Mechanism design with two alternatives in quasi-linear environments

Abstract

We study mechanism design in quasi-linear private values environments when there are two alternatives. We show that under a mild range condition, every implementable allocation rule is a generalized utility function maximizer. In unbounded domains, if we replace our range condition by an independence condition, then every implementable allocation rule is an affine maximizer. Our results extend Roberts’ affine maximizer theorem (Roberts, In: Laffont J-J (ed) The characterization of implementable choice rules, 1979) to the case of two alternatives.

Appendix

1.1 Proofs of Theorems 2 and 3

We prove Theorems 2 and 3 in this section. Before we do so, we comment on the methodology of the proof. The proof methodology is based on an ordering based approach of Mishra and Sen (2012) (M&S from now on). M&S provide an alternate proof of Roberts’ theorem when there are at least three alternatives. The general idea of their proof is to characterize weighted efficiency using neutrality and implementability. In the unrestricted domain, for every implementable allocation rule, there is another implementable allocation rule that satisfies neutrality. This new allocation rule can be obtained by translating the original allocation rule. One can then leverage the weighted efficiency characterization to get a characterization of affine maximization in the unrestricted domain.

Although, we employ this methodology, our proof is different in many aspects from M&S. This is mainly because we have two alternatives. Our characterization of weighted efficiency requires stronger condition than the neutrality condition of M&S. Further, our affine maximization characterization requires implementability and a new condition called independence, which M&S do not require if there are more than two alternatives.

1.2 Proof of Theorem 3

Like in M&S, we start by proving the characterization of weighted efficiency first, and then use this result to prove the affine maximizer characterization.

Fix an implementable allocation rule $f$. Consider the binary relation $R^f$ over $U$ defined by $x R^f y$ iff $a_{1} \in C^{f}(v)$, with $v(a_{1}) = x$ and $v(a_{2}) = y$. Let $P^f$ and $I^f$ respectively denote the asymmetric and symmetric part of $R^f$. They are well-behaved (in a sense made precise in Lemma 4) if $f$ satisfies a neutrality condition.

Definition 10

An allocation rule $f$ is neutral if for every pair of valuations $v,v' \in V$ such that $v(a_{1})=v'(a_{2})$ and $v(a_{2}) = v'(a_{1})$ we have

$$\begin{aligned} C^{f}(v) = \left\{ \begin{array}{ll} C^{f}(v') &{} \quad \text {if } \quad C^{f}(v) = A \\ A \setminus C^{f}(v) &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

The usual definition of neutrality will require that for every pair of valuations $v, v' \in V$ such that $v(a ) = v'(-a )$ and $v(-a ) = v'(a )$ with $v \ne v'$ we have $\{f(v')\} = A \setminus \{f(v)\}$. One can verify that this version of neutrality implies our version of neutrality if the allocation rule is implementable—see Mishra and Sen (2012) for a proof.

Lemma 4

Suppose $f$ is neutral and implementable. Then $R^{f}$ is reflexive and complete. Further, if $v(a_{1}) = x$ and $v(a_{2}) = y$, then

• $C^{f}(v) = \{a_{1}\}$ implies $xP^{f}y$ and $C^{f}(v) = \{a_{2}\}$ implies $yP^{f}x$, and

• $C^{f}(v) = A$ implies $xI^{f}y$.

Proof

$R^{f}$ is reflexive. For any $x \in U$, consider the valuation profile $v$ where $v(a_{1}) = v(a_{2}) = x$. Since $C^{f}(v)$ is non-empty and $f$ is neutral, $C^{f}(v) = A$. Hence, $x R^f x$.

$R^{f}$ is complete. For every $x, y \in U$, we can construct a valuation profile $v$ with $v(a_{1}) = x$ and $v(a_{2}) = y$. If $a_{1} \in C^{f}(v)$, then $x R^{f} y$. If $a_{1} \notin C^{f}(v)$, then $a_{2} \in C^{f}(v)$. Then, by neutrality, $a_{1} \in C^{f}(v')$, with $v'(a_{1}) = y$ and $v'(a_{2}) = x$. Therefore, $y R^{f} x$.

We now show that $C^{f}(v) = \{a_{1}\}$ implies $xP^{f}y$. Suppose $C^{f}(v) = \{a_{1}\}$. This clearly implies $xR^{f}y$. Assume for contradiction that we also have $y R^{f} x$. This implies that $a_{1} \in C^{f}(v')$, with $v'(a_{1}) = y$ and $v'(a_{2}) = x$. Then, by neutrality, $a_2 \in C^{f}(v)$, which gives us a contradiction.

A similar reasoning ensures that $C^{f}(v) = \{a_{2}\}$ implies $yP^{f}x$.

Finally, we show that $C^{f}(v) = A$ implies $x I^{f} y$. Suppose $C^{f}(v) = A$. This clearly implies $x R^{f} y$. Neutrality implies that $C^{f}(v') = A$, with $v'(a_{1}) = y$ and $v'(a_{2}) = x$. So, $y R^{f} x$, and hence, $x I^f y$. $\square$

Lemma 5

Suppose $f$ is an implementable and transitive allocation rule. Then, $f$ is unanimous if and only if it is neutral.

Proof

Suppose $f$ is neutral and implementable. Consider $x,y \in U$ such that $x_i > y_i$ for all $i \in N$. Then, due to neutrality, $C^f(v)=A$ if $v(a)=y$ for all $a \in A$. By PAD, $C^f(v')=\{a\}$ if $v'(a)=x$ and $v'(-a)=y$. Hence, $f$ is unanimous.

Now, suppose $f$ is unanimous and transitive. Assume for contradiction that $f$ is not neutral. Then, for some $x,y \in U$, we consider $v$ and $v'$ such that $v(a_1)=x=v'(a_2)$ and $v(a_2)=y=v'(a_1)$. We consider two cases.

Case 1. Assume for contradiction $C^f(v)=A$ but $C^f(v')=\{a_1\}$ (the argument does not change if $C^f(v')=\{a_2\}$). Since $a_2 \notin C^f(v')$, there is some $\epsilon \in \mathbb {R}^n_{++}$ such that $f(v'(a_1),v'(a_2)+2\epsilon )=a_1$. This implies that $C^f(v'(a_1),v'(a_2)+\epsilon )=\{a_1\}$. Choose $\epsilon ' \in \mathbb {R}^n_{++}$ such that $\epsilon '_i < \epsilon _i$ for all $i \in N$. Since $C^f(v)=A$, by PAD, $f(v(a_1)+\epsilon ,v(a_2)+\epsilon ')=a_1$. Moreover, by PAD, $C^f(v(a_1)+\epsilon , v(a_2)+\epsilon ')=\{a_1\}$. Now, consider the valuation profile $v''$ such that $v''(a_1)=v'(a_1)=y$ and $v''(a_2)=v(a_2)+\epsilon ' = y+\epsilon '$. By transitivity, $f(v'')=a_1$. But this contradicts the fact that $f$ is unanimous.

Case 2. Assume for contradiction $C^f(v)=\{a_1\}$ but $C^f(v') \ne \{a_2\}$ (the argument is unchanged if $C^f(v)=\{a_2\}$). If $C^f(v')=A$, then we can apply the argument in Case 1 to reach a contradiction (by interchanging the roles of $v$ and $v'$). Now, assume for contradiction $C^f(v')=\{a_1\}$. Since $a_2 \notin C^f(v)$, there is some sufficiently small $\epsilon \in \mathbb {R}^n_{++}$ such that $C^f(v(a_1),v(a_2)+\epsilon )=\{a_1\}$. Also, there is some $\epsilon '$ such that $\epsilon '_i < \epsilon _i$ for all $i \in N$ such that $f(v'(a_1)+\epsilon ,v'(a_2)+\epsilon ')=a_1$. Moreover, by PAD, $C^f(v'(a_1)+\epsilon ,v'(a_2)+\epsilon ')=\{a_1\}$. Consider a valuation profile $v''$ such that $v''(a_1)=v(a_1)=x$ and $v''(a_2)=v'(a_2)+\epsilon '=x+\epsilon$. By transitivity, $f(v'')=a_1$. But this contradicts the fact that $f$ is unanimous. $\square$

Finally, we show that if $f$ is implementable, transitive, and unanimous, then $R^f$ is transitive.

Lemma 6

If an implementable allocation rule $f$ is transitive and unanimous, then $R^f$ is an ordering.

Proof

By Lemmas 4 and 5, if $f$ is an implementable allocation rule that is transitive and unanimous, then $R^f$ is a well-behaved binary relation. We need to show that $R^f$ is transitive. We will show that $P^f$ and $I^f$ are each transitive, and this in turn will imply that $R^f$ is transitive.

$P^f$ is transitive. Consider $x,y,z \in U$ such that $x P^f y$ and $y P^f z$. Fix any $\epsilon \in \mathbb {R}^n_{++}$. By definition, if $v(a_1)=x$ and $v(a_2)=y$, then $f(v(a_1)+2\epsilon ,v(a_2)+\epsilon )=a_1$. Moreover, by PAD, $C^f(v(a_1)+2\epsilon ,v(a_2)+\epsilon )=\{a_1\}$. Similarly, if $v'(a_1)=y$ and $v'(a_2)=z$, then $C^f(v'(a_1)+\epsilon ,v'(a_2))=\{a_1\}$. Consider the valuation profile $v''$ such that $v''(a_1)=x$ and $v''(a_2)=z$. By transitivity, $f(v''(a_1)+2\epsilon ,v''(a_2))=a_1$. Hence, $a_1 \in C^f(v'')$.

Also, for some $\epsilon \in \mathbb {R}^n_{++}$, we have $C^f(v(a_1),v(a_2)+\epsilon )=\{a_1\}$ and for some $\epsilon ' \in \mathbb {R}^n_{++}$, we have $C^f(v'(a_1)+\epsilon , v'(a_2)+\epsilon ') = \{a_1\}$. Again, by transitivity, $f(v''(a_1),v''(a_2)+\epsilon ')=a_1$. Hence, $a_2 \notin C^f(v'')$. This shows that $x P^f z$.

$I^f$ is transitive. Consider $x,y,z \in U$ such that $x I^f y$ and $y I^f z$. Fix some $\epsilon \in \mathbb {R}^n_{++}$. By definition, if $v(a_1)=x$ and $v(a_2)=y$, then $C^f(v(a_1)+2\epsilon ,v(a_2)+\epsilon )=\{a_1\}$. Similarly, if $v'(a_1)=y$ and $v'(a_2)=z$, then $C^f(v'(a_1)+\epsilon ,v'(a_2))=\{a_1\}$. Consider the valuation profile $v''$ such that $v''(a_1)=x$ and $v''(a_2)=z$. By transitivity, $f(v''(a_1)+2\epsilon ,v''(a_2))=a_1$. Hence, $a_1 \in C^f(v'')$. A similar argument shows $a_2 \in C^f(v'')$. Hence, $x I^f z$. $\square$

An ordering $R$ on $U$ satisfies weak Pareto if for any $x, y \in U$ if $x_i > y_i$ for all $i \in N$, then $x P y$.

An ordering $R$ on $U$ satisfies translation invariance (tr-invariance) if for any $x,y \in U$ and $z \in \mathbb {R}^n$ such that $x+z,y+z \in U$, we have $x P y$ implies $(x+z) P (y+z)$ and $x I y$ implies $(x+z) I (y+z)$.

An ordering $R$ on $U$ satisfies continuity if for every $x \in U$, the sets $\{y \in U: x R y\}$ and $\{y \in U: y R x\}$ are closed in $U$.

Lemma 7

If $f$ is an implementable allocation rule such that $R^f$ is an ordering, then $R^f$ satisfies weak Pareto, tr-invariance, and continuity.

Proof

Since $f$ is unanimous, it is clear that $R^f$ satisfies weak Pareto.

Further, since $f$ satisfies PAD (by Lemma 3), $R^f$ satisfies tr-invariance. To see this, pick $x,y \in U$ and $z \in \mathbb {R}^n$ such that $x+z,y+z \in U$. Suppose $x P^f y$. Then, if $v(a_1)=x$ and $v(a_2)=y$ for every $\epsilon \in \mathbb {R}^n_{++}$, $f(v(a_1)+\epsilon ,v(a_2))=a_1$. Choose such an $\epsilon$. By PAD, for every $\epsilon ' \in \mathbb {R}^n_{++}$ such that $\epsilon '_i > \epsilon _i$ for all $i \in N$, we have $f(v(a_1)+z+\epsilon ',v(a_2)+z)=a_1$. Hence, $a_1 \in C^f(v(a_1)+z,v(a_2)+z)$. We also know that for some $\epsilon \in \mathbb {R}^n_{++}$, we have $f(v(a_1),v(a_2)+2\epsilon )=a_1$. By PAD, $f(v(a_1)+z,v(a_2)+z+\epsilon )=a_1$. Hence, $a_2 \notin C^f(v(a_1)+z,v(a_2)+z)$. This shows that $(x+z) P^f (y+z)$. A similar argument shows that $x I^f y$ implies $(x+z) I^f (y+z)$. Hence, $R^f$ satisfies tr-invariance.

We now show that $R^f$ satisfies continuity. To see this consider $x \in U$. We will first show that $\{y \in U: y R^f x\}$ is closed. Consider a sequence of points $\{x^k\}_k$ such that $x^k R^f x$ and the limit of this sequence is $z \in U$. Assume for contradiction that $x P^f z$. Hence, if $v(a_1)=x$ and $v(a_2)=z$, then $f(v(a_1),v(a_2)+\epsilon )=a_1$ for some $\epsilon \in \mathbb {R}^n_{++}$. Hence, $x R^f (z+\epsilon )$. Since the sequence converges to $z$, there is a point $z'$ in the sequence arbitrarily close to $z$ such that $z' R^f x$. Since $z'$ is arbitrarily close to $z$, we know that $(z+\epsilon ) P^f z'$. Hence, by transitivity of $R^f$, $(z+\epsilon ) P^f x$. This is a contradiction.

Next, we show that $\{y \in U: x R^f y\}$ is closed. Consider a sequence of points $\{x^k\}_k$ such that $x R^f x^k$ and the limit of this sequence is $z \in U$. Assume for contradiction that $z P^f x$. Interchanging the role of $x$ and $z$ in the previous argument, we get that $z R^f (x+\epsilon )$ for some $\epsilon \in \mathbb {R}^n_{++}$. Since the sequence converges to $z$, there is a point in the sequence $z'$ arbitrarily close to $z$ such that $x R^f z'$. Since $z'$ is arbitrarily close to $z$, by weak Pareto, $(x+\epsilon ) P^f z'$. This is a contradiction. $\square$

Proof of Theorem 3

Proof

Suppose $f$ is an implementable allocation rule that is unanimous and transitive. By Lemmas 6 and 7, the relation $R^f$ is an ordering on $U$ satisfying weak Pareto, tr-invariance, and continuity. Since $U$ is open and convex, by Mishra and Sen (2012), there exists $\lambda _1,\ldots ,\lambda _n \ge 0$ with $\lambda _i > 0$ for some $i \in N$, such that for every $x,y \in U$, $x R^f y$ if and only if $\sum _{i \in N}\lambda _i x_i \ge \sum _{i \in N}\lambda _i y_i$.

Now, consider any valuation profile $v$. Since $f(v) \in C^f(v)$, we know that $v(f(v)) R^f v(a)$ for all $a \in A$. Hence, $\sum _{i \in N}\lambda _i v_i(f(v)) \ge \sum _{i \in N}\lambda _i v_i(a)$. So, $f$ is a weighted efficient allocation rule.

Clearly, a weighted efficient allocation rule is transitive and unanimous. It is well known that if a weighted efficient allocation rule satisfies UIA, then it is implementable (Mishra and Sen 2012). $\square$

1.3 Proof of Theorem 2

We now use Theorem 3 to give a proof of Theorem 2. Before, we go into the details of the proof, we highlight the richness assumption of our domain. We assume that for every $i \in N$, the range of values for every alternative lies in an open interval $L_i$, which is unbounded from above. This implies that for every $i \in N$, $D_i=\mathbb {R}$,⁴ a fact which we will use extensively in our proofs. Denote by $D=D_1 \times \cdots \times D_n$, and note that $D=\mathbb {R}^n$.

We will use the standard range condition of Roberts (1979) for the proof.

Definition 11

An allocation rule $f$ satisfies non-imposition if for every $a \in A$, there exists $v \in V$ such that $f(v)=a$.

Fix an implementable allocation rule $f$. Suppose $f$ satisfies independence. We first observe that the choice set only depends on differences of valuations.

Lemma 8

Suppose $f$ is implementable. Then, for every pair of valuation profiles, $v,v'$ such that $\partial v_i = \partial v'_i$ for all $i \in N$, we have $C^f(v)=C^f(v')$.

Proof

Choose $v,v'$ such that $\partial v_i = \partial v'_i$ for all $i \in N$. Pick $a \in C^f(v)$ and $\epsilon \in \mathbb {R}^n_{++}$. By definition, $f(v(a)+\frac{\epsilon }{2},v(-a))=a$. By PAD and using the fact that $\partial v_i = \partial v'_i$ for all $i \in N$, we have $f(v'(a)+\epsilon ,v'(-a))=a$. Hence, $a \in C^f(v')$. Switching the role of $v$ and $v'$, we can show that if $a \in C^f(v')$, then $a \in C^f(v)$. As a result, $C^f(v)=C^f(v')$. $\square$

As a consequence of Lemma 8, we will define a mapping $c^f:D \rightarrow \{S \subseteq A: S \ne \emptyset \}$, such that for every $x \in D$, $c^f(x)=C^f(v)$, where $v$ is such that $\partial v_i = x_i$ for all $i \in N$.

Now, define $\kappa ^f$ as follows. For every $\alpha \in \mathbb {R}$, denote by $1_{\alpha }$ the vector in $\mathbb {R}^n$ such that each component of $1_{\alpha }$ has value $\alpha$. By our assumption on $D$, $1_0 \in D$. If $a_1 \in c^f(1_0)$, then let

$$\begin{aligned} \kappa ^f&= -\sup \{\alpha \in \mathbb {R}_+: a_1 \in c^f(1_{-\alpha })\}. \end{aligned}$$

If $a_1 \notin c^f(1_0)$, then let

$$\begin{aligned} \kappa ^f&= \inf \{\alpha \in \mathbb {R}_+: a_1 \in c^f(1_{\alpha })\}. \end{aligned}$$

Lemma 9

If $f$ is an implementable allocation rule satisfying non-imposition, then $\kappa ^f$ is a well defined real number.

Proof

Suppose $a_1\in c^f(1_0)$. By non-imposition (and using Lemma 8), we get that there is some $\beta \in \mathbb {R}$ such that $a_2 \in c^f(1_{-\beta })$. Since $a_1 \in c^f(1_0)$, by PAD, $\beta > \sup \{\alpha \in \mathbb {R}_+: a_1 \in c^f(1_{-\alpha })\} \ge 0$. This shows that $\kappa ^f$ exists since the set $\{\alpha \in \mathbb {R}_+: a_1 \in c^f(1_{-\alpha })\}$ is bounded. So, $\kappa ^f$ is a real number. A similar proof works if $a_1 \notin c^f(1_0)$. $\square$

The next lemma proves another property of $c^f$.

Lemma 10

If $f$ is an implementable allocation rule satisfying non-imposition, then $c^f(1_{\kappa ^f})=A$.

Proof

By our assumption on $D$, $1_{\kappa ^f} \in D$. First, we show that $a_1 \in c^f(1_{\kappa ^f})$. Assume for contradiction that $a_1 \notin c^f(1_{\kappa ^f})$. In that case, for all $v \in V$ with $\partial v_i = \kappa ^f$, we have $a_1 \notin C^f(v)$. This implies that there is some $\epsilon \in \mathbb {R}^n_{++}$ such that $f(v(a_1)+\epsilon ,v(a_2)) \ne a_1$. Hence, $a_1 \notin c^f(1_{\kappa ^f}+\frac{\epsilon }{2})$. But, by definition of $\kappa ^f$, for any $\epsilon ' \in \mathbb {R}^n_{++}$, $a_1 \in c^f(1_{\kappa ^f}+\epsilon ')$, and this is a contradiction.

Next, we show that $a_2 \in c^f(1_{\kappa ^f})$. Again, assume for contradiction that $a_2 \notin c^f(1_{\kappa ^f})$. As in the previous case, there is some $\epsilon \in \mathbb {R}^n_{++}$ and $v \in V$ such that $\partial v_i = \kappa ^f - \epsilon$ and $f(v) \ne a_2$. Hence, $a_2 \notin c^f(1_{\kappa ^f} - \frac{\epsilon }{2})$. But, by definition of $\kappa ^f$, for any $\epsilon ' \in \mathbb {R}^n_{++}$, $a_1 \notin c^f(1_{\kappa ^f}-\epsilon ')$. Since for any $\epsilon ' \in \mathbb {R}^n_{++}$, $c^f(1_{\kappa ^f}-\epsilon ')$ is non-empty, $a_2 \in c^f(1_{\kappa ^f}-\epsilon ')$. This is a contradiction. $\square$

Now, let $f$ be an implementable allocation rule satisfying non-imposition. Define a new allocation rule $\bar{f}$ as follows. For every $v \in V$, define the valuation profile $v^{tr}$ as follows: $\partial v_i^{tr} = \partial v_i + \kappa ^f$ for all $i \in N$. Note that by our assumption of $D$, $v^{tr} \in V$. Now, the allocation rule $\bar{f}$ is defined as:

$$\begin{aligned} \bar{f}(v)&= f(v^{tr}). \end{aligned}$$

We now establish an important lemma.

Lemma 11

If $f$ is an implementable allocation rule satisfying independence and non-imposition, then $\bar{f}$ is implementable, unanimous, and transitive.

Proof

Suppose $f$ is an implementable allocation rule satisfying independence and non-imposition. Let $(p_1,\ldots ,p_n)$ be the payments that implement $f$. For every $i \in N$ and for every $v_{-i}$, let $\bar{p}_i(v_i,v_{-i})=p_i(v^{tr}_i,v^{tr}_{-i}) - \kappa ^f$ if $f(v_i,v_{-i})=a_1$ and $\bar{p}_i(v_i,v_{-i})=p_i(v^{tr}_i,v^{tr}_{-i})$ if $f(v_i,v_{-i})=a_2$. We will show that $(\bar{p}_1,\ldots ,\bar{p}_n)$ implement $\bar{f}$. To see this, consider $i \in N$ and $v_{-i}$. Also, consider $v_i,v'_i$ such that $\bar{f}(v_i,v_{-i})=a_1$ and $\bar{f}(v'_i,v_{-i})=a_2$ (a similar proof works if $\bar{f}(v_i,v_{-i})=a_2$ and $\bar{f}(v'_i,v_{-i})=a_1$). Now,

$$\begin{aligned} v_i(a_1) - \bar{p}_i(v_i,v_{-i})&= v^{tr}_i(f(v^{tr}_i,v^{tr}_{-i})) - p_i(v^{tr}_i,v^{tr}_{-i}) \\&\ge v^{tr}_i(f(v'^{tr}_i,v^{tr}_{-i})) - p_i(v'^{tr}_i,v^{tr}_{-i}) \\&= v_i(\bar{f}(v'_i,v_{-i})) - \bar{p}_i(v'_i,v_{-i}). \end{aligned}$$

Hence, $(\bar{p}_1,\ldots ,\bar{p}_n)$ implement $\bar{f}$.

We show that $\bar{f}$ is unanimous. Consider a valuation profile $v$ such that $v(a_1)=x$, $v(a_2) = y$, and $x_i > y_i$ for all $i \in N$. We need to show that $\bar{f}(v)=a_1$. To see this, consider the valuation profile $v'$ such that $v'(a_1)=y=v'(a_2)$. But $c^{\bar{f}}(1_0)=c^f(1_{\kappa ^f})=A$. Hence, $C^{\bar{f}}(v')=A$, and using PAD, we get that $\bar{f}(v)=a_1$.

Finally, we show that $\bar{f}$ is transitive. For this, we consider $x,y,z \in D$ and $v,v',v''$ such that $v(a_1)=x=v''(a_1)$, $v(a_2)=y=v'(a_1)$, and $v'(a_2)=z=v''(a_2)$.

Suppose $C^{\bar{f}}(v)=\{a_1\}$ and $C^{\bar{f}}(v')=\{a_1\}$. We will show that $\bar{f}(v'')=a_1$. Note that since $C^{\bar{f}}(v')=\{a_1\}$, there is some $\epsilon \in \mathbb {R}^n_{++}$ such that $\bar{f}(v'(a_1)-\epsilon ,v'(a_2))=a_1$. To see this, suppose for all $\epsilon \in \mathbb {R}^n_{++}$, we have $\bar{f}(v'(a_1)-\epsilon ,v'(a_2))=a_2$. We know that for some $\epsilon ' \in \mathbb {R}^n_{++}$, we have $\bar{f}(v'(a_1),v'(a_2)+\epsilon ')=a_1$ (since $a_2 \notin C^{\bar{f}}(v')$). By PAD, $\bar{f}(v'(a_1)-\frac{\epsilon '}{2},v'(a_2))=a_1$. This is a contradiction. Similarly, there is an $\epsilon ' \in \mathbb {R}^n_{++}$ such that $f(v(a_1)-\epsilon ',v(a_2))=a_1$.

Now, choose an $\epsilon '' \in \mathbb {R}^n_{++}$ such that $\bar{f}(v'(a_1)-\epsilon '',v'(a_2))=a_1$ and $\bar{f}(v(a_1)-\frac{\epsilon ''}{2},v(a_2))=a_1$—note that such an $\epsilon ''$ can be chosen. In that case, by independence, either $\bar{f}(v(a_1),v'(a_2))=a_1$ or $\bar{f}(v'(a_1)-\frac{\epsilon ''}{2},v(a_2))=a_1$. Since $v'(a_1)=v(a_2)=y$ and $\bar{f}$ is unanimous, the latter is not possible. Hence, $\bar{f}(v'')=\bar{f}(v(a_1),v'(a_2))=a_1$.

A similar argument shows if $C^{\bar{f}}(v)=\{a_2\}$ and $C^{\bar{f}}(v')=\{a_2\}$, then $\bar{f}(v'')=a_2$. $\square$

This leads to the proof of Theorem 2.

Proof of Theorem 2

Proof

Suppose $f$ is an implementable allocation rule. If $f$ does not satisfy non-imposition, then clearly it is an affine maximizer. Now, suppose $f$ satisfies non-imposition and independence. Then, by Lemma 11, $\bar{f}$ is an implementable allocation rule which is unanimous and transitive. By Theorem 3, there exists non-negative weights $\lambda _1,\ldots ,\lambda _n$ such that for all $v$, if $\sum _{i \in N}\lambda _i \partial v_i > 0$, then $\bar{f}(v)=a_1$ and if $\sum _{i \in N}\lambda _i \partial v_i < 0$, then $\bar{f}(v)=a_2$. Furthermore, we can choose these weights, without loss of generality, such that $\sum _{i \in N}\lambda _i=1$.

Now, using the definition of $\bar{f}$, we get that if $\sum _{i \in N}\lambda _i \partial v_i > \kappa ^f$, then $f(v)=a_1$ and if $\sum _{i \in N}\lambda _i \partial v_i < \kappa ^f$, then $f(v)=a_2$. Setting $\gamma (a_1)=\kappa ^f$ and $\gamma (a_2)=0$, we get that $f$ is an affine maximizer.

For the converse, Lemma 2 shows that an affine maximizer satisfies independence. It is well known that an affine maximizer is implementable by generalized Groves payments if it satisfies UIA. $\square$

1.4 Independence of Axioms used in Theorem 1

We give three examples below to illustrate the requirement of the conditions in Theorem 1. The three examples below refer to the version of Theorem 1 with condition Ca, but can be easily adapted for the version with condition Cb.

Example 3

This example illustrates that there are implementable affine maximizers when $D_i$s are interval which violate agent sovereignty. Hence, the agent sovereignty condition is necessary for our characterization.

Let $D_{i}=[0,1] \ \forall i \in N$ with $n>2$ and $f(v) = \arg \max _{a \in A} \sum _{i \in N} v_{i}(a)$, with ties broken in favor of $a_1$. This allocation rule violates agent sovereignty. Indeed, fix $i \in N$ and note that if $v_{j}(a_1)=0, v_{j}(a_2)=1$ for all $j \ne i$, then $\not \exists ~v_{i} : f(v_i,v_{-i})=a_1$. However, $f$ is clearly monotone and, hence, implementable.

Example 4

This example illustrates that there are implementable allocation rules satisfying agent sovereignty when $D_i$s are not intervals. Hence, the requirement that $D_i$s are intervals is necessary for our characterization.

Let $D_{i}= \mathbb {N}~\forall i \in N$ with $n>2$, where $\mathbb {N}$ is the set of non-negative integers. Let $f(v) = \arg \max _{a \in A} \sum _{i \in N} v_{i}(a)$, with ties broken in favor of $a_1$. This allocation rule is clearly monotone, and hence, implementable. It also satisfies agent sovereignty. But the domain is not an interval.

Example 5

The following example illustrates that there are non-implementable allocation rules satisfying agent sovereignty in domains where $D_i$s are intervals. Hence, agent sovereignty does not imply implementability in such domains.

Let $D_{i}=(0,\infty ) \ \forall i \in N$ with $n>2$ and $f(v) = \arg \max _{a \in A} \Pi _{i \in N} v_{i}(a)$, with ties broken in favor of $a_1$. Allocation rule $f$ satisfies agent sovereignty because the domain is unbounded above. Monotonicity (and hence Implementability) is violated. Indeed, suppose $n=2$. Fix $v_2(a_1)=9,v_2(a_2)=1$. Consider $v_1(a_1)=0.1,v_1(a_2)=1$ and $v'_1(a_1)=1,v'_1(a_2)=2$. Note that $f(v_1,v_2)=a_2$ and $v_1(a_2)-v_1(a_1) = 0.9 < 1 = v'_1(a_2) - v'_1(a_1)$. However, $f(v'_1,v_2)=a_2$, violating monotonicity.