Efficient and strategy-proof social choice when preferences are single-dipped

Abstract

We study the problem of locating a single public good along a segment when agents have single-dipped preferences. We ask whether there are unanimous and strategy-proof rules for this model. The answer is positive and we characterize all such rules. We generalize our model to allow the set of alternatives to be unbounded. If the set of alternatives does not have a maximal and a minimal element, there is no meaningful notion of efficiency. However, we show that the range of every strategy-proof rule has a maximal and a minimal element. We then characterize all strategy-proof rules.

Appendices

Appendices

1.1 The Pareto-set

The following lemmas pertain to the nature of the Pareto set.

Lemma 14

(Domination by extremes) For each $R\in {\mathcal {R}}^N$ and each $x \notin P(R)$, either $0 \mathrel {\bar{P}} x$ or  $T \mathrel {\bar{P}} x$.

Proof

Let $R\in {\mathcal {R}}^N$. If $x\notin P(R)$, then there is $y\in [0,T]$ such that $y\mathrel {\bar{P}} x$. If $x < y$, then $y = \text {med}\{x,y,T\}$. Since $y \mathrel {\bar{P}} x$, $y \mathrel R_{N} x$. By single-dippedness of preferences, $T \mathrel P_N y$. Therefore, $T \mathrel {\bar{P}} x$.

If $x > y$, an analogous argument shows that $0 \mathrel {\bar{P}} x$. $\square$

Lemma 15

(Shape of the Pareto set) For each $R\in {\mathcal {R}}^N$, $P(R)$ has one of the following configurations:

• 1. $\{0\}$ if $n_0(R) > 0$ and $n_T(R) = 0$.

• 2. $\{T\}$ if $n_0(R) = 0$ and $n_T(R) > 0$.

• 3. $\{0,T\}\cup ({\underline{x}}(R), {\overline{x}}(R))$ if $n_0(R) > 0$, $n_T(R) > 0$, ${\underline{x}}(R) \in L(R_{{\underline{i}}(R)},0)$, ${\overline{x}}(R) \in L(R_{{\overline{I}}(R)},0)$, and ${\underline{x}}(R) < {\overline{x}} (R)$

• 4. $\{0,T\}\cup [{\underline{x}}(R), {\overline{x}}(R))$ if $n_0(R) > 0$, $n_T(R) > 0$, ${\underline{x}}(R) \notin L(R_{\underline{i}(R)},0)$, ${\overline{x}}(R) \in L(R_{{\overline{I}}(R)},0)$, and ${\underline{x}}(R) < {\overline{x}} (R)$

• 5. $\{0,T\}\cup [{\underline{x}}(R), {\overline{x}}(R)]$ if $n_0(R) > 0$, $n_T(R) > 0$, ${\underline{x}}(R) \notin L(R_{\underline{i}(R)},0)$, ${\overline{x}}(R) \notin L(R_{{\overline{I}}(R)},0)$, and ${\underline{x}}(R) < {\overline{x}} (R)$

• 6. $\{0,T\}\cup ({\underline{x}}(R), {\overline{x}}(R)]$ if $n_0(R) > 0$, $n_T(R) > 0$, ${\underline{x}}(R) \in L(R_{{\underline{i}}(R)},0)$, ${\overline{x}}(R) \notin L(R_{{\overline{I}}(R)},0)$, and $\underline{x}(R) < {\overline{x}} (R)$

• 7. $\{0,T\}$ if $n_0(R) > 0$, $n_T(R) > 0$, and $\underline{x}(R) > {\overline{x}} (R)$, or $N_{0T} = N$.

See Fig. 6.

Fig. 6

Five profiles of preferences showing the Pareto sets described in cases 3 through 7 of Lemma 15

Proof

Case 1: :

$n_0(R) > 0, n_T(R) = 0$. For each $x\in (0,T], 0 \mathrel {\bar{P}} x$. Thus, $P(R) = \{0\}$.

Case 2: :

$n_T(R) > 0, n_0(R) = 0$. For each $x\in [0,T), T \mathrel {\bar{P}} x$. Thus, $P(R) = \{T\}$.

Case 3: :

$N_{0T} = N$. For each $x\in (0,T)$, $0\mathrel {\bar{P}} x$ and $0\mathrel I_N T$. Thus, $P(R) = \{0,T\}$.

Case 4: :

$n_0(R) > 0, n_T(R) > 0$. By definition of ${\underline{i}}(R)$ and ${\overline{I}}(R)$, $d_{{\underline{i}}(R)} \le {\underline{x}}(R)$ and $d_{{\overline{I}}(R)} \le {\overline{x}}(R)$.

Step 1: For each $y\in (0,{\underline{x}}(R))$, $0 \mathrel {\bar{P}} y$. First, $0 \mathrel P_{N_0(R)} T$. Thus, $0 \mathrel P_{N_0(R)} y \equiv \text {med}\{0,y,T\}$. Second, $0 \mathrel P_{N_{0T}(R)} y$. Finally, $0 {\mathrel {R}}_{N_T(R)} y$. Suppose not. Then there is $j\in N_T(R)$ such that $y \mathrel P_j 0$. By definition ${\underline{x}}(R)= \displaystyle \min _{i\in N_T(R)}\{\max ({\overline{L}}(R_i, 0)\}.$ Thus, for each $i \in N_T(R)$, $y\in {\overline{L}}(R_i, 0)$. Thus for each $i\in N_T(R)$, $0 \mathrel P_i y$.

Step 2: For each $y\in ({\overline{x}}(R), T)$, $T \mathrel {\bar{P}} y$. The proof is analogous to that of Step 1.

Step 3: If ${\underline{x}}(R) < {\overline{x}}(R)$, then $({\underline{x}}(R),{\overline{x}}(R))\subset P(R)$. Let $y \in ({\underline{x}}(R),{\overline{x}}(R))$. Then $y \mathrel P_{\overline{i}(R)} T$ and $y \mathrel P_{{\underline{i}}(R)} 0$. Thus, neither $0 \mathrel {\bar{P}} y$ nor $T\mathrel {\bar{P}} y$. By domination by extremes, $y\in P(R)$.

Step 4: $\{0,T\} \subseteq P(R)$. For each $i \in N_0(R),$ $\tau ([0,T], R_i) = \{0\}$ and, for each $j\in N_T(R), \tau ([0,T], R_j) = \{T\}$. Thus, $\{0,T\} \subseteq P(R)$.

Step 5: If ${\underline{x}}(R) < {\overline{x}}(R)$, and ${\underline{x}}(R) \notin L(R_{{\underline{i}}(R)}, 0)$ then $\underline{x}(R) \in P(R)$. Since ${\underline{x}}(R) \notin L(R_{\underline{i}(R)},0)$, ${\underline{x}}(R) \mathrel P_{{\underline{i}}(R)} 0$. Since ${\underline{x}}(R) < {\overline{x}}(R)$, ${\underline{x}}(R) \mathrel P_{{\overline{I}}(R)} T$. By domination by extremes, $\underline{x}(R) \in P(R)$.

Step 6: If ${\underline{x}}(R) < {\overline{x}}(R)$, and ${\overline{x}}(R) \notin L(R_{{\overline{I}}(R)}, 0)$ then $\overline{x}(R) \in P(R)$. The proof is analogous to that of Step 5. $\square$