Matching with restricted trade

Abstract

Motivated by various trade restrictions in real-life object allocation problems, we introduce an object allocation with a particular class of trade restrictions model. The set of matchings that can occur through a market-like process under such restrictions is defined, and each such matching is called feasible. We then introduce a class of mechanisms, which we refer to as “Restricted Trading Cycles” (RTC). Any RTC mechanism is feasible, constrained efficient, and respects endowments. An axiomatic characterization of RTC is obtained, with feasibility, constrained efficiency, and a new property that we call hierarchically mutual best. In terms of strategic issues, feasibility, constrained efficiency, and respecting endowments together turns out to be incompatible with strategy-proofness. This in particular implies that no RTC mechanism is strategy-proof. Lastly, we consider a probabilistically restricted trading cycles (PRTC) mechanism, which is obtained by introducing a certain randomness to the RTC class. While PRTC continues to be manipulable, compared to RTC, it is more robust to truncations and reshufflings.

Appendix

Proof of Theorem 1

Let $\psi$ be a RTC mechanism. We first show that it terminates in a finite time. The first stage of $\psi$ iteratively matches agents with their top choices once they are endowed with them and removes the assigned agents as well as their assignments from the problem. It terminates whenever there is no such agent left. From here, because there are finitely many agents, we conclude that Step 1 terminates in a finitely many round and gives us a submatching. Then, the second stage of $\psi$ considers the reduced problem, emerging after removing the assigned agents along with their assignments in the first stage. To show that the second stage gives us a matching in the reduced problem, it is enough to show that there always exists an implementable collection of cycles and null-pairs in the reduced economy. This is what we show in the following.

Let $N'$ and $\mu ^{N'}$ be the set of assigned agents and the associated submatching in Step 1 of $\psi$. Let us first enumerate the remaining agents $N{\setminus } N'=\{i_1,\ldots ,i_n\}$. Then, start with agent $i_1$. If his favorite object in $I_{i_1}(\mu ^{N'})$ is the null-object, then let him form a null-pair. Otherwise, let him form a cycle with his favorite object (note that the null-object always belongs to $I_{i_1}(\mu ^{N'})$). Let $\mu ^{N'\cup \{i_1\}}$ be the obtained submatching after implementing the null-pair or the cycle over $\mu ^{N'}$. Next, let us consider agent $i_2$ and his favorite object in $I_{i_2}(\mu ^{N'\cup \{i_1\}})$, and repeat the same arguments for him, and so on. If we consider the permutation that orders the collection of these cycles and null-pairs as the same as the associated agents’ indexes, then the collection becomes implementable under this permutation. Therefore, in Step 2 of $\psi$, there always exists an implementable collection of cycles and null-pairs, hence a matching in the reduced problem is always obtained. This, along with our finding above, shows that $\psi$ terminates in a finitely many round, with producing a matching.

Next, we show that $\psi$ is feasible. Consider a problem P, and suppose that the set of agents that are assigned within Step 1 of $\psi$ is non-empty. Among them, consider the ones that are matched in the first substep of Step 1. By its definition, if i is such an agent, then he is assigned to his top object, say c, and $c\in I_i(\mu ^{\emptyset })$, which means that $i\in \Gamma _c(\mu ^{\emptyset })$. Hence, the agent-object pair (i, c) constitutes a cycle that is viable after $\mu ^{\emptyset }$. Let us now suppose that there exists another such agent-object pair, say $(j,c')$. That is, agent j is assigned object $c'$ in the first substep of Step 1. As the same as before, it implies that $j\in \Gamma _{c'}(\mu ^{\emptyset })$; hence $(j,c')$ constitutes a cycle that is viable after $\mu ^{\emptyset }$. On the other hand, by our supposition, agent j continues to keep his object $c'$ endowment even after the submatching formed by the agent i’s object c assignment. Therefore, each of such cycles continues to be viable even after the implementation of others in any order.

Let $N'$ be the set of agents that are assigned in the first substep of Step 1, and $\mu ^{N'}$ be the associated submatching. Then, in the second substep of Step 1 (if it takes place), only the agents who inherit their top choices are assigned. That is, if i is an assigned agent to some object c, then c is his top choice, and moreover, $c\in I_i(\mu ^{N'})$. That is, $i\in \Gamma _c(\mu ^{N'})$. Hence, (i, c) constitutes a cycle that is viable after $\mu ^{N'}$. Moreover, by the same reasoning as above, any such cycle continues to be viable even after the implementation of the others in any order.

The same arguments for the other substeps of Step 1 show that Step 1 of $\psi$ does not violate feasibility. Once Step 1 is finalized, in Step 2, an implementable collection of cycles and null-pairs is selected. By Definition 3, it is immediate to verify that Step 2 is compatible with feasibility as well. All of these show that $\psi$ is feasible.

Because of the feasibility of $\psi$ and our supposition as to $\Gamma$, for any agent i, if $c\in I_i(\mu ^{\emptyset })$, then $c\in I_i(\mu ^{N'})$ where $i\notin N'$ and $\mu ^{N'}$ is any submatching that occurs in the course of $\psi$. This, along with the definition of implementable collection of cycles and null-pairs, implies that no agent i is assigned with an object which is worse than any object in $I_i(\mu ^{\emptyset })$, which means that $\psi (P)$ respects endowments.

All the agents that are assigned in Step 1 receive their top objects. Moreover, by construction, the selected implementable collection of cycles and null-pairs in Step 2 of $\psi$ produces a matching in the associated reduced problem such that no other such collection produces a matching that dominates the former. These two facts together implies that $\psi (P)$ is a constrained efficient matching, which finishes the proof. $\square$

Proof of Theorem 2

“If” Part. Let $\psi$ be a RTC mechanism. By the definition of Step 1, $\psi$ satisfies hierarchically mutual best. From Theorem 3, $\psi$ is both feasible and constrained efficient.

“Only If” Part. Let $\psi$ be a mechanism that is feasible, constrained efficient, and that satisfies hierarchically mutual best. Consider a problem P. As there is no multiplicity in Step 1 of the RTC, any RTC mechanism’s Step 1 outcome is the same. By its definition, moreover, for any Substep 1.k ($k\in \{1,\ldots ,|N|\}$), any agent $i\in \cup _{l=1}^{|N|} Z_l$ is matched with his top object in Step 1. As $\psi$ satisfies hierarchically mutual best, any such agent is matched with his top choice at $\psi$ as well.

Because of the feasibility and constrained efficiency of $\psi$, the assignments of the rest of the agents, that is, $N{\setminus } \cup _{k=1}^{k=|N|} Z_k$, under $\psi$ can be obtained by implementing a collection of cycles and null-pairs such that it is implementable in the reduced problem after the agents in $\cup _{k=1}^{k=|N|} Z_k$ receive their top choices, and its induced matching is not dominated by that of any other such collection in the reduced problem. But then, it means that the assignments of the agents in $N{\setminus } \cup _{k=1}^{k=|N|} Z_k$ at $\psi (P)$ coincide with the Step 2 outcome of some RTC mechanism at P. This, along with the above observation, shows that $\psi$ is a RTC mechanism, which finishes the proof. $\square$

The Independence of the Theorem 2 Axioms Let $N=\{i,j\}$ and $O=\{a\} \cup \{\emptyset \}$, with $q_a=1$. Object a has a ranking order, which generates $\Gamma _a$. Let the preferences and ranking order be as follows:

$$\begin{aligned} P_i=P_j:\ a, \emptyset ; \succ _a:\ i, j. \end{aligned}$$

Suppose that there is no trade restriction, that is, $\tau _k(a)=N$ for any $k\in N$. Consider a mechanism $\psi$ such that at this problem P, $\psi _i(P)=\emptyset$ and $\psi _j(P)=a$. Suppose that at any other problem, $\psi$ gives the same outcome as a RTC mechanism. Consequently, $\psi$ is feasible and constrained efficient. Yet, it does not satisfy hierarchically mutual best (indeed, it does not respect endowments because $a\in I_i(\mu ^{\emptyset })$, yet $a P_i \psi _i(P)$).

With the same set of agents, let us now consider $O=\{a,b\} \cup \{\emptyset \}$, with $q_a=q_b=1$. The preferences and object ranking orders are as follows:

$$\begin{aligned}&P_i:\ a, b, \emptyset ; P_j:\ b, a, \emptyset .\\&\qquad \succ _a:\ j, i; \succ _b:\ i, j. \end{aligned}$$

There is no trade restriction (that is, $\tau _k(c)=N$ for any $k\in N$ and $c\in O$). Consider a mechanism $\psi$ such that $\psi _i(P)=b$ and $\psi _j(P)=a$. Suppose that at any other problem, $\psi$ gives the same outcome as a RTC mechanism. Hence, $\psi$ is feasible and satisfies hierarchically mutual best, yet it is not constrained efficient.

Let us now introduce a trade restriction to the above problem by letting $\tau _{i}(a)=\{i\}$. Consider a mechanism $\psi$ such that $\psi _i(P)=a$ and $\psi _j(P)=b$. Suppose that at any other problem, $\psi$ gives the same outcome as a RTC mechanism. In this case, $\psi$ is constrained efficient and satisfies hierarchically mutual best, yet it is not feasible.

Proof of Proposition 5

For ease of notation, let us write $\psi$ for PRTC. We first show that $\psi$ is not strongly manipulable via truncation at any problem P. Let $P'_i$ be a truncation of $P_i$ and $P'=(P'_i,P_{-i})$, and let $\sigma =\psi (P)$ and $\sigma '=\psi (P')$. Suppose that $\sigma '$ first order stochastically dominates $\sigma$ with respect to $P_i$.

First of all, at problem P, if agent i receives his assignment in Step 1 of $\psi$, then this implies that he is assigned to his top object with probability one. Hence, in this case, he cannot be better off via truncating his preferences.

Let us suppose that it is not the case, that is, at problem P, agent i receives his assignment in Step 2 of $\psi$. Here, we have two cases. Consider that at $P'$, agent i receives his assignment, say c, in Step 1 of $\psi$. This implies that he inherits (or is initially endowed with) his assigned object c in some substep of Step 1. As $P_{-i}=P'_{-i}$, he inherits object c at P in some substep of Step 1 as well. But then, by the definitions, any implementable collection of cycles and null-pairs that arises in Step 2 of $\psi$ at P gives agent i an object that is not worse than object c. This in turn implies that he does not receive any object that is worse than object c with a positive probability under $\sigma$, which shows that he cannot benefit from reporting $P'_i$.

Let us now consider the other case where at $P'$, agent i receives his assignment in Step 2 of $\psi$. If the sets of implementable collection of cycles and null-pairs that arise in Step 2 of $\psi$ at P and $P'$ are the same, then agent i’s assignment is the same at both problems. Suppose it is not the case. Let us first observe that because $P_{-i}=P'_{-i}$, and agent i is not assigned in Step 1 of $\psi$ at both problems, the Step 1 assignments of $\psi$ are the same at both problems; hence so are the reduced problems in Step 2 of $\psi$.

Next, observe that in Step 2, any implementable collection of cycles and null-pairs at P that matches agent i with an object in $Ac(P'_i)\cup \{\emptyset \}$ will continue to arise in Step 2 at $P'$. However, any such collection that matches agent i with an object in $Ac(P_i){\setminus } Ac(P'_i)$ will no longer arise in Step 2 at $P'$. This is because agent i, under $P'_i$, finds the null-object better than any object in $Ac(P_i){\setminus } Ac(P'_i)$. Moreover, as $Ac(P'_i) \subset Ac(P_i)$, any implementable collection of cycles and null-pairs that occurs in Step 2 of $\psi$ at $P'$ and that assigns agent i to an object in $Ac(P'_i)$ appears at P as well. All of these show that in order for $P'_i$ to be beneficial, there has to exist an implementable collection of cycles and null-pairs that arises at P, but not at $P'$ (because, otherwise, the assignment of agent i would be the same at both P and $P'$ under $\psi$). By our observations, any such collection gives agent i an object in $Ac(P_i){\setminus } Ac(P'_i)$, and as stated above, the only reason for it not to appear at $P'$ is that agent i finds the null-object better than any object in $Ac(P_i){\setminus } Ac(P'_i)$. All of these observations imply that $\sigma '_{i,\emptyset } > \sigma _{i,\emptyset }$. This, along with the fact that no agent receives an unacceptable object of himself with a positive probability under $\psi$,¹⁶ implies that for the least preferred acceptable object (with respect to $P_i$), say c, we have $\sum _{c'\in O:\ c' R_i c} \sigma _{i,c'} > \sum _{c'\in O:\ c' R_i c} \sigma '_{i,c'}$. This contradicts our supposition that $\sigma '$ first order stochastically dominates $\sigma$ with respect to $P_i$.

Let us now show that $\psi$ is not strongly manipulable via reshuffling. First, we need to introduce some notations. For any object c, $U(P_i,c)=\{c'\in O: c' R_i c\}$ and $SU(P_i,c)=U(P_i,c){\setminus } \{c\}$. Assume now that $P'_i$ is a reshuffling of $P_i$, and let $\sigma =\psi (P)$ and $\sigma '=\psi (P'_i,P_{-i})$. Assume for a contradiction that $\sigma '$ first order stochastically dominates $\sigma$ with respect to $P_i$.

By the same arguments above, if agent i is matched in Step 1 of $\psi$ at either (or both) P or $P'$, then we have the result. Hence, let us consider the case where he receives his assignment in Step 2 of $\psi$ at both problems P and $P'$. As the same as above, it implies that the reduced problems in Step 2 of $\psi$ are the same at both P and $P'$.

Let us first observe that swapping the places of a pair of objects at $P_i$ can only be beneficial only if it decreases the number of implementable collection of cycles and null-pairs that arise in Step 2 of $\psi$. To see this, let us suppose that $P'_i$ is such that (i) $Ac(P_i)=Ac(P'_i)$, (ii) for a particular pair of objects $c,c'$, $c' P'_i c$ whereas $c P_i c'$, and (iii) the ranking of every object remains the same. Any implementable collection of cycles and null-pairs that arises in Step 2 of $\psi$ at P that gives agent i an object $d\in U(P'_i,c')$ continues to arise in Step 2 of $\psi$ at $P'$. However, any such collection that gives an agent i an object in $U(P'_i,c){\setminus } U(P'_i,c')$ may disappear. The only reason for it to disappear at $P'$ is that agent i reports object $c'$ better than any object in $U(P'_i,c){\setminus } U(P'_i,c')$; thereby it may no longer produce a constrained efficient matching in the reduced problem. Moreover at $P'$, some new implementable collections of cycles and null-pairs that give object $c'$ to agent i may arise. These, along with the fact that each implementable collection of cycles and null-pairs is chosen with the equal probability in Step 2 of $\psi$, implies that the only way for $P'_i$ to be profitable to agent i is to have a fewer such collections so that the probabilities of getting objects in $U(P_i,c)$ may increase.

Let us now suppose that $P'_i$ is any reshuffling of $P_i$. As $P'_i$ can be obtained by iteratively swapping the places of objects at $P_i$, the above reasoning can be applied iteratively to conclude that the same is true for $P'_i$. That is, in order for $P'_i$ to be beneficial for agent i, the number of implementable collection of cycles and null-pairs in Step 2 of $\psi$ has to decrease at $P'$.

Let W be the set of objects such that $c\in W$ whenever there exists an implementable collection of cycles and null-pairs in Step 2 of $\psi$ at P that assigns agent i to object c, yet it does not constitute such a collection in Step 2 of $\psi$ at $P'$. By our observation above, $W\ne \emptyset$. Let $c'\in W$ such that it is the worst ranked one with respect to $P_i$. This implies that any implementable collection of cycles and null-pairs in Step 2 of $\psi$ at P and assigns an object which is worse than $c'$ (with respect to $P_i$) continues to arise in Step 2 of $\psi$ at $P'$.

Suppose that there are x many implementable collections of cycles and null-pairs in Step 2 of $\psi$ at P that give agent i an object that is not worse than $c'$. Moreover, suppose that there are in total $n_1$ many implementable collections of cycles and null-pairs arising in Step 2 of $\psi$ at P. Then, $\sum _{c R_i c'} \sigma _{i,c}=x/n_1$. On the other hand, let k many such collections disappear at $P'$. Then, we have $\sum _{c R_i c'} \sigma '_{i,c}=(x-k)/(n_1-k)$. It is immediate to verify that $x/n_1-(x-k)/(n_1-k) \ge 0$. If it is strict, then we are done as it yields a contradiction to our supposition that $\sigma '$ first order stochastically dominates $\sigma$ with respect to $P_i$. Otherwise, that difference is exactly equal to zero if and only if $x=n_1$. But then, it means that agent i does not receive any object that is worse than object $c'$ with a positive probability under $\sigma$, which implies that $\sum _{c R_i c'} \sigma _{i,c}=1$. However, at $P'$, some implementable collection of cycles and null-pairs that gives object $c'$ to agent i disappears. The only reason for it is that agent i reports an object that is actually worse than object $c'$ as a better alternative at $P'$. This, along with the fact that $c'$ is the least preferred object in W (with respect to $P_i$), implies that under $\sigma '$, agent i receives an object that is worse than object $c'$ with some positive probability. Hence, $\sum _{c R_i c'} \sigma '_{i,c} < 1$. Therefore, $\sum _{c R_i c'} \sigma _{i,c} > \sum _{c R_i c'} \sigma '_{i,c}$, showing that $\psi$ is not strongly manipulable via reshuffling.

In the rest of the proof, we will show that $\psi$ is weakly manipulable via truncation and reshuffling. To see this, let us consider a problem instance where $N=\{i_1,i_2,i_3,i_4,i_5,i_6\}$ and $O=\{c_1,c_2,c_3,c_4,c_5\}$, each with unit capacity, except $q_{c_4}=2$. The preferences and object ranking orders are as follows:

$$\begin{aligned}&P_{i_1}:\ c_1, c_2, c_3, c_4, \emptyset ; P_{i_2}:\ c_1, c_3, \emptyset ; P_{i_3}:\ c_2, c_3, c_1, \emptyset ; P_{i_4}=P_{i_5}:\ c_5, c_4, \emptyset ; P_{i_6}:\ c_4, c_5, \emptyset .\\&\quad \quad \quad \qquad \qquad \succ _{c_1}:\ i_3,\ldots ; \succ _{c_2}:\ i_1,\ldots ; \succ _{c_3}:\ i_2,\ldots ; \succ _{c_4}:\ i_1,i_4,i_5..; \succ _{c_5}:\ i_6,\ldots \end{aligned}$$

Suppose that for any agent-object pair (i, c), $\tau _i(c)=N$. That is, there is no trade restriction. Then, at the true preference profile P, agent $i_1$’s assignment is as follows: $\psi (P)_{i_1,c_1}=1/2$ and $\psi (P)_{i_1,c_2}=1/2$. Now, consider the false preferences $P'_{i_1}:\ c_1, c_4, c_2, c_3, \emptyset$; and let $P'=(P'_{i_1},P_{-i_1})$. Note that $P'_{i_1}$ is a reshuffling of $P_{i_1}$. At $P'$, $\psi (P')_{i_1,c_1}=2/3$ and $\psi (P')_{i_1,c_4}=1/3$. Hence, $\psi$ is weakly manipulable via reshuffling.

For the weak manipulation via truncation, consider the same set of agents and objects, but each with unit capacity now. Let the preferences and ranking orders be as follows:

$P_{i_1}:\ c_1, c_2, \emptyset$; $P_{i_2}:\ c_1, c_3, \emptyset$; $P_{i_3}:\ c_2, c_3, c_1, \emptyset$; $P_{i_4}:\ c_2, c_4, c_5, c_6, \emptyset$; $P_{i_5}:\ c_6, c_4, \emptyset$; $P_{i_6}:\ c_6, c_5, \emptyset$.

$\succ _{c_1}:\ i_3,\ldots$; $\succ _{c_2}:\ i_1,i_4,\ldots$; $\succ _{c_3}:\ i_2,\ldots$; $\succ _{c_4}:\ i_5,\ldots$; $\succ _{c_5}:\ i_6,\ldots$; $\succ _{c_6}:\ i_4,i_5,\ldots$

There is no trade restriction, as above. Under the true preference profile, $\psi (P)_{i_1,c_1}=1/2$ and $\psi (P)_{i_1,c_2}=1/2$. Let us consider the following truncation: $P'_{i_1}:\ c_1, \emptyset$, and let $P'=(P'_i,P_{-i})$. Then, under $P'$ : $\psi (P')_{i_1,c_1}=2/3$ and $\psi (P')_{i_1,\emptyset }=1/3$. Hence, $\psi$ is weakly manipulable via truncation. $\square$