Partially dominant choice

Abstract

This paper proposes and analyzes a model of context-dependent choice with stable but incomplete preferences that is based on the idea of partial dominance: an alternative is chosen from a menu if it is not worse than anything in the menu and is also better than something else. This choice procedure provides a simple explanation of the attraction/decoy effect. It reduces to rational choice when preferences are complete in two ways that are made precise. Some preference identification and choice consistency properties associated with this model are analyzed, and certain ways in which its predictions differ from those of other recently proposed models of the attraction effect are also discussed.

Appendices

Appendix 1: Proofs

Proof of Proposition 1

It is well known that (a) $\Rightarrow$ (b) and (d) $\Rightarrow$ (a). It will be shown that (b) $\Rightarrow$ (c) $\Rightarrow$ (d).

(b) $\Rightarrow$ (c). Suppose $a$-WARP is violated. Then $C(\{x,y\})=x$, $x\in A$ and $y\in C(A)$ for some $x,y\in X$ and $A\in {\mathcal {M}}$. Since $y\not \in C(\{x,y\})$, this obviously violates Property $\alpha$. Thus, $a$-WARP is satisfied. Suppose $b$-WARP is not. There are $A\in {\mathcal {M}}$ and $x,y\in A$ such that $x\in C(A)$, $y\in A\!{\setminus }\!C(A)$ and $y\in C(\{x,y\})$. Since Property $\alpha$ holds and $x\in C(A)$, it follows that $x\in C(\{x,y\})$ too. Thus, $C(\{x,y\})=\{x,y\}$. Since $A\supset \{x,y\}$ and Property $\beta$ holds, $C(\{x,y\})=\{x,y\}$ and $x\in C(A)$ implies $y\in C(A)$, a contradiction. Therefore, $b$-WARP is also satisfied.

(c) $\Rightarrow$ (d). Define the relation $\succsim$ on $X$ by $x\succsim y$ if $x\in C(\{x,y\})$. Since all binary menus are included in ${\mathcal {M}}$ and $C$ is nonempty-valued, $\succsim$ is complete. Suppose $x\succsim y$ and $y\succsim z$ and assume to the contrary that $x\not \succsim z$. Completeness implies $z\succ x$, i.e., $C(\{x,z\})=z$. By assumption, $A:=\{x,y,z\}\in {\mathcal {M}}$ and $C(A)\ne \emptyset$. It follows from a-WARP that $x\not \in C(A)$. Suppose $y\in C(A)$. Since $x\in A\!{\setminus }\!C(A)$ and $y\in C(A)$, it follows from b-WARP that $C(\{x,y\})=y$. This contradicts the postulate $x\succsim y$. Similarly, $z\not \in C(A)$. Thus, $x,y,z\not \in C(A)$, which is impossible. This establishes that $x\succsim z$ and therefore that $\succsim$ is also transitive, hence a weak order.

Now let $x\in C(A)$ and suppose there exists $y\in A$ such that $y\succ x$. Since $C(\{x,y\})=y$, $y\in A$ and $x\in C(A)$, this contradicts a-WARP. Thus, $x\succsim y$ for all $y\in A$. Conversely, suppose $x\succsim y$ for all $y\in A$ and suppose $x\not \in C(A)$. It holds that $z\in C(A)$ for some $x\ne z\in A$. Since $z\in C(A)$, $x\in A\!{\setminus }\!C(A)$ and $x\in C(\{x,z\})$ by assumption, b-WARP is contradicted. Thus, $\succsim$ rationalizes $C$. Finally, since all binary menus are included in ${\mathcal {M}}$, $\succsim$ is unique. $\square$

Proof of Proposition 2

It is straightforward that (a) $\Rightarrow$ (b), (b) $\Rightarrow$ (c) (cf Observation 1) and (d) $\Rightarrow$ (a). It will be shown that (c) $\Rightarrow$ (d). Suppose $C$ satisfies CIR and $a$-WARP. Define $\succ$ by $x\succ y$ if $C(\{x,y\})=x$. This is asymmetric by definition. From CIR and the full-domain assumption, it is also complete. Suppose $x\succ y$, $y\succ z$ and $x\nsucc z$. From CIR, $x\nsucc z$ implies $z\succ x$. Since $C$ is nonempty-valued, $C(\{x,y,z\})\ne \emptyset$. Suppose $x\in C(\{x,y,z\})$. Since $a$-WARP holds, this contradicts the postulate $z\succ x$. Therefore, $x\not \in C(\{x,y,z\})$. Similarly, $y,z\not \in C(\{x,y,z\})$. Thus, $C(\{x,y,z\})=\emptyset$, a contradiction. It follows then that $x\succ z$. Thus, $\succ$ is also transitive and hence a strict linear order.

Suppose $x\in C(A)$. It follows from $a$-WARP that $y\nsucc x$ for all $y\in A$. Now suppose the converse is true. From CIR this implies $x\succ y$ for all $y\in A\!{\setminus }\!\{x\}$. Suppose $x\not \in C(A)$. From nonempty-valuedness, $w\in C(A)$ for some $w\in A$. Since $x\succ w$ from above, and since $a$-WARP holds, this is a contradiction. Thus, $C(A)=\{x\in A: y\nsucc x$ for all $y\in A\}$. Finally, since all binary menus are included in ${\mathcal {M}}$, $\succ$ is unique. $\square$

Proof of Proposition 3

It is straightforward that (b) implies (a). Conversely, define $\succ$ by $x\succ y$ if $C(\{x,y\})=x$. Let $A\in {\mathcal {M}}$ and $x\in A$ be such that $y\nsucc x$ for all $y\in A$. From nonempty-valuedness, $x\in C(\{x,y\})$ for all $y\in A$. From Property $\gamma$, $x\in C(A)$. Now let $x\in C(A)$ and suppose, per contra, that $y\succ x$ for some $y\in A$. Since this is equivalent to $C(\{x,y\})=y$ and a-WARP holds by assumption, this is a contradiction. Finally, since all binary menus are included in ${\mathcal {M}}$, $\succ$ is unique. $\square$

Proof of Proposition 4

Suppose there exists an acyclic relation $\succ$ on $X$ such that (2) holds. Assume, per contra, that a-WARP is violated. There exist $x,y\in X$ and $A\in {\mathcal {M}}$ such that $C(\{x,y\})=x$, $x\in A$ and $y\in C(A)$. From $C(\{x,y\})=x$ and (2b) it follows that $x\succ y$. Since $y\in C(A)$ it also follows from (2a) or (2b) (if $C(A)=A$ or $C(A)\subset A$, respectively) that $x\nsucc y$. This is a contradiction.

Now let $A_1,\ldots ,A_k\in {\mathcal {M}}$ be such that $A_1\ne \{x\}$, $x=C(A_1)$ and $x\in C(A_i)$, $2\le i\le k$. Define $A:=\bigcup _{i=1}^kA_i$. It is implied by (2b) and $C(A_1)=x$ that $z\nsucc x$ for all $z\in A_1$ and $x\succ y$ for some $y\in A_1$. Moreover, $x\in C(A_i)$ and (2) imply $z\nsucc x$ for all $z\in A_i$. Thus, $z\nsucc x$ for all $z\in A$ and $x\succ y$ for some $y\in A$. From (2b), this implies $x\in C(A)$. Hence, Weak Property $\gamma$ is satisfied.

Finally, suppose $x\in C(A)$ and there exists $z\in A\!{\setminus }\!C(A)$. This implies $C(A)\subset A$. Hence, (2b) and $x\in C(A)$ implies $x\succ y$ and therefore $C(\{x,y\})=x$ for some $y\in A$. From (2b) and $x\succ y$, it also follows that $y\not \in C(A)$. Thus, Partial b-WARP is also satisfied.

Conversely, assume that $C$ is nonempty-valued and satisfies a-WARP, Weak Property $\gamma$ and Partial b-WARP. Define the asymmetric relation $\succ$ on $X$ by $x\succ y$ if $C(\{x,y\})=x$. Suppose $\succ$ is not acyclic. There exist $x_1,x_2,\ldots ,x_k\in X$ such that $x_1\succ x_2\succ \ldots \succ x_k\succ x_1$. By assumption, $B:=\{x_1,x_2,\ldots ,x_k\}\in {\mathcal {M}}$, while $C(B)\ne \emptyset$ is also true by assumption. Thus, $x_i\in C(B)$ for some $i\le k$. From the $\succ$-cycle above, it follows that there is $x_j\in B$ such that $x_j\succ x_i$, i.e., $C(\{x_i,x_j\})=x_j$. This contradicts a-WARP.

Suppose $A\in {\mathcal {M}}$ is such that $x\nsucc y$ and $y\nsucc x$ for all $x,y\in A$. Let $x\in C(A)$ and $C(A)\subset A$. If $|A|=2$, then $A=\{x,y\}$ for some $y\in X$, and $x\succ y$ obviously holds, a contradiction. Suppose $|A|>2$. From Partial b-WARP, there exists $y\in A\!{\setminus }\!C(A)$ such that $x\succ y$, which is a contradiction too. It follows, therefore, that $C(A)=A$. In the other direction, let $C(A)=A$ and suppose $x\succ y$ for some $x,y\in A$. Since $y\in C(A)$ by assumption, this is a violation of a-WARP. This establishes (2a).

To establish (2b), let $C(A)\subset A$ for some $A\in {\mathcal {M}}$. Suppose $x\in C(A)$. If $z\succ x$ for some $z\in A$, then a-WARP is violated. Moreover, $x\in C(A)$, $C(A)\subset A$ and Partial b-WARP together imply $x\succ y$ for some $y\in A\!{\setminus }\!C(A)$. Thus, $x\in C(A)$ and $C(A)\subset A$ implies $z\nsucc x$ for all $z\in A$ and $x\succ y$ for some $y\in A$. In the other direction, suppose there is $x\in A$ such that $z\nsucc x$ for all $z\in A$ and $x\succ y$ for some $y\in A$. This implies $C(\{x,y\})=x$ for some $y\in A$ and, since $C$ is nonempty-valued, $x\in C(\{x,z\})$ for all $z\in A$. Let $F:=\{x,y\}$ and label all other elements in $A$ by $z_1,\ldots ,z_k$. Also, let $G_i:=\{x,z_i\}$. Since $x=C(F)$, $x\in \bigcap _{i=1}^kC(G_i)$ and $\bigcup _{i=1}^kG_i\cup F=A$, it follows from Weak Property $\gamma$ that $x\in C(A)$. Finally, suppose $C(A)$ consists of all $x\in A$ with the above two properties and assume to the contrary that $C(A)=A$. Since there exist $x,y\in A$ such that $C(\{x,y\})=x$ by assumption, while $y\in C(A)=A$ also holds by assumption, a-WARP is violated. Hence, $C(A)\subset A$. As above, uniqueness is straightforward. $\square$

Proof of Proposition 5

It is obviously true that (a) always implies (b). To show that the converse implication holds under the stated conditions fix an arbitrary integer $n>2$ and assume that $x\in A$ implies $y\not \in C(A)$ for all $A\in {\mathcal {N}}$. Then, ${\mathcal {N}}$ includes all subsets with $n$ elements and there are $n-2$ distinct alternatives that are incomparable to everything in $X$. Let $S$ be the set of all these $n-2$ alternatives. Also, let $T=S\cup \{x,y\}$. Since $T$ has $n$ elements, it follows that $T\in {\mathcal {N}}$. Suppose $w\in C(T)$ for some $w\in S$. Since $w$ is universally incomparable and $C$ is PD, the above can happen if and only if $C(T)=T$. In this case, $y\in C(T)$ also holds, which contradicts (b). Thus, $w\not \in C(T)$ for all $w\in S$. Moreover, since $x\in T$, it follows from (b) that $y\not \in C(T)$. Hence, $C(T)=x$. Since $C(T)\subset T$, it follows from (2b) that $x\succ z$ for some $z\in T$. Since $x\nsucc w$ for all $w\in S$ by assumption, it follows that $x\succ z$ for some $z\in T\!{\setminus }\! {S}=\{x,y\}$. Since $\succ$ is irreflexive, this implies $x\succ y$. $\square$

Proof of Observation 3

Suppose first that $w,x,y,z$ are such that $w\succ x$, $y\succ z$ and no comparison is possible in all other pairs. It holds that $C(\{w,x,y,z\})=\{w,y\}$ and $C(\{w,y,z\})=y$, in violation of Property $\alpha$. If $x\succ z$ is also true, then $C(\{x,y,z\})=\{x,y\}$ and $C(\{w,x,y,z\})=\{w,y\}$, in violation of Property $\beta$. These preferences finally lead to the PD choices $C(\{w,z\})=\{w,z\}$, $C(\{w,y\})=\{w,y\}$ and $C(\{w,y,z\})=y$, in violation of Property $\gamma$.

Next, let $A=\{x,y,z\}$, $B=\{x,z\}$ and suppose $x\succ y$ and that all other options are incomparable. Then, $C(A)=x$, $\{x\}\subset B\subset A$ and $C(B)=\{x,z\}$, but $z\not \in C(A)=x$, showing that Aizerman’s axiom is not satisfied. Now suppose $A=\{x,y,w,z\}$, $B=\{x,y,w\}$, $w\succ y$, $x\succ y$, $x\succ z$, $z\succ w$ and all other pairs consist of incomparable options. These preferences imply $C(\{x,y\})=x=C(A)$ but $C(B)=\{x,w\}$, in violation of Weak WARP. Next, consider $A=\{w,x,y\}$, $B=\{x,y,z\}$, $x\succ w$, $y\succ z$ and let all other alternatives in $A,B$ be incomparable. It holds that $C(A)=x$ and $C(B)=y$, in violation of Weakened WARP.

To verify that Reference Consistency is not satisfied in general, suppose $x\succ z$, $w\succ y$ and that incomparability applies to all other pairs of options derived from these four. It holds that $C(\{w,x,y,z\})=\{w,x\}$. Moreover, $C(\{w,z\})=\{w,z\}$ and $C(\{x,y\})=\{x,y\}$, the menu $\{w,x,y,z\}$ is $C$-awkward, the menus $\{x,y\}$ and $\{w,z\}$ form a $C$-cover of $\{w,x,y,z\}$, and $C(\{w,x,y,z\})\cap \{w,z\}\ne C(\{w,z\})$, $C(\{w,x,y,z\})\cap \{x,y\}\ne C(\{x,y\})$.

Now suppose $x,y\in C(A)$. Since this implies $x\nsucc y$ and $y\nsucc x$, it follows that $C(\{x,y\})=\{x,y\}$. Therefore, a PD $C$ satisfies Rationality of Indifference. Finally, to see that such a $C$ also conforms with ERR notice that $w\in C(A\cup \{x\})$ and $w\in C(A\cup \{y\})$ implies $z\nsucc w$ for all $z\in A\cup \{x,y\}$ and $w\succ z',z''$ for some $z'\in A\cup \{x\}$ and $z''\in A\cup \{y\}$. Hence, $w\succ z$ for some $z\in A\cup \{x,y\}$, which ensures that $x\in C(A\cup \{x,y\})$. $\square$

Appendix 2: Axiom independence

The tightness of the axiomatic system of Proposition 4 is established here by means of counterexamples. Let $X=\{w,x,y,z\}$ and ${\mathcal {M}}=\{A:A\ne \emptyset ,\; A\subseteq X\}$. Each of the following cases provides an example where choices satisfy all but one axiom (singletons are ignored for brevity):

Not a-WARP

$$\begin{aligned} C(\{w,x\})= & {} w,\;\; C(\{w,y\})=\{w,y\},\;\; C(\{w,z\})=\{w,z\}, \\ C(\{x,y\})= & {} \{x,y\},\;\; C(\{x,z\})=x,\;\; C(\{y,z\})=y \\ C(\{w,x,y\})= & {} C(\{w,y,z\})=\{w,y\},\;\; C(\{w,x,z\})=\{w,x\},\;\; C(\{x,y,z\})=\{x,y\} \\ C(X)= & {} \{w,x,y\} \end{aligned}$$

Not Weak Property $\gamma$

$$\begin{aligned} C(\{w,x\})= & {} w,\;\; C(\{w,y\})=\{w,y\},\;\; C(\{w,z\})=\{w,z\}, \\ C(\{x,y\})= & {} \{x,y\},\;\; C(\{x,z\})=x,\;\; C(\{y,z\})=y \\ C(\{w,x,y\})= & {} C(\{w,y,z\})=C(\{w,x,z\})=w,\;\; C(\{x,y,z\})=y \\ C(X)= & {} w \end{aligned}$$

Not Partial b-WARP

$$\begin{aligned} C(\{w,x\})= & {} w,\;\; C(\{w,y\})=\{w,y\},\;\; C(\{w,z\})=\{w,z\}, \\ C(\{x,y\})= & {} \{x,y\},\;\; C(\{x,z\})=x,\;\; C(\{y,z\})=\{y,z\} \\ C(\{w,x,y\})= & {} C(\{w,y,z\})=\{w,y\},\;\; C(\{w,x,z\})=w,\;\; C(\{x,y,z\})=y \\ C(X)= & {} \{w,y\} \end{aligned}$$