A modification aimed at reducing the manipulability and inefficiency of the Boston school choice mechanism

Abstract

The Boston mechanism (BOS) is widely used for the assignment of students to schools. Yet, BOS is highly manipulable and, therefore, may lead to Pareto inferior assignments. We propose a new indirect matching mechanism (${{ NBOS }}$) that is a slight variant of BOS. ${{ NBOS }}$ is less manipulable than BOS in two important ways: (i) students always have a best reply featuring truthful reported preference, (ii) rational students avoid particular re-rankings that are typical in BOS. Most importantly, the lower manipulability of ${{ NBOS }}$ helps reaching more efficient assignments than those reached by BOS. We show that each equilibrium of BOS is associated to a set of equilibria of ${{ NBOS }}$, all of which are Pareto superior. ${{ NBOS }}$ generalizes to a class of mechanisms whose members are even less manipulable than ${{ NBOS }}$.

Appendix A

1.1 A.1 The DA and ${{ NBOS }}$ algorithms

We first describe the (student-proposing) DA.

• Round 1: Every student applies to the school she reported as her most preferred acceptable school (if any). Every school that receives more applications than its capacity definitively rejects the applicants with lowest pre-existing priorities, up to the point where it meets its capacity. All other applicants are temporarily accepted at the schools they applied to (this means they could still be rejected in a later round).

• $\vdots$

• Round l: Every student who was rejected in round $l-1$ applies to her next acceptable school (if any). Every school considers the new applicants of round l together with the students it temporarily accepted. If needed, each school definitely rejects the students with lowest pre-existing priorities, up to the point where it meets its capacity. All other applicants are temporarily accepted at the schools they applied to (this means they could still be rejected in a later round).

The algorithm terminates when all acceptable schools of the reported profile have been considered, or when every student is assigned to a school.

We then describe ${{ NBOS }}$. Each school has a standard queue and a priority queue.

• Round 1: Every student applies to the school she reported as her most preferred acceptable school (if any). If this school is her neutralized school, she goes to the priority queue, otherwise she goes to the standard queue. Every school that receives more applications than its capacity definitively rejects from its standard queue the applicants with lowest pre-existing priorities, up to the point where it meets its capacity or where its standard queue is empty. In the latter case, the school definitively rejects from its priority queue the applicants with lowest pre-existing priorities, up to the point where it meets its capacity. All other applicants are temporarily accepted at the schools they applied to (this means they could still be rejected in a later round).

• $\vdots$

• Round l: Every student who was rejected in round $l-1$ applies to her next acceptable school (if any). If this school is her neutralized school, she goes to the priority queue, otherwise she goes to the standard queue. Every school considers the new applicants of round l together with the students it temporarily accepted. If needed, each school definitely rejects some applicants from its two queues, up to the point where it meets its capacity. To do so, the school first rejects from its standard queue the applicants who have applied (in previous rounds) to the more numerous standard queues of other schools. In case of ties, the school rejects the applicants with lowest pre-existing priorities. If the school does not meet its capacity while rejecting applicants from the standard queue, it rejects the applicants in its priority queue with lowest pre-existing priorities. All other applicants are temporarily accepted at the schools they applied to (this means they could still be rejected in a later round).

The algorithm terminates when all acceptable schools of the reported profile have been considered, or when every student is assigned to a school.

1.2 A.2 Proof Lemma 1

Lemma 1 is a direct corollary of Lemmas 2 and 3 that we present below.

Strategies affect the implicit priority profile of ${{ NBOS }}$. For any two priority profiles F and $F'$, we say that the priority of $t_i$ at s is as high in $F'$ as in F if $t_i~F_s~t_j~\Rightarrow ~t_i~F_s'~t_j$ for all $t_j \ne t_i$. The following lemma is immediate and its proof is omitted.

Lemma 2

(As high priority as)

Take any two strategies $\sigma _i$ and $\sigma _i'$ and any profile $\sigma _{-i}$. Let $F^{N}$ and $F^{N'}$ be the implicit priority profiles respectively associated in ${{ NBOS }}$ to $(\sigma _i,\sigma _{-i})$ and $(\sigma _i',\sigma _{-i})$. If $r^{N}_s(\sigma _i')\le r^{N}_s(\sigma _i)$, then the priority of $t_i$ at s is as high in $F^{N'}$ as in $F^{N}$.

Lemma 3 presents how a student’s assignment under DA is affected by a particular variation of the priority profile. We have not found an equivalent of this result in the literature.

Lemma 3

(Higher priority at accessible school guarantees this school in DA)

Take any two strategies $Q_i$ and $Q_i'$, any profile $Q_{-i}$ and any two priority profiles F and $F'$. If (i) $DA_i((Q_i,Q_{-i}),F)=s$, (ii) the priority of $t_i$ at s is as high in $F'$ as in F, and (iii) $t_j~F_{s'}~t_k~\Leftrightarrow ~t_j~F_{s'}'~t_k$ for all $s' \in S$ and $t_j,t_k \ne t_i$, then we have $DA_i((Q_i',Q_{-i}),F')~Q_i'~s$.

Proof

First, we show for the ordinal strategy $Q_i^s:~s~t_i$ that $DA_i((Q_i^s,Q_{-i}),F)=s$ using the IR Monotonicity property of DA (Kojima and Manea 2010).

A reported preference ordering $Q_i'$ is an individually rational monotonic transformation (i.r.m.t.) of $Q_i$ at $\mu _i\in S\cup \{t_i\}$ if any school ranked above both $\mu _i$ and $t_i$ under $Q_i'$ is ranked above $\mu _i$ under $Q_i$, which is

$$\begin{aligned} s~Q_i'~\mu _i \quad \text {and} \quad s~Q_i'~t_i\quad \Rightarrow \qquad s~Q_i~\mu _i \quad \text {for all } s\in S. \end{aligned}$$

A reported profile $Q'$ is an IR monotonic transformation of Q at an assignment $\mu$ if $Q_i'$ is an i.r.m.t. of $Q_i$ at $\mu _i$ for all $t_i \in T$. A direct mechanism M satisfies IR Monotonicity if we have that

$$\begin{aligned} Q'~i.r.m.t.~Q~ \text {at } M(Q,F)\quad \Rightarrow \quad M_i(Q',F)~Q_i'~M_i(Q,F) \text { for all } t_i\in T. \end{aligned}$$

By assumption (i) and the construction of $Q_i^s$, we have that $(Q_i^s,Q_{-i})~i.r.m.t.~(Q_i,Q_{-i})$ at $DA((Q_i,Q_{-i}),F)$. As DA satisfies IR monotonicity (Kojima and Manea 2010), we have $DA_i((Q_i^s,Q_{-i}),F)~Q_i^s~DA_i((Q_i,Q_{-i}),F)$. As $t_i$ reports no school above s in $Q_i^s$, we must have $DA_i((Q_i^s,Q_{-i}),F)=s$.

Second, we show that $DA_i((Q_i^s,Q_{-i}),F')=s$. Consider the contradiction assumption for which $DA_i((Q_i^s,Q_{-i}),F')\ne s$. This assumption implies that $t_i$ is rejected from s over the course of DA under $(Q_i^s,Q_{-i})$ and $F'$. By assumption (iii), the course of DA under $(Q_i^s,Q_{-i})$ and $F'$ is identical as the course of DA under $(Q_i^s,Q_{-i})$ and F at least until the round at which $t_i$ is rejected from s. If $t_i$ is rejected from s, then at least $q_s$ students $t\in T$ with $t~F_s'~t_i$ have applied to s up to that round. However, by assumption (ii), this implies that $t~F_s~t_i$ for all these $q_s$ students. Given that the course of DA under $(Q_i^s,Q_{-i})$ and F and the course of DA under $(Q_i^s,Q_{-i})$ and $F'$ are identical up to that round, this implies that $t_i$ is also rejected from s over the course of DA under $(Q_i^s,Q_{-i})$ and F, a contradiction.

Finally, if we have that $s~Q_i'~DA_i((Q_i',Q_{-i}),F')$, then we obtain a contradiction to the strategy-proofness of DA (Dubins and Freedman 1981) because $DA_i((Q_i^s,Q_{-i}),F')= s$. Therefore, we have $DA_i((Q_i',Q_{-i}),F')~Q_i'~s$. $\square$

1.3 A.3 Proof Proposition 1

Take any best reply $\sigma _i^*$ to $\sigma _{-i}$. Let $s^*={{ NBOS }}_i(\sigma _{i}^*,\sigma _{-i})$ be the school assigned to $t_i$ under $(\sigma _{i}^*,\sigma _{-i})$. Consider the strategy $\sigma _i=(R_i,x_i)$ with $x_i=s^*$ and let ${\hat{F}}$ and ${\hat{F}}^*$ be the implicit priority profiles respectively associated in ${{ NBOS }}$ to $(\sigma _{i},\sigma _{-i})$ and $(\sigma _{i}^*,\sigma _{-i})$. As $x_i=s^*$, Lemma 2 implies that the priority of $t_i$ at $s^*$ is as high in ${\hat{F}}$ as in ${\hat{F}}^*$. Lemma 3 applies and we therefore have ${{ NBOS }}_i(\sigma _{i},\sigma _{-i})~R_i~s^*$. As $\sigma _i^*$ is a best reply to $\sigma _{-i}$, we must have ${{ NBOS }}_i(\sigma _{i},\sigma _{-i})=s^*$, showing that $\sigma _i$ is a best reply to $\sigma _{-i}$.

1.4 A.4 Proof Proposition 2

We only provide the proof for part (ii).

Let $\sigma _i'=(Q_i^{s,s'},s)$ and let ${\hat{F}}$ and ${\hat{F}}'$ be the implicit priority ordering respectively associated to $(\sigma _i,\sigma _{-i})$ and $(\sigma _i',\sigma _{-i})$.

The proof builds on the fact that priority orderings ${\hat{F}}$ and ${\hat{F}}'$ are identical. In effect, strategies $\sigma _i$ and $\sigma _i'$ report the same schools, neutralize the same school s and rank non-neutralized schools in the same order. As a result, we have $r^{N}_{s''}(\sigma _i)=r^{N}_{s''}(\sigma _i')$ for any school $s''$ reported in these strategies.

Take any $\sigma _{-i}$. The proof that strategy $\sigma _{i}'$ dominates $\sigma _{i}$ is in three parts.

Part 1::

If ${{ NBOS }}_i(\sigma _{i},\sigma _{-i})\ne s$, then ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})={{ NBOS }}_i(\sigma _{i},\sigma _{-i})$.

Case 1.1::

${{ NBOS }}_i(\sigma _{i},\sigma _{-i})~Q_i~s$.

By construction, both $Q_i$ and $Q_i^{s,s'}$ rank the first $r^{N}_{s'}(\sigma _{i})-1$ reported schools in the same order. As ${\hat{F}}$ and ${\hat{F}}'$ are identical, Lemma 3 implies that ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})={{ NBOS }}_i(\sigma _{i},\sigma _{-i})$.

Case 1.2::

$s'~Q_i~{{ NBOS }}_i(\sigma _{i},\sigma _{-i})$ and ${{ NBOS }}_i(\sigma _{i},\sigma _{-i})\ne s'$.

As ${\hat{F}}$ and ${\hat{F}}'$ are identical, Lemma 3 implies that we cannot have ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})~Q_i~s'$ in this subcase. Hence, we have $s'~Q_i~{{ NBOS }}_i(\sigma _{i}',\sigma _{-i})$ and ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})\ne s'$. By construction, both $Q_i$ and $Q_i^{s,s'}$ rank the schools reported below s and $s'$ in the same order. Then, Lemma 3 implies that ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})={{ NBOS }}_i(\sigma _{i},\sigma _{-i})$.

Case 1.3::

${{ NBOS }}_i(\sigma _{i},\sigma _{-i})=s'$.

By Lemma 3, we have ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})~Q_i^{s,s'}~s'$. If ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})$ is a school $s''$ ranked above $s'$ in $Q_i^{s,s'}$, then Lemma 3 implies that ${{ NBOS }}_i(\sigma _{i},\sigma _{-i})=s''$, a contradiction. This implies that ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})=s'$.

Part 2::

If ${{ NBOS }}_i(\sigma _{i},\sigma _{-i})= s$, then ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})~R_i~s$.

By Lemma 3, we must have ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})~Q_i^{s,s'}~s$. School ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})$ cannot be a school $s''$ ranked above $s'$ in $Q_i^{s,s'}$, because Lemma 3 would imply that ${{ NBOS }}_i(\sigma _{i},\sigma _{-i})=s''$, a contradiction. Then, we must have either ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})=s$ or ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i})=s'$. This yields the result as both s and $s'$ are as desirable as s.

Part 3::

There exists $\sigma _{-i}'$ such that ${{ NBOS }}_i(\sigma _{i},\sigma _{-i}')=s$ and ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i}')=s'$.

Profile $\sigma _{-i}'$ is constructed as follows:

$\bullet$:

for any school $s''$ reported in $Q_i$ above s, take $q_{s''}$ students $t_k$ and let $\sigma _k'=(Q_k',s'')$ with $Q_k':~s''~t_k$,

$\bullet$:

no student other than $t_i$ reports s or $s'$.

Given that s is assignable under $\sigma _i$, there are enough students to construct profile $\sigma _{-i}'$.

Under $(\sigma _{i},\sigma _{-i}')$ and $(\sigma _{i}',\sigma _{-i}')$, $t_i$ is rejected from any school $s''$ reported in $Q_i$ above s since $q_{s''}$ have higher priority at $s''$ than $t_i$. Given that $t_i$ is the only student to report s or $s'$, we have ${{ NBOS }}_i(\sigma _{i},\sigma _{-i}')=s$ and ${{ NBOS }}_i(\sigma _{i}',\sigma _{-i}')=s'$, the desired result.

1.5 A.5 Proof Theorem 1

Statement 1: $\mu ^0 \in {\mathcal {M}}^0$.

Take any NE of BOS $Q^{0}$ supporting $\mu ^{0}$. Consider the strategy profile $\sigma ^{N}=(Q_i^{0},Q_i^{0}(1))_{t_i \in T}$, where the ordinal ranking of any $t_i$ is the one she uses in the NE of BOS and the neutralized school is the school reported first in $Q_i^{0}$. By construction, we have $F^{N}=F^{0}$, where $F^{N}$ and $F^{0}$ are the implicit priority structure respectively associated to $\sigma ^{N}$ in ${{ NBOS }}$ and $Q^{0}$ in BOS. As $F^{N}=F^{0}$ and the profile of ordinal rankings is equal to $Q^{0}$, we have that ${{ NBOS }}(\sigma ^{N})=\mu ^{0}$.

There remains to show that $\sigma ^{N}$ is a NE of ${{ NBOS }}$. We show for any student $t_i$ and any school $s'$ with $s'~P_i~{{ NBOS }}_i(\sigma ^{N})$ that there exists no strategy $\sigma _i'$ such that ${{ NBOS }}_i(\sigma _i',\sigma ^{N}_{-i})=s'$. Assignment $\mu ^{0}$ is stable since it is a NE-outcome of BOS (Theorem 1 in Ergin and Sönmez (2006)). Therefore, there are $q_{s'}$ students $t_k$ assigned to $s'$ in $\mu ^{0}$ such that $t_k~F_{s'}~t_i$ and $Q_k^{0}(1)=s'$. If not, then $BOS_i(Q_i',Q^{0}_{-i})=s'$ for $Q_i':~s'~t_i$, a contradiction to $Q^{0}$ being a NE of BOS. Finally, as $x_k=Q_k^{0}(1)$ for these $q_{s'}$ students, they have higher implicit priority at $s'$ than $t_i$ for all strategies $\sigma _i'$. Thus there is no strategy $\sigma _i'$ such that ${{ NBOS }}_i(\sigma _i',\sigma ^{N}_{-i})=s'$, the desired result.

Statement 2: If $\mu ^0$ is not Pareto efficient, then there exists $\mu ^1 \in {\mathcal {M}}^0$ such that $\mu ^1\ne \mu ^0$.

Let ${\mathcal {M}}^d$ be the set of assignments that Pareto dominate $\mu ^0$. By assumption, this set is non-empty. For any $\mu ^a \in {\mathcal {M}}^d$, let $T^a$ be the set of students who have the same assignment under $\mu ^a$ and $\mu ^0$, i.e. $T^a=\{t_i \in T~|~\mu _i^a= \mu _i^0\}$. Let $S^a$ be the set of schools whose set of assigned students is different under $\mu ^a$ and $\mu ^0$, i.e.

$$\begin{aligned} S^a=\{s \in S|~ \text { there is }t_i\text { for whom } \mu ^a_i= s \text { and }\mu ^0_i\ne s \}. \end{aligned}$$

We first show that all schools in $S^a$ have no vacant seat under both $\mu ^a$ and $\mu ^0$. If this holds, then assignment $\mu ^a$ is obtained from $\mu ^0$ by a Pareto improving exchange cycle of seats at schools in $S^a$. Take any school $s^*\in S^a$. As $\mu ^a$ Pareto dominates $\mu ^0$, we have $\mu _i^a~P_i~\mu _i^0$ for all $t_i \in T\backslash _{T^a}$. By Theorem 1 in Ergin and Sönmez (2006), $\mu ^0$ is stable. Therefore $\mu ^0$ is non-wasteful, which implies that school $s^*$ has no vacant seat in $\mu ^0$. Given that $\mu ^a$ Pareto dominates $\mu ^0$, all students assigned to a school under $\mu ^0$ must be assigned to a school under $\mu ^a$. By non-wastefulness again, no student in $T\backslash _{T^a}$ can be assigned to a seat that is vacant under $\mu ^0$. Together, school $s^*$ has no vacant seat in $\mu ^a$.

We construct a finite sequence of assignments $(\mu ^g)_{g \in \{2,3,\ldots ,N\}} \in {\mathcal {M}}^d$ and show that the last assignment in the sequence belongs to ${\mathcal {M}}^0$. Let $\mu ^2$ be any assignment in ${\mathcal {M}}^d$ such that $\#T^2 \ge \#T^a$ for all $\mu ^a \in {\mathcal {M}}^d$. Take any $g \in \{2,3,\ldots ,N\}$. The following three steps check whether $\mu ^g\in {\mathcal {M}}^0$; and if $\mu ^g\notin {\mathcal {M}}^0$, construct the next element $\mu ^{g+1}$.

• Step 1: Construct a profile $\sigma ^g=(Q^g_i,x_i)_{t_i \in T}$ for which ${{ NBOS }}(\sigma ^g)=\mu ^g$.

  • $Q_i^g:~\mu _i^0~t_i$ and $x_i=\mu _i^0$ for all $t_i \in T^g$,

  • $Q_i^g:~\mu _i^g~\mu _i^0~t_i$ and $x_i=\mu _i^0$ for all $t_i \notin T^g$.

  Observe that this construction implies that students unassigned in $\mu ^0$ report no school. We have shown above that all schools in $S^g$ have no vacant seat under both $\mu ^g$ and $\mu ^0$. Therefore, for any school s the construction of $Q^g$ is such that the number of students $t_i$ for whom $Q_i^g(1)=s$ is exactly equal to the number of students assigned to s under $\mu ^g$. As a result, no student is rejected after the first round of ${{ NBOS }}$ under $\sigma ^g$ and we obtain by construction that ${{ NBOS }}(\sigma ^g)=\mu ^g$.

• Step 2: Construct $Q^{g'}$, a NE of BOS such that $BOS(Q^{g'})=\mu ^0$ and whose implicit priority profile in BOS is the same as that of $\sigma ^g$ in ${{ NBOS }}$.

  • $Q_i^{g'}:~\mu _i^0~t_i$ for all $t_i \in T^g$,

  • $Q_i^{g'}:~\mu _i^0~\mu _i^g~t_i$ for all $t_i \notin T^g$.

  We have by construction that $BOS(Q^{g'})=\mu ^0$. Profile $Q^{g'}$ is a NE of BOS because $Q^{g'}_i(1)=\mu ^0_i$ for all $t_i$ and $\mu ^0$ is stable (Theorem 1 in Ergin and Sönmez (2006)). By construction, the implicit priority profiles respectively associated to $Q^{g'}$ in BOS and to $\sigma ^g$ in ${{ NBOS }}$ are the same.¹³

• Step 3: If $\sigma ^g$ is not a NE of ${{ NBOS }}$, then there exists another $\mu ^{g+1} \in {\mathcal {M}}^d$ with the following properties: $\mu _i^{g+1}=\mu _i^g$ for all $t_i\ne t_j,t_k$; $\#T^{g+1}= \#T^g$; $\mu _j^{g+1}=\mu _k^g$ and $\mu _k^{g+1}=\mu _j^g$; $\mu _j^g=\mu ^0_k=\mu ^0_j$; and $t_j~F_{\mu _k^g}~t_k$.

  If $\sigma ^g$ is not a NE of ${{ NBOS }}$, then there a exists $t_j$ for whom ${{ NBOS }}_j(\sigma _j',\sigma _{-j}^g)=s^*$ for some $s^*~P_j~\mu _j^g$ and some $\sigma _j'$. This implies that ${{ NBOS }}_j(\sigma _j'',\sigma _{-j}^g)=s^*$ for the particular $\sigma _j''=(Q_j'',s^*)$ with $Q_j'':~s^*~t_j$. We show that $\mu ^{g+1}={{ NBOS }}(\sigma _j'',\sigma _{-j}^g)$ has the desired properties. First, we show that $\mu ^{g+1} \in {\mathcal {M}}^d$.

  As $\mu ^0$ is a stable assignment, there are by construction at least $q_{s^*}$ students t with $t~F_{s^*}~t_j$ for whom $\mu _t^0=s^*$. By construction, all students $t_l\ne t_j$ neutralize in $\sigma ^g_{-j}$ the school they are assigned in $\mu ^0$.¹⁴ Hence, all students $t_l\ne t_j$ have top-priority at $\mu ^0_l$ in the implicit priority profile associated to $(\sigma _j'',\sigma _{-j}^g)$. As a result, we have by Lemma 3 that ${{ NBOS }}_l(\sigma _j'',\sigma _{-j}^g)~Q_l^g~\mu _l^0$ for all $t_l\ne t_j$. By construction, this implies

  $$\begin{aligned} NBOS_l(\sigma _j'',\sigma _{-j}^g)~R_l~\mu _l^0 \qquad \text {for all } t_l\ne t_j, \end{aligned}$$

  (4)

  which shows that $\mu ^{g+1} \in {\mathcal {M}}^d$, as ${{ NBOS }}_j(\sigma _j',\sigma _{-j}^g)=s^*~P_j~\mu _j^0$.

  We then show that $\#T^{g+1}= \#T^g$ and that there is a unique $t_k\ne t_j$ for whom $\mu ^{g+1}_k\ne \mu ^g_k$. As ${{ NBOS }}_i(\sigma ^g)=Q_i^g(1)$ for all $t_i$, we have that only $t_j$ prefers her assignment under $\mu ^{g+1}$ to her assignment under $\mu ^g$. This implies that there is no student $t_o\ne t_j$ for whom $t_o \in T^g$ and $t_o \notin T^{g+1}$. As ${{ NBOS }}_j(\sigma _j'',\sigma _{-j}^g)=s^*$ even if $q_{s^*}$ students different than $t_j$ have top-priority at $s^*$ in the implicit priority profile associated to $(\sigma _j'',\sigma _{-j}^g)$, there is at least one student $t_k$ for whom $\mu _k^g=s^*$ and $\mu _k^{g+1}\ne s^*$. As ${{ NBOS }}_l(\sigma _j'',\sigma _{-j}^g)~Q_l^g~\mu _l^0$ for all $t_l\ne t_j$, this implies $\mu _k^{g+1}= \mu _k^0$, and hence, $t_k\notin T^g$ and $t_k \in T^{g+1}$. Together, we have that $\#T^{g+1}\ge \#T^g$. By definition of $\mu ^2$, there exists no $\mu ^a \in {\mathcal {M}}^d$ such that $\#T^a>\#T^g= \#T^2$, which yields $\#T^{g+1}= \#T^g$. Therefore, there is a unique $t_k\ne t_j$ for whom $\mu ^{g+1}_k\ne \mu ^g_k$.

  Finally, we show that $\mu _k^{g+1}=\mu _j^g$; $\mu _j^g=\mu ^0_k=\mu ^0_j$; and $t_j~F_{\mu _k^g}~t_k$. Given that $\mu ^g$ and $\mu ^{g+1}$ are both obtained from $\mu ^0$ by a Pareto improving exchange cycle of seats, we have that $\mu ^0_k, \mu ^0_j \in S$. It also implies that $\mu ^{g+1}$ is obtained from $\mu ^g$ by an exchange cycle of seats between $t_j$ and $t_k$ (though not a Pareto improving one). As $\mu _j^{g+1}=\mu _k^g$ and given that $t_k$ is the only student other than $t_j$ for whom $\mu ^{g+1}_k\ne \mu ^g_k$, we must have $\mu _k^{g+1}=\mu _j^g$. By construction of $\sigma ^g$, this is only possible if $\mu _j^g=\mu ^0_k=\mu ^0_j$ and $t_j~F_{\mu _k^g}~t_k$.

If $\sigma ^g$ is a NE of ${{ NBOS }}$, then $\mu ^g=\mu ^N\in {\mathcal {M}}^0$ and the proof is complete. There remains to show that the sequence of assignments $(\mu ^g)_{g \in \{2,3,\ldots ,N\}}$ is finite. If the sequence is finite, then for some $N \in {\mathbb {N}}$ there is no assignment $\mu ^{N+1}$ with the properties listed in Step 3. In that case, Step 3 implies that $\sigma ^N$ is a NE of ${{ NBOS }}$, and $\mu ^N\in {\mathcal {M}}^0$, the desired result.

We show that there can only exist a finite number of assignments with the properties listed in Step 3. Consider the set of seats that are exchanged during the Pareto improving exchange cycle leading from $\mu ^0$ to $\mu ^2$. The properties listed in Step 3 imply that this set of seat coincides with the set of seats exchanged during the Pareto improving exchange cycle leading from $\mu ^0$ to $\mu ^g$, for any g. Consider $T\backslash _{T^g}$, i.e. the set of students who hold these seats in $\mu ^g$ and who prefer their assignment under $\mu ^g$ over that under $\mu ^0$. Crucially, the priority of students in $T\backslash _{T^g}$ at these seats never decrease with g. What is more, the priority of $t_j$ at $\mu _k^{g}$ is strictly higher than the priority at $\mu _k^{g}$ of the student $t_k$ from whom $t_j$ takes the seat in $\mu _k^{g}$ when moving from $\mu ^g$ to $\mu ^{g+1}$. As the number of students and seats is finite, this sequence must therefore be finite, the desired result.

Statement 3: If $\mu ^1 \in {\mathcal {M}}^0$ and $\mu ^1\ne \mu ^0$, then $\mu ^1$ Pareto dominates $\mu ^0$

As $\mu ^{1}$ is associated to $\mu ^{0}$, there exist a NE of ${{ NBOS }}$ $\sigma ^{1}$ supporting $\mu ^{1}$ and a NE of BOS $Q^{0}$ supporting $\mu ^{0}$ such that $F^{1}=F^{0}$, where $F^{1}$ and $F^{0}$ are the respective implicit priority profiles.

As $\mu ^1\ne \mu ^0$, it is sufficient to show that $\mu ^{1}_i~R_i~\mu ^{0}_i$ for all $t_i$. Consider the contradiction assumption according to which $\mu ^{0}_j~P_j~\mu ^{1}_j$ for some $t_j$. As $\sigma ^{1}$ is a NE of ${{ NBOS }}$, there is no strategy $\sigma _j'$ such that ${{ NBOS }}_j(\sigma _j',\sigma ^{1}_{-j})=\mu ^{0}_j$. Hence, letting $s'=\mu ^{0}_j$, there are $q_{s'}$ students $t_k$ with $t_k~F_{s'}~t_j$ such that $x_k=s'$. As $F^{1}=F^{0}$, this implies that there are $q_{s'}$ students $t_k$ with $t_k~F_{s'}~t_j$ such that $Q^{0}_k(1)=s'$. This is a contradiction to $BOS_j(Q^{0})=\mu ^{0}_j=s'$.

1.6 A.6 NE-outcome of ${{ NBOS }}$ not associated to any NE-outcome of BOS

This example is directly adapted from Example 6.2 in Haeringer and Klijn (2009). There are three students and three schools, each endowed with one seat. The only stable assignment is shown in bold in the preference profile. We present a NE of ${{ NBOS }}$ $(Q_i,x_i)_{t_i \in T}$ whose outcome is boxed.

$$\begin{aligned} \begin{matrix} R_1 &{} R_2 &{} R_3 \\ \hline \mathbf {s_1} &{} \mathbf {s_3} &{} s_3 \\ s_3^* &{} s_1 &{} \mathbf {s_2^*} \\ s_2 &{} s_2 &{} s_1^* \end{matrix} \quad \begin{matrix} {F_1} &{} {F_2} &{} {F_3} \\ \hline t_3 &{} t_3 &{} t_1 \\ t_1&{} t_1 &{} t_2 \\ t_2 &{} t_2 &{} t_3 \end{matrix} \qquad \begin{matrix} Q_1 &{} Q_2 &{} Q_3 \\ \hline \boxed {s_1} &{} s_1 &{} \boxed {s_3} \\ {\hat{s}}_3^* &{} s_3 &{} {\hat{s}}_1^* \\ s_2 &{} \boxed {{\hat{s}}_2} &{} s_2^* \end{matrix} \end{aligned}$$

(5)

Student $t_2$ prefers her assignment under the only stable assignment than the school she is assigned under this NE-outcome. Yet, if $t_2$ neutralizes $s_3$ instead of $s_2$, she does not obtain $s_3$ because of the presence of the following rejection cycle:

• $t_3$ is rejected from $s_3$ and applies to $s_1$,

• $t_1$ is then rejected from $s_1$ and applies to $s_3$,

• $t_2$ is rejected from $s_3$ and is still assigned to $s_2$.

The particularity of this equilibrium is that $t_3$ creates a rejection chain by re-ranking $s_2$ and $s_1$ and neutralizing $s_1$. But observe that $t_3$ has top-priority both at $s_1$ and $s_2$, and prefers the latter. So, $t_3$ could guarantee a seat at $s_2$ by playing $(R_3,s_2)$ instead of $\sigma _3$, thereby securing a better assignment in the case she does not obtain $s_3$.

1.7 A.7 Proof Proposition 4

Step 1: Take any $w\ge 1$ and any student $t_i$, if $t_i$ has no more than w acceptable schools, then strategy $\sigma _i^*=(R_i,W_i)$ where all acceptable schools are neutralized is dominant in $w\text {-}{{ BOS }}$.

Take any strategy $\sigma _i'=(Q_i',W_i')$ with $\#W_i\le w$, we show that strategy $\sigma _i^*=(R_i,W_i)$ where $W_i$ contains all acceptable schools (weakly) dominates $\sigma _i'$ in $w\text {-}{{ BOS }}$. For any $\sigma _{-i}'$, let ${F}^{W^*}$ and ${F}^{W'}$ denote the implicit priority profiles respectively associated to the strategy profiles $(\sigma _i^*,\sigma _{-i}')$ and $(\sigma _i',\sigma _{-i}')$. Take any acceptable school s. As $s\in W_i$, the priority of $t_i$ at s is as high under ${F}^{W^*}$ as under ${F}^{W'}$. By Lemma 3, if $w\text {-}{{ BOS }}_i(\sigma _i',\sigma _{-i}')=s$, then we have $w\text {-}{{ BOS }}_i(\sigma _i^*,\sigma _{-i}')~R_i~s$. We have just shown that $w\text {-}{{ BOS }}_i(\sigma _i^*,\sigma _{-i}')~R_i~w\text {-}{{ BOS }}_i(\sigma _i',\sigma _{-i}')$ for all $\sigma _{-i}'$ for which $w\text {-}{{ BOS }}_i(\sigma _i',\sigma _{-i}')=s$ is acceptable. The desired result follows from the fact that $w\text {-}{{ BOS }}_i(\sigma _i^*,\sigma _{-i}')$ cannot be an unacceptable school.

Step 2: All students have a (truthful) dominant strategy in m-BOS.

As students cannot have more than m acceptable schools, Step 1 directly implies the result.

Step 3: The assignment associated to the dominant strategy equilibrium in m-BOS is the assignment associated to the dominant strategy equilibrium in DA.

By Step 2, any student $t_i$ has a dominant strategy in m-BOS of the form $\sigma _i^*=(R_i,W_i)$ where all acceptable schools are neutralized. This strategy is strategically equivalent to $\sigma _i'=(R_i,W_i')$ where all schools are neutralized. Let $\sigma '$ be the dominant strategy profile where all students play their strategy $\sigma _i'$. As all students neutralize all schools, the implicit priority profile ${F}^{W'}$ associated to the strategy profiles $\sigma _i'$ is equal to the pre-existing priority profile F. Then, by the definition of $w\text {-}{{ BOS }}$, we have $w\text {-}{{ BOS }}(\sigma ')=DA(R)$, where R is a dominant strategy profile of DA (Dubins and Freedman 1981).