Abstract
We consider the problem of allocating sets of objects to agents and collecting payments. Each agent has a preference relation over the set of pairs consisting of a set of objects and a payment. Preferences are not necessarily quasilinear. Nonquasilinear preferences describe environments where the wealth effect is nonnegligible: the payment level changes agents’ willingness to pay for swapping sets. We investigate the existence of efficient and strategyproof rules. A preference relation is unitdemand if given a payment level, for each set of objects, the most preferred one in the set is at least as good as the set itself; it is multidemand if given a payment level, when an agent receives an object, receiving some additional object(s) makes him better off. We show that if a domain contains enough variety of unitdemand preferences and at least one multidemand preference relation, and if there are more agents than objects, then no rule satisfies efficiency, strategyproofness, individual rationality, and no subsidy for losers on the domain.
Introduction
We consider an object assignment problem with money. Each agent receives a (possibly empty) set of objects and, possibly, pays money for the set. He has a preference relation over the set of pairs consisting of a set of objects and a payment. An allocation specifies how the objects are allocated and how much each agent pays. An (allocation) rule is a mapping from a class of admissible preference profiles, which we call a “domain,” to the set of allocations. An allocation is efficient if, without reducing the total payment, no other allocation makes all agents at least as well off and at least one agent better off. A rule is efficient if it always selects an efficient allocation. A rule is strategyproof if, for each agent, it is a weakly dominant strategy to report his true preferences. We investigate the existence of efficient and strategyproof rules.
Our model can be viewed as a multiobject auction model. Much of the literature on auction theory assumes preferences to be “quasilinear.” This means that the valuations over sets of objects are not affected by payment level. On the quasilinear domain, the socalled “VCG rules” (Vickrey 1961; Clarke 1971; Groves 1973) are efficient and strategyproof, and they are the only rules satisfying these properties (Holmström 1979).
As Marshall (1920) demonstrates, preferences are approximately quasilinear if payments are sufficiently low. However, in important applications of auction theory such as spectrum license allocation, house allocation, etc., prices are often equal to or exceed agents’ annual revenues. Excessive payments for objects may impair an agent’s ability to purchase complements for an effective use of the objects, and thus may influence the benefit the agent derives from the objects. Another reason why preferences may not be quasilinear is that an agent may need a loan to be able to pay high prices, and typically financial costs are nonlinear in borrowing.^{Footnote 1} \(^{\text {,}}\) ^{Footnote 2}
Another common assumption is the “unitdemand” property.^{Footnote 3} It says that given a payment level, for each set of objects, the most preferred one in the set is at least as good as the set itself. For unitdemand preferences, there exists the minimum Walrasian equilibrium price, which is lower than any other Walrasian equilibrium price. Thus, on the “unitdemand domain,” minimum price Walrasian (MPW) rules are welldefined. A minimum price Walrasian (MPW) rule always selects an allocation associated with the minimum price Walrasian equilibria for each preference profile. The MPW rules are strategyproof on the unitdemand domain (Demange and Gale 1985). It is straightforward to see that the MPW rules satisfy the following additional two properties: One is individual rationality, which says that each agent finds his assignment at least as desirable as getting no object and paying nothing. The other is no subsidy for losers, which says that the payment of an agent who receives no object is nonnegative. On the unitdemand domain, when there are more agents than objects, the MPW rules are the only rules satisfying efficiency, strategyproofness, individual rationality, and no subsidy for losers (Morimoto and Serizawa 2015).^{Footnote 4}
Although the unitdemand assumption is suitable in some important cases such as house allocation, etc., in many other cases, some agents may well wish to receive more than one object, and indeed, many authors have analyzed such situations.^{Footnote 5}
Now, a natural question arises. On a domain that is neither quasilinear nor unitdemand, do efficient and strategyproof rules exist? This is the question we address. To state our result, we need an additional property of preferences. A preference relation satisfies the multidemand property if given a payment level, when an agent receives an object, receiving some additional object(s) makes him better off. We show that when there are more agents than objects, on any domain that contains enough variety of unitdemand preferences and at least one multidemand preference relation, no rule satisfies efficiency, strategyproofness, individual rationality, and no subsidy for losers. In most impossibility results in the literature on strategyproofness, the incompatibility of a list of properties of rules is established on a fixed domain.^{Footnote 6} On the other hand, our result is more general in the sense that the incompatibility of our properties holds on any domain containing enough variety of unitdemand preferences and some multidemand preferences.
This article is organized as follows. In Sect. 2, we introduce the model and basic definitions. In Sect. 3, we introduce the unitdemand model and the richness condition. In Sect. 4, we define the minimum price Walrasian rule. In Sect. 5, we state our result and show the sketch of the proof. Sect. 6 concludes. All the proofs appear in the Appendix.
The model and definitions
There are \(n\ge 2\) agents and \(m\ge 2\) objects. We denote the set of agents by \(N\equiv \{1,\dots ,n\}\) and the set of objects by \(M\equiv \{1,\dots ,m\}\). Let \(\mathcal {M}\) be the power set of M. With abuse of notation, for each \(a\in M\), we may write a to mean \(\{a\}\). Each agent receives a subset of M and pays some amount of money. Thus, the agents’ common consumption set is \(\mathcal {M}\times \mathbb {R}\) and a generic (consumption) bundle for agent i is a pair \(z_i=(A_i,t_i)\in \mathcal {M}\times \mathbb {R}\). Let \(\varvec{0}\equiv (\emptyset ,0)\).
Each agent i has a complete and transitive preference relation \(R_i\) over \(\mathcal {M}\times \mathbb {R}\). Let \(P_i\) and \(I_i\) be the strict and indifference relations associated with \(R_i\). A typical class of preferences is denoted by \(\mathcal {R}\). We call \(\mathcal {R}^n\) a domain. The following are standard conditions of preferences.
Money monotonicity: For each \(A_i\in \mathcal {M}\) and each pair \(t_i,t'_i\in \mathbb {R}\) with \(t_i<t'_i\), \((A_i,t_i)\mathrel {P_i}(A_i,t'_i)\).
First object monotonicity: For each \((\{a\},t_i)\in \mathcal {M}\times \mathbb {R}\), \((\{a\},t_i)\mathrel {P_i}(\emptyset ,t_i)\).
Possibility of compensation: For each \((A_i,t_i)\in \mathcal {M}\times \mathbb {R}\) and each \(A'_i\in \mathcal {M}\), there are \(t'_i,t''_i\in \mathbb {R}\) such that \((A_i,t_i)\mathrel {R_i}(A'_i,t'_i)\) and \((A'_i,t''_i)\mathrel {R_i}(A_i,t_i)\).
Continuity: For each \(z_i\in \mathcal {M}\times \mathbb {R}\), the upper contour set at \(z_i\), \(UC_i(z_i)\equiv \{z'_i\in \mathcal {M}\times \mathbb {R}:z'_i\mathrel {R_i}z_i\}\), and the lower contour set at \(z_i\), \(LC_i(z_i)\equiv \{z'_i\in \mathcal {M}\times \mathbb {R}:z_i\mathrel {R_i}z'_i\}\), are both closed.
Free disposal: For each \((A_i,t_i)\in \mathcal {M}\times \mathbb {R}\) and each \(A'_i\in \mathcal {M}\) with \(A'_i\subseteq A_i\), \((A_i,t_i)\mathrel {R_i}(A'_i,t_i)\).
Definition 1
A preference relation is classical if it satisfies money monotonicity, first object monotonicity, possibility of compensation, and continuity.
Let \(\mathcal {R}^C\) be the class of classical preferences. We call \((\mathcal {R}^{C})^n\) the classical domain. Let \(\mathcal {R}^C_+\) be the class of classical preferences satisfying free disposal. Obviously, \(\mathcal {R}^C_+\subsetneq \mathcal {R}^C\).
Lemma 1 holds for classical preferences. The proof is relegated to the Appendix.
Lemma 1
Let \(R_i\in \mathcal {R}^C\) and \(A_i,A'_i\in \mathcal {M}\). There is a continuous function \(V_i(A'_i;(A_i,\cdot )):\mathbb {R}\rightarrow \mathbb {R}\) such that for each \(t_i\in \mathbb {R}\), \((A'_i,V_i(A'_i;(A_i,t_i)))\mathrel {I_i}(A_i,t_i)\).
For each \(R_i\in \mathcal {R}^C\), each \(z_i\in \mathcal {M} \times \mathbb {R}\), and each \(A_i\in \mathcal {M}\), we call \(V_i(A_i;z_i)\) the valuation of \(A_i\) at \(z_i\) for \(R_i\). By money monotonicity, for each \(R_i\in \mathcal {R}^C\) and each pair \((A_i,t_i),(A'_i,t'_i)\in \mathcal {M}\times \mathbb {R}\), \((A_i,t_i)\mathrel {R_i}(A'_i,t'_i)\) if and only if \(V_i(A'_i;(A_i,t_i))\le t'_i\).
Definition 2
A preference relation \(R_i\in \mathcal {R}^C\) is quasilinear if for each pair \((A_i,t_i),(A_i',t'_i)\in \mathcal {M}\times \mathbb {R}\) and each \(t''_i\in \mathbb {R}\), \((A_i,t_i)\mathrel {I_i}(A_i',t'_i)\) implies \((A_i,t_i+t''_i)\mathrel {I_i}(A_i',t'_i+t''_i)\).
Let \(\mathcal {R}^Q\) be the class of quasilinear preferences. We call \((\mathcal {R}^Q)^n\) the quasilinear domain. Obviously, \(\mathcal {R}^Q\subsetneq \mathcal {R}^C\).
Remark 1
Let \(R_i\in \mathcal {R}^Q\). Then,

(i)
there is a valuation function \(v_i:\mathcal {M}\rightarrow \mathbb {R}_{+}\) such that \(v_i(\emptyset )=0\), and for each pair \((A_i,t_i),(A_i',t'_i)\in \mathcal {M}\times \mathbb {R}\), \((A_i,t_i)\mathrel {R_i}(A_i',t'_i)\) if and only if \(v_i(A_i')t'_i\le v_i(A_i)t_i\), and

(ii)
for each \((A_i,t_i)\in \mathcal {M}\times \mathbb {R}\) and each \(A'_i\in \mathcal {M}\), \(\textit{V}_i(A'_i;(A_i,t_i))t_i=v_i(A'_i)v_i(A_i)\).
Now we define important classes of preferences. The following property formalizes the notion that given a payment level, an agent desires to consume at most one object.
Definition 3
A preference relation \(R_i\in \mathcal {R}^C\) satisfies the unitdemand property if for each \((A_i,t_i)\in \mathcal {M}\times \mathbb {R}\) with \(A_i>1\), there is \(a\in A_i\) such that \((a,t_i)\mathrel {R_i}(A_i,t_i)\).^{Footnote 7} \(^{\text {,}}\) ^{Footnote 8}
The condition means that given a payment level, for each set of objects, the most preferred one in the set is at least as good as the set itself. Note that it is possible that when an agent with a unitdemand preference relation receives an object and his payment is fixed, an additional object makes him better off. However, this occurs only when he prefers the additional object to the original one. Figure 1 illustrates a unitdemand preference relation.
Let \(\mathcal {R}^U\) be the class of unitdemand preferences. We call \((\mathcal {R}^U)^n\) the unitdemand domain. Obviously, \(\mathcal {R}^U\subsetneq \mathcal {R}^C\).
We also consider a property that formalizes the notion that given a payment level, an agent desires to consume several objects.
Definition 4
A preference relation \(R_i\in \mathcal {R}^C\) satisfies the multidemand property if for each \((\{a\},t_i)\in \mathcal {M}\times \mathbb {R}\), there is \(A_i\in \mathcal {M}\) such that \(a\in A_i\) and \((A_i,t_i)\mathrel {P_i}(\{a\},t_i)\).
The condition says that given a payment level, when an agent receives an object, receiving some additional object(s) makes him better off. Note that given a payment level, even if an agent has a multidemand preference relation, when he receives a set consisting of several objects, he may find it worse than each object in the set. Figure 2 illustrates a multidemand preference relation.
Let \(\mathcal {R}^M\) be the class of multidemand preferences. We call \((\mathcal {R}^M)^n\) the multidemand domain. The following are examples of preferences satisfying the multidemand property.
Example 1:
kobjectdemand preferences. Given \(k\in \{1,\dots , m\}\), a preference relation \(R_i\in \mathcal {R}^C\) satisfies the kobjectdemand property if (i) for each \((A_i,t_i)\in \mathcal {M}\times \mathbb {R}\) with \(A_i<k\), and each \(a\in M{\setminus } A_i\), \((A_i\cup \{a\},t_i)\mathrel {P_i}(A_i,t_i)\), and (ii) for each \((A_i,t_i)\in \mathcal {M}\times \mathbb {R}\) with \(A_i\ge k\), there is \(A'_i\subseteq A_i\) with \(A'_i= k\) such that \((A'_i,t_i)\mathrel {R_i}(A_i,t_i)\).^{Footnote 9} Clearly, for each \(k\in \{2,\dots , m\}\), preferences satisfying the kobjectdemand property satisfy the multidemand property.
Example 2:
Substitutes and complements. Suppose that the set of objects are divided into two nonempty sets K and L, and agent i with a preference relation \(R_i\) views objects a and b as substitutes if both a and b are in the same set, and as complements if a and b are in different sets. For example, objects in K can be pens and objects in L can be notebooks. Formally, \(R_i\) satisfies the following property: For each \(A_i\in \mathcal {M}\) with \(A_i> 1\) and each \(t_i\in \mathbb {R}\), if \(A_i\subseteq K\) or \(A_i\subseteq L\), then there is \(a\in A_i\) such that \((A_i,t_i)\mathrel {I_i}(a,t_i)\), and otherwise, for each \(a\in A_i\), \((A_i,t_i)\mathrel {P_i}(a,t_i)\). Clearly, this preference relation \(R_i\) satisfies the multidemand property.
Some preferences in \(\mathcal {R}^C\) violate both of the unitdemand property and the multidemand property.
Example 3:
(Fig. 3). A preference relation violating the unitdemand property and the multidemand property. Let \(R_i\in \mathcal {R}^C\) be such that for each \(a\in M\) and each \(t_i\in \mathbb {R}\), \(V_i(a;(\emptyset ,t_i))=t_i+5\), and for each \(A_i\in \mathcal {M}\) with \(A_i>1\), and each \(t_i\in \mathbb {R}\),
Then, for each pair \(a,b\in M\) and each \(t_i\in \mathbb {R}\) with \(t_i<5\), \(V_i(\{a,b\};(\emptyset ,t_i))=\frac{1}{2}(t_i+5)>t_i+5=V_i(a;(\emptyset , t_i))=V_i(b;(\emptyset ,t_i))\), and thus, we have \((\{a,b\}, t_i+5)\mathrel {P_i}(a,t_i+5)\mathrel {I_i}(b,t_i+5)\). Thus, \(R_i\) does not satisfy the unitdemand property. Moreover, for each \(a\in M\), each \(A_i\in \mathcal {M}\) with \(a\in A_i\), and each \(t_i\in \mathbb {R}\) with \(t_i\ge 5\), \(V_i(A_i;(\emptyset ,t_i))=t_i+5=V_i(a;(\emptyset ,t_i))\), and thus, we have \((A_i,t_i+5)\mathrel {I_i}(a,t_i+5)\). Thus, \(R_i\) does not satisfy the multidemand property.
An object allocation is an ntuple \(A\equiv (A_1,\ldots ,A_{n})\in \mathcal {M}^n\) such that \(A_i\cap A_j=\emptyset \) for each \(i,j\in N\) with \(i\ne j\). We denote the set of object allocations by \(\mathcal {A}\). A (feasible) allocation is an ntuple \(z\equiv (z_1,\dots ,z_{n})\equiv ((A_1,t_1),\dots ,(A_{n},t_{n}))\in (\mathcal {M}\times \mathbb {R})^n\) such that \((A_1,\dots ,A_{n})\in \mathcal {A}\). We denote the set of feasible allocations by Z. Given \(z\in Z\), we denote the object allocation and the agents’ payments at z by \(A\equiv (A_1,\dots ,A_n)\) and \(t\equiv (t_1\dots , t_n)\), respectively, and we also write \(z=(A,t)\).
A preference profile is an ntuple \(R\equiv (R_1,\ldots R_n)\in \mathcal {R}^n\). Given \(R\in \mathcal {R}^n\) and \(i\in N\), let \(R_{i}\equiv (R_j)_{j\ne i}\).
An allocation rule, or simply a rule on \(\mathcal {R}^n\) is a function \(f: \mathcal {R}^n\rightarrow Z\). Given a rule f and \(R\in \mathcal {R}^n \), we denote the bundle assigned to agent i by \(f_i(R)\) and we write \(f_i(R)=(A_i(R),t_i(R))\).
Now, we introduce standard properties of rules. The efficiency notion here takes the planner’s preferences into account and assumes that he is only interested in his revenue. Formally, an allocation \(z\equiv ((A_i,t_i))_{i\in N}\in Z\) is (Pareto)efficient for \(R\in \mathcal {R}^n\) if there is no feasible allocation \(z'\equiv ((A_i',t'_i))_{i\in N}\in Z\) such that \((\text {i})\text { for each }i\in N,\;z_i'\mathrel {R_i}z_i\text {, }(\text {ii})\text { for some }j\in N,z_j'\mathrel {P_i}z_j, \text { and }(\text {iii})\sum _{i\in N}t'_i\ge \sum _{i\in N}t_i.\)
The first property states that for each preference profile, a rule chooses an efficient allocation.
Efficiency: For each \(R\in \mathcal {R}^n\), f(R) is efficient for R.
Remark 2
By money monotonicity and Lemma 1, the efficiency of allocation z is equivalent to the property that there is no allocation \(z'\equiv ((A'_i,t'_i))_{i\in N}\in Z\) such that
(i\('\)) for each \(i\in N\), \(z_i'\mathrel {I_i}z_i\), and (ii\('\)) \(\sum _{i\in N}t'_i>\sum _{i\in N}t_i\).
The second property states that no agent benefits from misrepresenting his preferences.
Strategyproofness: For each \(R\in \mathcal {R}^n\), each \(i\in N\), and each \(R'_i\in \mathcal {R}\), \(f_i(R)\,R_i \,f_i(R'_i,R_{i})\).
The third property states that an agent is never assigned a bundle that makes him worse off than he would be if he had received no object and paid nothing.
Individual rationality: For each \(R\in \mathcal {R}^n\) and each \(i\in N\), \(f_i(R)\mathrel {R_i}\varvec{0}\).
The fourth property states that the payment of each agent is always nonnegative.
No subsidy: For each \(R\in \mathcal {R}^n\) and each \( i\in N\), \(t_i(R)\ge 0\).
The final property is a weaker variant of the fourth: If an agent receives no object, his payment is nonnegative.
No subsidy for losers: For each \(R\in \mathcal {R}^n\) and each \(i\in N\), if \(A_i(R)=\emptyset \), \(t_i(R)\ge 0\).
Unitdemand model and rich domains
In our model, potentially each agent can receive several objects. However, some authors study a model in which no agent can receive more than one object owing to some reason, say by regulations, or by physical reasons.^{Footnote 10} We call this model the unitdemand model, and refer to our model as the multidemand model. Some important results are established in the unitdemand model, and they are related to our main result. Some of such results continue to hold in the multidemand model when preferences are unitdemand, and others continue to hold only when domains include enough variety of unitdemand preferences. In this section, we introduce “richness” of domains in our model, which guarantees that a domain of the multidemand model includes enough variety of unitdemand preferences.
In the unitdemand model, preferences are defined over \(M\cup \{0\}\times \mathbb {R}\), where 0 means not receiving any object in M and is called null object. To distinguish classes of preferences in the multidemand model and those in the unitdemand model, we denote a typical class of preferences in the unitdemand model by \(\mathscr {R}\).
Money monotonicity, first object monotonicity, possibility of compensation, and continuity are defined in the unitdemand model in the same manner as defined in our model. Thus, in the unitdemand model, classical preferences are defined in the same manner.
Definition 5
A preference relation \(R_i\) over \(M\cup \{0\}\times \mathbb {R}\) is classical if it satisfies money monotonicity, first object monotonicity, possibility of compensation, and continuity.
Let \(\mathscr {R}^C\) be the class of classical preferences in the unitdemand model.
In the unitdemand model, a feasible allocation is an ntuple \( z=((x_i,t_i))_{i\in N}\in (M\cup \{0\}\times \mathbb {R})^{n}\) such that for each pair \(i,j\in N\), \(x_i=x_j\) implies \(x_i=x_j=0\). As we mentioned, in the unitdemand model, no agent can receive more than one object. Other notions such as rules, properties of rules, etc., are defined in the same manner as defined in the multidemand model.
To define the richness, we introduce the following notions, which connect preferences in the multidemand model to those in the unitdemand model.
Definition 6
A preference relation \(R_i\) in the multidemand model induces a preference relation \(R_i^{\prime }\) over \(M\cup \{0\}\times \mathbb {R}\) if for each pair \((a,t_i),(b,t_i^{\prime })\in M\cup \{0\}\times \mathbb {R}\), \((a,t_i)\mathrel {R'_i}(b,t_i^{\prime })\) if and only if \((A_i,t_i)\mathrel {R_i}(A_i^{\prime },t_i^{\prime })\), where
Definition 7
A class of preferences \(\mathcal {R}\) in the multidemand model induces a class of preferences \(\mathscr {R}\) over \(M\cup \{0\}\times \mathbb {R}\) if (i) for each \(R_i^{\prime }\in \mathscr {R}\), there is \( R_i\in \mathcal {R}\) that induces \(R_i^{\prime }\), and (ii) for each \( R_i\in \mathcal {R}\), there is \(R_i^{\prime }\in \mathscr {R}\) that is induced by \(R_i\).
Remark 3
Each of \(\mathcal {R}^{U}\), \(\mathcal {R}_{+}^{U}\), and \(\mathcal { R}^{U}{\setminus } \mathcal {R}_{+}^{U}\) induces \(\mathscr {R}^{C}\).
Now, we introduce the richness of domain, which guarantees that a domain of the multidemand model includes enough variety of unitdemand preferences so that they induce the class of classical preferences in the unitdemand model.
Definition 8
A class of preferences \(\mathcal {R}\) is rich if \(\mathcal {R\,}\cap \mathcal {\,R}^{U}\) induces \(\mathscr {R}^{C}\).
By Remark 3, \(\mathcal {R}^U\), \(\mathcal {R}^U_+\), and \(\mathcal {R}^U{\setminus }\mathcal {R}^U_+\) are rich.
Minimum price Walrasian rules
In this section we define the minimum price Walrasian rules and state several facts related to them.
Let \(p\equiv (p^1,\ldots ,p^M)\in \mathbb {R}_{+}^m\) be a price vector. The budget set at p is defined as \(B(p)\equiv \{(A_i,t_i)\in \mathcal { M}\times \mathbb {R}: t_i=\sum _{a\in A_i}p^a\}\). Given \(R_i\in \mathcal {R}\), the demand set at p for \(R_i\) is defined as \(D(R_i,p)\equiv \{z_i\in B(p):\text {for each }z^{\prime }_i\in B(p),\,z_i\mathrel {R_i} z^{\prime }_i\}\).
Lemma 2
Let \(R_i\in \mathcal {R}^U\) and \(p\in \mathbb {R}^m_+\). (i) Suppose \(p\in \mathbb {R}^m_{++}\). Then, for each \((A_i,t_i)\in D(R_i,p)\), \( A_i \le 1\). (ii) Let \(A_i\in \mathcal {M}\) be such that \((A_i,\sum _{a\in A_i}p^a)\mathrel {R_i}(A^{\prime }_i,\sum _{a\in A^{\prime }_i}p^a)\) for each \( A^{\prime }_i\in \mathcal {M}\) with \(A^{\prime }_i\le 1\). Then, \(( A_i,\sum _{a\in A_i}p^a)\in D(R_i,p)\).
Definition 9
Let \(R\in \mathcal {R}^n\). A pair \(((A,t),p)\in Z\times \mathbb {R}_{+}^m\) is a Walrasian equilibrium (WE) for R if
Condition Wi says that each agent receives a bundle that he demands. Condition Wii says that an object’s price is zero if it is not assigned to anyone. Given \(R\in \mathcal {R}^n\), let W(R) and P(R) be the sets of Walrasian equilibria and prices for R, respectively.
Lemma 3
Let \(R\in (\mathcal {R}^U)^n\) and \(p\in P(R)\). (i) If \(n>m\), then \( p^a>0 \) for each \(a\in M\). (ii) There is \(((A,t),p)\in W(R)\) such that \( A_i\le 1\) for each \(i\in N \).
Let \(R\in (\mathcal {R}^{U})^{n}\) and \(R^{\prime }\) be preference profiles of the multidemand and unitdemand models respectively such that for each \( i\in N\), \(R_i^{\prime }\) is induced by \(R_i\). For each \(p\in P(R)\), by (ii) of Lemma 3, there is an allocation \(((\{a_i\},t_i))_{i\in N}\) such that \((((\{a_i\},t_i))_{i\in N},p)\in W(R)\), and thus, \( (((a_i,t_i))_{i\in N},p)\) is a WE for \(R^{\prime }\). On the other hand, for each \(p\in \mathbb {R}_{+}^M\), if p is a WE price vector for \( R^{\prime }\), then there is an allocation \(((a_i,t_i))_{i\in N}\) such that \((((a_i,t_i))_{i\in N},p)\) is a WE for \(R^{\prime }\), and thus by (ii) of Lemma 2, \((((\{a_i\},t_i))_{i\in N},p)\in W(R)\). Therefore, the set of WE price vectors for R coincides with the set of WE price vectors for \(R^{\prime }\).
In the unitdemand model, several results on Walrasian equilibrium are established. By the preceding argument, the same results continue to hold in our model for preferences satisfying the unitdemand property.
Fact 1
(Alkan and Gale 1990)^{Footnote 11} \(^{\text {,}}\) ^{Footnote 12} For each \(R\in (\mathcal {R}^U)^n\), a Walrasian equilibrium for R exists.
Fact 2
(Demange and Gale 1985) For each \(R\in (\mathcal {R}^U)^n\), there is a unique minimum Walrasian equilibrium price vector, i.e., a vector \(p\in P(R)\) such that for each \( p^{\prime }\in P(R)\), \(p\le p^{\prime }\).^{Footnote 13}
Given \(R \in {\mathcal {R}}^n\), a minimum price Walrasian equilibrium (MPWE) for R is a Walrasian equilibrium for R whose price is minimum. Given \(R\in \mathcal {R}^n\), let \( p_{\min }(R)\) be the minimum Walrasian equilibrium price for R, and \(Z_{\min }^W(R)\) be the set of Walrasian equilibrium allocations associated with \(p_{\min }(R)\). Although there might be several minimum price Walrasian equilibria, they are indifferent for each agent, i.e., for each \(R\in \mathcal {R}^n\), each pair \(z,z^{\prime }\in Z_{\min }^W(R)\), and each \(i\in N\), \(z_i\mathrel {I_i}z^{\prime }_i\).
Definition 10
A rule f on \(\mathcal {R}^n\) is a minimum price Walrasian (MPW) rule if for each \(R\in \mathcal {R}^n\), \(f(R)\in Z_{\min }^W(R)\).
It is easy to show that the MPW rules on \((\mathcal {R}^{U})^{n}\) satisfy efficiency, individual rationality, and no subsidy. Demange and Gale (1985) show that the MPW rules are strategyproof on the classical domain in the unitdemand model. Our arguments above allow us to convert each MPWE allocation in the unitdemand model into an MPWE for a unitdemand profile in the multidemand model. Moreover, all the minimum price Walrasian equilibria are indifferent for each agent. Thus, the result by Demange and Gale (1985) implies that for each rich class of preferences \(\mathcal {R}\subseteq \mathcal {R}^{U}\), the MPW rules on \( \mathcal {R}^{n}\) also satisfy strategyproofness in multidemand model.
Morimoto and Serizawa (2015) shows that in the unitdemand model, when \(n>m\), only the MPW rules satisfy efficiency, strategyproofness, individual rationality, and no subsidy for losers on \((\mathscr {R}^{C})^{n}\). The following lemma states that in the multidemand model, when \(n>m\), efficient rules never assign more than one object to agents whose preferences satisfy the unitdemand property.
Lemma 4
(Single object assignment) Let \(n>m\). Let \(\mathcal {R}\subseteq \mathcal {R}^C\) and f be an efficient rule on \(\mathcal {R}^n\). Let \(R\in \mathcal {R}^n\) and \( i\in N\). If \(R_i\in \mathcal {R}^U\), \(A_i(R)\le 1\).
By Lemma 4, when \(n>m\), for each \(\mathcal {R}\subseteq \mathcal {R}^{U}\) that is rich, and each rule on \(\mathcal {R}^{n}\) satisfying efficiency, it always assigns each agent at most one object. Thus, when \(n>m \), for each \(\mathcal {R}\subseteq \mathcal {R}^{U}\) that is rich, and each rule on \(\mathcal {R}^{n}\) satisfying efficiency, strategyproofness, individual rationality, and no subsidy for losers, there is a corresponding rule in the unitdemand model, and moreover, it is easy to see that the corresponding rule also satisfies the four properties. Thus, the result by Morimoto and Serizawa (2015) continues to hold in our model.
Fact 3
(Demange and Gale 1985 for (i); Morimoto and Serizawa 2015 for (ii)) Let \(\mathcal {R}\subseteq \mathcal {R}^{U}\). (i) The minimum price Walrasian rules on \(\mathcal {R}^{n}\) satisfy efficiency, strategyproofness, individual rationality and no subsidy. (ii) Let \(n>m\), and \(\mathcal {R}\) be rich. Then, the minimum price Walrasian rules are the only rules on \( \mathcal {R}^{n}\) satisfying efficiency, strategyproofness, individual rationality and no subsidy for losers.
Main result
In this section, first we state the main theorem. Next, we explain how we prove the theorem.
Impossibility result
We consider rich domains containing some multidemand preferences and we investigate whether efficient and strategyproof rules still exist on such domains. In marked contrast to Fact 3 in Sect. 3, the results are negative. Namely, if there are more agents than objects, and if the domain is rich and contains even a single multidemand preference relation, then no rule on the domain satisfies efficiency, strategyproofness, individual rationality and no subsidy for losers.
Theorem
Let \(n>m\). Let \(R_{0}\in \mathcal {R}^M\) and \( \mathcal {R}\) be a rich class of preferences such that \(R_{0}\in \mathcal {R}\). Then, no rule on \(\mathcal {R}^{n}\) satisfies efficiency, strategyproofness, individual rationality and no subsidy for losers.
Corollary 1
Let \(n>m\). Let \(\mathcal {R}=\mathcal {R}^{U}\cup \mathcal {R}^M \). Then, no rule on \(\mathcal {R}^{n}\) satisfies efficiency, strategyproofness, individual rationality and no subsidy for losers.
Remark 4
The Corollary 1 is a standard form of impossibility results on strategyproofness in that since Gibbard (1973) and Satterthwaite (1975), many impossibility results on strategyproofness of this form are established. In such results, a domain is fixed and incompatibility of some properties of rules is established on this domain. Results of this form cannot be applied unless all the preferences in the fixed domain are deemed plausible. For example, the Corollary 1 cannot be applied unless all the preferences in \( \mathcal {R}^M\) in addition to \(\mathcal {R}^{U}\) are deemed plausible. On the other hand, our Theorem can be applied as soon as in addition to a rich class of preferences \(\mathcal {R}\), just one preference relation \(R_{0}\) arbitrarily chosen from \(\mathcal {R}^M\) is deemed plausible. Accordingly our Theorem can be applied to more variety of environments than Corollary 1. For example, consider an environment where \(n=40\) , \(m=20\) and there are only the preferences satisfying kobjectdemand property for \(k\in \{1,2,3,4,5\} \). The Theorem can be applied to this environment, but the Corollary 1 cannot be.
By Remark 3, we also have the following corollaries. These corollaries demonstrate the wide applicability of our results even more. In this paper, we do not maintain free disposal. However, it is a standard assumption for preferences. Corollary 2 states that our conclusion holds even if free disposal is assumed.
Corollary 2
Let \(n>m\). Let \(R_{0}\in \mathcal {R}^M\) and \( \mathcal {R}\equiv \mathcal {R}_{+}^{U}\cup \{R_{0}\}\). Then, no rule on \(\mathcal {R} ^{n}\) satisfies efficiency, strategyproofness, individual rationality and no subsidy for losers.
Free disposal is not a suitable assumption in some environment. Corollary 3 states that our conclusion holds even in such an environment.
Corollary 3
Let \(n>m\). Let \(R_{0}\in \mathcal {R}^M\) and \( \mathcal {R}\equiv (\mathcal {R}^{U}{\setminus } \mathcal {R}_{+}^{U})\cup \{R_{0}\}\). Then, no rule on \(\mathcal {R}^{n}\) satisfies efficiency, strategyproofness, individual rationality and no subsidy for losers.
Remark 5
In this paper, we assume that preferences are drawn from a common class of \(\mathcal {R}\). If preferences of each agent are drawn from a class \(\mathcal {R}_i\) that depends on the identity of the agent, our theorem can be strengthened as follows: Suppose that, for each \(i\in N\), \(\mathcal {R}_i\) is rich, and there are \(j\in N\) and \(R_j\in \mathcal {R}^M\) such that \(R_j\in \mathcal {R}_j\). Then, when \(n>m\), no rule on \(\prod _{i\in N} \mathcal {R}_i\) satisfies efficiency, strategyproofness, individual rationality and no subsidy for losers.
Sketch of the proof
Preliminary results
We state seven lemmas which we use in the sketch of the proof and in the formal proof. The proof of each lemma is relegated to the Appendix, or is omitted if it is straightforward.
Let \(\mathcal {R}\subseteq \mathcal {R}^C\) be rich. Let f be a rule on \(\mathcal {R}^n\) satisfying efficiency, strategyproofness, individual rationality and no subsidy for losers. Lemma 5 states that if an agent receives no object, then his payment is zero. This is immediate from individual rationality and no subsidy for losers. Thus we omit the proof.
Lemma 5
(Zero payment for losers) Let \(R\in \mathcal {R}^n\) and \(i\in N\). If \(A_i(R)=\emptyset \), \(t_i(R)=0\).
Lemma 6 states that for each agent, his payment is at most the valuation, at \(\varvec{0}\), of the set of objects that he receives. This is immediate from individual rationality. Thus, we omit the proof.
Lemma 6
For each \(R\in \mathcal {R}^n\) and each \(i\in N\), \(t_i(R)\le V_i(A_i(R);\varvec{0})\).
Lemma 7 states that each object is assigned to some agent. This follows from efficiency, \(n>m\), and first object monotonicity. We omit the proof.
Lemma 7
(Full object assignment) Let \(n>m\). For each \(R\in \mathcal {R}^n\) and each \(a\in M\), there is \(i\in N\) such that \(a\in A_i(R)\).
Lemma 8 states a necessary condition for efficiency.
Lemma 8
(Necessary condition for efficiency) Let \(R\in \mathcal {R}^n\) and \(i,j\in N\) with \(i\ne j\). Let \(A_i,A_j\in \mathcal {M}\) be such that \(A_i\cap A_j=\emptyset \) and \(A_i\cup A_j\subseteq A_i(R)\cup A_j(R)\). Then, \(V_i(A_i;f_i(R))+V_j(A_j;f_j(R)) \le t_i(R)+t_j(R)\).
Although no subsidy for losers itself tells us nothing about payment levels for nonempty sets of objects, Lemma 9 states that for each nonempty set of objects, there is a lower bound of the payment level for the set.
Lemma 9
(Payment lower bound) Let \(n>m\). Let \(R\in \mathcal {R}^n\) and \(i\in N\). Let \(R'_i\in \mathcal {R}\cap \mathcal {R}^U\) be such that for each \(a\in M\) and each \(t_i\in \mathbb {R}\), \(V'_i(a;(\emptyset ,t_i))t_i<\min _{j\in N{\setminus } \{i\}}V_j(a;\varvec{0})\).^{Footnote 14} Then, \(t_i(R)\ge V'_i(A_i(R);\varvec{0})\).
By first object monotonicity, Lemma 9 implies that for each \(R\in \mathcal {R}^n\) and each \(i\in N\), if \(A_i(R)=1\), then \(t_i(R)\ge 0\).
Lemma 10 states that f coincides with an MPW rule on \((\mathcal {R}\cap \mathcal {R}^U)^n\). This is immediate from (ii) of Fact 3. Thus we omit the proof.
Lemma 10
Let \(n>m\). For each \(R\in (\mathcal {R}\cap \mathcal {R}^U)^n\), \(f(R)\in Z^W_{\min }(R)\).
Given \(i\in N\) and \(R_{i}\in \mathcal {R}^{n1}\), we define the option set of agent i for \(R_{i}\) by
Lemma 11 states that (i) the option set does not contain more than one bundle with the same set of objects, and (ii) each agent receives one of the most preferred bundles in his option set. This is straightforward from strategyproofness. Thus, we omit the proof.
Lemma 11
Let \(i\in N\) and \(R_{i}\in \mathcal {R}^n\). (i) For each pair \((A_i,t_i),(A'_i,t'_i)\in o_i(R_{i})\), if \(A_i=A'_i\), then \(t_i=t'_i\). (ii) For each \(R_i\in \mathcal {R}\) and each \(z_i\in o_i(R_{i})\), \(f_i(R_i,R_{i})\mathrel {R_i}z_i\).
Threeagent and twoobject example
Since the proof of the Theorem is very complicated, we relegate it to the Appendix. Here we demonstrate the ideas and techniques of the proof by applying them to a particular example in a threeagent and twoobject setting. Let \(M=\{a,b\}\) and \(N=\{1,2,3\}\). In the formal proof, the preference relation \(R_0\) is an arbitrary element of \(\mathcal {R}^M\), but here we pick \(R_0\) from \(\mathcal {R}^{Q}\cap \mathcal {R}^M\). For concreteness, let
However, the idea of our proof does not depend on \(R_0\in \mathcal {R}^{Q}\). We assume \(R_0\in \mathcal {R}^{Q}\) only for simplicity of expression.
Let \(\mathcal {R}\subseteq \mathcal {R}^{C}\) satisfy the richness and contain \(R_0\). For example, let \(\mathcal {R} \supseteq \mathcal {R}^U\cup \{R_0\}\). We suppose that there is a rule f on \(\mathcal {R}^{3}\) satisfying efficiency, strategyproofness, individual rationality and no subsidy for losers, and derive a contradiction.
Step A: Constructing a preference profile.
Let \(R_1=R_0\). We construct \(R_2\in \mathcal {R}^{U}\) and \(R_{3}\in \mathcal {R}^{U}\) depending on \(R_1\) so that a contradiction is derived. We define \(R_2\) satisfying \(V_2(a;\varvec{0})>v_1(\{a,b\})\) and^{Footnote 15}
For example, let \(R_2\in \mathcal {R}^{U}\) be such that for each \(A_2\in \mathcal {M} {\setminus } \{\emptyset \}\),
We define \(R_{3}\) satisfying \(R_{3}\in \mathcal {R}^U\cap \mathcal {R}^{Q}\), and
For example, let \(v_{3}(a)=v_{3}(b)=9\). Note that \(V_2(a;\varvec{0} )>v_{3}(\{a,b\})\). Let \(R\equiv (R_1,R_2,R_{3})\). Figure 4 illustrates R . \(\square \)
Step B: \(A_2(R)\ne \emptyset \) .
Suppose by contradiction that \(A_2(R)=\emptyset \). By Lemma 5, \( f_2(R)=\varvec{0}\). By Lemma 7, there is \(i\ne 2\) such that \(a\in A_i(R)\).
Let \(A_i=\emptyset \) and \(A_2=\{a\}\). Note that \(A_i\cap A_2=\emptyset \) and \(A_i\cup A_2\subseteq A_i(R)\cup A_2(R)\). If \(i=1\), then \(V_1(\emptyset ; f_1(R))\ge t_1(R)40\). If \(i=3\), then \(V_3(\emptyset ;f_3(R))=t_3(R)9\). Thus, \(V_i(\emptyset ;f_i(R))\ge t_i(R)40\). Since \(V_2(a;\varvec{0})=41\),
Thus, by Lemma 8, efficiency is violated, a contradiction. \(\square \)
Step C: \(A_1(R)=a\).
Substep C1: \((a,9)\in o_1(R_{1})\).
Let \(R^{\prime }_1\in \mathcal {R}^U\) be such that
For example, let \(R^{\prime }_1\in \mathcal {R}^U\) be such that
Since \((R^{\prime }_1,R_{1})\in (\mathcal {R}^U)^3\), by Lemma 10, \( f(R^{\prime }_1,R_{1})\in Z_{\min }^{W}(R^{\prime }_1,R_{1})\). Let \(z\in Z\) be such that
Figure 5 illustrates \((R^{\prime }_1,R_{1})\) and z.
Let \(p\equiv (9,9)\). Then, \(D(R^{\prime }_1,p)=\{\{a\}\}\), \( D(R_2,p)=\{\{b\}\}\), and \(D(R_3,p)=\{\emptyset ,\{a\},\{b\}\}\). Thus \( (z,p)\in W(R^{\prime }_1,R_{1})\), implying \(p_{\min }(R^{\prime }_1 ,R_{1})\le p\). If \(p_{\min }^{a}(R^{\prime }_1,R_{1})<9\) or \(p_{\min }^{b}(R^{\prime }_1,R_{1})<9\), then \(0\notin D(R^{\prime }_1,p_{\min }(R^{\prime }_1,R_{1}))\) and for each \(i\in \{2,3\}\), \(0\notin D(R_i,p_{\min }(R^{\prime }_1,R_{1}))\), which implies \(p_{\min }(R^{\prime }_1,R_{1})\notin P(R^{\prime }_1,R_{1})\), a contradiction. Thus, \( p_{\min }(R^{\prime }_1,R_{1})=(9,9)\). Moreover, z is the only WE allocation supported by \(p_{\min }(R^{\prime }_1,R_{1})\). Thus, \( f_1(R^{\prime }_1,R_{1})=z\), and hence, \(f_1(R^{\prime }_1,R_{1})=(a,9)\in o_1(R_{1})\).
Substep C2: \((b,10)\in o_1(R_{1})\).
Let \(R^{\prime \prime }_1\in \mathcal {R}^U\) be such that
For example, let \(R^{\prime \prime }_1\in \mathcal {R}^U\) be such that
Since \((R^{\prime \prime }_1,R_{1})\in (\mathcal {R}^U)^3\), by Lemma 10, \(f(R^{\prime \prime }_1,R_{1})\in Z_{\min }^{W}(R^{\prime \prime }_1,R_{1})\). Let \(z^{\prime }\in Z\) be such that
Figure 6 illustrates \((R^{\prime \prime }_1,R_{1})\) and \(z^{\prime }\).
Let \(p^{\prime }\equiv (9,10)\). Then, \(D(R^{\prime \prime }_1,p^{\prime })=\{\{b\}\}\), \(D(R_2,p^{\prime })=\{\{a\},\{b\}\}\), and \(D(R_3,p^{\prime })=\{\emptyset ,\{a\}\}\). Thus \((z^{\prime },p^{\prime })\in W(R^{\prime \prime }_1,R_{1})\), implying \(p_{\min }(R^{\prime \prime }_1,R_{1})\le p^{\prime }\). If \(p_{\min }^{a}(R^{\prime \prime }_1,R_{1})<9\), then \( 0\notin D(R^{\prime \prime }_1,p_{\min }(R^{\prime \prime }_1,R_{1}))\) and for each \(i\in \{2,3\}\), \(0\notin D(R_i,p_{\min }(R^{\prime \prime }_1,R_{1}))\), which implies \(p_{\min }(R^{\prime \prime }_1,R_{1})\notin P(R^{\prime \prime }_1,R_{1})\), a contradiction. Thus, \(p_{\min }^{a}(R^{ \prime \prime }_1,R_{1})=9\). If \(p_{\min }^{b}(R^{\prime \prime }_1,R_{1})<10\), then we have \(D(R^{\prime \prime }_1,p_{\min }(R^{\prime \prime }_1,R_{1}))=\{\{b\}\}\) and \(D(R_2,p_{\min }(R^{\prime \prime }_1,R_{1}))=\{\{b\}\}\), which further implies \(p_{\min }(R^{\prime \prime }_1,R_{1})\notin P(R^{\prime \prime }_1,R_{1})\), a contradiction. Thus, \( p_{\min }(R^{\prime \prime }_1,R_{1})=(9,10)\). Moreover, \(z^{\prime }\) is the only WE allocation which is supported by \(p_{\min }(R^{\prime \prime }_1,R_{1})\). Thus, \(f(R^{\prime \prime }_1,R_{1})=z^{\prime }\), and hence, \(f_1(R^{\prime \prime }_1,R_{1})=(b,10)\in o_1(R_{1})\).
Substep C3: \(A_1(R)=a\).
Since \(A_2(R)\ne \emptyset \) by Step B, \(A_1(R)\le 1\). If \( A_1(R)=\emptyset \), then by Lemma 5, we have \(f_1(R)=\varvec{0}\), and thus, \(f_1(R^{\prime }_1,R_{1})\mathrel {P_1}f_1(R)\), which contradicts strategyproofness. Thus, \(A_1(R)=a\) or b, and therefore, by (i) of Lemma 11, \(f_1(R)=(a,9)\) or (b, 10). Since \((a,9)\mathrel {P_1} (b,10)\), (ii) of Lemma 11 implies \(f_1(R)=(a,9)\). \(\square \)
Step D: f(R) is not efficient for R.
By \(A_2(R)\ne \emptyset \) and \(A_1(R)=a\), \(A_2(R)=b\). By Lemma 6, \( t_2(R)\le V_2(b;\varvec{0})=12\). By Lemma 9, \(t_2(R)\ge 0\).
Let \(z\equiv ((A_i,t_i))_{i\in N}\in Z\) be such that
Figure 7 illustrates z.
Since \(V_1(\{a,b\};f_1(R))=V_1(\{a,b\};(a,9))=29\), it is easy to see that \( z_1\mathrel {I_1}f_1(R)\) and \(z_3\mathrel {I_3}f_3(R)\). Also by \(t_2(R)\ge 0\) and \(A_2(R)=b\), \(z_2=(\emptyset ,3)\mathrel {I_2}(b,0) \mathrel {R_2}f_2(R)\). Moreover, by \(t_1(R)=9\) and \(t_2(R)\le 12\),
implying that f(R) is not efficient for R, a contradiction. \(\square \)
We emphasize the difference between a (direct) proof of the Corollary 1 that one might write, and the proof of the Theorem that we have shown. To prove the Corollary 1 directly, we can freely pick preference profiles in \(\mathcal { R}^U\cup \mathcal {R}^M\) to derive a contradiction. On the other hand, in the proof of the Theorem, we may only choose preferences from \(\mathcal {R} ^U\cup \{R_0\}\). Moreover, the preference relation \(R_0\), which could be anything in \(\mathcal {R}^M\), forces us to construct profiles depending on \( R_0\), further complicating the process.
In the above sketch, \(R_0\in \mathcal {R}^M\) is assumed to be quasilinear, but the basic logic of the sketch works even in the case \(R_0\in \mathcal {R} ^M\backslash \mathcal {R}^{Q}\).
In the formal proof in the Appendix, we have six steps. Steps A, B, C, and D correspond to Steps 1, 3, 4, and 6, respectively, of the formal proof. Steps 2 and 5 of the formal proof are necessary only for the more general case, so they do not appear in the above sketch.
Concluding remarks
In this article, we have considered an object assignment problem with money where each agent can receive more than one object. We focused on domains that contain enough variety of unitdemand preferences and some multidemand preferences. We studied allocation rules satisfying efficiency, strategyproofness, individual rationality, and no subsidy for losers, and showed that if the domain contains enough variety of unitdemand preferences and at least one multidemand preference relation, and if there are more agents than objects, then no rule satisfies the four properties. As discussed in Sect. 1, we have been motivated by the search for efficient and strategyproof rules on a domain which is not quasilinear or unitdemand. Our result establishes the difficulty of designing efficient and strategyproof rules on such a domain. We state three remarks on our result.
Maximal domain. Some literature on strategyproofness investigates the existence of maximal domains on which there are rules satisfying desirable properties.^{Footnote 16} A domain \(\mathcal {R}^n\) is a maximal domain for a list of properties of rules if there is a rule on \(\mathcal {R}^n\) satisfying the properties, and for each \(\mathcal {R}'\supsetneq \mathcal {R}\), no rule on \((\mathcal {R}')^n\) satisfies the properties. Our result is rather closer to maximal domain results than impossibility results of the form of the Corollary 1. However, our result does not imply that the unitdemand domain is a maximal domain for the four properties in the Theorem, since we add only multidemand preferences to the rich domains and derive the nonexistence of rules satisfying the four properties. In fact, what domains including \((\mathcal {R}^U)^n\) are maximal domains for the four properties is an open question.
However, we are sure that \((\mathcal {R}^U)^n\) is not a maximal domain for the four properties. For example, consider \(R_i\) in Example 3 and let \(\mathcal {R}\equiv \mathcal {R}^U\cup \{R_i\}\). Since \(R_i\) does not satisfy the unitdemand property, \(\mathcal {R}\supsetneq \mathcal {R}^U\). Note that for each \(R_j\in \mathcal {R}\), each \(A_j\in \mathcal {M}\) with \(A_j>1\) and each \(t_j\in \mathbb {R}\), if \(t_j\ge 0\), then there is \(a\in A_j\) such that \((a,t_j)\mathrel {I_j}(A_j,t_j)\). Thus, since each agent never pays negative amount of money under the minimum price Walrasian rules, and since they satisfy the four properties on the unitdemand domain, they also satisfy the four properties on \(\mathcal {R}^n\). Hence, \((\mathcal {R}^U)^n\) is not a maximal domain for the four properties.
Although we do not find maximal domains for the four properties, the multidemand class includes most of natural preferences outside the unitdemand class. Thus, our result implies that on most of natural domains properly including the unitdemand domain, if there are more agents than objects, we have an impossibility of designing rules satisfying the four properties.
Other properties. Efficiency is not the only property studied in the literature on auction theory. For example, some authors study strategyproof and individually rational rules that achieve as much revenue as possible. Since efficiency takes the auctioneer’s revenue into account, efficiency is closely related to maximizing the auctioneer’s revenue. However, there may exist strategyproof and individually rational rules that is not efficient but achieve as much revenue as possible.^{Footnote 17}
While efficiency takes the auctioneer’s revenue into account, some authors study another efficiency notion that takes only agents’ preferences into account.^{Footnote 18} An allocation is efficient with no deficit if (i) the sum of payments is nonnegative, and (ii) no other allocation with nonnegative sum of payments makes each agent at least as well off and at least one agent better off. Notice that efficiency is implied by efficiency with no deficit.^{Footnote 19} Thus, the Theorem holds even if we replace efficiency by efficiency with no deficit.
Identical objects. Some literature on object assignment problems also study the case in which the objects are identical.^{Footnote 20} In this paper, we do not make this assumption. When objects are not identical, the domain includes a greater variety of preference profiles than when objects are identical. This variety plays an important role in our proof. Therefore, our theorem does not exclude the possibility that when objects are identical, multidemand preferences can be added to the unitdemand domain without preventing the existence of rules satisfying the four properties.
Notes
 1.
See Saitoh and Serizawa (2008) for numerical examples.
 2.
 3.
 4.
Note that the unitdemand domain contains nonquasilinear preferences, and thus the result by Holmström (1979) does not apply
 5.
 6.
 7.
Given a set X, X denotes the cardinality of X.
 8.
Gul and Stacchetti (1999) define the unitdemand property for quasilinear preferences. In their model, a preference relation \(R_i\in \mathcal {R}^Q\) satisfies the unitdemand property if for each \(A_i\in \mathcal {M}\) with \(A_i>1\), \(v_i(A_i)=\max _{a\in A_i}v_i(a)\).
 9.
In Gul and Stacchetti (1999), this notion is called \(k\) satiation.
 10.
 11.
Precisely, Alkan and Gale (1990) show the nonemptiness of the core in a twosided matching model. However, the twosided matching model includes the unitdemand model, and in the unitdemand model, nonemptiness of the core is equivalent to the existence of a Walrasian equilibrium.
 12.
 13.
For each \(p,p^{\prime }\in \mathbb {R}^m\), \(p\le p^{\prime }\) if and only if for each \(i\in \{1,\dots ,m\}\), \(p^i\le p^{\prime i}\).
 14.
Notice that in each class of preferences satisfying the richness, there exists such a preference relation.
 15.
In this sketch, we assume that \(M=\{a,b\}\) and \(R_0\in \mathcal {R}^{Q}\) for the simplicity of expression. For a general multidemand preference relation \(R_0\), the RHS of the first inequality is set as the maximal difference between various \(t_1\) in \([0,V_1(\{a,b\};\varvec{0})]\) and the valuation of empty set at \((\{a,b\},t_1)\) for \(R_1\), i.e.,
$$\begin{aligned} \max _{t_1\in [ 0,V_1(\{a,b\};\varvec{0})]}\{t_1 V_1(\emptyset ;(\{a,b\},t_1))\}. \end{aligned}$$For a general set M, the RHS is defined as \(\overline{t}_1\) in the Appendix.
For a general multidemand preference relation \(R_0\), the RHS of the second inequality is set as the minimum value of the marginal valuations of the second object, i.e.,
$$\begin{aligned} \min \left\{ \min _{t_1\in [ 0,V_1(\{a\};\varvec{0})]} \{V_1(\{a,b\};(a,t_1))t_1\},\min _{t_1\in [ 0,V_1(\{b\};\varvec{0})]}\{V_1(\{a,b\};(b,t_1))t_1\}\right\} . \end{aligned}$$For a general set M of objects, the RHS of the second inequality is defined as \(\underline{t}_1\) in the Appendix.
 16.
 17.
For the single object case with quasilinear preferences, Myerson type rules are not necessarily efficient but maximize the auctioneer’s revenue. See Myerson (1981).
 18.
For example, see Sprumont (2013).
 19.
The proof is as follows. Let \(R\equiv (R_i)_{i\in N}\) be a preference profile. Suppose there is an allocation \(z\equiv ((A_i,t_i))_{i\in N}\) that is efficiency with no deficit for R but not efficient for R. Then, there is an allocation \(z'\equiv ((A'_i,t'_i))_{i\in N}\) such that
$$\begin{aligned} \text {for each }i\in N,\ z'_i\mathrel {R_i}z_i,\text { for some }j\in N,\ z'_j\mathrel {P_j}z_j,\text { and }\sum _{i\in N}t'_i\ge \sum _{i\in N}t_i. \end{aligned}$$By condition (i) of efficiency with no deficit, we have \(\sum _{i\in N}t_i\ge 0\). Thus, \(\sum _{i\in N}t'_i\ge 0\). However, this implies that z is not efficient with no deficit for R, a contradiction.
 20.
 21.
See Debreu (1959).
 22.
See Berge (1963).
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Acknowledgments
The authors are grateful to the editor, the associate editor and two anonymous referees for their many detailed and helpful comments. The preliminary version of this article was presented at the IDGP 2015 Workshop, the 2015 Conference on Economic Design, and the II MOMA Meeting. The authors thank participants at those conferences for their comments. They are also grateful to seminar participants at Shanghai University of Finance and Economics, University of Calfornia, Berkeley, and Waseda University for their comments. The authors specially thank Yuichiro Kamada, Fuhito Kojima, James Schummer, William Thomson, and Ryan Tierney for their detailed comments. This research was supported by the Joint Usage/Research Center at ISER, Osaka University. The authors acknowledge the Japan Society for the Promotion of Science (Kazumura, 14J05972; Serizawa, 15H03328 and 15H05728).
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Appendix: Proofs
Appendix: Proofs
A Proofs of Lemmas
Proof of Lemma 1:
First, we define \(V_i(A'_i;(A_i,t_i))\) for each \(t_i\in \mathbb {R}\). Take any \(t_i\in \mathbb {R}\). Let \(B\equiv \{t'_i\in \mathbb {R}:(A_i',t'_i)\mathrel {R_i}(A_i,t_i)\}\) and \(W\equiv \{t'_i\in \mathbb {R}:(A_i,t_i)\mathrel {R_i}(A_i',t'_i)\}\). By possibility of compensation, \(B\ne \emptyset \) and \(W\ne \emptyset \). By continuity, B and W are both closed. By money monotonicity, B is bounded above and W is bounded below. Thus, \(\overline{t}_i\equiv \max B\) and \(\underline{t}_i\equiv \min W\) exist.
Suppose \(\overline{t}_i>\underline{t}_i\). Then, by money monotonicity, \((A_i',\underline{t}_i)\mathrel {P_i}(A_i',\overline{t}_i)\mathrel {R_i}(A_i,t_i)\). This implies \(\underline{t}_i\notin W\). This contradicts \(\underline{t}_i= \min W\). Thus, \(\overline{t}_i\le \underline{t}_i\).
Suppose \(\overline{t}_i<\underline{t}_i\). Let \(t'_i\in (\overline{t}_i,\underline{t}_i)\). By \(t'_i<\underline{t}_i\), \(t'_i\notin W\), that is, \((A_i,t_i)~R_{i}~(A_{i}^{\prime },t_{i}^{\prime })\) does not hold. By \(t'_i>\overline{t}_i\), \(t'_i\notin B\), that is, \((A_{i}^{\prime },t_{i}^{\prime })~R_{i}~(A_i,t_i)\) does not hold. This contradicts completeness. Thus, \(\overline{t}_i=\underline{t}_i\).
Let \(V_i(A'_i;(A_i,t_i))\equiv \overline{t}_i=\underline{t}_i\). Then, \(B\cap W=\{V_i(A'_i;(A_i,t_i))\}\) and so \((A_i',V_i(A'_i;(A_i,t_i)))\mathrel {R_i}(A_i,t_i)\) and \((A_i,t_i)\mathrel {R_i}(A_i',V_i(A'_i;(A_i,t_i))),\) which implies \((A_i',V_i(A'_i;(A_i,t_i)))\mathrel {I_i}z_i\). Since \(B\cap W\) is a singleton, such \(V_i(A'_i;(A_i,t_i))\) is unique.
Finally, we show that the function \(V_{i}(A_{i}^{\prime };(A_{i},\cdot ))\) is continuous. Note that it is sufficient to show that for each \(s_{i}\in \mathbb {R}\), the function’s inverse images of \((\infty ,s_{i}]\) and \([s_{i},+\infty )\) are both closed.^{Footnote 21} Take any \(s_{i}\in \mathbb {R}\). Let \(t_{i}\equiv V_{i}(A_{i};(A_{i}^{\prime },s_{i}))\). Then, by money monotonicity and continuity of \(R_{i}\), the inverse images of \((\infty ,s_{i}]\) and \([s_{i},+\infty )\) are \((\infty ,t_{i}]\) and \([t_{i},+\infty )\) respectively, which are closed. Thus, \(V_{i}(A_{i}^{\prime };(A_{i},\cdot ))\) is continuous. \(\square \)
Proof of (i) of Lemma 2:
Suppose that there is \((A_i,t_i)\in D(R_i,p)\) such that \(A_i>1\). By \(R_i\in \mathcal {R}^U\), there is \(a\in A_i\) such that \((a,t_i)\mathrel {R_i}(A_i,t_i)\). By \((A_i,t_i)\in B(p)\), \(t_i=\sum _{b\in A_i}p^b\). By \(p\in \mathbb {R}^m_{++}\) and \(A_i>1\), \(p^a<\sum _{b\in A_i}p^b=t_i\). Thus, by first object monotonicity,
which contradicts \((A_i,t_i)\in D(R_i,p)\). \(\square \)
Proof of (ii) of Lemma 2:
Let \(A'_i\in \mathcal {M}\). If \(A'_i\le 1\), then by the def. of \(A_i\), \((A_i,\sum _{a\in A_i}p^a)\mathrel {R_i}(A'_i,\sum _{a\in A'_i}p^a)\). Suppose \(A'_i>1\). By \(R_i\in \mathcal {R}^U\), there is \(a\in A'_i\) such that \((a,\sum _{b\in A'_i}p^b)\mathrel {R_i}(A_i,\sum _{b\in A'_i}p^b)\). By \(p\in \mathbb {R}^m_{+}\) and money monotonicity, \((a,p^a)\mathrel {R_i}(a,\sum _{b\in A'_i}p^b)\mathrel {R_i}(A'_i,\sum _{b\in A'_i}p^b)\). Thus, by the def. of \(A_i\),
Thus, \((A_i,\sum _{b\in A_i}p^b)\in D(R_i,p)\). \(\square \)
Proof of (i) of Lemma 3:
By contradiction, suppose that \(n>m\) and \(p^a=0\) for some \(a\in M\). Then, by first object monotonicity, for each \(i\in N\), \((a,p^a)\mathrel {P_i}\varvec{0}\). Thus, for each \(i\in N\), \(\varvec{0}\notin D(R_i,p)\), which implies that for each \(((A,t),p)\in W(R)\), \(A_i\ne \emptyset \). However, this contradicts \(n>m\).\(\square \)
Proof of (ii) of Lemma 3:
By \(p\in P(R)\), there is \((A,t)\in Z\) such that \(((A,t),p)\in W(R)\).
Let \(N^*=\{i\in N: A_i>1\}\) and \(i\in N^*\). By \(R_i\in \mathcal {R}^U\), there is \(a\in A_i\) such that \((a,t_i)\mathrel {R_i}(A_i,t_i)\). By \((A_i,t_i)\in B(p)\), \(t_i=\sum _{b\in A_i}p^b\). By \(p\in \mathbb {R}^m_+\), \(p^a\le \sum _{b\in A_i}p^b=t_i\). Thus, by money monotonicity, \((a,p^a)\mathrel {R_i}(a,t_i)\mathrel {R_i}(A_i,t_i)\), which implies \((a,p^a)\in D(R_i,p)\) and \(p^{b}=0\) for each \(b\in A_{i}\ \{a\}\). Hence, for each \(i\in N^*\), there is \(a_i\in A_i\) such that \((a_i,p^{a_i})\in D(R_i,p)\) and \(p^{b}=0\) for each \(b\in A_{i}\backslash \{a_{i}\}\).
Let \(z'\in Z\) be such that for each \(i\in N^*\), \(z'_i=(a_i,p^{a_i})\), and for each \(i\in N{\setminus } N^*\), \(z'_i=(A_i,t_i)\). Then, each agent receives at most one object at \(z'\), and clearly, \((z',p)\in W(R)\).\(\square \)
Proof of Lemma 4:
Suppose by contradiction that \(R_i\in \mathcal {R}^U\) and \(A_i(R)>1\). Then, there is \(a\in A_i(R)\) such that \((a,t_i(R))\mathrel {R_i}f_i(R)\). By \(A_i(R)>1\), there is \(b\in A_i(R)\) such that \(b\ne a\).
By \(n>m\), there is \(j\in N{\setminus } \{i\}\) such that \(A_j(R)=\emptyset \). Let \(z\equiv ((A_k,t_k))_{k\in N}\in Z\) be such that
Clearly, \(\sum _{k\in N}t_k=\sum _{k\in N}t_k(R)\), and for each \(k\in N {\setminus }\{i,j\}\), \(z_k\mathrel {I_k}f_k(R)\). Moreover, \(z_i=(a,t_i(R))\mathrel {R_i}f_i(R)\), and by first object monotonicity, \(z_j=(b,t_j(R))\mathrel {P_j}f_j(R)\). This contradicts efficiency. \(\square \)
Proof of Lemma 8:
Suppose by contradiction that \(V_i(A_i;f_i(R))+V_j(A_j;f_j(R))> t_i(R)+t_j(R)\). Let \(z'\in Z\) be such that \(z'_i=(A_i,V_i(A_i;f_i(R)))\), \(z'_j=(A_j,V_j(A_j;f_j(R)))\), and for each \(k\in N{\setminus }\{i,j\}\), \(z'_k=f_k(R)\). Then \(z'_k\mathrel {I_k}f_k(R)\) for each \(k\in N\). Moreover, \(V_i(A_i;f_i(R))+V_j(A_j;f_j(R))+\sum _{k\ne i,j}t_k(R) >\sum _{k\in N}t_k(R)\). By Remark 2, this contradicts efficiency. \(\square \)
Proof of Lemma 9:
(Fig. 8) Suppose by contradiction that \(t_i(R)<V'_i(A_i(R);\varvec{0})\). If \(A_i(R)=\emptyset \), then \(t_i(R)<V'_i(\emptyset ;\varvec{0})=0\), which contradicts no subsidy for losers. Hence, \(A_i(R)\ne \emptyset \).
Next, we show \(A_i(R'_i,R_{i})\ne \emptyset \). Suppose not. Then, by Lemma 5, \(f_i(R'_i,R_{i})=\varvec{0}\). By \(t_i(R)<V'_i(A_i(R);\varvec{0})\), \(f_i(R)\mathrel {P'_i}\varvec{0}=f_i(R'_i,R_{i})\), which contradicts strategyproofness. Hence \(A_i(R'_i,R_{i})\ne \emptyset \).
By \(R'_i\in \mathcal {R}^U\), \(A_i(R'_i,R_{i})\ne \emptyset \), and Lemma 4, there is \(a\in M\) such that \(A_i(R'_i,R_{i})=a\). Since \(n>m\) and \(A_i(R'_i,R_{i})\ne \emptyset \), there is \(j\in N{\setminus } \{i\}\) such that \(A_j(R'_i,R_{i})=\emptyset \). By Lemma 5, \(f_j(R'_i,R_{i})=\varvec{0}\). Thus, letting \(s_i\equiv V'_i(\emptyset ;f_i(R'_i,R_{i}))\),
This contradicts Lemma 8. \(\square \)
B Proof of Theorem
The proof of the Theorem has six steps.
Step 1: Constructing preferences.
Let \(R_1\equiv R_{0}\). For each \(a\in M\), let \(\mathcal {M}_a\equiv \{A_1\in \mathcal {M}:a\in A_1\}\).
Claim 1: There is \(\underline{t}_1\in \mathbb {R}\) such that \(\underline{t}_1>0\) and
Proof
For each \(a\in M\), let \(g^a\) be a function on \(\mathcal {M}_a\times \mathbb {R}\) such that for each \(A_1\in \mathcal {M}_a\) and each \(t_1\in \mathbb {R}\), \(g^a(A_1,t_1)=V_1(A_1;(a,t_1))t_1\), and let \(g^a_{\max }\) be a function on \(\mathbb {R}\) such that for each \(t_1\in \mathbb {R}\), \(g^a_{\max }(t_1)=\max _{A_1\in \mathcal {M}_a}g^a(A_1,t_1)\).
Note that by Lemma 1, for each \(a\in M\), and each \(A_1\in \mathcal {M}_a\), \(g^a(A_1,\cdot )\) is continuous in \(\mathbb {R}\). Thus, Berge’s maximum theorem implies that for each \(a\in M\), \(g^a_{\max }(\cdot )\) is also continuous in \(\mathbb {R}\).^{Footnote 22} For each \(a\in M\), since \([0,V_1(a;\varvec{0})]\) is compact, there is \(\hat{t}_1^a\in [0,V_1(a;\varvec{0})]\) such that \(g^a_{\max }(\hat{t}_1^a)=\min _{t_1\in [0,V_1(a;\varvec{0})]}g^a_{\max }(t_1)\). For each \(a\in M\), let \(\underline{t}_1^a\equiv g^a_{\max }(\hat{t}_1^a)\). Since M is finite, \(\min \{\underline{t}_1^a:a\in M\}\) exists. Let \(\underline{t}_1\equiv \min \{\underline{t}_1^a:a\in M\}\). Then,
Next, we show \({\underline{t}}_1>0\). Let \(a\in M\) be such that \(\underline{t}_1=\max _{A_1\in \mathcal {M}_a}\{V_1(A_1;(a,\hat{t}_1^a))\hat{t}_1^a\}\). By \(R_1\in \mathcal {R}^M\), there is \(\hat{A}_1\in \mathcal {M}_a\) such that \((\hat{A}_1,\hat{t}_1^a)\mathrel {P_1}(a,\hat{t}_1^a)\), that is, \(V_1(\hat{A}_1;(a,\hat{t}_1^a))\hat{t}_1^a>0\). Thus,
\(\square \)
By Claim 1, there is \(d_*\in \mathbb {R}_{++}\) such that \(5(m1)d_*<\underline{t}_1\) and for each \(a\in M\), \((a,3d_*)\mathrel {P_1}\varvec{0}\). Let \(a^*\in M\) be such that for each \(a\in M{\setminus } \{a^*\}\), \((a^*,3d_*)\mathrel {R_1}(a,3d_*)\). Without loss of generality, assume \(a^*=1\).
Since \(\mathcal {R}\) is rich, there is \(R'_1\in \mathcal {R}\cap \mathcal {R}^U\) such that for each \(a\in M\) and each \(t_1\in \mathbb {R}\), \(V'_1(a;(\emptyset ,t_1))=\frac{d_*}{2}+t_1\). Let
Note that by first object monotonicity, \(\mathcal {M}(R_1,s^*_1)\ne \emptyset \).
Claim 2: There is \(\overline{t}_1\in \mathbb {R}\) such that \(\overline{t}_1\ge \underline{t}_1\) and
Proof
For each \(A_1\in \mathcal {M}(R_1,s^*_1)\), let \(g^{A_1}\) be a function on \(\mathbb {R}\) such that for each \(t_1\in \mathbb {R}\), \(g^{A_1}(t_1)=t_1V_1(\emptyset ;(A_1,t_1))\). Note that by Lemma 1, for each \(A_1\in \mathcal {M}(R_1,s^*_1)\), \(g^{A_1}\) is continuous in \(\mathbb {R}\). For each \(A_1\in \mathcal {M}(R_1,s^*_1)\), since \([s^*_1,V_1(A_1;\varvec{0})]\) is compact, there is \(\hat{t}^{A_1}_1\in [s^*_1,V_1(A_1;\varvec{0})]\) such that \(g^{A_1}(\hat{t}^{A_1}_1)=\max _{t_1\in [s^*_1,V_1(A_1;\varvec{0})]}g^{A_1}(t_1)\). For each \(A_1\in \mathcal {M}(R_1,s^*_1)\), let \(\overline{t}^{A_1}=g^{A_1}(\hat{t}^{A_1}_1)\). Since \(\mathcal {M}(R_1,s^*_1)\) is finite, \(\max \{\overline{t}^{A_1}:A_1\in \mathcal {M}(R_1,s^*_1)\}\) exists. Let \(\overline{t}_1\equiv \max \{\overline{t}^{A_1}:A_1\in \mathcal {M}(R_1,s^*_1)\}\). Then,
Next, we show \(\overline{t}_1\ge \underline{t}_1\). Let \(a\in M \), \(t_1\in [ 0,V_1(a;\varvec{0})]\) and \(A_1\in \mathcal {M}_a\) be such that \(\underline{t}_1=V_1(A_1;(a,t_1))t_1\). Let \(\hat{t}_1\equiv V_1(A_1;(a,t_1))\). Since \(V_1(\emptyset ;(A_1,\hat{t}_1))=V_1(\emptyset ;(a,t_1))\) and since first object monotonicity implies \(t_1V_1(\emptyset ;(a,t_1))> 0\),
By \(t_1\le V_1(a;\varvec{0})\), \(\hat{t}_1\le V_1(A_1;\varvec{0})\). By \(t_1\ge 0\), \(\underline{t}_1>0\), and \(s^*_1\le 0\), \(\hat{t}_1=\underline{t}_1+t_1> 0\ge s^*_1\). Thus, \(s^*_1\le \hat{t}_1\le V_1(A_1;\varvec{0})\). This implies \(A_1\in \mathcal {M}(R_1,s^*_1)\), and thus,
\(\square \)
Let \(d^*\in \mathbb {R}_{++}\) be such that \(d^*>\overline{t}_1\). Note that by Claim 2, \(5(m1)d_*<\underline{t}_1\le \overline{t}_1<d^*\). Since \(\mathcal {R}\) is rich, for each \(i\in \{2,\dots , m\}\), there is \(R_i\in \mathcal {R}\cap \mathcal {R}^U\) satisfying the following conditions:
Since \(\mathcal {R}\) is rich, for each \(i\in N{\setminus } \{1,\dots , m\}\), there is \(R_i\in \mathcal {R}\cap \mathcal {R}^U\) such that for each \(a\in M\) and each \(t_i\in \mathbb {R}\),
Denote \(R\equiv (R_1,\dots , R_n)\). Figure 9 illustrates \(R_i\) and \(R_j\), where \(i\in \{2,\dots , m\}\) and \(j\in \{m+1,\dots , n\}\).
Notice that for each \(a\in M\) and each \(t_1\in \mathbb {R}\),
Thus, by Lemma 9, \(t_1(R)\ge V'_1(A_1(R);\varvec{0})\ge s^*_1\). \(\blacksquare \)
Step 2: For each \(i\in \{2,\dots ,m\}\), \(V_i(\emptyset ;f_i(R))\ge d_*\).
Let \(i\in \{2,\dots ,m\}\). If \(A_i(R)=\emptyset \), then by Lemma 5, \(V_i(\emptyset ;f_i(R))=t_i(R)=0>d_*\). Suppose \(A_i(R)\ne \emptyset \). By \(R_i\in \mathcal {R}^U\) and Lemma 4, there is \(a\in M\) such that \(A_i(R)=a\). Thus, Lemma 9 implies \(t_i(R)\ge 0\).
If \(a=i\), then by money monotonicity, \((i,0)\mathrel {R_i}f_i(R)\), and thus, by ib, \(V_i(\emptyset ;f_i(R))\ge V_i(\emptyset ;(i,0))=d_*\). If \(a\ne i\), then by ic,
which implies \((i,0)\mathrel {P_i}f_i(R)\), and thus, by ib \(V_i(\emptyset ;f_i(R))> V_i(\emptyset ;(i,0))=d_*\). \(\blacksquare \)
Step 3: For each \(i\in \{2,\dots ,m\}\), \(A_i(R)\ne \emptyset \).
Suppose by contradiction that there is an agent \(i\in \{2,\dots , m\}\) such that \(A_i(R)=\emptyset \). By Lemma 5, \(t_i(R)=0\). By Lemma 7, there is an agent \(j\in N{\setminus } \{i\}\) such that \(i1\in A_j(R)\). We show the following claim.
Claim 1: \(t_j(R)V_j(\emptyset ;f_j(R))<d^*\).
Proof
We have three cases.
Case 1: \(j=1\). By Lemma 6 and Lemma 9, \(s^*_1\le t_1(R)\le V_1(A_1(R);\varvec{0})\), implying \(A_1(R)\in \mathcal {M}(R_1,s^*_1)\). Thus, by the def. of \(\overline{t}_1\),
Case 2: \(j\in \{2,\dots , m\}\). By \(R_j\in \mathcal {R}^U\) and Lemma 4, \(A_j(R)=i1\). By Lemma 6, \(t_j(R)\le V_j(i1;\varvec{0})\). Since \(j1\ne i1\), \(V_j(i1;\varvec{0})=4d_*\) or \(d_*\). Thus, \(V_j(i1;\varvec{0})\le 4d_*\). Moreover, by Step 2, \(V_j(\emptyset ;f_j(R))\ge d_* \). Therefore, by \(5d_*<5(m1)d_*<d^*\),
Case 3: \(j\in \{m+1,\dots , n\}\). By \(R_j\in \mathcal {R}^U\) and Lemma 4, \(A_j(R)=i1\). Let \(t^*_j\equiv V_j(\emptyset ;f_j(R))\). By the def. of \(R_j\) and \(3d_*<5(m1)d_*<d^*\),
\(\square \)
By \(f_i(R)=\varvec{0}\), \(V_i(i1;f_i(R))=V_i(i1;\varvec{0})=d^*\). Thus, by Claim 1,
This contradicts Lemma 8. \(\blacksquare \)
Step 4: \(A_1(R)=1\).
First we show the following claim.
Claim 1: For each \(a\in M\), there is \(t(a)\in \mathbb {R}\) such that \((a,t(a))\in o_1(R_{1})\) and
Proof
Let \(a\in M\). Since \(\mathcal {R}\) is rich, there is \(R''_1\in \mathcal {R}\cap \mathcal {R}^U\) such that
Figure 10 illustrates \(R''_1\).
By \((R''_1,R_{1})\in (\mathcal {R}^U)^n\) and Lemma 10, \(f(R''_1,R_{1})\in Z_{\min }^W(R''_1,R_{1})\). Let \(p\equiv p_{\min }(R''_1,R_{1})\). By \((R''_1,R_{1})\in (\mathcal {R}^U)^n\) and Lemma 4, for each \(i\in N\), \(A_i(R''_1,R_{1})\le 1\). In the following three paragraphs, we show \((a,p^a)\in o_1(R_{1})\).
First, suppose \(A_1(R''_1,R_{1})=\emptyset \). By Lemma 5, \(f_1(R''_1,R_{1})=\varvec{0}\). By \(A_1(R''_1,R_{1})=\emptyset \) and Lemma 7, there is an agent \(i\in N{\setminus } \{1\}\) such that \(A_i(R''_1,R_{1})=a\). Since \(f(R''_1,R_{1})\in Z_{\min }^W(R''_1,R_{1})\), \(t_i(R''_1,R_{1})=p^a\). By Lemma 6, \(p^a=t_i(R''_i,R_{i})\le V_i(a;\varvec{0})\). Since \(f(R''_1,R_{1})\in Z_{\min }^W(R''_1,R_{1})\), \(f_1(R''_1,R_{1})=\varvec{0}\in D(R''_1,p)\), and thus, \(\varvec{0}\mathrel {R''_1}(a,p^a)\). Therefore,
This contradicts \(1a''\). Hence, \(A_1(R''_1,R_{1})\ne \emptyset \).
Next, suppose that for some \(b\in M{\setminus } \{a\}\), \(A_1(R''_1,R_{1})=b\). Since \(f(R''_1,R_{1})\in Z_{\min }^W(R''_1,R_{1})\), \(t_1(R''_1,R_{1})=p^b\). By Lemma 6, \(p^b=t_1(R''_1,R_{1})\le V''_1(b;\varvec{0})\). By \(A_1(R''_1,R_{1})\ne \emptyset \) and \(n>m\), there is an agent \(i\in N{\setminus } \{1\}\) such that \(A_i(R''_1,R_{1})=\emptyset \). By Lemma 5, \(f_i(R''_1,R_{1})=\varvec{0}\). Since \(f(R''_1,R_{1})\in Z_{\min }^W(R''_1,R_{1})\), \(f_i(R''_1,R_{1})=\varvec{0}\in D(R_i,p)\), and thus, \(\varvec{0}\mathrel {R_i}(b,p^b)\). Therefore,
This contradicts \(1b''\). Thus for each \(b\in M{\setminus } \{a\}\), \(A_1(R''_1,R_{1})\ne b\).
By \(A_1(R''_1,R_{1})\ne \emptyset \), \(A_1(R''_1,R_{1})\ne b\) for each \(b\in M{\setminus }\{a\}\), and \(A_1(R''_1,R_{1})\le 1\), we conclude that \(A_1(R''_1,R_{1})=a\). Since \(f(R''_1,R_{1})\in Z_{\min }^W(R''_1,R_{1})\), \(t_1(R''_1,R_{1})=p^a\). Hence, \((a,p^a)\in o_1(R_{1})\).
Next, we show that \(p^a\le 3d_*\) if \(a=1\), and \(p^a> 3d_*\) otherwise.
Case 1: \(a=1\). Let \(z\in Z\) be such that for each \(i\in N\),
Let \(\hat{p}\in \mathbb {R}^m_+\) be such that for each \(b\in M\), \(\hat{p}^b=3d_*\). We show that \(z_1\in D(R''_1,\hat{p})\), and for each \(i\in N{\setminus } \{1\}\), \(z_i\in D(R_i,\hat{p})\). This implies that \((z,\hat{p})\in W(R''_1,R_{1})\), and thus, by \(p=p_{\min }(R''_1,R_{1})\), we conclude that \(p^1\le \hat{p}^1=3d_*\).
Note that by \(\hat{p}\in \mathbb {R}^m_{++}\) and (i) of Lemma 2, for each \((A_1,t_1)\in D(R''_1,\hat{p})\), \(A_1\le 1\), and for each \(i\in N{\setminus } \{1\}\) and each \((A_i,t_i)\in D(R_i,\hat{p})\), \(A_i \le 1\).
Let \(i\in N\). We have three subcases.
Subcase 11: \(i=1\). By \(1c''\) and \(\hat{p}\in \mathbb {R}^m_{++}\), for each \(b\in M{\setminus }\{1\}\),
Thus for each \(b\in M{\setminus } \{1\}\), \((1,\hat{p}^1)\mathrel {P'_1}(b,\hat{p}^b)\). Also by \(1c''\) and first object monotonicity, \(V''_1(\emptyset ;(1;\hat{p}^1))=V''_1(\emptyset ;(1,3d_*))<0\), which implies \((1,\hat{p}^1)\mathrel {P'_1}\varvec{0}\). Thus, \(z_1=(1,\hat{p}^1)\in D(R''_1,\hat{p})\).
Subcase 12: \(i\in \{2,\dots ,m\}\). By ic and \(\hat{p}\in \mathbb {R}^m_{++}\), for each \(b\in M{\setminus } \{i\}\),
Thus, for each \(b\in M{\setminus } \{i\}\), \((i,\hat{p}^i)\mathrel {P_i}(b,\hat{p}^b)\). Also by ic and first object monotonicity, \(V_i(\emptyset ;(i,\hat{p}^i))=V_i(\emptyset ;(i,3d_*))<0\), which implies \((i,\hat{p}^i)\mathrel {P_i}\varvec{0}\). Thus, \(z_i=(i,\hat{p}^i)\in D(R_i,\hat{p})\).
Subcase 13: \(i\in \{m+1,\dots ,n\}\). For each \(b\in M\), \(V_i(b;\varvec{0})=3d_*=\hat{p}^b\). This implies \(\varvec{0}\mathrel {R_i}(b,\hat{p}^b)\). Thus, \(z_i=\varvec{0}\in D(R_i,\hat{p})\).
Case 2: \(a\in \{2\dots ,m\}\). Let \(i=a\). Suppose by contradiction that \(p^a\le 3d_*\). Note that by \(n<m\) and (i) of Lemma 3, \(p\in \mathbb {R}_{++}^m\). Thus by (i) of Lemma 2, for each \((A_i,t_i)\in D(R_i,p)\), \(A_i\le 1\). By \(p^a\le 3d_*\), \(i=a\), ic, and \(p\in \mathbb {R}^m_{++}\), for each \(b\in M{\setminus } \{a\}\),
Thus for each \(b\in M{\setminus } \{a\}\), \((a,p^a)\mathrel {P_i}(b,p^b)\). Also by \(i=a\), \(p^a\le 3d_*\), ic, and first object monotonicity, \(V_i(\emptyset ;(a,p^a))\le V_i(\emptyset ;(i,3d_*))<0\), which implies \((a,p^a)\mathrel {P_i}\varvec{0}\). Therefore, \(D(R_i,p)=\{(a,p^a)\}\). Since \(f(R''_1,R_{1})\in Z_{\min }^W(R''_1,R_{1})\), \(A_i(R''_1,R_{1})=a\). This contradicts \(A_1(R''_1,R_{1})=a\). \(\square \)
Recall that \((1,3d_*)\mathrel {P_1}\varvec{0}\) and for each \(a\in M{\setminus } \{1\}\), \((1,3d_*)\mathrel {R_1}(a,3d_*)\). By Claim 1, \(t(1)\le 3d_*\) and for each \(a\in M{\setminus } \{1\}\), \(t(a)>3d_*\). Thus,
and for each \(a\in M{\setminus }\{1\}\),
Therefore, (ii) of Lemma 11 implies \(A_1(R)\ne \emptyset \) and for each \(a\in M{\setminus } \{1\}\), \(A_1(R)\ne a\). By Step 3, \(A_1(R)\le 1\), because otherwise there exists an agent \(i\in \{2,\dots , m\}\) such that \(A_i(R)=\emptyset \), which contradicts Step 3. Hence, \(A_1(R)=1\). \(\blacksquare \)
Step 5: For each \(i\in \{2,\dots , m\}\), \(A_i(R)=i\).
We show that for each \(i\in \{2,\dots , m\}\), \(A_i(R)\subseteq \{i,i1\}\). Then, since \(A_1(R)=1\) by Steps 3 and 4, we conclude that for each \(i\in \{2,\dots , m\}\), \(A_i(R)=i\).
Suppose by contradiction that there is \(i\in \{2,\dots , m\}\) such that \(A_i(R)\not \subseteq \{i1, i\}\). By Lemma 4, there is \(a\in M{\setminus } \{i1, i\}\) such that \(A_i(R)=a\). By \(a\in M{\setminus } \{i1,i\}\) and ia, \(V_i(a;\varvec{0})=d_*\). Thus, by Step 2 and Lemma 6,
Let \(j\in \{m+1,\dots , n\}\). By Step 3 and Step 4, \(A_j(R)=\emptyset \), and thus, by Lemma 5, \(f_j(R)=\varvec{0}\). Thus, by the def. of \(R_j\),
Therefore, by (1), (2), and \(t_j(R)=0\),
This contradicts Lemma 8. \(\blacksquare \)
Step 6: Completing the proof.
For each \(i\in \{2,\dots , m\}\), by Step 2, \(V_i(\emptyset ;f_i(R))\ge d_*\), by ia, \(V_i(i;\varvec{0})=4d_*\), and by Lemma 6 and Step 5, \(t_i(R)\le V_i(A_i(R);\varvec{0})=V_i(i;\varvec{0})\). Thus, for each \(i\in \{2,\dots , m\}\),
By \(A_1(R)=1\), there is \(A_1\in \mathcal {M}\) such that \(1\in A_1\) and
By \(A_1(R)=1\), Lemma 6, and Lemma 9, \(t_1(R)\in [0,V_1(1;\varvec{0})]\). Thus, by the def. of \(\underline{t}_1\),
Let \(N'\equiv \{i\in \{2,\dots ,m\}: A_i(R)\subseteq A_1\}\). Let \(z'\equiv ((A'_i,t'_i))_{i\in N}\in Z\) be such that
Figure 11 is an illustration of \(z'\) when \(A_1=\{1,2,3\}\).
Clearly, for each \(i\in N\), \(z'_i\mathrel {I_i}f_i(R)\). By Step 5 and \(1\in A_1\), \(N'=A_11\le m1\). Thus, \(5N'd_*\le 5(m1)d_*<\underline{t}_1\), Therefore, by (3), (4), and \(5N'd_*<\underline{t}_1\),
This contradicts efficiency. \(\blacksquare \)
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Kazumura, T., Serizawa, S. Efficiency and strategyproofness in object assignment problems with multidemand preferences. Soc Choice Welf 47, 633–663 (2016). https://doi.org/10.1007/s0035501609868
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Keywords
 Strategyproofness
 Efficiency
 Multidemand preferences
 Unitdemand preferences
 Nonquasilinear preferences
 Minimum price Walrasian rule