A necessary and sufficient condition for stable matching rules to be strategy-proof

Abstract

We study one-to-one matching problems and analyze conditions on preference domains that admit the existence of stable and strategy-proof rules. In this context, when a preference domain is unrestricted, it is known that no stable rule is strategy-proof. We introduce the notion of the no-detour condition, and show that under this condition, there is a stable and group strategy-proof rule. In addition, we show that when the men’s preference domain is unrestricted, the no-detour condition is also a necessary condition for the existence of stable and strategy-proof rules. As a result, under the assumption that the men’s preference domain is unrestricted, the following three statements are equivalent: (i) a preference domain satisfies the no-detour condition, (ii) there is a stable and group strategy-proof rule, (iii) there is a stable and strategy-proof rule.

Appendix: Proof of theorem

To prove the Theorem, we use three lemmata. The first one is know as the blocking Lemma.

Lemma 1

(Hwang, c.f., Lemma 3.5 in Roth and Sotomayor 1990) Given $P \in \fancyscript{D}$, let $\mu =DA(P)$ and $\mu ' \in \fancyscript{M}^{\textit{IR}}(P)$. Let $M^{\mu '}(P) \equiv \{m \in M: \mu '(m) \ P_m \ \mu (m)\}$. Then, if $M^{\mu '}(P) \ne \emptyset$, there is a pair $(m, w) \in M\setminus M^{\mu '}(P) \times \mu '\left( M^{\mu '}(P)\right)$ who blocks $\mu$ under $P$.⁹

The second lemma shows that the NDC is sufficient for the DA to be group strategy-proof. Note that, again, to prove this lemma, we do not need the assumption that the set of preferences for each man is unrestricted.

Lemma 2

If $\fancyscript{D}$ satisfies the NDC, then the DA rule is group strategy-proof on $\fancyscript{D}$.

Proof

Suppose that $\fancyscript{D}$ satisfies the NDC. We prove that $DA \in GSP(\fancyscript{D} )$ by contradiction. Suppose that $DA$ is manipulated by a group $C \subseteq M \cup W$ by misstating $P'_C$ instead of $P_C$. To simplify the notations, let $P' \equiv (P'_C, P_{-C}), \mu \equiv DA(P)$ and $\mu ' \equiv DA(P')$. Then $\mu '(x) \ P_x \ \mu (x)$ for all $x \in C$.

Claim 1

$\mu ' \in \fancyscript{M}^{\textit{IR}}(P)$.

Proof of Claim 1

Suppose not. Then there exists an agent $z \in M \cup W$ such that $z$ prefers being unmatched to $\mu '(z)$ under $P_z$. If $z \not \in C$, then $P_z=P'_z$, which immediately violates $\mu ' \in \fancyscript{M}^{\textit{IR}}(P')$. Therefore, $z \in C$. Then $z$ prefers being unmatched to $\mu (z)$, which violates $\mu \in \fancyscript{M}^{\textit{IR}}(P)$. (Claim 1 ends.)

Claim 2

$C \subseteq W$.

Proof of Claim 2

Suppose that $M^{\mu '}(P) = \{m \in M: \mu '(m) \ P_m \ \mu (m)\} \ne \emptyset$. By Lemma 1, there is $(m, w) \in M \setminus M^{\mu '}(P) \times \mu '\left( M^{\mu '}(P)\right)$ such that $w \ P_m \ \mu '(m)$ and $m \ P_w \ \mu '(w)$. Since $m \in M \setminus M^{\mu '}(P), \mu (m) \ R_m \ \mu '(m)$. Therefore, $m \not \in C$. If $w \not \in C$, then $P_m=P'_m$ and $P_w=P'_w$. This implies that $(m, w)$ also blocks $\mu '$ under $P'$, which contradicts $\mu ' \in \fancyscript{M}^{S}(P')$. Therefore, $w \in C$.

Let $\tilde{m} \equiv \mu '(w)$. Since $w \in \mu '\left( M^{\mu '}(P)\right)$,

$$\begin{aligned} w=\mu '(\tilde{m}) \ P_{\tilde{m}} \ \mu (\tilde{m}). \end{aligned}$$

Since $w \in C$,

$$\begin{aligned} \tilde{m}=\mu '(w) \ P_w \ \mu (w). \end{aligned}$$

Then $(\tilde{m}, w)$ blocks $\mu$ under $P$, which violates $\mu \in \fancyscript{M}^{S}(P)$. Therefore, $M^{\mu '}(P)=\emptyset$, and $C \subseteq W$ (Claim 2 ends).

Step 1: Construction of a cycle. By Claim 2, there is $w\in W\cap C$ such that $\mu '(w) \ P_w \ \mu (w)$. Suppose that $w_1$ is such a woman; namely

$$\begin{aligned} \mu '(w_1) \ P_{w_1} \ \mu (w_1). \end{aligned}$$

(1)

We construct a sequence of women and men $\{m_2, w_2, m_3, w_3, \ldots \}$ by means of the following algorithm:

Algorithm (Fig. 4)

• (A) Initialize $k$, i.e., let $k=0$.

• (B) Increase $k$ by one. Define $m_{k+1} \equiv \mu '(w_k)$ and $w_{k+1} \equiv \mu (m_{k+1})$. If $w_{k+1} \ne w_1$, then go back to the beginning of (B). If not, then the algorithm terminates.

We show that the algorithm is well-defined. We first check if $m_2 \in M$ is well-defined. If $\mu '(w_1) =m_0$, then by (1), $\mu \not \in \fancyscript{M}^{\textit{IR}}(P)$, which violates $\mu \in \fancyscript{M}^{S}(P)$. Therefore $\mu '(w_1) \in M$ and we can define $m_2 \in M$ by $m_2 \equiv \mu '(w_1)$.

To check if $w_2$ is well-defined, we show that $\mu (m_2) \ P_{m_2} \ \mu '(m_2)$. Suppose not, then $\mu '(m_2) \ R_{m_2} \ \mu (m_2)$. Since $w_1=\mu '(m_2)$ and $\mu '(w_1) \ne \mu (w_1)$ by (1), $\mu '(m_2) \ne \mu (m_2)$. Therefore, $w_1 \ P_{m_2} \ \mu (m_2)$. Since we have $m_2 \ P_{w_1} \ \mu (w_1)$ by (1), it must be that $(m_2, w_1)$ blocks $\mu$ under $P$. This is a contradiction. Therefore, $\mu (m_2) \ P_{m_2} \ \mu '(m_2)$. Since $P_{m_2}=P'_{m_2}, \mu '(m_2) \ne w_0$ and $\mu ' \in \fancyscript{M}^{IR}(P')$,

$$\begin{aligned} \mu (m_2) \ P_{m_2} \ \mu '(m_2) \ P_{m_2} \ w_0. \end{aligned}$$

(2)

Therefore $\mu (m_2) \in W$ and we can define $w_2 \equiv \mu (m_2) \in W$. Note that $w_2 \ne w_1$ by (2).

Fig. 4

The flow chart of the algorithm in the proof of Lemma 2

To check if $m_3$ is well-defined, we show that $\mu '(w_2) \ P_{w_2} \ \mu (w_2)$. Since $\mu '(m_2)=w_1 \ne w_2=\mu (m_2), \mu '(w_2) \ne \mu (w_2)$. In the case that $w_2 \in C$, we have $\mu '(w_2) \ P_{w_2} \ \mu (w_2)$. In the other case, we have

$$\begin{aligned} m_2=\mu (w_2) \ P_{w_2} \ \mu '(w_2). \end{aligned}$$

Since $w_2, m_2 \not \in C$, we have $P_{w_2}=P'_{w_2}$ and $P_{m_2}=P'_{m_2}$. Therefore, $(m_2, w_2)$ blocks $\mu '$ under $P'$, which contradicts $\mu ' \in \fancyscript{M}^{S}(P')$. Therefore, since $\mu \in \fancyscript{M}^{\textit{IR}}(P)$ and $\mu (w_2)\ne m_0$,

$$\begin{aligned} \mu '(w_2) \ P_{w_2} \ \mu (w_2) \ P_{w_2} \ m_0. \end{aligned}$$

(3)

By repeating the argument above, we can show that if $m_k$ is defined for some $k \ge 2$, then

• 1. $w_k$ is defined by $w_k \equiv \mu (m_k)$, and if $w_k \ne w_1$ then

  $$\begin{aligned} \mu '(w_k) \ P_{w_k} \ \mu (w_k) \ P_{w_k} \ m_0. \end{aligned}$$

Moreover, if $w_k \ne w_1$, then

• 2. $m_{k+1}$ is defined by $m_{k+1} \equiv \mu '(w_k)$, and

  $$\begin{aligned} \mu (m_{k+1}) \ P_{m_{k+1}} \ \mu '(m_{k+1}) \ P_{m_{k+1}} \ w_0. \end{aligned}$$

We next check if all the men defined by following the algorithm are distinct and so are all the women. To this end, suppose that $m_i$ and $m_k (i<k)$ are defined and $m_i= m_k$. Then, $w_{i-1}=\mu '(m_i)=\mu '(m_k)=w_{k-1}$ and $m_{i-1}= \mu (w_{i-1})= \mu (w_{k-1})=m_{k-1}$. By repeating this procedure, $w_{1}=w_{k-(i-1)}$ and $k-(i-1) \ne 1$. The algorithm should terminate at the step where $w_{1}=w_{k-(i-1)}$. Therefore it terminates before defining $w_k$, which is a contradiction. Therefore, the men defined by following the algorithm are all distinct, which immediately implies that so are the women.

By the finiteness of $M$ and $W$, this algorithm terminates in finitely many steps. Hence, there is $\ell +1 \ge 2$ such that $w_{\ell +1}=w_1$. Define $m_1$ by $m_1 \equiv m_{\ell +1}$.¹⁰ Since $\mu \in \fancyscript{M}^{\textit{IR}}(P)$ and $\mu (w_1)=\mu (w_{\ell +1})=m_{\ell +1}=m_1 \ne m_0$, together with (1),

$$\begin{aligned} \mu '(w_1) \ P_{w_1} \ \mu (w_1) \ P_{w_1} \ m_0. \end{aligned}$$

(4)

Then $(P_{w_1}, P_{w_2}, \ldots , P_{w_{\ell }}; m_1, m_2, m_3, \ldots , m_{\ell })$ is a cycle. (Step 1 ends.)

Figure 5 illustrates the connection of agents and the structure of preferences with respect to the cycle. Note that $w_1$ is chosen arbitrarily from the women who are members of $C$.

Fig. 5

The connection of the agents and the structure of their preferences in Step 1 in the proof of Lemma 2 (any man–woman pair conned by a line means that the man is matched to the woman at $\mu$. Similarly, any man–woman pair connected by a double line means that the man is matched to the woman at $\mu '$).

Step 2: Checking if there is a cycle with a detour. We consider two cases:

Case 1 $\mu \not \in \fancyscript{M}^{\textit{IR}}(P')$. There exists an agent $z$ whose assignment at $\mu$ is not acceptable under $P'_z$. Since $\mu \in \fancyscript{M}^{\textit{IR}}(P)$, it must be that $P_z \ne P'_z$, that is, $z \in C$. Since $z \in W$ by Claim 2, we rename $z$ as $w_1 \in W$. By Step 1, we can construct a cycle, $(P_{w_1}, P_{w_2}, \ldots , P_{w_{\ell }}; m_1, m_2, m_3, \ldots , m_{\ell })$, where $m_1 \equiv \mu (w_1)$.

We show that the cycle has a detour from $m_1$ with $x=m_0$. Since $\mu '(w_1) \ne m_0$ by (4) and $\mu ' \in \fancyscript{M}^{\textit{IR}}(P')$,

$$\begin{aligned} m_2=\mu '(w_1) \ P'_{w_1} \ m_0 \ P'_{w_1} \ \mu (w_1)=m_1. \end{aligned}$$

Since $\mu \in \fancyscript{M}^{\textit{IR}}(P)$,

$$\begin{aligned} m_1=\mu (w_1) \ P_{w_1} \ m_0. \end{aligned}$$

Therefore, the cycle has a detour from $m_1$.

Case 2 $\mu \in \fancyscript{M}^{\textit{IR}}(P')$. Let $M^{\mu }(P') \equiv \{m \in M: \mu (m) \ P'_m \ \mu '(m)\}$. Note that again, since $C \subseteq W$ by Claim 2, for each $m \in M$, we have $P'_m=P_m$. Therefore, $M^{\mu }(P')=M^{\mu }(P)$. By Step 1, it is obvious that $M^{\mu }(P) \ne \emptyset$. By Lemma 1, there is $(\widehat{m}, \widehat{w}) \in M \setminus M^{\mu }(P) \times \mu \left( M^{\mu }(P)\right)$ such that $(\widehat{m}, \widehat{w})$ blocks $\mu$ under $P'$.

If $\widehat{w} \not \in C$, then $P_{\widehat{w}}=P'_{\widehat{w}}$. Then, since $\widehat{m} \not \in C$, the pair also blocks $\mu$ under $P$, which violates $\mu \in \fancyscript{M}^{S}(P)$. Therefore,

$$\begin{aligned} \widehat{w} \in C. \end{aligned}$$

(5)

By Claim 2, $M^{\mu '}(P)=\emptyset$. Therefore, $\widehat{m} \in M \setminus M^{\mu }(P)$ implies $\mu (\widehat{m})=\mu '(\widehat{m})$. Since $(\widehat{m}, \widehat{w})$ blocks $\mu$ under $P'$ and $P_{\widehat{m}}=P'_{\widehat{m}}$,

$$\begin{aligned} \widehat{w} \quad P_{\widehat{m}} \quad \mu (\widehat{m})=\mu '(\widehat{m}) \end{aligned}$$

(6)

and

$$\begin{aligned} \widehat{m} \quad P'_{\widehat{w}} \quad \mu (\widehat{w}). \end{aligned}$$

(7)

Since $\mu \in \fancyscript{M}^{S}(P)$ and $\mu ' \in \fancyscript{M}^{S}(P')$, it must be that

$$\begin{aligned} \mu (\widehat{w}) \quad R_{\widehat{w}} \quad \widehat{m} \quad \mathrm{and} \quad \mu '(\widehat{w}) \quad R'_{\widehat{w}} \quad \widehat{m}. \end{aligned}$$

(8)

Since $\mu (\widehat{m})\ne \widehat{w}$ by (6), $\mu (\widehat{w})\ne \widehat{m}$. Since $\mu (\widehat{m})=\mu '(\widehat{m})$, we have $\mu '(\widehat{w}) \ne \widehat{m}$. Therefore, by (8),

$$\begin{aligned} \mu (\widehat{w}) \quad P_{\widehat{w}} \quad \widehat{m} \end{aligned}$$

(9)

and

$$\begin{aligned} \mu '(\widehat{w}) \quad P'_{\widehat{w}} \quad \widehat{m}. \end{aligned}$$

(10)

Here, rename $\widehat{w}$ as $w_1$. Since $w_1 \in C$, by applying Step 1, we can construct a cycle $(P_{w_1}, P_{w_2}, \ldots , P_{w_{\ell }}; m_1, m_2, \ldots , m_{\ell })$, where $m_1 \equiv \mu (w_1)$.

We show that the cycle has a detour from $m_1$ with $x=\widehat{m}$. Note that since $\widehat{m} \in M \setminus M^{\mu }(P)$ and $m_1, m_2, \ldots , m_\ell \in M^{\mu }(P)$, we have $\widehat{m} \in \left( M \cup \{m_0\} \right) \setminus \{m_1, m_2, \ldots , m_\ell \}$.

By (7) and (10), we have

$$\begin{aligned} m_2=\mu '(w_1) \ P'_{w_1} \ \widehat{m} \ \ P'_{w_1} \ \mu (w_1)=m_1. \end{aligned}$$

(11)

Since $\mu \in \fancyscript{M}^{IR}(P')$ in this case, $\mu (w_1) \ R'_{w_1} \ m_0$. By (11),

$$\begin{aligned} \widehat{m} \ P'_{w_1} \ m_0. \end{aligned}$$

(12)

By (9),

$$\begin{aligned} m_1=\mu (w_1) \ P_{w_1} \ \widehat{m}. \end{aligned}$$

(13)

Hence, by (11), (12) and (13), the cycle has a detour from $m_1$.

In both cases, there exists a cycle with a detour. This contradicts the assumption that $\fancyscript{D}$ satisfies the NDC. Therefore, $DA \in GSP(\fancyscript{D} )$. $\square$

The third lemma shows that the NDC is a necessary condition for the DA to be strategy-proof under an additional assumption.

Lemma 3

Suppose that the set of preferences for each man is unrestricted. If the DA rule is strategy-proof, then $\fancyscript{D}$ satisfies the NDC.

Proof

Suppose that $\fancyscript{D} _{m}=\fancyscript{P}_M$ for all $m \in M$. We prove that $\fancyscript{D}$ satisfies the NDC by contradiction. Let $(P_{w_1}, P_{w_2}, \ldots , P_{w_{\ell }}; m_1, m_2, m_3, \ldots , m_{\ell })$ be a cycle on $\fancyscript{D}$, that is, for each $k \in \{1, 2, \ldots , \ell \}$,

$$\begin{aligned} m_{k+1} \ P_{w_k} \ m_k \ P_{w_k} \ m_0, \end{aligned}$$

where $m_{\ell +1} \equiv m_1$. Suppose that this cycle has a detour, say from $m_1$. Then there is $x \in \left( M \cup \{m_0\} \right) \setminus \{m_1, m_2, \ldots , m_{\ell }\}$, and $P'_{w_1} \in \fancyscript{D} _{w_1}$ such that

• 1. $m_2 \ P'_{w_1} \ x \ P'_{w_1} \ m_1$,

• 2. $x \ R'_{w_1} \ m_0$, and

• 3. $m_1 \ P_{w_1} \ x$.

Case 1 $x \ne m_0$. Rename $x$ as $x \equiv \widehat{m} \in M$. Then $\widehat{m} \ne m_1, m_2, \ldots , m_{\ell }$. Let $P_{m_1}, P_{m_2}, \ldots , P_{m_{\ell }}, P_{\widehat{m}}$ be defined as in Table 1.¹¹ Let the preferences of the remaining men, if any, be such that the option of being unmatched is most preferred. For each woman $w \in W \setminus \{w_1, w_2, \ldots , w_{\ell }\}$, let $P_{w}$ be an arbitrary preference in $\fancyscript{D} _w$.

Table 1 Preferences of men in Case 1 in the proof of Lemma 3

It is easy to see that

$$\begin{aligned} \textit{DA}(P)=\begin{pmatrix} m_1 &{} m_2 &{} \ldots &{} m_{\ell -1} &{}m_{\ell } \\ w_1 &{} w_2 &{} \ldots &{} w_{\ell -1} &{}w_{\ell } \end{pmatrix}. \end{aligned}$$

On the other hand, if $w_1$ misrepresents her preference as $P'_{w_1}$ instead of $P_{w_1}$, then

$$\begin{aligned} \textit{DA}\left( P'_{w_1}, P_{-{w_1}}\right) = \begin{pmatrix} m_1 &{} m_2 &{} \ldots &{} m_{\ell -1} &{} m_{\ell } \\ w_\ell &{} w_1 &{} \ldots &{} w_{\ell -2} &{} w_{\ell -1} \end{pmatrix}. \end{aligned}$$

Therefore, since

$$\begin{aligned} \textit{DA}\left( P'_{w_1}, P_{-{w_1}}\right) ({w_1}) = m_2 \ \ P_{w_1} \ \ m_1=DA(P)({w_1}), \end{aligned}$$

$w_1$ can manipulate ${\textit{DA}}$ by misrepresenting her preference as $P'_{w_1}$ when her true preference is $P_{w_1}$.This contradicts ${\textit{DA}} \in {\textit{SP}}(\fancyscript{D} )$.

Case 2 $x=m_0$. Define $P_{m_1}, P_{m_2}, \ldots , P_{m_{\ell }}$ as in Table 2 and let the preferences of the remaining men be such that being unmatched is most preferred.

Table 2 Preferences of men in Case 2 in the proof of Lemma 3

In this case,

$$\begin{aligned} \textit{DA}(P)=\begin{pmatrix} m_1 &{} m_2 &{} \ldots &{} m_{\ell -1} &{}m_{\ell } \\ w_1 &{} w_2 &{} \ldots &{} w_{\ell -1} &{}w_{\ell } \end{pmatrix}. \end{aligned}$$

On the other hand, if $w_1$ represents her preference as $P'_{w_1}$ instead of $P_{w_1}$, then

$$\begin{aligned} \textit{DA}(P'_{w_1}, P_{-{w_1}})=\begin{pmatrix} m_1 &{} m_2 &{} \ldots &{} m_{\ell -1} &{} m_{\ell } \\ w_\ell &{} w_1 &{} \ldots &{} w_{\ell -2} &{} w_{\ell -1} \end{pmatrix}. \end{aligned}$$

Therefore, since

$$\begin{aligned} \textit{DA}(P'_{w_1}, P_{-{w_1}})({w_1}) = m_2 \ \ P_{w_1} \ \ m_1=DA(P)({w_1}), \end{aligned}$$

$w_1$ can manipulate ${\textit{DA}}$ by misrepresenting her preference as $P'_{w_1}$ when her true preference is $P_{w_1}$. This contradicts ${\textit{DA}} \in {\textit{SP}}(\fancyscript{D} )$. $\square$

By using Lemmas 2 and 3, we provide a Proof of Theorem.

Proof of Theorem

Suppose that $\fancyscript{D}$ satisfies the NDC. By Lemma 2, the DA is group strategy-proof. Therefore, there is a stable and strategy-proof rule. To prove the converse, suppose that there exists a stable and strategy-proof rule. By Fact, it is the DA rule. Since the DA is now strategy-proof, by Lemma 3, the NDC is satisfied. $\square$