Downstream competition and upstream labor market matching

Abstract

This study investigates how externalities from downstream competition shape sorting in upstream labor markets. I model this as a two-stage game: A first stage of simultaneous one-to-one matching between firms and managers and a second stage of Cournot competition among matched pairs. If a firm’s technology and human capital are strategic complements, it is rational for each firm-manager pair to expect that the remaining agents will form a positive assortative matching (PAM), and the PAM on the grand market is a stable matching under rational expectations. The PAM remains stable even when they are strategic substitutes but the substitutive effect is moderate. However, if the substitutive effect is sufficiently strong, a negative assortative matching is stable. Social welfare induced by stable matchings is discussed.

Appendices

Appendix

Cournot competition

Lemma A.1

Suppose $[q\cdot D'(q)]'<0$.

• (i) The inverse function $G^{-1}$ exists, and it is decreasing.

• (ii) $[D\circ G^{-1}]'\in (0,\frac{1}{n})$.

• (iii) For all $x\ge 0$, $\frac{\partial ^2 q\cdot D(q+x)}{\partial q \partial x}<0$ and $\frac{\partial ^2 q\cdot D(q+x)}{\partial q^2}<0$.

• (iv) For all $x\ge 0$, $\frac{\frac{\partial ^2 q\cdot D(q+x)}{\partial q \partial x}}{\frac{\partial ^2 q\cdot D(q+x)}{\partial q^2}}\in (0,1)$.

Proof

First, because $G'(q)=nD'(q)+[q\cdot D'(q)]'<0$, the inverse function $G^{-1}$ exists and it is decreasing.

Second, $[D\circ G^{-1}(c)]'=\frac{D'(q)}{G'(q)}=\frac{D'(q)}{nD'(q)+[qD'(q)]'}$. The value of the last term is in between 0 and 1 / n, because $D'(q)<0$ and $[qD'(q)]'<0$.

Third, since $[q\cdot D'(q)]'<0$ can be rewritten as $D'(q)+q\cdot D''(q)<0$ and $D''(q)>0$ almost everywhere, the second order derivatives of $q\cdot D(q+x)$ are

$$\begin{aligned} \frac{\partial ^2 q\cdot D(q+x)}{\partial q \partial x}&=D'(q+x)+q\cdot D''(q+x) \nonumber \\&=[D'(q+x)+(q+x)\cdot D''(q+x)]-x\cdot D''(q+x) \nonumber \\&<0\text { for all }x\ge 0;\end{aligned}$$

(3)

$$\begin{aligned} \frac{\partial ^2 q\cdot D(q+x)}{\partial q^2}&=2D'(q+x)+q\cdot D''(q+x)\nonumber \\&=[D'(q+x)+(q+x)\cdot D''(q+x)]+D'(q+x)-x\cdot D''(q+x) \nonumber \\&<0\text { for all }x\ge 0. \end{aligned}$$

(4)

Fourth, following from Eqs. (3) and (4), $\frac{\frac{\partial ^2 q\cdot D(q+x)}{\partial q \partial x}}{\frac{\partial ^2 q\cdot D(q+x)}{\partial q^2}}$ is larger than 0 but less than 1. $\square$

Proof of Lemma 1.1

In the Cournot game, given that agent pairs other than $(i,j)\in \mu$ produce in total an amount of $Q^{-ij}\ge 0$, (i, j) choose an output level $q\ge 0$ to maximize

$$\begin{aligned} qD(q+Q^{-ij})-qc^{ij}. \end{aligned}$$

This objective function is strictly concave, because, by Lemma A.1(iii), the second order condition is satisfied:

$$\begin{aligned}&\frac{\partial ^2 qD'(q+Q^{-ij})}{\partial q^2} <0. \end{aligned}$$

The first order condition for this problem is:

$$\begin{aligned}&\frac{\partial qD(q+Q^{-ij})}{\partial q}-c^{ij}=0 \\&\quad \Leftrightarrow D(q+Q^{-ij})+ qD'(q+Q^{-ij})-c^{ij}=0.\nonumber \end{aligned}$$

(5)

Hence, in equilibrium, the output of pair (i, j) is

$$\begin{aligned} q^{ij}_{\mu }&=\max \left\{ 0,~-\frac{D( Q_{\mu })-c^{ij}}{D'(Q_{\mu })}\right\} , \end{aligned}$$

where the total output under matching $\mu$, $Q_{\mu }$, satisfies:

$$\begin{aligned} Q_{\mu }&=-\frac{1}{D'(Q_{\mu })} \cdot \sum _{(i,j)\in \mu } \max \left\{ 0,~D(Q_{\mu })-c^{ij}\right\} \nonumber \\&\Leftrightarrow Q_{\mu }D'(Q_{\mu })+\sum _{(i,j)\in \mu } \max \left\{ 0,~D(Q_{\mu })-c^{ij}\right\} =0. \end{aligned}$$

(6)

Following from Assumption 1.1, the term on the left hand side of Eq. (6) is strictly decreasing in the value of $Q_{\mu }$, and it is strictly positive when $Q_{\mu }$ approached 0 and negative when $Q_{\mu }$ approached d. Hence, following from the intermediate value theorem, there is a unique value of $Q_{\mu }$ that solves the above equation. Equation (6) also implies

$$\begin{aligned} Q_{\mu }D'(Q_{\mu })+\sum _{(i,j)\in \mu } D(Q_{\mu })-c^{ij}\le 0 \Leftrightarrow G(Q_{\mu })\le C_{\mu } \Leftrightarrow Q_{\mu }\le G^{-1}(C_{\mu }), \end{aligned}$$

(7)

where the equalities hold if each firm produces a positive amount.

Last, I prove that in the unique equilibrium the output produced by each firm-manager pair is strictly positive. It is sufficient to show $D(Q_{\mu })-c^{ij}>0$ for all $(i,j)\in \mu$. Because both D and $G^{-1}$ are strictly decreasing, $D\circ G^{-1}$ is strictly increasing and

$$\begin{aligned} D(Q_{\mu })-c^{ij}&\ge D\circ G^{-1}(C_{\mu })-c^{ij}\\&\ge D\circ G^{-1}(c^{ij})-c^{ij}\\&=[D\circ G^{-1}(1)-1]-\int _{c^{ij}}^{1}[D\circ G^{-1}(c)-c]'dc\\&>[D\circ G^{-1}(1)-1]-\int _{c^{ij}}^{1}\frac{1}{n}-1 dc\\&>0, \end{aligned}$$

where the third inequality follows from Lemma A.1(ii).

Hence, in the unique equilibrium, the total output is

$$\begin{aligned} Q_{\mu }&=G^{-1}(C_{\mu }), \end{aligned}$$

the output of pair (i, j) is

$$\begin{aligned} q^{ij}_{\mu }&=-\frac{D( Q_{\mu })-c^{ij}}{D'(Q_{\mu })}, \end{aligned}$$

(8)

and the profit of (i, j), obtained by substituting $Q_{\mu }$ and $q^{ij}_{\mu }$ back into the objective function, is

$$\begin{aligned} \pi ^{ij}_{\mu }&=-\frac{\left[ D( Q_{\mu })-c^{ij}\right] ^2}{D'(Q_{\mu })}. \end{aligned}$$

$\square$

Corollary A.1

For any pair (i, j), $\frac{d}{dc} \left( \frac{D\circ G^{-1}(c)-c^{ij}}{D'\circ G^{-1}(c)}\right) <0$ for all $c\in [c^{ij},n-1+c^{ij}]$.

Proof

It is equivalent to prove that the derivative $\frac{d}{dQ} \left( \frac{D(Q)-c^{ij}}{D'(Q)}\right) >0$ for all $Q\in [G^{-1}(n-1+c^{ij}), G^{-1}(c^{ij})]$.

Denote $Q^{-ij}$ as the total output of all firm-manager pairs other than (i, j) and $q^{ij}(Q^{-ij})$ as (i, j)’s best response. Restrict attention to the set of $Q^{-ij}>0$ such that $q^{ij}(Q^{-ij})>0$. First, applying the implicit function theorem and Lemma A.1, to the first order condition (Eq. (5)), we have

$$\begin{aligned} \frac{dq^{ij}}{d Q^{-ij}}&=-\frac{\frac{\partial ^2 q^{ij}D(q^{ij}+Q^{-ij})}{\partial q^{ij}\partial Q^{-ij}}}{\frac{\partial ^2 q^{ij}D(q^{ij}+Q^{-ij})}{\partial (q^{ij})^2}}\in (-1,0). \end{aligned}$$

Therefore, $q^{ij}(Q^{-ij})+Q^{-ij}$ is strictly increasing in $Q^{-ij}$ for $Q^{-ij}\ge 0$, with the infimum achieved when $Q^{-ij}=0$ and the supremum achieved when $q^{ij}=0$.¹¹ In the Cournot equilibrium under any matching, every firm-manager pair produces a positive amount. Hence, the values of $G^{-1}(n-1+c^{ij})$ and $G^{-1}(c^{ij})$ lie between the infimum and supremum.

Second, taking the derivative with respect to $Q^{-ij}$ on both sides of Eq. (8) yields

$$\begin{aligned} \frac{dq^{ij}}{dQ^{-ij}} =&-\frac{d\left[ \frac{ D(q^{ij}+Q^{-ij})-c^{ij}}{ D'(q^{ij}+Q^{-ij}) }\right] }{d(q^{ij}+Q^{-ij})}\cdot \frac{d (q^{ij}+Q^{-ij})}{Q^{-ij}}\\ =&-\frac{d\left[ \frac{ D(q^{ij}+Q^{-ij})-c^{ij}}{ D'(q^{ij}+Q^{-ij}) }\right] }{d(q^{ij}+Q^{-ij})}\cdot \left( \frac{d q^{ij}}{dQ^{-ij}}+1\right) <0. \end{aligned}$$

Hence, $\frac{d\left[ \frac{ D(q^{ij}+Q^{-ij})-c^{ij}}{ D'(q^{ij}+Q^{-ij}) }\right] }{d(q^{ij}+Q^{-ij})}>0$ for all $q^{ij}(Q^{-ij})+Q^{-ij}\in [G^{-1}(n-1+c^{ij}),G^{-1}(c^{ij})]$. This yields the desired result. $\square$

Stable matchings

1.1 Linear programming

Proof of Lemma 1.2

The proof uses Shapley and Shubik (1971) linear programming method:

$$\begin{aligned} \begin{aligned}&{\text {(Primal)}}\\&\max _x z= \sum _{i\in I_s} \sum _{j\in J_s} \pi ^{ij}_{\varphi _{\mu ^{M_1}}^{M_2}(i,j)}\cdot x_{ij}\\&s.t.~\textstyle \sum _{i\in I_2} x_{ij}\le 1 ,~~\forall j\in J_2\\&\textstyle \sum _{j\in J_2} x_{ij}\le 1 ,~~\forall i\in I_2\\&x_{ij}\ge 0 \end{aligned} \quad \begin{aligned}&{\text {(Dual)}}\\&\min _{u,v} w=\sum _{i\in I_2} u_i +\sum _{j\in J_2} v_j\\&s.t.~u_i+v_j\ge \pi ^{ij}_{\varphi _{\mu ^{M_1}}^{M_2}(i,j)},~~\forall i\in I_2 ,\forall j\in J_2\\&u_i\ge 0 \\&v_j\ge 0 \end{aligned} \end{aligned}$$

The maximum value $z_{max}$ of the primal problem is attained when $x_{ij}=0$ or 1 for all i and j. By strong duality, both the primal and the dual have the same optimal value, i.e., $z_{max}=w_{min}$. By complementary slackness, if $x^*_{ij}=1$ then the corresponding dual constraint must be tight which implies $u_i+v_j= \pi ^{ij}_{\varphi _{\mu ^{M_1}}^{M_2}(i,j)}$.

Taking an extreme solution $x^*$ to the primal problem, I construct a matching $\mu ^{M_2}$ such that it satisfies that

$$\begin{aligned} (i,j)\in \mu ^{M_2}\quad \textit{ if and only if }\quad x^*_{ij}=1. \end{aligned}$$

(Sufficiency). Take a solution $(u^*,v^*)^{M_2}$ to the dual problem. If $\mu ^{M_2}$ is admissible, then the triplet $(\mu ^{M_2},(u^*,v^*)^{M_2})$ satisfies the conditions in Definition 1.1, and thus $\mu ^{M_1}\cup \mu ^{M_2}$ is a $\varphi _{\mu ^{M_1}}^{M_2}$-stable matching.¹²$^,$¹³

(Necessity). Take a matching $\mu ^{M_2}$. If condition ii of Lemma 1.2 is not satisfied, the matching is not stable. If condition i is not satisfied, then $V(\mu ^{M_2}| \varphi _{\mu ^{M_1}}^{M_2})<z_{max}=w_{min}$. Therefore, the feasibility condition of Definition 1.1 cannot be satisfied, and thus $\mu ^{M_2}$ is not stable. $\square$

1.2 Stable matchings under rational beliefs

Due to externalities from the downstream, it is quite involving to prove Theorem 2.1. In particular, in case i PAM results in the highest industry production efficiency and thus the lowest Cournot equilibrium market price, which is a counterforce of stability (because it requires the highest total expected profit). This conflict makes the proof difficult. Similarly, this difficulty also exists in case (iii). Case ii has no such conflict and thus is easier to prove.

To prove cases (i) and (iii), I provide a series of technical lemmas below. The general idea is to prove these two cases in a recursive method, a series of pairwise permutations (Lemma B.7). I want to show that the total expected profit under a permuted matching is greater than that under the original one. However, the profit of a firm-manager pair is proportional to the square of their unit profit, and I need to make comparisons across the square terms. In order to do so, I factorize differences in the square terms (Lemmas B.5–B.6) and make comparisons across the unit profits (Lemmas B.2–B.4).

Since an expected profit crucially depends on its associated expected industry production efficiency, the first step is to make comparisons across expected industry production efficiencies (Lemma B.1).

Now, consider a partition $[M_1,M_2]$ with $\mu ^{M_1}$ on market $M_1$. Denote

$$\begin{aligned} M_2=&I_2\cup J_2:=\{i_1,\ldots ,i_{t}\}\cup \{j_1,\ldots ,j_{t}\}. \end{aligned}$$

Denote $\mu _P^{M_2\backslash \{i,j \}}$ as the positively ordered matching (based on indicators) on market $M_2\backslash \{i,j \}$ such that for all $(i_x,j_z),(i_{x'},j_{z'})\in \mu _P^{M_2\backslash \{i,j \}}, x>x'\Leftrightarrow z>z'$. In short, let us denote $C^{ij}:=C_{\mu ^{M_1}\cup (i,j)\cup \mu _P^{M_2\backslash \{i,j \}}}$. Focus on the two scenarios below for Lemmas B.1–B.7:

• Scenario 1: $\frac{\partial ^2 c}{\partial f \partial s}\le 0$; $f_{i_k}>f_{i_{k+1}}~\text {and}~s_{i_k}>s_{i_{k+1}}$ for $k=1,\ldots ,t-1$.

• Scenario 2: $\frac{\partial ^2 c}{\partial f \partial s}> 0$; $f_{i_k}>f_{i_{k+1}}~\text {and}~s_{i_k}<s_{i_{k+1}}$ for $k=1,\ldots ,t-1$.

Lemma B.1

Take three numbers, k, p, q, such that $1\le k<p, q\le t$. In Scenarios 1 and 2, the four industry production efficiency differences, $C^{i_kj_k}-C^{i_kj_q}$, $C^{i_pj_q}-C^{i_pj_k}$, $C^{i_pj_q}-C^{i_pj_k}$, and $C^{i_pj_q}-C^{i_kj_q}$, are all non-positive.

Proof

Let us first consider the case in which $p\le q$. The industrial production efficiencies under four matchings are:

$$\begin{aligned} C^{i_kj_k}&=C_{\mu ^{M_1}\cup (i_k,j_k)\cup \mu _P^{M_2\backslash \{i_k,j_k \}}} =\sum _{l=1}^{t}c^{i_lj_l} +C_{\mu ^{M_1}};\\ C^{i_kj_q}&=C_{\mu ^{M_1}\cup (i_k,j_q)\cup \mu _P^{M_2\backslash \{i_k,j_q \}}} =\sum _{l=1}^{k-1}c^{i_lj_l} +c^{i_kj_q} +\sum _{l=k+1}^{q}c^{i_{l}j_{l-1}} \\&\quad +\sum _{l=q+1}^{t}c^{i_lj_l} +C_{\mu ^{M_1}}; \\ C^{i_pj_q}&=C_{\mu ^{M_1}\cup (i_p,j_q)\cup \mu _P^{M_2\backslash \{i_p,j_q \}}} =\sum _{l=1}^{p-1}c^{i_lj_l} +c^{i_pj_q} +\sum _{l=p+1}^{q}c^{i_{l}j_{l-1}} \\&\quad +\sum _{l=q+1}^{t}c^{i_lj_l} +C_{\mu ^{M_1}} ; \\ C^{i_pj_k}&=C_{\mu ^{M_1}\cup (i_p,j_k)\cup \mu _P^{M_2\backslash \{i_p,j_k \}}} = \sum _{l=1}^{k-1}c^{i_lj_l} +c^{i_pj_k}\\&\quad +\sum _{l=k+1}^{p}c^{i_{l-1}j_{l}} +\sum _{l=p+1}^{t}c^{i_lj_l} +C_{\mu ^{M_1}}. \end{aligned}$$

In both scenarios:

$$\begin{aligned} C^{i_kj_k}-C^{i_kj_q}&=\sum _{l=1}^{t}c^{i_lj_l} -\left[ \sum _{l=1}^{k-1}c^{i_lj_l} +c^{i_kj_q} +\sum _{l=k+1}^{q}c^{i_{l}j_{l-1}} +\sum _{l=q+1}^{t}c^{i_lj_l} \right]&\nonumber \\&=(c^{i_kj_k}-c^{i_kj_q}) +\sum _{l=k+1}^{q}(c^{i_{l}j_{l}}- c^{i_{l}j_{l-1}})&\end{aligned}$$

(9)

$$\begin{aligned}&=\int ^{s_{j_k}}_{s_{j_q}}\frac{\partial c(f_{i_k},s)}{\partial s}ds -\sum _{l=k+1}^{q}\int ^{s_{j_{l-1}}}_{s_{j_l}}\frac{\partial c(f_{i_{l}},s)}{\partial s}ds \nonumber \\&=\sum _{l=k+1}^{q}\int ^{s_{j_{l-1}}}_{s_{j_l}}\left[ \frac{\partial c(f_{i_k},s)}{\partial s}-\frac{\partial c(f_{i_{l}},s)}{\partial s}\right] ds&\nonumber \\&=\sum _{l=k+1}^{q}\int ^{s_{j_{l-1}}}_{s_{j_l}}\int _{f_{i_l}}^{f_{i_k}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dfds&\nonumber \\&\le 0; \end{aligned}$$

(10)

$$\begin{aligned} C^{i_pj_q} - C^{i_pj_k}&= (c^{i_pj_q}-c^{i_pj_k}) + \sum _{l=k+1}^{p}(c^{i_{l-1}j_{l-1}}-c^{i_{l-1}j_{l}} ) + \sum _{l=p+1}^{q}(c^{i_{l}j_{l-1}}-c^{i_{l}j_{l}}) \end{aligned}$$

(11)

$$\begin{aligned}&=-\int _{s_{j_q}}^{s_{j_k}}\frac{\partial c(f_{i_p},s)}{\partial s}ds + \sum _{l=k+1}^{p}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial c(f_{i_{l-1}},s)}{\partial s}ds \nonumber \\&\quad + \sum _{l=p+1}^{q}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial c(f_{i_l},s)}{\partial s}ds \nonumber \\&= \sum _{l=k+1}^{p}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial c(f_{i_{l-1}},s)}{\partial s}-\frac{\partial c(f_{i_p},s)}{\partial s}ds \nonumber \\&\quad + \sum _{l=p+1}^{q}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial c(f_{i_l},s)}{\partial s}-\frac{\partial c(f_{i_p},s)}{\partial s}ds \nonumber \\&= \sum _{l=k+1}^{p}\int _{s_{j_l}}^{s_{j_{l-1}}}\int _{f_{i_p}}^{f_{i_{l-1}}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dfds\nonumber \\&\quad + \sum _{l=p+1}^{q}\int _{s_{j_l}}^{s_{j_{l-1}}}\int _{f_{i_p}}^{f_{i_{l}}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dfds\nonumber \\&\le 0; \end{aligned}$$

(12)

$$\begin{aligned} C^{i_kj_k}-C^{i_pj_k}&=\sum _{l=k}^pc^{i_lj_l}-\left[ c^{i_pj_k}+ \sum _{l=k+1}^pc^{i_{l-1}j_l} \right] \nonumber \\&=(c^{i_kj_k}-c^{i_pj_k})+\frac{1}{n}\sum _{l=k+1}^p(c^{i_lj_l}-c^{i_{l-1}j_l}) \end{aligned}$$

(13)

$$\begin{aligned}&=\int _{f_{i_p}}^{f_{i_k}}c(f,s_{j_k})df -\sum _{l=k+1}^p\int _{f_{i_l}}^{f_{i_{l-1}}}c(f,s_{j_l})df \nonumber \\&=\sum _{l=k+1}^p\int _{f_{i_l}}^{f_{i_{l-1}}}\int _{s_{j_l}}^{s_{j_k}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dsdf\nonumber \\&\le 0; \end{aligned}$$

(14)

$$\begin{aligned} C^{i_pj_q}-C^{i_kj_q}&=\left[ \sum _{l=k}^{p-1}c^{i_lj_l}+c^{i_pj_q}\right] -\left[ c^{i_kj_q}+\sum _{l=k+1}^{p}c^{i_lj_{l-1}}\right] \\&=-(c^{i_kj_q}-c^{i_pj_q})+\sum _{l=k+1}^{p}( c^{i_{l-1}j_{l-1}}-c^{i_lj_{l-1}}) \\&=-\int ^{f_{i_k}}_{f_{i_p}}\frac{\partial c(f,s_{j_q})}{\partial f}df+\sum _{l=k+1}^{p}\int ^{f_{i_{l-1}}}_{f_{i_l}}\frac{\partial c(f,s_{j_{l-1}})}{\partial f}df \\&=\sum _{l=k+1}^{p}\int ^{f_{i_{l-1}}}_{f_{i_l}}\left[ \frac{\partial c(f,s_{j_{l-1}})}{\partial f}-\frac{\partial c(f,s_{j_q})}{\partial f}\right] df \\&=\sum _{l=k+1}^{p}\int ^{f_{i_{l-1}}}_{f_{i_l}}\int _{s_{j_q}}^{s_{j_{l-1}}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dsdf. \\&\le 0. \end{aligned}$$

The proof for the case in which $p\ge q$ is just a mirror image and thus is omitted. $\square$

Denote the unit profit of (i, j) in a Nash equilibrium of the Cournot game, given a matching $\mu ^{M_1}\cup (i,j)\cup \mu _P^{M_2\backslash \{i,j \}}$, by

$$\begin{aligned} y^{ij}:=D\circ G^{-1}(C^{ij})-c^{ij}. \end{aligned}$$

Lemma B.2

In Scenario 1, the three unit profit differences, $y^{i_kj_k}-y^{i_kj_q}$, $y^{i_kj_k}-y^{i_pj_k}$, $y^{i_kj_q}-y^{i_pj_q}$, and $y^{i_pj_k}-y^{i_pj_q}$, are all strictly positive.

Proof

This proof is an application of Lemmas A.1(ii) and B.1. Let us first consider the case in which $p\le q$. First,

$$\begin{aligned} y^{i_kj_k}-y^{i_kj_q}&=D\circ G^{-1}(C^{i_kj_k}) -D\circ G^{-1}(C^{i_kj_q}) -(c^{i_kj_k}-c^{i_kj_q})\\&\ge \frac{1}{n}(C^{i_kj_k}-C^{i_kj_q} ) -(c^{i_kj_k}-c^{i_kj_q})\\&= -\frac{1}{n}\sum _{l=k+1}^{q}\int ^{s_{j_{l-1}}}_{s_{j_l}}\frac{\partial c(i_{l},s)}{\partial s}ds-\frac{n-1}{n}\int ^{s_{j_k}}_{s_{j_q}}\frac{\partial c(f_{i_k},s)}{\partial s}ds\\&>0, \end{aligned}$$

where the first inequality follows from Lemma A.1(ii) and $C_P^{i_kj_k}-C_P^{i_kj_q}\le 0$ and the second equation follows from Eqs. (9) and (10). Similarly,

$$\begin{aligned} y^{i_kj_k}-y^{i_pj_k}&=D\circ G^{-1}(C^{i_kj_k}) -D\circ G^{-1}(C^{i_pj_k}) -(c^{i_kj_k}-c^{i_pj_k})\\&\ge \frac{1}{n}(C^{i_kj_k}- C^{i_pj_k}) -(c^{i_kj_k}-c^{i_pj_k})\\&=-\frac{1}{n}\sum _{l=k+1}^p\int _{f_{i_l}}^{f_{i_{l-1}}}\frac{\partial c(f,s_{j_l})}{\partial f}df -\frac{n-1}{n}\int _{f_{i_p}}^{f_{i_k}}\frac{\partial c(f,s_{j_k})}{\partial f}df\\&>0, \end{aligned}$$

where the first inequality follows from Lemma A.1(ii) and $C_P^{i_kj_k}-C_P^{i_pj_k}\le 0$ and the second equation following from Eqs. (13) and (14). Third,

$$\begin{aligned} y^{i_kj_q}-y^{i_pj_q}&=D\circ G^{-1}(C^{i_kj_q})- D\circ G^{-1}(C^{i_pj_q}) -(c^{i_kj_q}-c^{i_pj_q})\\&\ge -(c^{i_kj_q}-c^{i_pj_q})\\&=-\int _{f_{i_p}}^{f_{i_k}}\frac{\partial c(f,s_{j_q})}{\partial f}df\\&>0, \end{aligned}$$

where the first inequality follows from Lemma A.1(ii) and $C_P^{i_pj_q}-C_P^{i_kj_q}\le 0$. Fourth,

$$\begin{aligned} y^{i_pj_k}-y^{i_pj_q}&=D\circ G^{-1}(C^{i_pj_k})- D\circ G^{-1}(C^{i_pj_q}) -(c^{i_pj_k}-c^{i_pj_q})\\&\ge -(c^{i_pj_k}-c^{i_pj_q})\\&=-\int _{s_{j_q}}^{s_{j_q}}\frac{\partial c(f_{i_p},s)}{\partial s}ds\\&>0, \end{aligned}$$

where the first inequality follows from Lemma A.1(ii) and $C_P^{i_pj_q}-C_P^{i_pj_k}\le 0$.

The proof for the case in which $p\ge q$ is just a mirror image and thus is omitted. $\square$

Lemma B.3

In Scenarios 1 and 2, for $n\ge 2$,

$$\begin{aligned} (y^{i_kj_k}-y^{i_kj_q})+(y^{i_pj_q}-y^{i_pj_k}) \ge -\frac{n-2}{n}\int ^{f_{i_k}}_{f_{i_p}}\int ^{s_{j_k}}_{s_{j_q}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dsdf \ge 0. \end{aligned}$$

Proof

$$\begin{aligned}&\left( y^{i_kj_k}-y^{i_kj_q}\right) +\left( y^{i_pj_q}-y^{i_pj_k}\right) \\&\quad = \left\{ \left[ D\circ G^{-1}(C^{i_kj_k})-c^{i_kj_k}\right] - \left[ D\circ G^{-1}(C^{i_kj_q})-c^{i_kj_q}\right] \right\} \\&\qquad +\left\{ \left[ D\circ G^{-1}(C^{i_pj_q})-c^{i_pj_q}\right] - \left[ D\circ G^{-1}(C^{i_pj_k})-c^{i_pj_k}\right] \right\} \\&\quad \ge \left[ \frac{1}{n}(C^{i_kj_k}-C^{i_kj_q} ) -(c^{i_kj_k}-c^{i_kj_q})\right] + \left[ \frac{1}{n}(C^{i_pj_q} - C^{i_pj_k}) - ( c^{i_pj_q}-c^{i_pj_k})\right] \\&\quad = -\frac{1}{n}\sum _{l=k+1}^{q}\int ^{s_{j_{l-1}}}_{s_{j_l}}\frac{\partial c(f_{i_{l}},s)}{\partial s}ds\\&\qquad +\frac{1}{n}\sum _{l=k+1}^{p}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial c(f_{i_{l-1}},s)}{\partial s}ds +\frac{1}{n}\sum _{l=p+1}^{q}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial c(f_{i_l},s)}{\partial s}ds\\&\qquad -\frac{n-1}{n}\int ^{s_{j_k}}_{s_{j_q}}\left[ \frac{\partial c(f_{i_k},s)}{\partial s} - \frac{\partial c(f_{i_p},s)}{\partial s} \right] ds \\&\quad =\frac{1}{n}\sum _{l=k+1}^{p}\int _{f_{i_{l}}}^{f_{i_{l-1}}}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dsdf -\frac{n-1}{n}\int ^{f_{i_k}}_{f_{i_p}}\int ^{s_{j_k}}_{s_{j_q}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dsdf\\&\quad \ge -\frac{n-2}{n}\int ^{f_{i_k}}_{f_{i_p}}\int ^{s_{j_k}}_{s_{j_q}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dsdf \\&\quad \ge 0 \text { for }n\ge 2, \end{aligned}$$

where the first inequality follows from Lemma A.1(ii), $C_P^{i_kj_k}-C_P^{i_kj_q}\le 0$, and $C^{i_pj_q}-C^{i_pj_k}\le 0$. $\square$

Lemma B.4

In Scenario 2, $y^{i_pj_q}-y^{i_pj_k}>0$.

Proof

For the case in which $p\le q$:

$$\begin{aligned} y^{i_pj_q}-y^{i_pj_k}&= D\circ G^{-1}\left( C^{i_pj_q}\right) - D\circ G^{-1}(C^{i_pj_k}) -\left( c^{i_pj_q}-c^{i_pj_k}\right) \\&\ge \frac{1}{n}\left[ C^{i_pj_q}-C^{i_pj_k}\right] -(c^{i_pj_q}-c^{i_pj_k})\\&=\frac{1}{n}\sum _{l=k+1}^{p}(c^{i_{l-1}j_{l-1}}-c^{i_{l-1}j_{l}} ) + \frac{1}{n}\sum _{l=p+1}^{q}(c^{i_{l}j_{l-1}}-c^{i_{l}j_{l}})\\&\quad +\frac{n-1}{n}\left( c^{i_pj_k} - c^{i_pj_q} \right) \\&= \frac{1}{n}\sum _{l=k+1}^{p}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial c(f_{i_{l-1}},s)}{\partial s}ds + \frac{1}{n}\sum _{l=p+1}^{q}\int _{s_{j_l}}^{s_{j_{l-1}}}\frac{\partial c(f_{i_l},s)}{\partial s}ds\\&\quad +\frac{n-1}{n}\int _{s_{j_q}}^{s_{j_k}}\frac{\partial c(f_{i_p},s)}{\partial s}ds\\&>0, \end{aligned}$$

where the first inequality follows from Lemma A.1(ii) and $C^{i_pj_q}-C^{i_pj_k}\le 0$ and the second equality follows from Eqs. (11) and (12). The proof for the case in which $p\ge q$ is the same as above except for that there is no summation from $p+1$ to q. $\square$

Lemma B.5

In Scenario 1,

$$\begin{aligned} (y^{i_kj_k})^2+(y^{i_pj_q})^2-(y^{i_kj_q})^2-(y^{i_pj_k})^2 > 0. \end{aligned}$$

Proof

$$\begin{aligned}&(y^{i_kj_k})^2+(y^{i_pj_q})^2-(y^{i_kj_q})^2-(y^{i_pj_k})^2\\&\quad =(y^{i_kj_k}-y^{i_kj_q})(y^{i_kj_k}+y^{i_kj_q})+(y^{i_pj_q}-y^{i_pj_k})(y^{i_pj_q}+y^{i_pj_k}) \\&\quad > \left[ (y^{i_kj_k}-y^{i_kj_q})+(y^{i_pj_q}-y^{i_pj_k})\right] (y^{i_pj_q}+y^{i_pj_k})\\&\quad \ge 0, \end{aligned}$$

where the first inequality follows from Lemma B.2 and the second inequality follows from Lemma B.3. $\square$

Lemma B.6

In Scenario 2, if $\frac{\partial ^2 c^2}{\partial f \partial s}/\frac{\partial ^2 c}{\partial f \partial s}\le \frac{n-2}{n}\cdot 2D\circ G^{-1}(0)$, then

$$\begin{aligned} (y^{i_kj_k})^2+(y^{i_pj_q})^2-(y^{i_kj_q})^2-(y^{i_pj_k})^2 \ge 0. \end{aligned}$$

Proof

First,

$$\begin{aligned}&(y^{i_kj_k})^2-(y^{i_kj_q})^2\\&\quad =\left( \left[ D\circ G^{-1}(C^{i_kj_k})-c^{i_kj_k}\right] -\left[ D\circ G^{-1}(C^{i_kj_q})-c^{i_kj_q}\right] \right) \\&\qquad \cdot \left( \left[ D\circ G^{-1}(C^{i_kj_k})-c^{i_kj_k}\right] +\left[ D\circ G^{-1}(C^{i_kj_q})-c^{i_kj_q}\right] \right) \\&\quad \ge ( y^{i_kj_k}-y^{i_kj_q})\cdot \left[ D\circ G^{-1}(C^{i_kj_k})+D\circ G^{-1}(C^{i_kj_q})\right] +\left[ (c^{i_kj_k})^2-(c^{i_kj_q})^2\right] , \end{aligned}$$

where the inequality follows from $C^{i_kj_k}-C^{i_kj_q}\le 0$. Second,

$$\begin{aligned}&(y^{i_pj_q})^2-(y^{i_pj_k})^2\\&\quad = \left( \left[ D\circ G^{-1}(C^{i_pj_q})-c^{i_pj_q}\right] -[D\circ G^{-1}(C^{i_pj_k})-c^{i_pj_k}]\right) \\&\qquad \cdot \left( \left[ D\circ G^{-1}(C^{i_pj_q})-c^{i_pj_q}\right] + \left[ D\circ G^{-1}(C^{i_pj_k})-c^{i_pj_k}\right] \right) \\&\quad \ge (y^{i_pj_q}-y^{i_pj_k})\cdot \left[ D\circ G^{-1}(C^{i_pj_q})+D\circ G^{-1}(C^{i_pj_k})\right] +\left[ (c^{i_pj_q})^2-(c^{i_pj_k})^2\right] \\&\quad \ge (y^{i_pj_q}-y^{i_pj_k})\cdot \left[ D\circ G^{-1}(C^{i_kj_k})+D\circ G^{-1}(C^{i_kj_q})\right] +\left[ (c^{i_pj_q})^2-(c^{i_pj_k})^2\right] , \end{aligned}$$

where the first inequality follows from $C^{i_pj_q}-C^{i_pj_k}\le 0$ and $c^{i_pj_q}<c^{i_pj_k}$ and the second inequality follows from equations $C^{i_kj_q}-C_P^{i_pj_q}\le 0$, $C^{i_kj_k}-C_P^{i_pj_k}\le 0$, and $y^{i_pj_q}-y^{i_pj_k}>0$ (Lemma B.4).

Sum up the two components:

$$\begin{aligned}&(y^{i_kj_k})^2+(y^{i_pj_q})^2-(y^{i_kj_q})^2-(y^{i_pj_k})^2\\&\quad \ge \left[ (y^{i_kj_k}-y^{i_kj_q})+(y^{i_pj_q}-y^{i_pj_k})\right] \cdot \left[ D\circ G^{-1}(C^{i_kj_k})+D\circ G^{-1}(C^{i_kj_q})\right] \\&\qquad +(c^{i_kj_k})^2-(c^{i_k\mu _k(i_k)})^2+(c^{i_pj_q})^2-(c^{i_pj_k})^2\\&\quad \ge -\frac{n-2}{n}\int ^{f_{i_k}}_{f_{i_p}}\int ^{s_{j_k}}_{s_{j_q}}\frac{\partial ^2 c(f,s)}{\partial f \partial s}dsdf \cdot 2D\circ G^{-1}(0)\\&\qquad +\int ^{f_{i_k}}_{f_{i_p}}\int ^{s_{j_k}}_{s_{j_q}}\frac{\partial ^2[c(f,s)]^2}{\partial f \partial s}(f,s)dsdf\\&\quad \ge 0~\text { if }\frac{\partial ^2 c^2}{\partial f \partial s}/\frac{\partial ^2 c}{\partial f \partial s}\le \frac{n-2}{n}\cdot 2D\circ G^{-1}(0), \end{aligned}$$

where the second inequality follows from Lemma B.3. $\square$

Lemma B.7

In Scenario 1,

$$\begin{aligned} \sum _{(i,j)\in \mu ^{M_2}_{P}} \left[ D\circ G^{-1}(C^{ij})-c^{ij}\right] ^2 - \sum _{(i,j)\in \mu ^{M_2}} \left[ D\circ G^{-1}(C^{ij})-c^{ij}\right] ^2> 0. \end{aligned}$$

In Scenario 2, if $\frac{\partial ^2 c^2}{\partial f \partial s}/\frac{\partial ^2 c}{\partial f \partial s}\le \frac{n-2}{n}\cdot 2D\circ G^{-1}(0)$,

$$\begin{aligned} \sum _{(i,j)\in \mu ^{M_2}_{P}} \left[ D\circ G^{-1}(C^{ij})-c^{ij}\right] ^2 - \sum _{(i,j)\in \mu ^{M_2}} \left[ D\circ G^{-1}(C^{ij})-c^{ij}\right] ^2\ge 0. \end{aligned}$$

Proof

First, define a sequence of permutated matchings of $\mu ^{M_2}$, $\{\mu _1,\mu _2,\ldots ,\mu _t\}$:

$$\begin{aligned} \mu _1&:=\mu ^{M_2};\\ \mu _{k+1}&:=\left\{ (i_k,j_k),(\mu _k(j_k),\mu _k(i_k)) \right\} \cup \mu _k\backslash \left\{ (i_k,\mu _k(i_k)),(\mu _k(j_k),j_k)\right\} \\&\quad \quad \text { for }\quad k=1,2,\ldots ,t-1;\\ \mu _t&:=\{(i_1,j_1),(i_2,j_2),\ldots ,(i_t,j_t) \}. \end{aligned}$$

Then,

$$\begin{aligned}&\sum _{(i,j)\in \mu ^{M_2}_{P}} \left[ D\circ G^{-1}(C^{ij})-c^{ij}\right] ^2 - \sum _{(i,j)\in \mu ^{M_2}} [D\circ G^{-1}(C^{ij})-c^{ij}]^2\\&\quad = \sum _{k=1}^{t-1}\left[ \sum _{(i,j)\in \mu ^{M_2}_{k+1}}[D\circ G^{-1}(C^{ij})-c^{ij}]^2 - \sum _{(i,j)\in \mu ^{M_2}_{k}}[D\circ G^{-1}(C^{ij})-c^{ij}]^2 \right] \\&\quad = \sum _{k=1}^{t-1}\left[ \left[ D\circ G^{-1}(C^{i_kj_k})-c^{i_kj_k}\right] ^2+\left[ D\circ G^{-1}(C^{\mu _k(j_k)\mu _k(i_k)})-c^{\mu _k(j_k)\mu _k(i_k)}\right] ^2\right. \\&\qquad \left. - \left[ D\circ G^{-1}(C^{i_k\mu _k(i_k)})-c^{i_k\mu _k(i_k)}\right] ^2 -\left[ D\circ G^{-1}(C^{\mu _k(j_k)j_k})-c^{\mu _k(j_k)j_k}\right] ^2 \right] \\&\quad = \sum _{k=1}^{t-1} \left( y^{i_kj_k}\right) ^2+\left( y^{\mu _{k}(j_k)\mu _k(i_k)}\right) ^2-\left( y^{i_k\mu _k(i_k)}\right) ^2-\left( y^{\mu _{k}(j_k)j_k}\right) ^2\\&\quad \ge 0~(>0 \text { in Scenario 1}) , \end{aligned}$$

where the last inequality follows from Lemmas B.5 and B.6 . $\square$

Proof of Theorem 2.1

To prove the theorem, it is sufficient to check that the conditions in Lemma 1.2 are satisfied. I only consider assortative matchings among available agents, and thus $\varphi$-admissibility is automatically satisfied. I only need to check condition i. Again, consider a partition $[M_1,M_2]$ with $\mu ^{M_1}$ on market $M_1$, where

$$\begin{aligned} M_2=&I_2\cup J_2:=\{i_1,\ldots ,i_{t}\}\cup \{j_1,\ldots ,j_{t}\}. \end{aligned}$$

Take a matching $\mu ^{M_2}$ on market $M_2$. For cases i and ii, the conditional beliefs that I propose are $\varphi _{\mu ^{M_1}}^{M_2}(i,j)=\mu ^{M_1}\cup \{ (i,j) \}\cup \mu _+^{M_1\backslash \{i,j\}}$, and it is sufficient to show that $V(\mu ^{M_2}_+|\varphi _{\mu ^{M_1}}^{M_2})>V(\mu ^{M_2}|\varphi _{\mu ^{M_1}}^{M_2})$ if $\mu ^{M_2}\ne \mu ^{M_2}_+$. For case iii, the conditional beliefs that I propose are $\varphi _{\mu ^{M_1}}^{M_2}(i,j)=\mu ^{M_1}\cup \{ (i,j) \}\cup \mu _-^{M_1\backslash \{i,j\}}$, and it is sufficient to show that $V(\mu ^{M_2}_-|\varphi _{\mu ^{M_1}}^{M_2})>V(\mu ^{M_2}|\varphi _{\mu ^{M_1}}^{M_2})$ if $\mu ^{M_2}\ne \mu ^{M_2}_-$.

In the following, I will prove cases i and iii first, because these two cases are similar, and prove case ii afterward.

Case (i) Take a matching $\mu ^{M_2}\ne \mu ^{M_2}_+$. Without loss of generality, let us focus on Scenario 1 as above.

$$\begin{aligned}&V\left( \mu ^{M_2}_+|\varphi _{\mu ^{M_1}}^{M_2}\right) -V\left( \mu ^{M_2}|\varphi _{\mu ^{M_1}}^{M_2}\right) \\&\quad =\sum _{(i,j)\in \mu ^{M_2}_{+}}\pi ^{ij}_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _+^{M_2\backslash \{i,j\}}}- \sum _{(i,j)\in \mu ^{M_2}}\pi ^{ij}_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _+^{M_2\backslash \{i,j\}}}\\&\quad = \sum _{(i,j)\in \mu ^{M_2}_{+}} \frac{[D\circ G^{-1}(C^{ij})-c^{ij}]^2}{-D'\circ G^{-1}(C^{ij})} - \sum _{(i,j)\in \mu ^{M_2}} \frac{[D\circ G^{-1}(C^{ij})-c^{ij}]^2}{-D'\circ G^{-1}(C^{ij})}\\&\quad \ge \sum _{(i,j)\in \mu ^{M_2}_{+}} \frac{[D\circ G^{-1}(C^{ij})-c^{ij}]^2}{-D'\circ G^{-1}\left( C_{\mu ^{M_1}\cup \mu _+^{M_2}}\right) } - \sum _{(i,j)\in \mu ^{M_2}} \frac{[D\circ G^{-1}(C^{ij})-c^{ij}]^2}{-D'\circ G^{-1}\left( C_{\mu ^{M_1}\cup \mu _+^{M_2}}\right) }\\&\quad >0. \end{aligned}$$

The first inequality follows from $C_{\mu ^{M_1}\cup \mu ^{M_2}_+}- C_{\mu ^{M_1}\cup \mu ^{M_2}}\le 0$, because $\frac{\partial ^2 c}{\partial f \partial s}\le 0$.¹⁴ The second inequality follows from Lemma B.7. Hence, the rational beliefs contain PAMs among available agents and $\mu _+$ is stable under rational beliefs.

Case (iii) Take a matching $\mu ^{M_2}\ne \mu ^{M_2}_-$. Without loss of generality, let us focus on Scenario 2 as above.

$$\begin{aligned}&V(\mu ^{M_2}_-|\varphi _{\mu ^{M_1}}^{M_2})-V(\mu ^{M_2}|\varphi _{\mu ^{M_1}}^{M_2})\\&\quad =\sum _{(i,j)\in \mu ^{M_2}_{-}}\pi ^{ij}_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _-^{M_2\backslash \{i,j\}}}- \sum _{(i,j)\in \mu ^{M_2}}\pi ^{ij}_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _-^{M_2\backslash \{i,j\}}}\\&\quad = \sum _{(i,j)\in \mu ^{M_2}_{-}} \frac{[D\circ G^{-1}(C^{ij})-c^{ij}]^2}{-D'\circ G^{-1}(C^{ij})} - \sum _{(i,j)\in \mu ^{M_2}} \frac{[D\circ G^{-1}(C^{ij})-c^{ij}]^2}{-D'\circ G^{-1}(C^{ij})}\\&\quad > \sum _{(i,j)\in \mu ^{M_2}_{-}} \frac{[D\circ G^{-1}(C^{ij})-c^{ij}]^2}{-D'\circ G^{-1}(C^{ij})} - \sum _{(i,j)\in \mu ^{M_2}} \frac{[D\circ G^{-1}(C^{ij})-c^{ij}]^2}{-D'\circ G^{-1}(C^{ij})}\\&\quad \ge 0. \end{aligned}$$

The first inequality follows from $C_{\mu ^{M_1}\cup \mu ^{M_2}_-}- C_{\mu ^{M_1}\cup \mu ^{M_2}}< 0$, because $\frac{\partial ^2 c}{\partial f \partial s}> 0$. The second inequality follows from Lemma B.7. Hence, the rational beliefs contain NAMs among available agents $\mu _-$ is stable under rational beliefs.

Case (ii) Again, suppose $\mu ^{M_2}\ne \mu ^{M_2}_+$, and without loss of generality, let us focus on Scenario 1 as above. First,

$$\begin{aligned} V\left( \mu ^{M_2}_+|\varphi _{\mu ^{M_1}}^{M_2}\right)&=\sum _{(i,j)\in \mu ^{M_2}_+}\pi ^{ij}_{\mu ^{M_1}\cup \mu _+^{M_2}}\\&=\sum _{(i,j)\in \mu ^{M_2}_+}-\frac{\left[ D\circ G^{-1}(C_{\mu ^{M_1}\cup \mu _+^{M_2}})-c^{ij}\right] ^2}{D'\circ G^{-1}(C_{\mu ^{M_1}\cup \mu _+^{M_2}})};\\ V\left( \mu ^{M_2}|\varphi _{\mu ^{M_1}}^{M_2}\right)&=\sum _{(i,j)\in \mu ^{M_2}}\pi ^{ij}_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _+^{M_2\backslash \{i,j\}}}\\&=\sum _{(i,j)\in \mu ^{M_2}}-\frac{\left[ D\circ G^{-1}(C_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _+^{M_2\backslash \{i,j\}}})-c^{ij}\right] ^2}{D'\circ G^{-1}(C_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _+^{M_2\backslash \{i,j\}}})}\\&<\sum _{(i,j)\in \mu ^{M_2}}-\frac{D\circ G^{-1}(C_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _+^{M_2\backslash \{i,j\}}})-c^{ij}}{D'\circ G^{-1}(C_{\mu ^{M_1}\cup \{(i,j)\}\cup \mu _+^{M_2\backslash \{i,j\}}})}\\&\quad \cdot \left[ D\circ G^{-1}(C_{\mu ^{M_1}\cup \mu _+^{M_2}})-c^{ij}\right] \\&<\sum _{(i,j)\in \mu ^{M_2}}-\frac{\left[ D\circ G^{-1}(C_{\mu ^{M_1}\cup \mu _+^{M_2}})-c^{ij}\right] ^2}{D'\circ G^{-1}(C_{\mu ^{M_1}\cup \mu _+^{M_2}})}. \end{aligned}$$

The first inequality follows from $C_{\mu ^{M_1}\cup \mu _+^{M_2}}>C_{\mu ^{M_1}\cup \mu ^{M_2}}$ for all $\mu ^{M_2}\ne \mu ^{M_2}_+$, because $\frac{\partial ^2 c}{\partial f \partial s}>0$. The second inequality follows from Corollary A.1. The condition of the corollary is satisfied because for all $1\le p,q\le t$,

$$\begin{aligned} (n-1)+c^{i_pj_q}\ge (n-1)+\min \left\{ c^{i_pj_p},c^{i_qj_q}\right\} \ge C_{\mu ^{M_1}}+\sum _{k=1}^tc^{i_kj_k}=C_{\mu ^{M_1}\cup \mu _+^{M_2}}. \end{aligned}$$

Second, the prescribed condition of case (ii) implies

$$\begin{aligned}&\frac{\partial ^2}{\partial f \partial s}\left[ D\circ G^{-1}(C_{\mu ^{M_1}\cup \mu _+^{M_2}})-c(f,s)\right] ^2>0. \end{aligned}$$

As a result in standard assignment games with no externality, this supermodularity implies

$$\begin{aligned} \sum _{(i,j)\in \mu ^{M_2}_+}\left[ D\circ G^{-1}(C_{\mu ^{M_1}\cup \mu _+^{M_2}})-c^{ij}\right] ^2>\sum _{(i,j)\in \mu ^{M_2}}\left[ D\circ G^{-1}(C_{\mu ^{M_1}\cup \mu _+^{M_2}})-c^{ij}\right] ^2 . \end{aligned}$$

Hence, $V(\mu ^{M_2}_+|\varphi _{\mu ^{M_1}}^{M_2})>V(\mu ^{M_2}|\varphi _{\mu ^{M_1}}^{M_2})$. Thus, again, the rational beliefs contain PAMs among available agents and $\mu _+$ is stable under rational beliefs. $\square$

Social welfare

Proof of Proposition 2.2

For a given matching $\mu$, the consumer surplus is

$$\begin{aligned} CS(\mu ):=\int _{0}^{Q_{\mu }}D(q)dq -D(Q_{\mu })Q_{\mu } \end{aligned}$$

and the producer surplus is

$$\begin{aligned} PS(\mu ):=D(Q_{\mu })Q_{\mu }-\sum _{(i,j)\in \mu }c^{ij}q^{ij}_{\mu }. \end{aligned}$$

Then, the total surplus is then $TS(\mu ):=CS(\mu )+PS(\mu )$.

Suppose $(i_1,j_1),(i_2,j_2)\in \mu$. Take another matching $\mu '$ such that $(i_1,j_2),(i_2,j_1) \in \mu '$ and $\mu '\backslash \{(i_1,j_2),(i_2,j_1) \}$=$\mu \backslash \{(i_1,j_1),(i_2,j_2) \}$. With out loss of generality, assume $i_1>i_2$ and $j_1>j_2$. Here, I only need to prove that $TS(\mu )>TS(\mu ')$ for cases 1 and 2 and that $TS(\mu )<TS(\mu ')$ for case 3.

Case 1 Let $M_1$, $M_2$, and $\mu ^{M_1}$ be $M\backslash \{i_1,i_2,j_1,j_2\}$, $\{i_1,i_2,j_1,j_2\}$, and $\mu \cap \mu '$, respectively. On the one hand, following from the proof of case (i) of Theorem 2.1, we have

$$\begin{aligned} PS(\mu )=&\sum _{(i,j)\in \mu ^{M_1}}\pi ^{ij}_{\mu ^{M_1}\cup \mu _+^{M_2}} +\sum _{(i,j)\in \mu ^{M_2}_{+}}\pi ^{ij}_{\mu ^{M_1}\cup \mu _+^{M_2}}\\ >&\sum _{(i,j)\in \mu ^{M_1}}\pi ^{ij}_{\mu ^{M_1}\cup \mu _-^{M_2}} +\sum _{(i,j)\in \mu ^{M_2}_-}\pi ^{ij}_{\mu ^{M_1}\cup \mu _-^{M_2}}=PS(\mu '). \end{aligned}$$

On the other hand, because $C_{\mu }<C_{\mu '}$ and, thus, $Q_{\mu }>Q_{\mu '}$, we have

$$\begin{aligned} CS(\mu )=&\int _{0}^{Q_{\mu }}D(q)dq -D(Q_{\mu })Q_{\mu }\\ =&\int _{0}^{Q_{\mu }}D(q)-D(Q_{\mu })dq \\>&\int _{0}^{Q_{\mu }}D(q)-D(Q_{\mu '})dq \\ >&\int _{0}^{Q_{\mu '}}D(q)-D(Q_{\mu '})dq =CS(\mu '). \end{aligned}$$

Hence, in this case $TS(\mu )>TS(\mu ')$.

Case 2 Here, $C_{\mu }>C_{\mu '}$ and thus $Q_{\mu }<Q_{\mu '}$. The difference between the total surpluses across the two matchings is

$$\begin{aligned}&TS(\mu )-TS(\mu ')\\&\quad =\int _{Q_{\mu '}}^{Q_{\mu }}D(q)dq -\left[ \sum _{(i,j)\in \mu }c^{ij}q^{ij} -\sum _{(i,j)\in \mu '}c^{ij}q^{ij} \right] \\&\quad>D(Q_{\mu '})[Q_{\mu }-Q_{\mu '}] -\left[ \sum _{(i,j)\in \mu }c^{ij}q^{ij} -\sum _{(i,j)\in \mu '}c^{ij}q^{ij} \right] \\&\quad =\sum _{(i,j)\in \mu }[D(Q_{\mu '})-c^{ij}]\cdot \frac{D\circ G^{-1}(C_{\mu })-c^{ij}}{-D'\circ G^{-1}(C_{\mu })}\\&\qquad -\sum _{(i,j)\in \mu '}[D(Q_{\mu '})-c^{ij}]\cdot \frac{D\circ G^{-1}(C_{\mu '})-c^{ij}}{-D'\circ G^{-1}(C_{\mu '})}\\&\quad >\sum _{(i,j)\in \mu }[D(Q_{\mu '})-c^{ij}]\cdot \frac{D\circ G^{-1}(C_{\mu '})-c^{ij}}{-D'\circ G^{-1}(C_{\mu '})}\\&\qquad -\sum _{(i,j)\in \mu '}\left[ D(Q_{\mu '})-c^{ij}\right] \cdot \frac{D\circ G^{-1}(C_{\mu '})-c^{ij}}{-D'\circ G^{-1}(C_{\mu '})}\\&\quad =\frac{\left[ D\circ G^{-1}(C_{\mu '})-c^{i_1j_1}\right] ^2+\left[ D\circ G^{-1}(C_{\mu '})-c^{i_2j_2}\right] ^2 }{-D'\circ G^{-1}(C_{\mu '})}\\&\qquad -\frac{\left[ D\circ G^{-1}(C_{\mu '})-c^{i_1j_2}\right] ^2+\left[ D\circ G^{-1}(C_{\mu '})-c^{i_2j_1}\right] ^2 }{-D'\circ G^{-1}(C_{\mu '})}\\&\quad =\frac{-2D'\circ G^{-1}(C_{\mu '})\left[ c^{ij}+c^{i'j'}-c^{i'j}-c^{ij'}\right] +\left[ (c^{ij})^2+(c^{i'j'})^2-(c^{i'j})^2-(c^{ij'})^2\right] }{-D'\circ G^{-1}(C_{\mu '})}\\&\quad =\frac{1}{-D'\circ G^{-1}(C_{\mu '})}\int _{f_{i'}}^{f_i}\int _{s_{j'}}^{s_j}-2D'\circ G^{-1}(C_{\mu '})\frac{\partial ^2 c(f,s)}{\partial f \partial s}+\frac{\partial ^2[c(f,s)]^2}{\partial f\partial s}dsdf\\&\quad \ge 0, \end{aligned}$$

where the second inequality follows from Corollary A.1. Hence, for any matching $\mu '\ne \mu _+$, there is a matching $\mu$ such that $TS(\mu )>TS(\mu ')$. Therefore, $\mu _+$ is efficient.

Case 3 This proof is the same as that of case 1 except that the inequalities hold in the opposite directions. In particular, $PS(\mu )<PS(\mu ')$ follows from the rationality requirement. Thus, $TS(\mu )<TS(\mu ')$. Therefore, the negative assortative matching is efficient. $\square$