Additive representation under idempotent attention

Abstract

This paper explores a scenario where a decision maker evaluates bundles by adding up the utility of the options that attract her attention. We introduce a novel attention rule called the “idempotent attention rule” and examine additive representations under this rule. With idempotent attention rules, we can narrow our focus to a subset of bundles to reveal attention rules and utility functions. As a generalization of attention filters, this rule sheds light on how alternatives interact in forming attention.

Appendices

Appendix A Proof of Theorem 1

For the only if part, suppose that ${\succcurlyeq }$ on ${{\mathcal {X}}}$ admits the AR-IA under $({\Gamma },u)$. Let $\left\{ S_n\right\} _{n=1}^m, \left\{ T_n\right\} _{n=1}^m$ be two collections of basic sets such that the DM prefers $S_n$ over $T_n$ for all $n\le m$, and strictly prefers $S_n$ to $T_n$ for some $n\le m$. The AR-IA implies that $\sum _{n=1}^m\sum _{s\in T_n}u(s)>\sum _{n=1}^m\sum _{t\in T_n}u(t)$ because basic sets catch full attention. However, if $\sum _{n=1}^m\mathbbm {1}_{S_n}(x)\le \sum _{n=1}^m\mathbbm {1}_{T_n}(x)$ for all $x\in X$, i.e., the collection of $S_n$ contains less x than the collection of $T_n$ for all $x\in X$, then $\sum _{n=1}^m\sum _{s\in T_n}u(s)<\sum _{n=1}^m\sum _{t\in T_n}u(t)$. Therefore, NR is necessary for the AR-IA.

We now move on to the sufficiency of WO and NR.

Step 1. Constructing the utility function by solving the system of linear inequalities induced by the ${\succcurlyeq }$ on ${{\mathcal {B}}}$. Let ${{\mathcal {B}}}=\{B_n\}_{n=1}^m$. If $B\sim B'$ for all $B, B'\in {{\mathcal {B}}}$, then it is obvious that the system of linear inequalities has a solution. We then can assume that there exists at least one pair of $B_n,B_{n'}\in {{\mathcal {B}}}$ such that $B_n\ {\succ }\ B_{n'}$. For any $B_{n}\ {\succ }\ B_{n'}$, we have $\sum _{x\in B_{n}}u(x)-\sum _{x\in B_{n'}}u(x)>0$. Similarly, $B_{n} \sim B_{n'}$ suggests that $\sum _{x\in B_{n}}u(x)-\sum _{x\in B_{n'}}u(x)=0$, and $B_{n}\ {\succcurlyeq }\ B_{n'}$ suggests that $\sum _{x\in B_{n}}u(x)-\sum _{x\in B_{n'}}u(x)\ge 0$. Moreover, we also need a restriction on u, i.e., $u(x)\ge 0$ for all $x\in X$. We then consider the solution to the above system of linear inequalities.

By Kraft et al. (1959), the above system of linear inequalities has a solution if and only if there are no $m_1>0$ pairs of $B_n\ {\succ }\ B_{n'}$, $m_2\ge 0$ pairs of $B_n \sim B_{n'}$, and $m_3\ge 0$ pairs of $B_n\ {\succcurlyeq }\ B_{n'}$ such that the collection of $B_n$ has less x than the collection of $B_{n'}$ for all $x\in X$. Hence, the system of linear inequalities has a solution if and only if ${\succcurlyeq }$ on ${{\mathcal {B}}}$ is a weak order and satisfies nonnegative remainders.

Step 2. Constructing ${\Gamma }$ by using basic sets. Let ${{\mathcal {E}}}{{\mathcal {S}}}(B_1)=\{S: B_1\subseteq S\ and \ B_1\sim S\}$ and ${{\mathcal {E}}}{{\mathcal {S}}}(B_i)=\{S: B_i\subseteq S, \ B_i\sim S,\ and\ S\notin {{\mathcal {E}}}{{\mathcal {S}}}_{B_{i-1}} \}$ for all $i>1$. For every $S\in {{\mathcal {E}}}{{\mathcal {S}}}(B_i)$, let ${\Gamma }(S)=B_i$. It’s clear that ${\Gamma }$ is idempotent attention.

Take any pair of $({\Gamma },u)$ constructed above. For any $S,T\in {{\mathcal {X}}}$ with $S\ {\succcurlyeq }\ T$, we know that there is a pair of basic sets $B_{n}$ and $B_{n'}$ such that $S\in {{\mathcal {E}}}{{\mathcal {S}}}(B_n)$ and $T\in {{\mathcal {E}}}{{\mathcal {S}}}(B_m)$. Therefore,

$$\begin{aligned} S\ {\succcurlyeq }\ T \iff B_n\ {\succcurlyeq }\ B_{n'} \iff \sum _{x\in B_n}u(x)\ge \sum _{y\in B_{n'}}u(y) \iff \sum _{x\in {\Gamma }(S)}u(x)\ge \sum _{y\in {\Gamma }(T)}u(y). \end{aligned}$$

Appendix B Proof of Proposition 2

Suppose that ${\succcurlyeq }$ on ${{\mathcal {X}}}$ admits AR-IAs. Let $\{({\Gamma }_i,u_i)\}_{i\in I_{\succcurlyeq }}$ be the collection of corresponding idempotent attentions and utility functions.

We first prove the only if part of this proposition. By contradiction, assume that there is a positive integer m and $\{S_n\}_{n=1}^m,\{T_n\}_{n=1}^m\subseteq {{\mathcal {B}}}$ where $S_n\ {\succcurlyeq }\ T_n$ for all n, and $S_n\ {\succ }\ T_n$ for some n, such that $\sum _{n=1}^m\mathbbm {1}_{S_n}(x)=\sum _{n=1}^m\mathbbm {1}_{T_n}(x)$ for all $x\notin MIN(X, {\succcurlyeq })$, and $\sum _{x\in MIN(X,{\succcurlyeq })}\sum _{n=1}^m\mathbbm {1}_{S_n}(x)>\sum _{x\in MIN(X,{\succcurlyeq })}\sum _{n=1}^m\mathbbm {1}_{T_n}(x)$. Then,

$$\begin{aligned} \sum _{n=1}^m\sum _{s\in S_n}u(s)&= \sum _{n=1}^m\sum _{s\in S_n\cap MIN(X,{\succcurlyeq })}u(s)+ \sum _{n=1}^m\sum _{s\in S_n\setminus MIN(X,{\succcurlyeq })}u(s)\\&>\sum _{n=1}^m\sum _{t\in T_n}u(t)\\&= \sum _{n=1}^m\sum _{t\in T_n\cap MIN(X,{\succcurlyeq })}u(t)+ \sum _{n=1}^m\sum _{t\in T_n\setminus MIN(X,{\succcurlyeq })}u(t). \end{aligned}$$

Therefore,$\sum _{n=1}^m\sum _{s\in S_n\cap MIN(X,{\succcurlyeq })}u(s) >\sum _{n=1}^m\sum _{t\in T_n\cap MIN(X,{\succcurlyeq })}u(t)$. We then have $u(x)>0$ for all $x\in MIN(X,{\succcurlyeq })$.

For the converse direction, suppose that there is a $u\in {{\mathcal {U}}}_{\succcurlyeq }$ such that $u(x)=0$ for all $x\in MIN(X,{\succcurlyeq })$. We then consider the system of linear inequalities we introduced in the proof of Theorem 1 combining with the restriction that $u(x)=0$ for all $x\in MIN(X,{\succcurlyeq })$. By Kraft et al. (1959), this linear system has a solution if and only if there are no $m_1> 0$ pairs of $B_n\ {\succ }\ B_{n'}$, $m_2\ge 0$ pairs of $B_n\sim B_{n'}$, $m_3\ge 0$ pairs of $B_n\ {\succcurlyeq }\ B_{n'}$, and $|MIN(X,{\succcurlyeq })|>0$ restrictions on $u(x)=0$ where $x\in MIN(X,{\succcurlyeq })$ such that the collection of $B_n$ only contains more dominated alternatives than the collection of $B_{n'}$. Hence, we have ENS.

Appendix C Proof of Theorem 2

Suppose that ${\succcurlyeq }$ on ${{\mathcal {X}}}$ admits an AR-AF under $({\Gamma },u)$. We first know that $({\Gamma },u)$ is also a pair of AR-IAs for ${\succcurlyeq }$ on ${{\mathcal {X}}}$. Therefore, we know that ${\succcurlyeq }$ on ${{\mathcal {X}}}$ satisfies WO and NR by Theorem 1.

We then only need to prove that ${\succcurlyeq }$ on ${{\mathcal {X}}}$ satisfies IB. Take any $S\in {{\mathcal {X}}}$, we can get ${\Gamma }(S)$ from the attention filter ${\Gamma }$. We first know that ${\Gamma }(Y)={\Gamma }(S)$ for all ${\Gamma }(S)\subseteq Y\subseteq S$. If ${\Gamma }(S)$ is basic, it is clear that ${\Gamma }(S)$ itself can serve as the corresponding T and B in IB, and ${\Gamma }(S){\setminus } {\Gamma }(S)=\emptyset \subseteq N$. Moreover, $\sum _{s\in {\Gamma }(S)}u(s)=\sum _{s\in {\Gamma }({\Gamma }(S))}u(s)$ which implies that $S\sim {\Gamma }(S)$. We then have IB.

If ${\Gamma }(S)$ is non-basic, we know that there is a corresponding basic set B of ${\Gamma }(S)$ such that $B\sim {\Gamma }(S)$. Hence, $\sum _{s\in {\Gamma }(S){\setminus } B}u(s)=0$. As a result, $u(s)=0$ for all $s\in {\Gamma }(S){\setminus } B$. By Proposition 2, we have ${\Gamma }(S)\setminus B\subseteq N$. We can let ${\Gamma }(S)$ be the corresponding T of S in IB, and the ${\succcurlyeq }$ on ${{\mathcal {X}}}$ satisfies IB.

We now suppose that ${\succcurlyeq }$ on ${{\mathcal {X}}}$ satisfies WO, NR, and IB. We know that ${\succcurlyeq }$ on ${{\mathcal {X}}}$ admits AR-IAs by Theorem 1. We then need to consider the construction of an attention filter. Let $\{T_i\}_{i=1}^m=\bigcup _{s\in {{\mathcal {X}}}}{{\mathcal {I}}{\mathcal {B}}}(S)$, and

$$\begin{aligned}{{\mathcal {E}}}{{\mathcal {B}}}(T, S):= {\left\{ \begin{array}{ll} \left\{ Y: T\subseteq Y \subseteq S\right\} &{} \text {if all}\ Y\ \text {inbetween}\ T\ \text {and}\ S\ \text {are indifferent,}\\ \emptyset &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

Let’s also define

$$\begin{aligned} {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_1):=\left\{ S: S\in {{\mathcal {X}}}\; such\; that\; \emptyset \ne {{\mathcal {E}}}{{\mathcal {B}}}(T_1,S)\right\} , \end{aligned}$$

and for all $i>1$, we denote

$$\begin{aligned} {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i):=\left\{ S: S\in {{\mathcal {X}}}\setminus \bigcup _{j<i}{{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(A_j) \; such\; that\; \emptyset \ne {{\mathcal {E}}}{{\mathcal {B}}}(T_i,S)\right\} . \end{aligned}$$

Claim 1

$\{{{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)\}_{i=1}^m$ forms a partition of ${{\mathcal {X}}}$.

Proof

We first show that for any set S, there is a $i\le m$ such that $S\in {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)$. Take any $S\in {{\mathcal {X}}}$, we know that there exists at least one of $T_i$ such that $T_i\in {{\mathcal {I}}{\mathcal {B}}}(S)$. We then can denote the ${{\mathcal {I}}{\mathcal {B}}}(S)$ as a finite subsequence $\{T_{i_j}\}_{j=1}^n$ of $\{T_i\}_{i=1}^m$. We then claim that $S\in {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_{i_1})$. To show this, we only need to show that $S\notin {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)$ where $i< i_1$. By contradiction, suppose that $S\in {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)$ for some $i< i_1$. Then, we know that $T_i\in {{\mathcal {I}}{\mathcal {B}}}(S)$ which is a contradiction.

We then show that for any $T_i,T_j\in \{T_i\}_{i=1}^m$ where $i\ne j$, ${{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)\cap {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_j)=\emptyset$. Without loss of generality, we can assume that $i>j$. By contradiction, suppose that there is a set S such that $S\in {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)\cap {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_j)$, we then know that $S\in {{\mathcal {E}}}{{\mathcal {B}}}(T_j, S)$ which implies that $S\notin {{\mathcal {X}}}\setminus \bigcup _{j<i}{{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_j)$. Hence, $S\notin {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)$.

In conclusion, $\{{{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)\}_{i=1}^m$ forms a partition of ${{\mathcal {X}}}$. $\square$

Let’s define a function ${\Gamma }:{{\mathcal {X}}}\rightarrow {{\mathcal {X}}}$ by ${\Gamma }(S)=T_i$ if $S\in {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_i)$. We then claim that ${\Gamma }$ is an attention filter. Take any $S\in {{\mathcal {X}}}$, and suppose that ${\Gamma }(S)=T_i$. Take any Y with $T_i\subseteq Y\subseteq S$, we want to show that ${\Gamma }(Y)=T_i$. We first know that $Y\sim S$. We then only need to show that $Y\notin {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_j)$ for any $j<i$. By contradiction, if $Y\in {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_j)$ for some $j<i$, then we know that $S\in {{\mathcal {U}}}{{\mathcal {E}}}{{\mathcal {B}}}(T_j)$. It is a contradiction. Therefore, ${\Gamma }$ is an attention filter.

We now want to show that there is a utility function u such that $S\ {\succcurlyeq }\ T\iff \sum _{s\in S}u(s)\ge \sum _{t\in T}u(t)$. Take any $S\ {\succcurlyeq }\ Y$, we know that there are two basic sets $T_S$ and $T_Y$ such that ${\Gamma }(S)=T_S$ and ${\Gamma }(Y)=T_Y$. By IB, there is a basic set $B_1$ where $B_1\subseteq T_S$ and $B_1\sim T_S$ such that $T_S{\setminus } B_1\subset N$. Similarly, we can get a basic set $B_2$ for $T_Y$. For all u that consist of AR-IAs, we have

$$\begin{aligned} S\ {\succcurlyeq }\ Y&\iff T_S\ {\succcurlyeq }\ T_Y\\&\iff B_1\ {\succcurlyeq }\ B_2\\&\iff \sum _{b\in B_1}u(b)\ge \sum _{b\in B_2}u(b)\\&\iff \sum _{s\in {\Gamma }(S)}u(s)\ge \sum _{y\in {\Gamma }(Y)}u(y). \end{aligned}$$

Appendix D Independence between WO, NR and IB

Completeness and NR imply transitivity on ${{\mathcal {B}}}$ (see Fishburn 1992). However, they can not characterize the AR-IA.

Example 5

Let $X=xy$, and the ${\succcurlyeq }$ on ${{\mathcal {X}}}$ is $x\ {\succ }\ y$, $xy\sim x$, $y\ {\succ }\ xy$. This preference is complete and nontransitive. Also, ${{\mathcal {B}}}=\{x,y\}$, and it satisfies NR. However, if it admits AR-IA under $({\Gamma },u)$, then ${\Gamma }(xy)=x$ or xy. When ${\Gamma }(xy)=x$, the AR-IA suggests that $xy\ {\succ }\ y$. When ${\Gamma }(xy)=xy$, it suggests that $xy\ {\succ }\ xy$.

The Example 2 provides a preference relation ${\succcurlyeq }$ on ${{\mathcal {X}}}$ that admits AR-IAs but not AR-AFs. To see the ${\succcurlyeq }$ provided in Example 2 violates IB, let us consider xyz. First, we know that $N=\emptyset$. xyz is not basic suggests that ${{\mathcal {I}}{\mathcal {B}}}(xyz)=\emptyset$.

We now want to show that WO and IB together cannot characterize AR-AFs.

Example 6

Let $X=xy$, and the ${\succcurlyeq }$ on ${{\mathcal {X}}}$ is $x\ {\succ }\ xy\ {\succ }\ y$. We know that $N=\emptyset$, and all the sets are basic. Then it satisfies WO and IB. However, it does not admit AR-AFs because $x\ {\succ }\ xy\ {\succ }\ y\iff u(y)<0$.