Matching markets and cultural selection

Abstract

We study the implications of two different matching mechanisms: bargaining in match (BIM) and binding agreement in the matching market (BAMM) in the context of cultural selection. Under BIM, matching across cultures is not operated through a well-regulated market, where agents’ payoffs are not determined by the market but through negotiation with their matched partners. Under BAMM, matching across cultures is operated through a formal market in which agents make binding agreements. We show that BIM results in inefficiency through cultural selection and possibly leads to assimilation, while BAMM restores efficiency and ensures cultural heterogeneity. The findings highlight the importance of regulating the matching market for multi-cultural societies.

Appendix

Proof of Lemma 1

The Nash bargaining problem is equivalent to

$$\begin{aligned} \max _{h(i, j)}(h(i, j)-g(i))(f(i, j)-h(i, j)-g(j)), \text { for } i, j \in \{a, b\}. \end{aligned}$$

(3)

The first order condition implies that $h(i, j)=\frac{f(i,i)+g(i)-g(j)}{2}$. $h(j, i)=f(i, j)-h(i, j)=\frac{f(i, j)-g(j)+g(i)}{2}$. $\square$

Proof of Proposition 1

(i) When we have $h(a, a)<h(a ,b)$ and $h(b, b)<h(b, a)$, any agent in the population strictly prefers being matched with a different type agent. If there exists a matching outcome in which the proportion of cross-type matches is smaller than that in the one described in the proposition, then there must exist a positive proportion of both $(a-a)$ matches and $(b-b)$ matches and these agents have an incentive to terminate their current matches and form cross-type matches instead. Therefore, the matching outcome described in the proposition must be an equilibrium because it induces the maximum proportion of cross-type matches.

(ii) When we either have $h(a, a)>h(a ,b)$ or $h(b, b)>h(b, a)$. If there exists a positive proportion of cross-type matches, then those agents who prefer to be matched with their own type members have an incentive to terminate their current matches and form same-type matches instead. Therefore, the matching outcome described in the proposition must be an equilibrium because there exists no cross-type match. $\square$

Proof of Proposition 2

When $h(a, a)<h(a, b)$ and $h(b, b)<h(b, a)$, the matching equilibrium induces the maximum proportion of cross-type matches as indicated in Proposition 1 (i). The average payoffs are given as follows. (A) If $x_t \ge \frac{1}{2}$, then

$$\begin{aligned}&F_a(x_t)=\frac{2x_t-1}{x_t}\times h(a, a)+\frac{1-x_t}{x_t}\times h(a, b); \end{aligned}$$

(4)

$$\begin{aligned}&F_b(x_t)=h(b, a). \end{aligned}$$

(5)

(B) If $x_t <\frac{1}{2}$, then

$$\begin{aligned}&F_a(x_t)=h(a, b); \end{aligned}$$

(6)

$$\begin{aligned}&F_b(x_t)=\frac{1-2x_t}{1-x_t}\times h(b, b)+\frac{x_t}{1-x_t}\times h(b, a). \end{aligned}$$

(7)

(i) If in addition $h(a, a)>h(b, a)$, then $F_a(x_t)>F_b(x_t)$ for any $x_t \in [0, 1]$. Hence, $x=1$ is stable.

(ii) If $h(a, a)<h(b, a)$, then we have a uniquely interior state

$$\begin{aligned} x= {\left\{ \begin{array}{ll} \frac{h(a, b)-h(b, b)}{f(a, b)-f(b, b)}, &{} \text {if } g(a) < g(b),\\ \frac{h(a, b)-h(a, a)}{f(a, b)-f(a, a)}, &{} \text {otherwise,} \end{array}\right. } \end{aligned}$$

Moreover, $F_a(x_t)>F_b(x_t)$ if $x_t<x$, and $F_b(x_t)>F_a(x_t)$ if $x_t>x$. Hence, x is stable.

(iii) If $h(a, a)>h(a, b)$ or $h(b, b)>h(b, a)$, then the matching equilibrium induces no cross-type match as indicated in Proposition 1 (ii). The average payoffs are $F_a(x_t)=h(a, a)=\frac{f(a, a)}{2}$ and $F_b(x_t)=h(b, b)=\frac{f(b, b)}{2}$, respectively. Since $F_a(x_t)>F_b(x_t)$ for any $x_t \in [0, 1]$, $x=1$ is stable. $\square$

Proof of Proposition 3

If there exist both a pair of $(a-a)$ match and a pair of $(b-b)$ match, there always exist a type-a agent and a type-b agent who have an incentive to form a new pair. Therefore, in any matching equilibrium, we must have the maximum proportion of cross-type matches.

(i) Suppose $x_t< \frac{1}{2}$. If there exists a pair of $(a-b)$, in which the type-b agent earns a wage w higher than $\frac{f(b, b)}{2}$, then the type-a agent has an incentive to form a new pair with another type$-b$ agent who earns $\frac{f(b, b)}{2}$ by offering the type-b agent a wage lower than w but higher than $\frac{f(b, b)}{2}$. Hence, in equilibrium, it must be the case that $w(b)=\frac{f(b, b)}{2}$ and $w(a)=f(a, b)-\frac{f(b, b)}{2}$.

(ii) Suppose $x_t< \frac{1}{2}$. The same logic applies and in equilibrium, $w(a)=\frac{f(a, a)}{2}$ and $w(b)=f(a, b)-\frac{f(a, a)}{2}$.

(iii) Suppose $x_t=\frac{1}{2}$. Each type-a agent is matched with a type-b agent. Therefore, in equilibrium, we must have $w(a)+w(b)=\frac{f(a, b)}{2}$. To prevent two same-type agents to form a pair, we also have $2w(a) \ge f(a, a)$ and $2w(b) \ge f(b, b)$. $\square$

Proof of Proposition 4

$F_a(x_t)=f(a, b)-\frac{f(b, b)}{2}>\frac{f(b, b)}{2}=F_b(x_t)$ if $x_t<\frac{1}{2}$, and $F_b(x_t)=f(a, b)-\frac{f(a, a)}{2}>\frac{f(a, a)}{2}=F_a(x_t)$ if $x_t>\frac{1}{2}$. Hence, $x=\frac{1}{2}$ is stable, provided that $F_a(\frac{1}{2})=F_b(\frac{1}{2})=\frac{f(a, b)}{2}$. $\square$

Lemma 2

In BIM with frictions,

• (i) If $x_t>\frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}$, then $\alpha ^*=\frac{h(a, b)-h(a, a)}{c(a)}\frac{1-x_t}{x_t}$ and $\beta ^*=1$.

• (ii) If $x_t<\frac{c(b)}{c(b)+h(b, a)-h(b, b)}$, then $\alpha ^*=1$ and $\beta ^*=\frac{h(b, a)-h(b, b)}{c(b)}\frac{x_t}{1-x_t}$.

• (iii) If $\frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)} \ge x_t \ge \frac{c(b)}{c(b)+h(b, a)-h(b, b)}$, then $\alpha ^*=\beta ^*=1$.

Proof of Lemma 2

First, we establish the fact that it is impossible to have an equilibrium in which $\alpha <1$ and $\beta <1$. In an equilibrium, we must have either $\alpha x_t\le \beta (1-x_t)$ or $\alpha x_t\ge \beta (1-x_t)$ or both. Suppose $\alpha x_t\le \beta (1-x_t)$, then all type-a agents in the mixed pool earn an expected payoff of $h(a, b)-c(a)$, which is larger than their payoff of h(a, a) in the homogeneous pool. Therefore, as long as $\alpha <1$, those type-a agents in the homogeneous pool have an incentive to enter the mixed pool, implying $\alpha <1$ cannot be in an equilibrium with $\alpha x_t\le \beta (1-x_t)$. Similarly, $\beta <1$ cannot be in an equilibrium with $\alpha x_t\ge \beta (1-x_t)$.

Now we proceed to discuss the cases in which at least one of $\alpha$ and $\beta$ equals 1. First, suppose $\alpha <1$, $\beta =1$. This can only happen if $\alpha x_t>1-x_t$. It also implies that an type-a agent is indifferent between entering the mixed pool or staying in the homogeneous pool, i.e,

$$\begin{aligned} \frac{\alpha x_t -(1-x_t)}{\alpha x_t} h(a, a)+\frac{1-x_t}{\alpha x_t} h(a, b)-c(a)=h(a,a). \end{aligned}$$

(8)

We get $\alpha =\frac{h(a, b)-h(a, a)}{c(a)}\frac{1-x_t}{x_t}$. To ensure $\alpha <1$, we need $x_t>\frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}$.

Second, consider the case $\alpha =1$, $\beta <1$. This can only happen if $x<\beta (1-x_t)$. It also implies that an type-b agent is indifferent between entering the mixed pool or staying in the homogeneous pool, i.e,

$$\begin{aligned} \frac{\beta (1- x_t)-x_t}{\beta (1-x_t)}h(b,b)+\frac{x_t}{\beta (1-x_t)} h(b,a)-c(b)=h(b, b). \end{aligned}$$

(9)

We get $\beta =\frac{h(b, a)-h(b, b)}{c(b)}\frac{x_t}{1-x_t}$. To ensure $\beta <1$, we need $x_t<\frac{c(b)}{c(b)+h(b, a)-h(b, b)}$.

Third, consider the case $\alpha =1$, $\beta =1$. First, suppose $x_t\le 1-x_t$. In this scenario, type-a agents do strictly better in the mixed pool than in the homogeneous pool and we need type-b agents to get an expected payoff in the mixed pool at least as high as that in the homogeneous pool. That is,

$$\begin{aligned} \frac{1-2x_t}{1-x_t}h(b,b)+\frac{x_t}{1-x_t} h(b,a)-c(b)\ge h(b, b). \end{aligned}$$

(10)

This requires $x_t\ge \frac{c(b)}{c(b)+h(b, a)-h(b, b)}$.

Second, suppose $x_t \ge 1-x_t$. In this scenario, type-b agents do strictly better in the mixed pool than in the homogeneous pool and we need type-a agents to get an expected payoff in the mixed pool at least as high as that in the homogeneous pool. That is,

$$\begin{aligned} \frac{2x_t-1}{x_t}h(a, a)+\frac{1-x_t}{x_t} h(a, b)-c(a)\ge h(a, a). \end{aligned}$$

(11)

This requires $x_t\le \frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}$.

In sum, when $x_t>\frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}$, $(\alpha ^*=\frac{h(a, b)-h(a, a)}{c(a)}\frac{1-x_t}{x_t},\beta ^*=1)$ is the unique equilibrium; when $x_t<\frac{c(b)}{c(b)+h(b, a)-h(b, b)}$, $(\alpha ^*=1, \beta ^*=\frac{h(b, a)-h(b, b)}{c(b)}\frac{x_t}{1-x_t})$ is the unique equilibrium; when $x_t \in [\frac{c(b)}{c(b)+h(b, a)-h(b, b)}, \frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}]$, $(\alpha ^*=\beta ^*=1)$ is the unique equilibrium. $\square$

Proof of Proposition 5

(1) When $x_t>\frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}$, by Lemma 2, we know a fraction of the type-a agents and all the type-b agents enter the mixed pool. Therefore, a type-a agent is indifferent between entering the mixed pool and staying in the homogeneous pool, implying that $F_a(x_t)=h(a, a)$ and $F_b(x_t)=h(b, a)-c(b)$. Hence, if $c(b)>h(b, a)-h(a, a)$, we have $F_a(x_t)>F_b(x_t)$, implying $x=1$ is the stable state in this region. If $c(b)\le h(b, a)-h(a, a)$, we have $F_a(x_t)\le F_b(x, t)$, implying no stable state in this region.

(2) When $x_t<\frac{c(b)}{c(b)+h(b, a)-h(b, b)}$, by Lemma 2, we know all the type-a agents and a fraction of the type-b agents enter the mixed pool. Therefore, a type-b agent is indifferent between entering the mixed pool and staying in the homogeneous pool, implying that $F_b(x_t)=h(b, b)$ and $F_a(x_t)=h(a, b)-c(a)$. Assumption 2 guarantees that $F_a(x_t)>F_b(x_t)$. Hence, there is no stable state in this region.

(3) When $x_t \in [\frac{c(b)}{c(b)+h(b, a)-h(b, b)}, \frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}]$, by Lemma 2, we know all agents enter the mixed pool. If $x_t\le \frac{1}{2}$. $F_a(x_t)=F_b(x_t)$ is equivalent to

$$\begin{aligned} h(a, b)-c(a)=\frac{1-2x_t}{1-x_t}h(b, b)+\frac{x_t}{1-x_t}h(b, a)-c(b), \end{aligned}$$

(12)

implying that $x_t={\hat{x}}=\frac{h(a, b)-h(b, b)-c(a)+c(b)}{f(a, b)-f(b, b)-c(a)+c(b)}$ is a stationary state. It is less than or equal to $\frac{1}{2}$ if $g(a)-c(a) \le g(b)-c(b)$. One can verify that when $x_t<{\hat{x}}$, $F_a(x_t)>F_b(x_t)$; and vice versa. Also, Assumption 2 ensures that ${\hat{x}}> \frac{c(b)}{c(b)+h(b, a)-h(b, b)}$. Hence, ${\hat{x}}$ exists and is stable.

On the contrary, if $x>\frac{1}{2}$, $F_a(x_t)=F_b(x_t)$ is equivalent to

$$\begin{aligned} \frac{2x_t-1}{x_t}h(a, a)+\frac{1-x_t}{x_t}h(a, b)-c(a)=h(b, a)-c(b), \end{aligned}$$

(13)

implying that $x_t=\tilde{x}=\frac{h(a, b)-h(a, a)}{f(a, b)-f(a, a)+c(a)-c(b)}$ is a stationary state. It is larger that $\frac{1}{2}$ if $g(a)-c(a) > g(b)-c(b)$. One can verify that when $x_t<\tilde{x}$, $F_a(x_t)>F_b(x_t)$; and vice versa. Also, as long as $c(b) < h(b, a)-h(a, a)$, $\tilde{x} < \frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}$. Hence, $\tilde{x}$ exists and is stable.

Note that when $c(b)\ge h(b, a)-h(a, a)$, we cannot have $g(a)-c(a) \le g(b)-c(b)$ at the same time because of Assumption 2. Hence, ${\hat{x}}$ cannot be a stable state in this case. Also, $\tilde{x}$ would be larger than or equal to $\frac{h(a, b)-h(a, a)}{c(a)+h(a, b)-h(a, a)}$ when $c(b) \ge h(b, a)-h(a, a)$, implying that $\tilde{x}$ is impossible to be a stable state in this case either. Therefore, when $c(b)>h(b, a)-h(a, a)$, $x=1$ is stable. When $c(b)< h(b, a)-h(a, a)$, there exists an interior stable state (either ${\hat{x}}$ or $\tilde{x}$). $\square$

Lemma 3

In BAMM with frictions,

• (i) If $x_t\le \frac{1}{2}$, then $\alpha ^*=1$ and $\beta ^*=\frac{x_t}{1-x_t}$.

• (ii) If $x_t>\frac{1}{2}$, then $\alpha ^*=\frac{1-x_t}{x_t}$ and $\beta ^*=1$.

Proof of Lemma 3

When $\alpha x_t < \beta (1-x_t)$, type-a agents are in the long side of the market in the mixed pool and they get $\frac{f(a, a)}{2}-c(a)$, which is lower that $\frac{f(a,a)}{2}$, the payoff they obtain in the homogeneous pool. Hence, they have an incentive to leave. Similarly, when $\alpha x_t > \beta (1-x_t)$, type-b agents in the mixed pool have an incentive to leave. Therefore, we should have $\alpha x_t=\beta (1-x_t)$ in equilibrium. Also, we cannot have $\alpha , \beta \in (0, 1)$ in equilibrium. Otherwise, given Assumption 3, those agents in the homogeneous pools have an incentive to enter the mixed pool. Therefore, we must have at least one of $\alpha$ and $\beta$ equals 1. When $x_t\le 1-x_t$, the only possibility is that $\alpha ^*=1$ and $\alpha ^* x_t=\beta ^* (1-x_t)$ implies that $\beta ^*=\frac{x_t}{1-x_t}$. Similarly, one can conclude that when $x_t>1-x_t$, $\alpha ^*=\frac{1-x_t}{x_t}$ and $\beta ^*=1$. $\square$

Proof of Proposition 6

According to Lemma 3, When $x_t\le \frac{1}{2}$, $\alpha ^*=1$ and $\beta ^*=\frac{x_t}{1-x_t}$, implying that $F_a(x_t)=\frac{f(a, b)}{2}-c(a)$ and $F_b(x_t)=\frac{1-2x}{1-x}\frac{f(b, b)}{2}+\frac{x}{1-x}(\frac{f(a, b)}{2}-c(b))$. If there exists a stationary state $x^*$ in this region, it satisfies $F_a(x^*)=F_b(x^*)$, which indicates $x^*=\frac{\frac{f(a, b)-f(a, a)}{2}-c(a)}{f(a, b)-f(a, a)-c(a)-c(b)}$. $x^*\le \frac{1}{2}$ if $c(a) \ge c(b)$. One can verify that when $x_t<x^*$, $F_a(x_t)>F_b(x_t)$; and vice versa. Hence, $x^*$ is stable. Similarly, when $x_t> \frac{1}{2}$, Lemma 3 states that $\alpha ^*=\frac{1-x_t}{x_t}$ and $\beta ^*=1$, implying that $F_a(x_t)=\frac{2x-1}{x}\frac{f(a, a)}{2}+\frac{1-x}{x}(\frac{f(a, b)}{2}-c(a))$ and $F_b(x_t)=\frac{f(a, b)}{2}-c(b)$. If there exists a stationary state $x^{**}$ in this region, it satisfies $F_a(x^{**})=F_b(x^{**})$, which indicates $x^{**}=\frac{\frac{f(a, b)-f(b, b)}{2}-c(a)}{f(a, b)-f(b, b)-c(a)-c(b)}$. $x^{**}> \frac{1}{2}$ if $c(a) <c(b)$. One can verify that when $x_t<x^{**}$, $F_a(x_t)>F_b(x_t)$; and vice versa. Hence, $x^{**}$ is stable. To summarize, $x^*$ is stable if $c(a) \ge c(b)$ and $x^{**}$ is stable if $c(a)<c(b)$. $\square$

Proof of Proposition 7

First, consider the case $c(a)-c(b)<\frac{g(a)-g(b)}{2}$. We need to analyze several different scenarios. Suppose $c(b) \ge h(b, a)-h(a, a)$.¹⁶ From Proposition 5 (i) we know that $x=1$ is stable under BIM when $c(b) > h(b, a)-h(a, a)$ and $Avg^{\text {BIM}}=h(a, a)=\frac{f(a, a)}{2}$.¹⁷ Under BAMM, if $c(a)<c(b)$, the stable state satisfies $x>\frac{1}{2}$, which implies that $Avg^{\text {BAMM}}=\frac{f(a, b)}{2}-c(b)=\frac{2x-1}{x}\frac{f(a, a)}{2}+\frac{1-x}{x}(\frac{f(a, b)}{2}-c(a))>\frac{f(a, a)}{2}=Avg^{\text {BIM}}$ (because $\frac{f(a, b)}{2}-c(a)>\frac{f(a, b)}{2}-c(b)$ and $\frac{f(a, b)}{2}-c(b)$ is a linear combination of $\frac{f(a, a)}{2}$ and $\frac{f(a, b)}{2}-c(a)$); if $c(a) \ge c(b)$, the stable state satisfies $x\le \frac{1}{2}$, which implies that $Avg^{\text {BAMM}}=\frac{f(a, b)}{2}-c(a)>\frac{f(a, a)}{2}=Avg^{\text {BIM}}$ by Assumption 3.

If instead $c(b) < h(b, a)-h(a, a)$, then according to Proposition 5 (ii), there must exist an interior stable state under BIM. When $c(a) \le c(b)$, we have $Avg^{\text {BAMM}}=\frac{f(a, b)}{2}-c(b)>Avg^{\text {BIM}}=h(b, a)-c(b)$. When $\frac{g(a)-g(b)}{2}+c(b)>c(a) >c(b)$, we have $Avg^{\text {BAMM}}=\frac{f(a, b)}{2}-c(a)>Avg^{\text {BIM}}=h(b, a)-c(b)$.

Second, consider the case $c(a)-c(b) \ge \frac{g(a)-g(b)}{2}$. In this case, we can rule out the possibility that $c(b) \ge h(b, a)-h(a, a)$ because it violates Assumption 3. Given $c(b) < h(b, a)-h(a, a)$, we know that there must exist an interior stable sate under BIM. When $g(a)-g(b)>c(a)-c(b) \ge \frac{g(a)-g(b)}{2}$, we have $Avg^{\text {BAMM}}=\frac{f(a, b)}{2}-c(a) \le Avg^{\text {BIM}}=h(b, a)-c(b)$. When $c(a)-c(b) \ge g(a)-g(b)$, we have $Avg^{\text {BAMM}}=\frac{f(a, b)}{2}-c(a) < Avg^{\text {BIM}}=h(a, b)-c(a)$. $\square$