Where should your daughter go to college? An axiomatic analysis

Abstract

We consider the problem of choosing a location in an interval so as to best take into account the preferences of two agents who will use this location. One has single-peaked preferences and the other single-dipped preferences. The most preferred location for the agent with single-peaked preferences is known and it is the least preferred location for the agent with single-dipped preferences. We show that the only efficient and strategy-proof rules are dictatorial.

Appendix

This appendix is devoted to the straightforward but tedious proofof Lemma 1.

Fig. 3

Step 1: Identifying the Pareto set in Case (i) of Lemma 1

Case (i): (Fig. 3). We first consider the location 0. We have already noted that $e_1(0)$ exists. We have $U^*_1(0)=]0,e_1(0)[$ and $L^*_1(0)=]e_1(0), C]$. Also, $U^*_2(0)=\emptyset$; if $0 \mathrel {P_2} C$, $L^*_2(0)=]0, C]$, and if $0 \mathrel {I_2} C$, $L^*_2(0)=]0, C[$. Making the father better off implies making the daughter worse off. The daughter cannot be made better off. Thus $0\in E(R)$.

Let $a\in ]0,h]$. Note that $e_1(a)$ exists. We have $U^*_1(a)=]a, e_1(a)[$ and $L^*_1(a)=]0, a]\cup ]e_1(0), C]$. If $e_2(a)$ does not exist, then $U^*_2(a)=[0,a[$ and $L^*_2(a)=]a,C]$. If $e_2(a)$ exists and is equal to C, then $U^*_2(a)=[0,a[$ and $L^*_2(a)=]a,C[$. Otherwise, as is the case in Fig. 3, $U^*_2(a)=[0,a[ \cup ]e_2(a), C]$ and $L^*_2(a)=]a,e_2(a)[$. A simple calculation, which we omit (and we will omit the similar calculations that are needed to settle the other cases) shows that $e_1(a) <e_2(a)$. In any case, making either the father or the daughter better off implies making the other agent worse off, so $a \in E(R)$.

Let $b\in ]h, e_1(0)]$. Then $e_1(b)$ exists and $U^*_1(b)=]e_1(b), b[$. Also, $e_2(b)$ exists and $U^*_2(b)=[0,e_2(b)[\cup ]b,C]$. Note that $e_1(b) <e_2(b)$. Any point of $]e_1(b), e_2(b)[$ Pareto-dominates b, so $b \notin E(R)$.

Let $c\in ]e_1(0), C[$. Then $U^*_1(c)=[0,c[$. Also, $U^*_2(c)=[0,e_2(c)[\cup ]c,C]$. Any point of $[0, e_2(c)[$ Pareto-dominates c, so $c\notin E(R)$.

Finally $U^*_1(C)=[0,C[$. If $0 \mathrel {P_2} C$, $U^*_2(c)=[0,e_2(c)[$ and $L^*_2(C)=]e_2(C),C[$. Any point of $[0, e_2(c)[$ Pareto-dominates c, so $c\notin E(R)$. If $0 \mathrel {I_2} C$, $U^*_2(c)=\emptyset$ and $L^*_2(C)=]0,C[$. The location 0 Pareto-dominates C, so $C\notin E(R)$.

Fig. 4

Step 1: Identifying the Pareto set in Case (ii) of Lemma 1

Case (ii) (Fig. 4). Let $a\in [0,h[$. Then $e_1(a)$ exists. We have $U^*_1(a)=]a,e_1(a)[$ and if $a=0$, $L^*_1(a)=]e_1(0),C]$ and otherwise $L^*_1(a)=[0,a[\cup ]e_1(0),C]$. Also $e_2(a)$ exists. If $a=0$, $U^*_2(a)=]e_2(0),C]$ and $L^*_2(a)=]0, e_2(0)[$, and otherwise $U^*_2(a)=[0,a[\cup ]e_2(a), C[$ and again $L^*_2(a)=]a, e_2(a)[$. Note that $e_1(a)<e_2(a)$. Making either the father or the daughter better off implies making the other agent worse off, so $a \in E(R)$.

Let $b\in ]h, e_1(0)]$. Then $e_1(b)$ exists and $U^*_1(b)= ]e_1(b),b[$. Also, $e_2(b)$ exists and $e_2(0)>0$; $U^*_2(b)= [0, e_2(b)[\cup ]b,C]$. Thus, 0 Pareto-dominates b, so $b \notin E(R)$.

Let $c\in ]e_1(0), e_2(0)]$. Then $U^*_1(c)=[0,c[$. Also, if $c<e_2(0)$, $U^*_2(c)= [0, e_2(c)[\cup ]c,C]$, and if $c=e_2(0)$, $U^*_2(c)= [0, e_2(c)[$. Again, 0 Pareto-dominates c, so $c \notin E(R)$.

Let $d\in \, ]e_2(0), C[$. Then $U^*_1(d)=[0,d[$ and $L^*_1(d) = ]d,C]$. Also, $U^*_2(c)=]d,C]$; if $d=e_2(0)$, $L^*_2(d)=]0,e_2(0)[$ and if $d>e_2(0)$, $L^*_2(d)=[0,e_2(0)[$. Making either the father or the daughter better off implies making the other agent worse off, so $d\in E(R)$. (Note that $e_2(0)$ itself is Pareto dominated by 0; thus the Pareto set is not closed.)

Finally, $U^*_1(C)=[0, C[$ and $L^*_1(C)=\emptyset$. Also, $U^*_2(C)=\emptyset$ and $L^*_2(C)=[0,C[$. Making the father better off implies making the daughter worse off. The daughter cannot be made better off. Thus, $C\in E(R)$.

Case (iii). The proof that $E(R) =[0,C]$ is simply because both agents have exactly opposite preferences.

Case (iv) (Fig. 5). Let $a<h$. Then $e_1(a)$ exists and $U^*_1(a)=]a,e_1(a)[$. Also, $e_2(a)$ exists. If $a=0$, $U^*_2(a) = ]e_2(a), C]$ and if $a>0$, $U^*_2(a)=[0,a[\cup ]e_2(a),C]$. Note that $e_2(a)<e_1(a)$. Any point of $]e_2(a), e_1(a)[$ Pareto-dominates a, so $a \notin E(R)$.

Let $b \in ]h, e_2(0)]$. Then $e_1(b)$ exists. We have $U^*_1(b)=]e_1(b), b[$ and $L^*_1(b)=[0,e_2(b)[\cup ]b,C]$. Also, $e_2(b)$ exists. If $b=e_2(0)$, then $U^*_2(b)= ]e_2(b), C]$ and $L^*_2(b)=]0,b[$; if $b<e_2(b)$, $U^*_2(b)=[0,e_2(b)[ \cup ]b, C]$ and $L^*_2(b)=]e_2(b),b[$. Making either the father or the daughter better off implies making the other agent worse off, so $b \in E(R)$.

Let $c \in ]e_2(0), e_1(0)]$. Then $e_1(c)$ exists. If $c<e_1(0)$, then $U^*_1(c)=]e_1(c), c[$ and $L^*_1(c)= [0, e_1(c)[\cup ]c,C]$; if $c=e_1(0)$, then $U^*_1(c)=]0, e_1(0)[$ and $L^*_1(c)= [e_1(0),C]$. Also, $U^*_2(c)=]c, C]$ and $L^*_2(c)=[0,c[$. Thus making either the father or the daughter better off implies making the other agent worse off, so $c \in E(R)$.

Let $d \in ]e_1(0), C[$. Then $U^*_1(d)=[0, d[$ and $L^*_1(d)=]d,C]$. Also, $U^*_2(d)= ]d, C[$ and $L^*_2(d)=[0,d[$. Making either the father or the daughter better off implies making the other agent worse off, so $d \in E(R)$.

Finally, $U^*_1(C)=[0,C[$ and $L^*_1(C)=\emptyset$. Also, $U^*_2(C)=\emptyset$ and $L^*_1(C)=[0,C[$. Making the father better off implies making the daughter worse off. The daughter cannot be made better off. Thus, $C \in E(R)$.

$\square$

Fig. 5

Step 1: Identifying the Pareto set in Case (iv) of Lemma 1