Conditional choice with a vacuous second tier

Abstract

This paper studies a generalization of rational choice theory. I briefly review the motivations that Helzner gives for his conditional choice construction (2013). Then, I focus on the important class of conditional choice functions with vacuous second tiers. This class is interesting for both formal and philosophical reasons. I argue that this class makes explicit one of conditional choice’s normative motivations in terms of an account of neutrality advocated within a certain tradition in decision theory. The observations recorded—several of which are generalizations of central results in the standard theory of rational choice—are intended to provide further insight into how conditional choice generalizes the standard account and are offered as additional evidence of the fruitfulness of the conditional choice framework.

Appendices

Appendix: Proofs

Proof of observation 1

Proof

Let $\mathcal {C}:\mathcal {X} \times \mathcal {E} \rightarrow \fancyscript{P}(X)$ be a conditional choice function.

$(\Leftarrow )$ Suppose that $\mathcal {C}$ satisfies $\nu$ and that $\succ _e = \emptyset$ for all $e \in E.$ Let $x \in \mathcal {C}(Y|e).$ Since $\mathcal {C}$ is a conditional choice function, it follows immediately that there is an $f \in E$ such that $e \sqsubseteq f$ and $x \in \mathcal {C}(Y|g)$ whenever $f \sqsubseteq g.$ For the other direction of $(ii'),$ assume that there is an $f \in E$ such that $e \sqsubseteq f$ and $x \in \mathcal {C}(Y|g)$ whenever $f \sqsubseteq g.$ By $\nu ,$ we have $\mathcal {C}(Y|e) \ne \emptyset .$ So there is some $y \in \mathcal {C}(Y|e).$ Suppose that $x \notin \mathcal {C}(Y|e).$ It follows that $y \succ _e x,$ which is a contradiction. So $x \in \mathcal {C}(Y|e).$ Hence, $\mathcal {C}$ satisfies $(ii').$

$(\Rightarrow )$ Suppose that $\mathcal {C}$ satisfies $(ii').$ Suppose that $\mathcal {C}(Y|e') \ne \emptyset$ for some $e' \in E$ such that $e \sqsubseteq e'.$ Then there is some $x \in \mathcal {C}(Y|e').$ By $(ii'),$ there is an $f \in E$ such that $e' \sqsubseteq f$ and $x \in \mathcal {C}(Y|g)$ whenever $f \sqsubseteq g.$ But $e \sqsubseteq f$ too, so by $(ii'),\ x \in \mathcal {C}(Y|e),$ so $\mathcal {C}(Y|e) \ne \emptyset .$ Hence, $\mathcal {C}$ satisfies $\nu .$ Now suppose that $\succ _e \ne \emptyset$ for some $e \in E.$ Then, for some $Y \in \mathcal {X},$ there are $x, y \in Y$ such that $x \in \mathcal {C}(Y|e),\ y \notin \mathcal {C}(Y|e),$ but $y \in \mathcal {C}(Y|f)$ for some $f \in E$ such that $e \sqsubseteq f.$ It follows from $(ii')$ that there is an $f' \in E$ such that $f \sqsubseteq f'$ and $y \in \mathcal {C}(Y|g)$ whenever $f' \sqsubseteq g.$ But since $e \sqsubseteq f',$ by $(ii'),\ y \in \mathcal {C}(Y|e),$ which is a contradiction. Thus, $\succ _e = \emptyset$ for all $e \in E.$ $\square$

Proof of observation 2

Proof

That $\mathcal {\tilde{C}}(Y|e) \subseteq \bigcup \{\mathcal {\tilde{C}}(Y|e') :e \sqsubseteq e'\}$ holds for all $Y \in \mathcal {X}$ and $e \in E$ is easy to see from the fact that $e \sqsubseteq e.$ For the other inclusion claim, let $x$ be in $\bigcup \{\mathcal {\tilde{C}}(Y|e') :e \sqsubseteq e'\}.$ By $\nu ,$ $\mathcal {\tilde{C}}(Y|e) \ne \emptyset .$ So there is some $y \in \mathcal {\tilde{C}}(Y|e).$ Suppose that $x$ is not in $\mathcal {\tilde{C}}(Y|e).$ It follows that $y \succ _e x,$ which contradicts the assumption that the second tier is vacuous. Hence, $x \in \mathcal {\tilde{C}}(Y|e).$ $\square$

Proof of observation 3

Proof

Immediate from condition (i) in the definition of conditional choice, since a conditional choice function is a choice function in its first argument. $\square$

Proof of observation 4

Proof

Let $C$ be a choice function on $X$. $C^{\star } :\mathcal {X} \times \mathcal {E} \rightarrow \fancyscript{P}(X)$ since, for all $Y \in \mathcal {X}$ and $e \in E$, $C^{\star }(Y|e)=C(Y)$. For condition (i), $C^{\star }(Y|e) \subseteq Y$ because $C(Y) \subseteq Y$ for all $Y \in \mathcal {X}$. As for condition (ii), first suppose that $x \in C^{\star }(Y|e)$. Since, for any $e' \in E$, $C^{\star }(Y|e)=C^{\star }(Y|e')$, it follows that there is an $f \in E$ such that $e \sqsubseteq f$ and $x \in C^{\star }(Y|g)$ whenever $f \sqsubseteq g$. Finally, for the other direction, suppose that there is an $f \in E$ such that $e \sqsubseteq f$ and $x \in C^{\star }(Y|g)$ whenever $f \sqsubseteq g$. Again, since $C^{\star }(Y|f)=C(Y)=C^{\star }(Y|e')$ for all $e' \in E$, it follows that $x \in C^{\star }(Y|e)$. $\square$

Proof of observation 5

Proof

Suppose that $C$ is a choice function on $X$ and satisfies $P$. Since $C^{\star }_e=C$ for all $e \in E$, it follows that $C^{\star }$ satisfies $P^{\star }$. For the other direction, assume that $C^{\star }$ satisfies $P^{\star }$. Let $e \in E$. By assumption, we have it that there is an $f \in E$ such that $e \sqsubseteq f$ and, for all $g \in E$ such that $f \sqsubseteq g, C^{\star }_g$ satisfies $P$. But this implies that $C^{\star }_f$ satisfies $P$ and $C^{\star }_f =C$, so $C$ satisfies $P$. $\square$

Proof of observation 6

Proof

Suppose that $\mathcal {\tilde{C}}$ satisfies $P^{\star }$. Let $x \in \mathcal {\tilde{C}}(Y|e)$. By definition, there is some $f \in E$ such that $e \sqsubseteq f$ and $x \in \mathcal {\tilde{C}}(Y|g)$ when $f \sqsubseteq g$. From the assumption, we have that there is some $f' \in E$ such that $f \sqsubseteq f'$ and $\mathcal {\tilde{C}}_{g'}$ satisfies $P$ whenever $f' \sqsubseteq g'$. It follows that $x \in \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'\ \text {and}\ \mathcal {\tilde{C}}_{e'}\ \text {satisfies}\ P\}$. Now assume that $x \in \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'\ \text {and}\ \mathcal {\tilde{C}}_{e'}\ \text {satisfies}\ P\}$. It follows immediately from Observation 2 that $x \in \mathcal {\tilde{C}}(Y|e)$. $\square$

Proof of observation 7

Proof

Let $\{C_i: i \in I\}$ be a collection of choice functions, $C_i: \mathcal {X} \rightarrow \fancyscript{P}(X)$, and define $C(Y) = \bigcup _{i \in I}C_i(Y)$, for all $Y \in \mathcal {X}$. It is clear that $C$ is a well-defined choice function. We check each axiom.

$(I)$ Assume that for all $i \in I$, $C_i$ satisfies $(I)$. Suppose that $Y \subseteq Y'$ and $x \in Y \cap C(Y')$. Since $x \in C(Y')$, it follows that $x \in C_i(Y')$ for some $i \in I$. By assumption $C_i$ satisfies $(I)$, so it follows that $x \in C_i(Y)$. Hence, $x \in C(Y)$, too. So, $C$ satisfies $(I)$.

$(I^-)$ Suppose that for all $i \in I$, $C_i$ satisfies $(I^-)$. Assume that $Y \subseteq Y'$, $C(Y') \subseteq Y$, and $x \in C(Y')$. Then, for some $i \in I$, $x \in C_i(Y')$. Since $C_i$ satisfies $(I^-)$, it follows that $x \in C_i(Y)$. Hence, $x \in C(Y)$. So, $C$ satisfies $(I^-)$.

$(I')$ Suppose that $C_i$ satisfies $(I')$ for all $i \in I$. Assume that $x \in C(Y \cup Y')$. So, for some $i \in I$, $x \in C_i(Y \cup Y')$. By $(I')$, $x \in [C_i(Y) \cup C_i(y')]$. In either case, it follows that $x \in [C(Y) \cup C(Y')]$. So $C$ satisfies $(I')$.

$(III)$ Let $C_i$ satisfy $(III)$ for all $i \in I$. Suppose that $Y \subseteq Y'$ and $C(Y') \subseteq Y$, and let $x$ be in $C(Y)$. It must then be the case that $C_i(Y') \subseteq Y$ for all $i \in I$. And, for some $i \in I$, $x \in C_i(Y)$. By $(III)$, $x \in C_i(Y')$. Hence, $x \in C(Y')$. It follows that $C$ satisfies $(III)$.

$(IV^+)$ Let $C_i$ satisfy $(IV^+)$ for all $i \in I$. Assume that $Y \subseteq Y'$ and $x \in C(Y)$. It follows that $x \in C_i(Y)$ for some $i \in I$. By $(IV^+)$, $x \in C_i(Y')$. Thus, $x \in C(Y')$. $C$ therefore satisfies $(IV^+)$.

(Faith1) Suppose that (Faith1) is satisfied by $C_i$ for all $i \in I$. Assume that $Y \cap B \ne \emptyset$ (where $B$ is the set of “absolutely satisfactory” options in $X$) and let $x$ be in $C(Y)$. It follows that there is an $i \in I$ such that $x \in C_i(Y)$. By (Faith1), $x \in B$. So, $C$ satisfies (Faith1).

(Faith2) Let $C_i$ satisfy (Faith2) for all $i \in I$. Assume that $x \in Y \cap B$. By (Faith2), $x \in C_i(Y)$ for all $i \in I$. Hence, $x \in C(Y)$. $C$ therefore satisfies (Faith2).

(Success) Suppose that $C_i$ satisfies (Success) for all $i \in I$. Consider some $Y \ne \emptyset$ in $\mathcal {X}$. By (Success), $C_i(Y) \ne \emptyset$ for all $i \in I$. Hence, $C(Y) \ne \emptyset$. So, $C$ satisfies (Success).

$(\emptyset 1)$ Let $C_i$ satisfy $(\emptyset 1)$ for all $i \in I$. Suppose that $Y \subseteq Y'$ and that $C(Y') = \emptyset$. Then, $C_i(Y') = \emptyset$ for all $i \in I$. By $(\emptyset 1)$, it follows that $C_i(Y) = \emptyset$ for all $i \in I$. Hence, $C(Y) = \emptyset$, too. So, $C$ satisfies $(\emptyset 1)$.

$(\emptyset 2)$ Suppose that $C_i$ satisfies $(\emptyset 2)$ for all $i \in I$. Assume that $Y \subseteq Y'$ and $C(Y) = \emptyset$. Then, $C_i(Y) = \emptyset$ for all $i \in I$. By $(\emptyset 2)$, we have that $C_i(Y') \cap Y = \emptyset$ for all $i \in I$. It follows that $C(Y') \cap Y = \emptyset$. $C$ therefore satisfies $(\emptyset 2)$. $\square$

Proof of observation 8

Proof

($\Leftarrow$) Since, by hypothesis, every local choice function of $\mathcal {\tilde{C}}$ satisfies property $P$, it follows that, for every $e \in E$ there is an $f \in E$ such that $e \sqsubseteq f$ and $\mathcal {\tilde{C}}_g$ satisfies $P$ for all $g \in E$ such that $f \sqsubseteq g$. That is, $\mathcal {\tilde{C}}$ satisfies $P^{\star }$.

($\Rightarrow$) Let $\mathcal {\tilde{C}}$ satisfy $P^{\star }$. By Observation 6, $\mathcal {\tilde{C}}(Y|e) = \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'\ \text {and}\ \mathcal {\tilde{C}}_{e'}\ \text {satisfies}\ P\}$ for all $Y \in \mathcal {X}$ and $e \in E$. For those constraints, $P$, that are preserved under unions (Observation 7), it follows that $\mathcal {\tilde{C}}_e$ satisfies $P$. Hence, every local choice function of $\mathcal {\tilde{C}}$ satisfies $P$. $\square$

Proof of observation 9

Proof

Assume that $\mathcal {\tilde{C}}: \mathcal {X} \times \mathcal {E} \rightarrow \fancyscript{P}(X)$ is a conditional choice with a vacuous second tier that satisfies $(\emptyset 1^{\star })$, and let $\mathcal {X} = \fancyscript{P}_{fin}(X)$.

$(\Rightarrow )$ Suppose that $\mathcal {\tilde{C}}$ satisfies $(I^{\star })$ and $(II^{\star })$. By Observation 8, every local choice function of $\mathcal {\tilde{C}}$ satisfies $(\emptyset 1)$ and $(I)$. From Observation 6 it follows that $\mathcal {\tilde{C}}(Y|e) = \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'$ and $\mathcal {\tilde{C}}_{e'}$ satisfies $(II)\}$ for all $Y \in \mathcal {X}$ and $e \in E$. So, $\mathcal {\tilde{C}}(Y|e) = \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'$ and $\mathcal {\tilde{C}}_{e'}$ satisfies $(\emptyset 1), (I),$ and $(II)\}$. By Theorem 1, $\mathcal {\tilde{C}}(Y|e) = \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'$ and $\mathcal {\tilde{C}}_{e'}$ is rationalizable $\} = \{x: xRy$ for all $y \in Y$ with respect to some rationalizing $R \in O_e\}$.

$(\Leftarrow )$ Suppose that, for all $x \in X, Y \in \mathcal {X}$, and $e \in \mathcal {E},\ \mathcal {\tilde{C}}(Y|e)=\{x :xRy$ for all $y \in Y$ wrt some rationalizing $R \in O_e\}$. Let $\mathcal {E}_e$ be the poset $\langle \{f|e \sqsubseteq f\}, \sqsubseteq \rangle$, where $\sqsubseteq$ is the relevant restriction of the partial order from $\mathcal {E}.$ By the assumption on $\mathcal {E}$, it follows that every chain in $\mathcal {E}_e$ has an upper bound in $\mathcal {E}_e$. Thus, by Zorn’s lemma, $\mathcal {E}_e$ has a maximal element. Let $f$ be such a maximal element of $\mathcal {E}_e$. Either (i) $\mathcal {\tilde{C}}(Y|f) \ne \emptyset$ for some $Y \in \mathcal {X}$ or (ii) $\mathcal {\tilde{C}}_f$ always returns the empty set. Suppose that (i). Then, for some $Y \in \mathcal {X},$ there is some $x \in \mathcal {\tilde{C}}(Y|f).$ By assumption, there is a rationalizing $R \in O_f$ such that $xRy$ for all $y \in Y.$ $f$ is maximal so $R_{\mathcal {\tilde{C}}_f}$ is rationalizing. Therefore, by Theorem 1, $\mathcal {\tilde{C}}_f$ satisfies $(I)$ and $(II)$. And because $f$ is maximal, $\mathcal {\tilde{C}}_g$ satisfies $(I)$ and $(II)$ for any $g \in E$ such that $f \sqsubseteq g$. Now, consider case (ii). $\mathcal {\tilde{C}}(Y|f) = \emptyset$ for all $Y \in \mathcal {X}.$ It is straightforward to verify that $\mathcal {\tilde{C}}_f$ vacuously satisfies constraints $(I)$ and $(II).$ Again, because $f$ is maximal, $\mathcal {\tilde{C}}_g$ satisfies $(I)$ and $(II)$ for any $g \in E$ such that $f \sqsubseteq g$. It follows that $\mathcal {\tilde{C}}$ satisfies $(I^{\star })$ and $(II^{\star })$. $\square$

Proof of observation 11

Proof

Assume that $\mathcal {\tilde{C}}: \mathcal {X} \times \mathcal {E} \rightarrow \fancyscript{P}(X)$ is a conditional choice with a vacuous second tier that satisfies $(\emptyset 1^{\star })$, and let $\mathcal {X} = \fancyscript{P}_{fin}(X)$.

$(\Rightarrow )$ Assume that $\mathcal {\tilde{C}}$ satisfies $(I^{\star })$ and $(IV^{\star }).$ By Observation 8, it follows from our assumptions that $\mathcal {\tilde{C}}$ satisfies $(\emptyset 1)$ and $(I)$ at each of its local choice functions. By Observation 6, $\mathcal {\tilde{C}}(Y|e) = \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'$ and $\mathcal {\tilde{C}}_{e'}$ satisfies $(IV)\}$ for all $Y \in \mathcal {X}$ and $e \in E$. Putting these together, it follows that $\mathcal {\tilde{C}}(Y|e) = \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'$ and $\mathcal {\tilde{C}}_{e'}$ satisfies $(\emptyset 1), (I),$ and $(IV)\}$ for all $Y \in \mathcal {X}$ and $e \in E$. By Theorem 2, this amounts to $\mathcal {\tilde{C}}(Y|e) = \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'$ and $\mathcal {\tilde{C}}_{e'}$ is weak order rationalizable$\}$ for all $Y \in \mathcal {X}$ and $e \in E$.

$(\Leftarrow )$ Now assume that for all $e \in E,\ \mathcal {\tilde{C}}_e$ is pseudo-rationalizable by a collection of local choice functions $\{\mathcal {\tilde{C}}_{e'} :e \sqsubseteq e'$ and $\mathcal {\tilde{C}}_{e'}$ is weak order rationalizable$\}.$ We establish that $\mathcal {\tilde{C}}$ satisfies $(I^{\star })$ and $(IV^{\star }).$ Let $\mathcal {E}_e$ be the poset $\langle \{f|e \sqsubseteq f\}, \sqsubseteq \rangle$, where $\sqsubseteq$ is the relevant restriction of the partial order from $\mathcal {E}.$ By the assumption on $\mathcal {E}$, it follows that every chain in $\mathcal {E}_e$ has an upper bound in $\mathcal {E}_e$. Thus, by Zorn’s lemma, $\mathcal {E}_e$ has a maximal element. Let $f$ be such a maximal element of $\mathcal {E}_e$. Either (i) $\mathcal {\tilde{C}}(Y|f) \ne \emptyset$ for some $Y \in \mathcal {X}$ or (ii) $\mathcal {\tilde{C}}_f$ always returns the empty set. Consider case $(i)$ first. Suppose that, for some $Y \in \mathcal {X}$, $x \in \mathcal {\tilde{C}}(Y|f)$. By the assumption of the pseudo-rationalizability of each local choice function (i.e., $\mathcal {\tilde{C}}(Y|e) = \bigcup \{\mathcal {\tilde{C}}(Y|e'): e \sqsubseteq e'$ and $\mathcal {\tilde{C}}_{e'}$ is weak order rationalizable$\}$ for all $Y \in \mathcal {X}$ and $e \in E$), it follows that there is a weak order rationalizable local choice function $\mathcal {\tilde{C}}_{f'}$ such that $f \sqsubseteq f'$ and $x \in \mathcal {\tilde{C}}_{f'}(Y).$ $f$ is maximal so ${\mathcal {\tilde{C}}_{f}}$ must itself be weak order rationalizable. Therefore, by Theorem 2, $\mathcal {\tilde{C}}_{f}$ satisfies $(I)$ and $(IV)$. And because $f$ is maximal, $\mathcal {\tilde{C}}_g$ satisfies $(I)$ and $(IV)$ for any $g \in E$ such that $f \sqsubseteq g$. Now consider case $(ii)$, $\mathcal {\tilde{C}}_f(Y) = \emptyset$ for all $Y \in \mathcal {X}$. $(I)$ and $(IV)$ are trivially satisfied. Because $f$ is maximal, $\mathcal {\tilde{C}}_g$ satisfies $(I)$ and $(IV)$ for any $g \in E$ such that $f \sqsubseteq g$. It follows that $\mathcal {\tilde{C}}$ satisfies $(I^{\star })$ and $(IV^{\star })$. $\square$