Preference for flexibility and dynamic consistency with incomplete preferences

Abstract

We generalize a previous result about dynamically consistent menu preferences to the case where preferences are not necessarily complete. We show that, as it is the case when preferences are complete, a subjective state space version of dynamic consistency is linked to a comparative theory of preference for flexibility. In words, an objective signal is interpreted as an event in the agent’s subjective state space and the agent acts in a dynamically consistent way after that if and only if we can attribute all the differences between the agent’s preferences before and after the signal to the fact that the agent values flexibility more before the signal than after.

Proofs

1.1 Proof of Theorem 2

[1 $\implies$ 2] Suppose $\succsim$ and $\succsim ^{*}$ are finite IPAEU preferences. Let S and $S^{*}$ be the unique subjective state spaces of $\succsim$ and $\succsim ^{*}$, respectively, and let $U:(S\cup S^{*})\times \Delta (X)\rightarrow \mathbb {R}$ be such that, for any $s\in S\cup S^{*}$, U(s, .) is an expected utility function that represents s. Now suppose there exist two menus A and B with

$$\begin{aligned} \max _{p\in A}U(s,p)=\max _{p\in B}U(s,p)\text { for all }s\in S{\setminus } S^{*}, \end{aligned}$$

(1)

but either $A\succsim ^{*}B$ and it is not true that $A\succsim B$ or $B\succsim A$ and it is not true that $B\succsim ^{*}A$. Without loss of generality, we may assume that $A,B\in \mathrm{int}(\mathcal {X})$.⁵ Now suppose C is a menu such that $A\cup B\cup C\sim ^{*}A\cup C$. By the representation of $\succsim ^{*}$, this can happen only if

$$\begin{aligned} \max _{p\in A\cup B\cup C}U(s,p)=\max _{p\in A\cup C}U(s,p)\text { for all }s\in S^{*}. \end{aligned}$$

Together with (1), this implies that

$$\begin{aligned} \max _{p\in A\cup B\cup C}U(s,p)=\max _{p\in A\cup C}U(s,p)\text { for all }s\in S, \end{aligned}$$

which, by the representation of $\succsim$, implies that $A\cup B\cup C\sim A\cup C$. That is, $\succsim$ and $\succsim ^{*}$ violate Flexibility Consistency II.

[2 $\implies$ 3] Now suppose statement 2 is true and let $(S,\mathcal {M},U)$ and $(S^{*},\mathcal {M}^{*},U^{*})$ be finite IPAEU representations of $\succsim$ and $\succsim ^{*}$, respectively. Fix any two menus A and B and let F be any sphere in $\mathrm{int}(\mathcal {X})$.⁶ Let C be the subset of F that includes only the maximizers of s for all $s\in S{\setminus } S^{*}$ and let D be the subset of F that includes only the maximizers of s for all $s\in S^{*}$.⁷ For each $\lambda \in (0,1)$, define $A_{\lambda }:=C\cup (\lambda D+(1-\lambda )A)$ and $B_{\lambda }:=C\cup (\lambda D+(1-\lambda )B)$. For $\lambda$ large enough, we have that $\max _{p\in A_{\lambda }}U(s,p)=\max _{p\in B_{\lambda }}U(s,p)=\max _{p\in C}U(s,p)$ for every $s\in S{\setminus } S^{*}$ and $\max _{p\in A_{\lambda }}U^{*}(s,p)-\max _{p\in B_{\lambda }}U^{*}(s,p)=(1-\lambda )( \max _{p\in A}U^{*}(s,p)-\max _{p\in B}U^{*}(s,p))$ for every $s\in S^{*}$. By statement 2 and the representations of $\succsim$ and $\succsim ^{*}$, we get that

$$\begin{aligned} A&\succsim ^{*}{B}\\&{\iff} \\ A_{\lambda }&\succsim ^{*}{B_{\lambda }}\\&{\iff} \\ A_{\lambda }&\succsim{B_{\lambda }}\\&{\iff} \\ \sum _{s\in S}\mu (s)\max _{p\in A_{\lambda }}U(s,p)&\ge \sum _{s\in S}\mu (s)\max _{p\in B_{\lambda }}U(s,p)\text { for every }\mu \in \mathcal {M}\\&{\iff} \\ \sum _{s\in S\cap S^{*}}\mu (s)\max _{p\in A_{\lambda }}U(s,p)&\ge \sum _{s\in S\cap S^{*}}\mu (s)\max _{p\in B_{\lambda }}U(s,p)\text { for every }\mu \in \mathcal {M}\\&{\iff} \\ \sum _{s\in S\cap S^{*}}\mu (s)\max _{p\in A}U(s,p)&\ge \sum _{s\in S\cap S^{*}}\mu (s)\max _{p\in B}U(s,p)\text { for every }\mu \in \mathcal {M}. \end{aligned}$$

Define $T:=S\cap S^{*}$.⁸ It is clear that $(T,\mathcal {M}_{T},U)$ is a finite IPAEU representation of $\succsim ^{*}$.

[3 $\implies$ 1] Let $\left( S,\mathcal {M},U\right)$ and $\left( T,\mathcal {M}_{T},U\right)$ be finite IPAEU representations of $\succsim$ and $\succsim ^{*}$, respectively, where $\mathcal {M}_{T}$ is the set with the Bayesian updates of all (possible) priors in $\mathcal {M}$ after the observation of T. Now suppose that $A\in \mathcal {X}$ and $B\in \mathrm{int}\left( \mathcal {X}\right)$ are such that $B\succsim A$, but it is not true that $B\succsim ^{*}A$, or $A\succsim ^{*}B$, but it is not true that $A\succsim B$. It is clear that this can happen only if there exists $s^{*}\in S{\setminus } S^{*}$ such that

$$\begin{aligned} \max _{p\in B}U\left( s^{*},p\right) >\max _{p\in A}U\left( s^{*},p\right) . \end{aligned}$$

We now show that this implies that there exists a menu C such that

$$\begin{aligned} \max _{p\in A\cup B\cup C}U\left( s,p\right) =\max _{p\in A\cup C}U\left( s,p\right) \text {, for every }s\in S^{*}\text {,} \end{aligned}$$

(2)

but

$$\begin{aligned} \max _{p\in A\cup B\cup C}U\left( s^{*},p\right) >\max _{p\in A\cup C}U\left( s^{*},p\right) . \end{aligned}$$

(3)

For each $s\in S^{*}$, define $q_{s}$ as follows: if $\max _{p\in A}U\left( s,p\right) \ge \max _{p\in B}U\left( s,p\right)$, let $q_{s}$ be any lottery in $\arg \max _{p\in A}U\left( s,p\right)$. If $\max _{p\in B}U\left( s,p\right) >\max _{p\in A}U\left( s,p\right)$ and there exists $q\in \arg \max _{p\in B}U\left( s,p\right)$ with $U\left( s^{*},q\right) <\max _{p\in B}U\left( s^{*},p\right)$, let $q_{s}:=q$. We are left with the case where $\max _{p\in B}U\left( s,p\right) >\max _{p\in A}U\left( s,p\right)$, but $U\left( s^{*},q\right) =\max _{p\in B}U\left( s^{*},p\right)$ for all $q\in \arg \max _{p\in B}U\left( s,p\right)$. We first note that this implies that it cannot be the case that $U\left( s,.\right)$ is cardinally equivalent to $-U\left( s^{*},.\right)$. If this was the case, we would necessarily have $\max _{p\in B}U\left( s^{*},p\right) <\max _{p\in A}U\left( s^{*},p\right)$. Therefore, there exist lotteries p and $p^{\prime }$ such that $U\left( s,p\right) =U\left( s,p^{\prime }\right)$, but $U\left( s^{*},p\right) <U\left( s^{*},p^{\prime }\right)$. Fix $q\in \arg \max _{p\in B}U\left( s,p\right)$ and let $\xi :=q+p-p^{\prime }$. Note that, for every $\lambda \in \left( 0,1\right)$, $U\left( s,\lambda q+\left( 1-\lambda \right) \xi \right) =U\left( s,q\right)$, but $U\left( s^{*},\lambda q+\left( 1-\lambda \right) \xi \right) <U\left( s^{*},q\right)$. Since $B\in \mathrm{int}\left( \mathcal {X}\right)$, we have $\lambda q+\left( 1-\lambda \right) \xi \in \Delta \left( X\right)$ when $\lambda$ is large enough. In this case, let $q_{s}:=\lambda q+\left( 1-\lambda \right) \xi$ for some $\lambda \in \left( 0,1\right)$ such that $\lambda q+\left( 1-\lambda \right) \xi \in \Delta \left( X\right)$. Now define $C:=\left\{ q_{s}:s\in S^{*}\right\}$. It is clear that (2) and (3) hold for such menu C and, consequently, $A\cup B\cup C\sim ^{*}A\cup C$, but $A\cup B\cup C\succ A\cup C$. That is, $\succsim$ and $\succsim ^{*}$ satisfy Flexibility Consistency II.

1.2 Proof of Proposition 1

Suppose $\{S_{1},\dots ,S_{I}\}$, the collection of subjective state spaces of the relations $\succsim _{1},\cdots ,\succsim _{I}$, is a partion of S, the subjective state space of $\succsim$. Fix menus A and B in $\mathrm{int}(\mathcal {X})$ with $B\subseteq A$ and $A\succ B$. Let $U:S\times \Delta (X)\rightarrow \mathbb {R}$ be such that, for any $s\in S$, $U\left( s,.\right)$ is an expected utility function that represents s. There must exist $s^{*}\in S$ such that $\max _{p\in A}U(s^{*},p)>\max _{p\in B}U(s^{*},p)$. Since $S_{1},\dots ,S_{I}$ are a partition of S, there exists a unique $i^{*}\in \{1,\dots ,I\}$ such that $s^{*}\in S_{i^{*}}$. Replicating the argument in the [3 $\implies$ 1] part of the proof of Theorem 2, we can find a menu C such that $\max _{p\in A\cup C}U(s,p)=\max _{p\in B\cup C}U(s,p)$ for every $s\in S{\setminus }\{s^{*}\}$, while $\max _{p\in A\cup C}U(s^{*},p)>\max _{p\in B\cup C}U(s^{*},p)$. This implies that $A\cup C\sim _{i}B\cup C$ for every $i\ne i^{*}$ and $A\cup C\succ _{i^{*}}B\cup C$. That is, $\succsim ,\succsim _{1},\cdots ,\succsim _{I}$ satisfy Flexibility Separability.

Conversely, suppose $\succsim ,\succsim _{1},\cdots ,\succsim _{I}$ satisfy Flexibility Separability. If there exists $s^{*}\in S{\setminus }\bigcup _{i=1}^{I}S_{i}$, then it is easy to find menus A and B in $\mathrm{int}(\mathcal {X})$ with $B\subseteq A$, $A\succ B$, but $A\sim _{i}B$ for $i=1,\dots ,I$, which contradicts Flexibility Separability. We conclude that $\bigcup _{i=1}^{I}S_{i}=S$. Now suppose there exist distinct $i,j\in \{1,\dots ,I\}$ with $S_{i}\cap S_{j}\ne \emptyset$. Pick any $s^{*}\in S_{i}\cap S_{j}$. It is easy to build menus A and B in $\mathrm{int}(\mathcal {X})$ with $B\subseteq A$, $\max _{p\in A}U(s,p)=\max _{p\in B}U(s,p)$ for every $s\in S{\setminus }\{s^{*}\}$ while $\max _{p\in A}U(s^{*},p)>\max _{p\in B}U(s^{*},p)$. Since $\bigcup _{i=1}^{I}S_{i}=S$, this implies that $A\succ B$. However, it is clear that, for any menu C, $A\cup C\sim _{i}B\cup C\iff A\cup C\sim _{j}B\cup C$, which again contradicts Flexibility Separability. We conclude that $\{S_{1},\dots ,S_{I}\}$ is a partition of S.