Abstract
This paper proposes and axiomatically characterizes a model of choice that builds on the criterion of partial dominance and allows for two types of avoidant behavior: choice deferral and status quo bias. These phenomena are explained in a unified way that allows for a clear theoretical distinction between them to be made. The model also explains the strengthening of the attraction effect that has been observed when deferral is permissible. Unlike other models of status quo biased behavior, the one analyzed in this paper builds on a unique, reference-independent preference relation that is acyclic and generally incomplete. When this relation is complete, the model reduces to rational choice.
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Notes
The object \(\diamond \) is defined as something not belonging to X in Masatlioglu and Ok (2005). This minor difference in the two definitions is unimportant mathematically.
Other interpretations have been adopted in the literature. For instance, Masatlioglu and Ok (2015) call \((A,\diamond )\) a problem without a status quo and (A, s) a problem with a status quo.
This is so because the relation \(\succ \) is acyclic. Thus, if (1a) is not satisfied in \((A,\diamond )\), it must be because \(x\nsucc y\) for all \(x,y\in A\).
Property \(\gamma \) is also commonly referred to as “Expansion” in the literature.
Evidence for this effect was first reported in Iyengar and Lepper (2000).
However, Ok et al. (2015) do analyze a model of choice in which certain feasible alternatives act as endogenous reference points that influence behavior by restricting the set of options over which the agent is maximizing his utility function.
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Acknowledgments
I thank Mark Dean, Robert Evans, Yusufcan Masatlioglu, Paola Manzini, Marco Mariotti, John Quah, Ran Spiegler, Jörg Stoye, audiences at Cambridge, Amsterdam, Queen Mary-London, St Andrews, EBIM 2010 (Bielefeld), EEA-ESEM 2011 (Oslo), SIRE-BIC 2012 (Edinburgh), and especially two referees of this journal for their very useful comments. Some of the material presented in this paper first appeared in a preliminary form in working papers titled “Rational indecisive choice” (2010) and “Asymmetric dominance, deferral and status quo bias in a model of choice with incomplete preferences” (2012). Partial financial assistance from the Scottish Institute for Research in Economics is gratefully acknowledged.
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Appendices
Appendix A: Proofs
Proof of Proposition 1
The proofs that the axioms are necessary and that the preference relation \(\succ \) in the EPD model is unique are straightforward and omitted. For sufficiency, define \(\succ \) by \(x\succ y\) if \(C(\{x,y\},\diamond )=x\). Suppose \(x_1\succ x_2,\ldots ,x_k\succ x_1\) for some \(x_1,\ldots ,x_k\in X\). Let \(A:=\{x_1,\ldots ,x_k\}\). Without loss of generality, consider \((A,x_1)\in {\mathcal {S}}\). By assumption, \(C(A,x_1)\ne \emptyset \). Suppose \(x_1\in C(A,x_1)\). Then, \(x_k\succ x_1\) together with A1 implies \(C(\{x_1,x_k\},x_1)=x_k\), which violates A2. Suppose instead that \(x_j\in C(A,x_1)\) for some \(j\ne 1\). It follows from A1 that \(x_j\in C(A,\diamond )\). But since \(x_{j-1}\succ x_j\) holds by assumption, this also violates A2. Hence, \(x_j\not \in C(A,x_1)\) for all \(x_j\ne x_1\). It follows then that \(C(A,x_1)=\emptyset \), which is impossible by assumption. Hence, \(\succ \) is acyclic. Moreover, since all binary problems \((\{x,y\},\diamond )\) are included in \({\mathcal {S}}\), \(\succ \) is unique.
Consider now \((A,\diamond )\in {\mathcal {S}}\) and suppose \(C(A,\diamond )\ne \emptyset \). Let \(x\in C(A,\diamond )\). From A2, \(z\nsucc x\) for all \(z\in A\). From A5, \(x\succ y\) for some \(y\in A\). Moreover, if \(x\in A\) is such that \(x\succ y\) for some \(y\in A\) and \(z\nsucc x\) for all \(z\in A\), then applying A4 as above leads to \(x\in C(A,\diamond )\). It is obvious, finally, that \(C(A,\diamond )\ne \emptyset \) if \(C(A,\diamond )\) is as in (1a). Hence, (1a) holds.
Next, consider \((A,s)\in {\mathcal {S}}\). Suppose \(x\in C(A,s)\) for some \(x\ne s\). Assume to the contrary that there exists \(y\in A\) such that \(y\succ x\). It follows from \(x\in C(A,s)\), \(x\ne s\) and A1 that \(x\in C(A,\diamond )\). But since \(y\succ x\), this contradicts A2. Thus, \(x\in C(A,s)\) and \(x\ne s\) implies \(y\nsucc x\) for all \(y\in A\). Moreover, it follows from A3 that \(C(\{x,s\},s)\subset \{x,s\}\). Suppose \(x=C(\{x,s\},s)\). Then, A1 implies \(x\in C(\{x,s\},\diamond )\). From A3, \(x=C(\{x,s\},\diamond )\), or \(x\succ s\). Conversely, if C(A, s) consists of all \(x\in A\) such that \(y\nsucc x\) for all \(y\in A\) and \(x\succ s\), then A2 ensures that \(C(A,s)\ne s\). This establishes (1b).
For completeness, we will show how the axioms also imply (2) and (3). For the former, consider \((A,\diamond )\in {\mathcal {S}}\) and let \(x\nsucc y\) and \(y\nsucc x\) for all \(x,y\in A\). Assume, per contra, that \(w\in C(A,\diamond )\). From A3, \(C(A,\diamond )\subset A\). From A5, \(C(\{w,z\},\diamond )=w\) for some \(z\in A\setminus C(A,\diamond )\). Since \(w\nsucc z\) by assumption, this is a contradiction. Thus, \(C(A,\diamond )=\emptyset \). Conversely, let \(C(A,\diamond )=\emptyset \). Since \(\succ \) is acyclic, there exists \(x\in A\) such that \(z\nsucc x\) for all \(z\in A\). Suppose \(x\succ y\) for some \(y\in A\). Let \(A:=\{x,y,z_1,\ldots ,z_k\}\). Then, \(C(\{x,y\},\diamond )=x\) and either \(x\in C(\{x,z_i\},\diamond )\) or \(C(\{x,z_i\},\diamond )=\emptyset \) for all \(i\le k\). A4 implies \(x\in C(A,\diamond )\), a contradiction. This establishes (2).
Finally, to establish (3) suppose \((A,s)\in {\mathcal {S}}\) is such that \(y\nsucc s\) for all \(y\in A\) and assume to the contrary that \(x\in C(A,s)\) for some \(x\ne s\). It follows from above that \(x\succ s\), a contradiction. Since \(C(A,s)\ne \emptyset \), this implies \(s=C(A,s)\). Conversely, suppose \(C(A,s)=s\) and let \(x\succ s\) for some \(x\in A\). A1 implies \(C(\{x,s\},s)=x\). A2 then also implies \(s\not \in C(A,s)\), a contradiction. \(\square \)
Proof of Proposition 2
The proof that (b) implies (a) is straightforward and omitted. For the converse, define the asymmetric relation \(\succ \) on X as in the preceding proof. It follows from A0 and A3 that \(x\succ y\) or \(y\succ x\) for all \(x,y\in X\). Hence, \(\succ \) is complete. Suppose \(x\succ y\), \(y\succ z\) and \(x\nsucc z\). Completeness implies \(z\succ x\), i.e., \(C(\{x,z\},\diamond )=z\). By assumption, \((\{x,y,z\},\diamond )\in Z\). Moreover, A0 implies \(C(\{x,y,z\},\diamond )\ne \emptyset \). If \(C(\{x,y,z\},\diamond )=x\), then \(z\succ x\) violates A2. The cases where \(C(\{x,y,z\},\diamond )=y\) and \(C(\{x,y,z\},\diamond )=z\) are similarly ruled out. Thus, \(x\succ z\) and \(\succ \) is also transitive, hence a strict linear order. As above, \(\succ \) is unique.
Consider \((A,\diamond )\in Z\). Suppose that \(C(A,\diamond )=x\) and let y be the \(\succ \)-maximal element of \(\succ \) in A. Assume to the contrary that \(x\ne y\). Completeness of \(\succ \) implies \(y\succ x\). Since \(C(A,\diamond )=x\) and \(y\in A\) by assumption, this violates A2. Thus, \(x=y\).
Next, let \((A,s)\in Z\) and define x and y as above. Consider the case where \(y\ne s\) first. By way of contradiction, suppose \(x\ne y\). Completeness of \(\succ \) and the definition of y imply \(y\succ x\) and \(y\succ s\). Let \(x\ne s\). From A1, \(x= C(A,s)\) implies \(x= C(A,\diamond )\). Since \(y\succ x\) and \(y\in A\) by assumption, \(x= C(A,\diamond )\) violates A2. Thus, \(x=s\). In this case, it follows from A1 and \(C(\{x,y\},\diamond )=y\) that \(C(\{s,y\},s)=y\). Since \(y\in A\) and \(s=x= C(A,s)\) by assumption, this violates A2. Thus, \(x=y\).
Consider finally the case where \(y=s\). Suppose \(x\ne y\). From the definition of y \((=s)\) one has \(s\succ x\). From the definition of x one also has \(C(A,s)=x\). Since \(x\ne s=y\), it follows from A1 and \(x=C(A,s)\) that \(x=C(A,\diamond )\). In view of \(s\succ x\) and \(s\in A\), this violates A2. Thus, \(x=y\) and (1) is established for all \((A,p)\in Z\). \(\square \)
Appendix B: Axiom independence
Let \(X=\{w,x,y\}\) and let \({\mathcal {S}}\) be the union of the following collections of sets:
Each of the examples below presents a set of choices that conforms with all but one of the axioms of Proposition 1. Axiom A1 is broken into A1a and A1b to reflect the former and the latter condition in the statement of the axiom, respectively.
Not A1a
Not A1b
Not A2
Not A3
Not A4
Not A5
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Gerasimou, G. Asymmetric dominance, deferral, and status quo bias in a behavioral model of choice. Theory Decis 80, 295–312 (2016). https://doi.org/10.1007/s11238-015-9499-7
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DOI: https://doi.org/10.1007/s11238-015-9499-7