\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E \cap F)}(g) = \delta$ and ${\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(g) = 1$. Using SEP and the choice of $\delta$, we must have

$$\begin{aligned} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }F} (g)= & {} {\sup }_{({\Pr },\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha (\Pr (E\cap F) \delta + \Pr (E^c \cap F) )\\\le & {} {\sup }_{\Pr \in {\mathcal {P}}^+|^{\chi }F} \alpha \Pr (E^c \cap F) {\mathrm {reg}}_{{M}}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(g). \end{aligned}$$

Clearly,

$$\begin{aligned} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }F} (g) \ge \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha \Pr (E^c \cap F) {\mathrm {reg}}_{{M}}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(g). \end{aligned}$$

Thus,

$$\begin{aligned} {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^{\chi }F} (g) = \sup _{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha \Pr (E^c \cap F) {\mathrm {reg}}_{{M}}^{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(g), \end{aligned}$$

violating the second condition of SEP. Therefore, by Theorem 4.3, Axiom 1 does not hold.

Finally, suppose that condition (c) in rectangularity does not hold. Then for some non-negative real vector $\theta \in {\mathbb {R}}^{|S|}$,

$$\begin{aligned}&{\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \left( {\alpha } \Pr (E) {\sup }_{({\Pr }_1,\alpha _1) \in {\mathcal {P}}^+|^{\chi }(E \cap F)} {\sum }_{s \in E\cap F} \alpha _1 {\Pr }_1(s|E) \theta (s)) \right. \nonumber \\&\qquad \left. + {\alpha } \Pr (E^c) {\sup }_{({\Pr }_2,\alpha _2) \in {\mathcal {P}}^+|^{\chi }(E^c \cap F)} {\sum }_{s\in E^c \cap F} \alpha _2 {\Pr }_2(s|E^c) \theta (s)) \right) \nonumber \\&\quad < {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F}{\alpha } {\sum }_{s \in F} \Pr (s) \theta (s). \end{aligned}$$

(7)

We construct a decision problem D based on $(S,\Sigma )$. D has two stages: in the first stage, nature chooses a state $s \in S$. In the second stage, the DM chooses an action from the set $M = \{f,g\}$, with utilities defined as follows:

$$\begin{aligned} u(g,s)= & {} -\theta (s) \ \ \ \hbox { for all } s \in S.\\ u(f,s)= & {} 0 \ \ \ \hbox { for all } s \in S. \end{aligned}$$

So we have

$$\begin{aligned}&{\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^pF} \alpha \left( \Pr (E\cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^p(E \cap F)}(g) + \Pr (E^c \cap F) {\mathrm {reg}}_{M}^{{\mathcal {P}}^+|^p(E^c \cap F)}(g)\right) \\&\quad = {\sup }_{(\Pr ,\alpha ) \in {\mathcal {P}}^+|^{\chi }F} \alpha \left( \Pr (E \cap F ) \overline{E}_{{\mathcal {P}}^+|^{\chi }(E \cap F)}(\theta ) + \Pr (E^c\cap F) \overline{E}_{{\mathcal {P}}^+|^{\chi }(E^c \cap F)}(\theta ) \right) \\&\quad < \overline{E}_{{\mathcal {P}}^+|^{\chi }F}(\theta ) \ \ \ \hbox {[by (7)]}\\&\quad = {\mathrm {reg}}_M^{{\mathcal {P}}^+|^pF}(g). \end{aligned}$$

This means that SEP, and hence Axiom 1, is violated, a contradiction. $\square$