Characterizations of the cumulative offer process

Abstract

In the matching with contracts setting, we provide new axiomatic characterizations of the “cumulative offer process” (${ COP}$) in the domain of hospital choice functions that satisfy “unilateral substitutes” and “irrelevance of rejected contracts.” We say that a mechanism is truncation-proof if no doctor can ever benefit from truncating his preferences. Our first result shows that the ${ COP}$ is the unique stable and truncation-proof mechanism. Next, we say that a mechanism is invariant to lower-tail preference change if no doctor’s assignment changes after he changes his preferences over the contracts that are worse than his assignment. Our second result shows that a mechanism is stable and invariant to lower-tail preference change if and only if it is the ${ COP}$. Lastly, by extending Kojima and Manea’s (Econometrica 78:633–653, 2010) result, we show that the ${ COP}$ is the unique stable and weakly Maskin monotonic mechanism.

Appendix

Proof of Lemma 1

• (i) Let $\psi$ be a mechanism that satisfies weak Maskin monotonicity. Let us consider a problem P, doctor d, and $P'_d$ such that $P_{d_{|_{U(P_d,\psi _d(P))}}}=P'_{d_{|_{U(P_d,\psi _d(P))}}}$. This implies that $P'_d$ is a monotonic transformation of $P_d$ at $\psi _d(P)$. Therefore, $P'=(P'_d,P_{-d})$ is a monotonic transformation of P at $\psi (P)$. By weak Maskin monotonicity, $\psi _d(P'_d,P_{-d}) R'_d \psi _d(P)$. This also implies that $P_d$ is a monotonic transformation of $P'_d$ at $\psi _d(P'_d,P_{-d})$. Therefore, P is a monotonic transformation of $(P'_d,P_{-d})$ at $\psi (P'_d,P_{-d})$. Then, by weak Maskin monotonicity, $\psi _d(P) R_d \psi _d(P'_d,P_{-d})$. These altogether shows that $\psi _d(P'_d,P_{-d})=\psi _d(P)$. Hence, $\psi$ is invariant to lower-tail preference change.

• (ii) Let $\psi$ be a mechanism that is both individually rational for doctors and invariant to lower-tail preference change.¹⁶ Assume for a contradiction that it is not truncation-proof. That is, there exist a problem instance P, doctor d, and truncation $P'_d$ of $P_d$ such that $\psi _d(P)=x$ and $\psi _d(P'_d,P_{-d})=x'$ with $x' P_d x$. Due to the individual rationality for doctors property of $\psi$ and the manipulation via truncation $P'_d$, there exists a contract $x''\ne x'$ with $x''_D=d$ and $x' P_d x'' P_d \emptyset$ (it may be that $x=x''$).

Let $P''_{d}$ be the truncation such that $\tilde{x}\notin Ac(P''_{d})$ if and only if $\tilde{x}_D=d$ and $x' P_d \tilde{x}$. Then, due to the invariance to lower-tail preference change property of $\psi$, we have $\psi _{d} (P''_{d},P_{-d})=x'$. This, along with $\psi _d(P)=x$, contradicts $\psi$ being invariant to lower-tail preference change. $\square$

Proof of Theorem 1

Due to Lemma 1, we only need to show that $(i)\ \Rightarrow \ (iv)$ and $(ii)\ \Rightarrow \ (i)$. Let us first show the former. That is, we want to show that under ${ US}$ and ${ IRC}$, the COP is stable and satisfies weak Maskin monotonicity. From Hatfield and Kojima (2010) and Aygün and Sönmez (2012), we already know that the COP is stable under ${ US}$ and ${ IRC}$. For weak Maskin monotonicity, let $P'$ be a monotonic transformation of P at ${ COP}(P)=X'$. By definition, $X'$ continues to be stable at $P'$. Let ${ COP}(P')=X''$. From Hatfield and Kojima (2010), we know that the COP produces the doctor-optimal stable allocation under ${ US}$ and ${ IRC}$. This, along with the stability of $X'$, implies that $X'' R'_d X'$ for any doctor $d\in D$. Hence, COP satisfies weak Maskin monotonicity.

Let us now show that $(ii)\ \Rightarrow \ (i)$. That is, we want to prove that if a mechanism is stable and truncation-proof, then it is nothing but the COP. Let $\psi$ be a stable and truncation-proof mechanism. Assume for a contradiction that $\psi \ne COP$, and let P be such that $\psi (P)\ne COP(P)$. For ease of notation, let $\psi (P)=X'$ and ${ COP}(P)=X''$. For any allocation $\tilde{X}$, similar to the doctor side, let $\tilde{X}_h$ be the set of hospital h’s contracts at $\tilde{X}$.

Let $S^0=\left\{ d\in D:\ X'' P_d X'\right\}$. From Hatfield and Kojima (2010), $X''$ is the doctor-optimal stable allocation. Thus, since $X'$ is also stable, we know that for every doctor $X''$ is at least as good as $X'$. This, along with $X'\ne X''$, implies that $S^0$ is a non-empty set. Let $d_1\in S^0$ and consider $P'_{d_1}$ that is the truncation of $P_{d_1}$ such that $x\notin Ac(P'_{d_1})$ if and only if $x_D=d_1$ and $X''_{d_1} P_{d_1} x$. Let $P^1=(P'_{d_1}, P_{-d_1})$. Note that as $X''_{d_1} P_{d_1} X'_{d_1} R_{d_1} \emptyset$ (the latter is due to the stability of $X''$), $(X''_{d_1})_H=h$ for some hospital h.

From Hatfield and Kojima (2010) and Aygün and Sönmez (2012), under ${ US}$ and ${ IRC}$, no previously rejected contract is accepted in a later step in the course of the COP. In other words, the COP coincides with the ${ DA}$. This simply implies that ${ COP}(P^1)=X''$. On the other hand, due to the truncation-proofness and the stability of $\psi$, we have $\psi _{d_1}(P^1)=\emptyset$.

Let $S^1=\left\{ d\in D:\ X'' P_d \psi (P^1)\right\} \setminus \left\{ d_1\right\}$. We now claim that $S^1$ is non-empty. As ${ COP}(P^1)=X''$ and the COP produces the doctor-optimal stable allocation under ${ US}$ and ${ IRC}$, we have $X'' R_d \psi (P^1)$ for any $d\in D$. If $S^1=\emptyset$, then by strictness of the doctor preferences, $X''_d=\psi _d(P^1)$ for any $d\in D\setminus \left\{ d_1\right\}$.¹⁷ From above, we also have $\psi _{d_1}(P^1)=\emptyset$. Let us now consider the set of contracts $\psi (P^1) \cup \left\{ X''_{d_1}\right\}$. As $X''_d=\psi _d(P^1)$ for any $d\in D\setminus \left\{ d_1\right\}$, $X''=\psi (P^1) \cup \left\{ X''_{d_1}\right\}$. From the stability of $X''$ and $X''_{d_1} P'_{d_1} \emptyset$, we have $X''_h=C_h(\psi (P^1) \cup \{X''_{d_1}\})\subseteq C_D\left( \psi (P^1) \cup \left\{ X''_{d_1}\right\} \right)$, contradicting the stability of $\psi (P^1)$ at $P^1$. Therefore, $S^1$ is non-empty.

Let $d_2\in S^1$ and, similar to above, consider $P'_{d_2}$ that is the truncation of $P_{d_2}$ such that $x\notin Ac(P'_{d_2})$ if and only if $x_D=d_2$ and $X''_{d_2} P_{d_2} x$. As the same as before, since $X'' P_{d_2} \psi (P^1) R_{d_2} \emptyset$, we have $(X''_{d_2})_H=h$ for some hospital h. Let us define $P^2=(P'_{d_2},P^1_{-d_2})$. By the same reason as above, ${ COP}(P^2)=X''$ and $\psi _{d_2}(P^2)=\emptyset$. Moreover, by the definition of $P^2$ and the fact that COP is the doctor-optimal stable mechanism, either $\psi _{d_1}(P^2)=\emptyset$ or $\psi _{d_1}(P^2)=X''_{d_1}$.

Let us now define $S^2=\left\{ d\in D:\ X'' P_d \psi (P^2)\right\} \setminus \left\{ d_1,d_2\right\}$. Similar to above, we claim that $S^2$ is non-empty. As ${ COP}(P^2)=X''$ and it is the doctor-optimal stable allocation, $X'' R_d \psi (P^2)$ for any $d\in D$. If $S^2=\emptyset$, then $X''_d=\psi _d(P^2)$ for any $d\in D\setminus \left\{ d_1,d_2\right\}$. From above, we also have $\psi _{d_2}(P^2)=\emptyset$. We have two cases to consider. Let us first consider the case where $\psi _{d_1}(P^2)=X''_{d_1}$. In this case, similar to above, consider the set of contracts $X''=\psi (P^2) \cup \left\{ X''_{d_2}\right\}$. Then, from the stability of $X''$ and the fact that $X''_{d_2} P'_{d_2} \emptyset$, we have $X''_h=C_h\left( \psi (P^2) \cup \left\{ X''_{d_2}\right\} \right) \subseteq C_D\left( \psi (P^2) \cup \left\{ X''_{d_2}\right\} \right)$, contradicting the stability of $\psi (P^2)$ at $P^2$.

Let us consider the other case of $\psi _{d_1}(P^2)=\emptyset$. If $(X''_{d_1})_H=h$, then $X''=\psi (P^2) \cup \left\{ X''_{d_1}\right\} \cup \left\{ X''_{d_2}\right\}$. Due to the same arguments as above, $X''_h=C_h\left( \psi (P^2)\cup \left\{ X''_{d_1}\right\} \cup \left\{ X''_{d_2}\right\} \right) \subseteq C_D\left( \psi (P^2)\cup \left\{ X''_{d_1}\right\} \cup \left\{ X''_{d_2}\right\} \right)$, contradicting the stability of $\psi (P^2)$ at $P^2$. If $(X''_{d_1})_H\ne h$, then let $\tilde{X}=\psi (P^2) \cup \left\{ X''_{d_2}\right\}$. Again by the same arguments above, $\tilde{X}_h=C_h\left( \psi (P^2) \cup \left\{ X''_{d_2}\right\} \right) \subseteq C_D\left( \psi (P^2) \cup \left\{ X''_{d_2}\right\} \right)$, contradicting the stability of $\psi (P^2)$ at $P^2$. Hence, $S^2$ is non-empty.

If we keep applying the same arguments above, each iteration would give us a different doctor. This, however, contradicts the finiteness of the doctor set, contradicting our starting supposition that $\psi \ne COP$. This finishes the proof. $\square$