Coordination and culture

Abstract

Culture constrains individual choice, rendering certain actions impermissible or taboo. While cultural constraints may regulate behavior within a group, they can have a pernicious effect in multicultural societies, inhibiting the emergence of unified social conventions. We analyze interactions between members of two cultural groups who are matched to play a coordination game with an arbitrary number of actions. Due to cultural constraints, miscoordination prevails despite strong incentives to coordinate behavior. In an application to identity-based conflict, exclusive ethnic and religious identities persist in poorer and more unequal societies. Occasional violation of cultural constraints can make miscoordination even more stable.

Appendix

Proof of Proposition 1.

Proof

We shall characterize the set of recurrent classes of the best response dynamic. As any finite Markov chain converges almost surely to one of its recurrent classes, this establishes the proposition.

Firstly, we claim that there are no recurrent classes of the best response dynamic other than the states of coordination and $\mathcal {M}$. To establish the claim it suffices to show that there is a positive probability of transiting from any state to a state of coordination or a state in $\mathcal {M}$ in a finite number of periods.

Consider an arbitrary time t and state $z^t$. Let x be a best response for group A members. Suppose that in period t, the set of revising players is $R_t=N_A$ and that each of them chooses x. In addition, suppose that in period $t+1, R_{t+1}= N_{B}$. All of this occurs with positive probability.

Case 1 $x\in X_{A}\cap X_{B}$. (By definition, this is impossible if the cultures are incompatible.) Then the expected payoff to each revising group B member from playing x is $\theta _{Bx} + \delta _{Bx}$ for all samples. The payoff from choosing $x'\ne x$ is at most $d_{B}$, which is less than the payoff from x by $\textit{ND}$. Hence each revising group B member chooses x and the dynamic transits with probability to a state of coordination in two periods.

Case 2 $x\notin X_{A}\cap X_{B}$. Then the payoff to each revising group B member from playing any $x'\in X_{B}$ is $\delta _{Bx'}$. Hence each revising group B member chooses an action in $\widetilde{X}_{B} \equiv argmax_{x\in X_{B}}\delta _{Bx}$. When the cultures are misaligned, every such action lies outside $X_{A}\cap X_{B}$, as does x. Hence the dynamic transits with positive probability to a state in $\mathcal {M}$ in two periods. When the cultures are aligned, we have either $\tilde{X}_A\cap X_B$ is nonempty or $\tilde{X}_B\cap X_A$ is nonempty. Thus the dynamic transits with positive probability in at most two periods to a state in which each group B member plays an action $x'\in X_{A}\cap X_{B}$. From there, the argument used in case 1 establishes that the dynamic transits to a state of coordination in finite time.

This establishes the claim.

The fact that the states of coordination are recurrent classes of the best response dynamic if and only if the cultures are compatible follows immediately from the argument in case 1. That $\mathcal {M}$ is a recurrent class if and only if the cultures are misaligned follows immediately from the argument in case 2. This establishes the proposition.

$\square$

Proposition 2. The proof of Proposition 2 employs two tree-surgery arguments. These arguments make use of the following three lemmas and corollary, which compare the resistance of transitions between recurrent classes, defined as follows. The cost of a path $(z^1,z^2,\ldots z^H)$ is the minimum number of errors required for the perturbed dynamic ($\varepsilon > 0$) to transit from state $z^1$ to $z^H$ along this path. The resistance $r(z^{1},z^H)$ is the minimum cost of a path starting at $z^1$ and ending in $z^H$. The definition can be extended so that for sets E and $E'$, the resistance $r(E,E')$ is the minimum cost of a path starting in some state in E and ending in some state in $E'$.

In addition, let $x_i(z)$ be the action played by i in state z. A direct path $(z^1,z^2,\ldots z^H)$ from state $z^1$ to $z^H$ is one in which $x_i(z^h)\in \{x_i(z^1), x_i(z^H)\}$ for all $h=1,2, \ldots H$ and $i\in N$.

Define $X_i(E) \equiv \{x_i(z): z\in E\}$. A direct path $(z^1,z^2,\ldots z^H)$ from set E to $E'$ is one in which $z^1\in E, z^H\in E'$ and $x_i(z^h)\in X_i(E)\cup X_i(E')$ for all $h=1,2, \ldots H$ and $i\in N$.

Lemma 1

Let $z_x$ denote the state of coordination on x. For all distinct pairs $(x,x') \in (X_A \cap X_B)^2$,

$$\begin{aligned} r(\mathcal {M}, z_x) \le r(z_{x'}, z_{x}). \end{aligned}$$

Proof

Let $r(z_{x'}, z_{x})=a$. This means that for some $k\in \{A,B\}$, after a errors by $k'$ members, x is a best response for k members, and hence a weakly better response than a mutually impermissible action. That is:

$$\begin{aligned} d_k\le & {} \frac{a}{n}\theta _k + \delta _{kx}. \end{aligned}$$

(10)

Let $\alpha _k$ be the minimum number of errors such that (10) holds:

$$\begin{aligned} \alpha _k = \bigg \lceil \frac{d_{k}-\delta _{kx}}{\theta _{k}}n\bigg \rceil \equiv \lceil D_{kx}n\rceil . \end{aligned}$$

(11)

We know $r(z_{x'}, z_{x})=a\ge \min _{k\in \{A,B\}}\alpha _k$. By inspection of (10), $\min _{k\in \{A,B\}}\alpha _k$ is the minimum cost of a direct path from any state in $\mathcal {M}$ to $z_x$. Hence $r(\mathcal {M}, z_x)\le \min _{k\in \{A,B\}}\alpha _k$. This establishes the Lemma. $\square$

Lemma 2

Suppose that in state $z, n_a$ members of group A and $n_b$ members of group B use strategy x. Define $z'$ as the state in which the same is true, with the remaining $n-n_a$ members of group A and $n-n_b$ members of group B using mutually impermissible actions.

Then $r(z',z_x)\le r(z,z_x)$.

Proof

The argument in the proof of Lemma 1 establishes the result. $\square$

Lemma 3

If miscoordination $\mathcal {M}$ is strictly risk dominant, then for n sufficiently large

$$\begin{aligned} r(z_{x}, \mathcal {M})<r(\mathcal {M}, z_x) \end{aligned}$$

for all $x \in X_A \cap X_B$.

Proof

We claim that a direct path from $\mathcal {M}$ to $z_x$ has minimum cost among all paths from $\mathcal {M}$ to $z_x$. Consider an arbitrary indirect path from $\mathcal {M}$ to $z_x$. Since the path is indirect, it runs through a state z in which some mutually permissible action $x'\ne x$ is played. The minimum cost of such a path is $r(\mathcal {M},z) + r(z,z_x)$.

In state z, replace every action by a k member not equal to x with an action in $X_{k}\backslash X_{k'}, k'\ne k$. Label this state $z'$. There is a direct path from $\mathcal {M}$ to $z_x$ through state $z'$. The minimum cost of such a path is $r(\mathcal {M},z') + r(z',z_x)$. Clearly, $r(\mathcal {M},z')<r(\mathcal {M},z)$. In addition, $r(z', z_x) \le r(z,z_x)$ by Lemma 2. Hence the direct path has lower cost than the indirect path, establishing the claim.

By (11), the minimum cost of a direct path is $\min _{k\in \{A,B\}}\lceil D_{kx} n\rceil$. Hence we need only find a path from $z_{x}$ to $\mathcal {M}$ with lower cost. Consider a direct path. Starting from $z_x, k'$ members make a errors. For $x''\in \widetilde{X}_{k}\backslash X_{k'}$ to be a best response for k members and to thereby reach a state in $\mathcal {M}$ without further errors, a must satisfy

$$\begin{aligned} \bigg (1 - \frac{a}{n}\bigg )\theta _{k} + \delta _{kx}\le d_{k}. \end{aligned}$$

That is,

$$\begin{aligned} a\ge \bigg (1 - \frac{d_{k}-\delta _{kx}}{\theta _{k}}\bigg )n \equiv (1-D_{kx})n. \end{aligned}$$

Thus the minimum cost of such a transition is

$$\begin{aligned} \min _{k\in \{A,B\}}\big \lceil (1-D_{kx})n \big \rceil . \end{aligned}$$

For n sufficiently large, this is less than $\min _{k\in \{A,B\}}\lceil D_{kx} n\rceil$ if and only if

$$\begin{aligned} \min _{k\in \{A,B\}}D_{kx}>&\min _{k\in \{A,B\}}\big (1 - D_{kx}\big )\nonumber \\ \min _{k\in \{A,B\}}D_{kx} + \max _{k\in \{A,B\}}D_{kx}>&1 \nonumber \\ D_{Ax} + D_{Bx}> & {} 1. \end{aligned}$$

(12)

The last inequality holds, because $\mathcal {M}$ strictly risk dominates $z_x$ [see (2)]. $\square$

The next result follows immediately from Lemmas 1 and 3:

Corollary 1

If miscoordination $\mathcal {M}$ is strictly risk dominant, then for n sufficiently large

$$\begin{aligned} r(z_x,\mathcal {M}) < r(z_{x'}, z_{x}) \end{aligned}$$

for all distinct pairs $(x,x') \in (X_A \cap X_B)^2$.

Proof of Proposition 2 (i) Suppose that $\mathcal {M}$ is strictly risk dominant. We shall show that for n sufficiently large, $\mathcal {M}$ is the unique stochastically stable class using the spanning-tree method of Young (1993a).

The following tree-surgery argument is used. An x-tree is a tree consisting of a set of $|X_A \cap X_B|$ directed edges, one emanating from each recurrent class (node) other than $z_x$. Consider a tree, say rooted at 0 and denoted by $T_0$, that has minimum resistance among x-trees. Through a series of operations on $T_0$ we will produce an $\mathcal {M}$-tree with lower resistance than $T_0$. Hence all trees with minimum resistance are $\mathcal {M}$-trees. By (Young 1993a, Theorem 2) then, $\mathcal {M}$ is the unique stochastically stable class.

Again let $T_0$ have minimum resistance among x-trees. By definition, there is a unique route from $\mathcal {M}$ to $z_{0}$ through $T_0$ (see Fig. 1a for example). Denote the nodes on this route by $z_{H}, z_{H-1}, \ldots z_1, z_{0}$. The edges composing this route are $(\mathcal {M}, z_{H}), (z_{H}, z_{H-1}), \ldots (z_{1}, z_{0})$.

Fig. 1

Beginning with the 0-tree in (a), steps 1–2 produce the $\mathcal {M}$-tree in (b)

• Step 1. Replace edge $(\mathcal {M}, z_{H})$ with $(z_{H},\mathcal {M})$.

• Step 2. Replace each edge $(z_{h}, z_{h-1})$ with $(z_{h-1}, \mathcal {M})$ for $h=1,\ldots H$.

Steps 1–2 produce an $\mathcal {M}$-tree (see Fig. 1b for example). The difference in resistance between the original 0-tree and this $\mathcal {M}$-tree is:

$$\begin{aligned} \big [r(\mathcal {M}, z_{H}) - r(z_{H}, \mathcal {M})\big ] + \sum _{h=1}^{H}\big [r(z_{h}, z_{h-1}) - r(z_{h-1}, \mathcal {M})\big ].\end{aligned}$$

(13)

For n sufficiently large, the first term is positive by Lemma 3 and the second term, if it exists, is positive by Corollary 1. Hence the $\mathcal {M}$-tree produced by steps 1–2 has lower resistance than the original x-tree, $T_0$. Since the original x-tree was chosen arbitrarily, all trees with minimum resistance are $\mathcal {M}$-trees, establishing the result.

(ii) Again a tree-surgery argument is employed. Consider an arbitrary $\mathcal {M}$-tree. There exists at least one edge $(z_{x}, \mathcal {M})$, for some $x\in X_A \cap X_B$. Replace this edge with $(\mathcal {M}, z_{x})$. This transforms the original $\mathcal {M}$-tree into an x-tree. See Fig. 2 for an illustration.

Fig. 2

Beginning with the $\mathcal {M}$-tree in (a), replace edge $(z_{x}, \mathcal {M})$ with $(\mathcal {M}, z_{x})$ to produce the x-tree in (b)

The difference in resistance between the original $\mathcal {M}$-tree and this x-tree is:

$$\begin{aligned} r(z_{x}, \mathcal {M}) - r(\mathcal {M}, z_{x}). \end{aligned}$$

We claim that this is nonnegative when payoffs are simple and $\mathcal {M}$ is not strictly risk dominant. Hence for every $\mathcal {M}$-tree there exists an x-tree with no greater resistance. By (Young 1993a, Theorem 2) then, $\mathcal {M}$ is not the unique stochastically stable class.

Let us now establish the claim. In the proof of Lemma 3, we showed that $r(\mathcal {M}, z_{x})=\min _{k\in \{A,B\}} \lceil D_{kx}n\rceil$ and the cost of a direct path from $z_{x}$ to $\mathcal {M}$ equals $\min _{k\in \{A,B\}}\lceil (1-D_{kx})n \rceil$.

An indirect path from $z_{x}$ to $\mathcal {M}$ involves at least b erroneous plays of some mutually permissible action $x'\ne x$ by $k'$ members, so that $x'$ is a best response for k members. Hence b satisfies:

$$\begin{aligned} \bigg (1-\frac{b}{n}\bigg )\theta _k + \delta _{kx}\le & {} \frac{b}{n}\theta _k + \delta _{kx'}\\ 2\frac{b}{n}\theta _k\ge & {} \theta _k + \delta _{kx} - \delta _{kx'}\\ b\ge & {} \frac{1}{2}n. \end{aligned}$$

The last line utilizes the fact that $\delta _{kx}= \delta _{kx'}$ because payoffs are simple. Thus the cost of an indirect path from $z_{x}$ to $\mathcal {M}$ is at least $\lceil \frac{1}{2}n\rceil$.

Hence when payoffs are simple, $r(z_{x}, \mathcal {M})\ge r(\mathcal {M}, z_{x})$ if n is sufficiently large and

$$\begin{aligned} \min \bigg \{\min _{k\in \{A,B\}}(1-D_{kx}), \frac{1}{2}\bigg \} \ge \min _{k\in \{A,B\}} D_{kx}.\end{aligned}$$

(14)

We shall now show that (14) is satisfied. By hypothesis, $\mathcal {M}$ is not strictly risk dominant. As payoffs are simple, this means that $\mathcal {M}$ is risk dominated by all states of coordination. By (12) this is equivalent to

$$\begin{aligned} \min _{k\in \{A,B\}}D_{kx} \le \min _{k\in \{A,B\}}\big (1 - D_{kx}\big ), \end{aligned}$$

(15)

which implies

$$\begin{aligned} \min _{k\in \{A,B\}}D_{kx}\le & {} \max _{k\in \{A,B\}}\big (1 - D_{kx}\big ) \nonumber \\ \min _{k\in \{A,B\}}D_{kx}\le & {} 1 - \min _{k\in \{A,B\}}D_{kx}\nonumber \\ \min _{k\in \{A,B\}}D_{kx}\le & {} \tfrac{1}{2}. \end{aligned}$$

(16)

Together (15) and (16) imply that (14) holds. This establishes the claim and indeed part (ii) of the proposition.

Proof of Proposition 3 The result follows immediately from the fact that 1 / x is strictly positive, strictly decreasing and strictly convex.

Proposition 4 When $\varepsilon =0$, there is no violation of cultural constraints. Hence the recurrent classes of the best response dynamic ($\varepsilon =0$) are still given by Proposition 1. These are the candidates for stochastic stability.

The only change in computation of resistances is to transitions from states of coordination to $\mathcal {M}$. Hence Lemmas 1 and 2 still apply.

We shall now introduce two new lemmas and a corollary. Lemma 4 and Corollary 2 are analogs of Lemma 3 and Corollary 1, respectively, for the case of unconstrained errors. We shall then employ the tree-surgery arguments used in Proposition 2 to establish the result.

Lemma 4

If (9) is satisfied, then for n sufficiently large

$$\begin{aligned} r(z_{x}, \mathcal {M})<r(\mathcal {M}, z_x) \end{aligned}$$

for all $x \in X_A \cap X_B$.

Proof

Recall that $r(\mathcal {M},z_x) = \min _{k\in \{A,B\}}\lceil D_{kx} n\rceil$.

Let us now compute $r(z_x,\mathcal {M})$. On the minimum cost path from $z_x$ to $\mathcal {M}$ one of two things must occur. Starting from $z_x$, through a series of errors, either (i) a mutually impermissible action becomes a best response, in which case the process transits to $\mathcal {M}$ without any further errors, or (ii) an action $x'\in X_A \cap X_B$ becomes a best response.

The minimum cost path conforming to case (i) is a direct path as follows. As errors are unconstrained, a transition from $z_x$ to $\mathcal {M}$ can occur with a erroneous plays of an action in $\widetilde{X}_k\setminus X_{k'}$ by $k'$ members, where a satisfies

$$\begin{aligned} \bigg (1 - \frac{a}{n}\bigg )\theta _k + \delta _{kx} \le \frac{a}{n}\theta _k + d_{k}. \end{aligned}$$

(17)

The value of a that equates (17) is $(1-D_{kx})\frac{n}{2}$. Hence the cost of such a path from $z_x$ to $\mathcal {M}$ is

$$\begin{aligned} \min _{k\in \{A,B\}} \bigg \lceil (1-D_{kx})\frac{n}{2}\bigg \rceil = \min _{k\in \{A,B\}} \bigg \lceil \bigg (1-\frac{d_{k}-\delta _{kx}}{\theta _k}\bigg )\frac{n}{2}\bigg \rceil . \end{aligned}$$

(18)

The minimum cost path conforming to case (ii) involves at least b erroneous plays of some mutually permissible action $x'\ne x$ by $k'$ members, so that $x'$ is a best response for k members. Hence b satisfies

$$\begin{aligned} \bigg (1 - \frac{b}{n}\bigg )\theta _k + \delta _{kx} \le \frac{b}{n}\theta _k + \delta _{kx'}. \end{aligned}$$

Thus the cost of such a path from $z_{x}$ to $\mathcal {M}$ is at least

$$\begin{aligned} \min _{k\in \{A,B\}} \bigg \lceil \bigg (1-\frac{\delta _{kx'}-\delta _{kx}}{\theta _k}\bigg )\frac{n}{2}\bigg \rceil . \end{aligned}$$

(19)

This is greater than (18) because $d_{k}>\delta _{kx'}$ by misalignment. Hence the minimum cost path from $z_x$ to $\mathcal {M}$ is direct and $r(z_x,\mathcal {M})$ equals (18).

Therefore, when errors are unconstrained and n is sufficiently large the following statements are equivalent:

$$\begin{aligned} r(\mathcal {M},z_x)>&r(z_x, \mathcal {M}),\nonumber \\ \min _{k\in \{A,B\}}D_{kx}>&\min _{k\in \{A,B\}}\tfrac{1}{2}\big (1 - D_{kx}\big ),\nonumber \\ 2\min _{k\in \{A,B\}}D_{kx} + \max _{k\in \{A,B\}}D_{kx}> & {} 1. \end{aligned}$$

(20)

The last line holds by hypothesis (9). $\square$

The next result follows immediately from Lemmas 1 and 4.

Corollary 2

If (9) is satisfied, then for n sufficiently large

$$\begin{aligned} r(z_x,\mathcal {M}) < r(z_{x'}, z_{x}) \end{aligned}$$

for all distinct pairs $(x,x') \in (X_A \cap X_B)^2$.

Lemma 5

$r(z_x, \mathcal {M}) \le r(z_x,z_{x'})$ for all distinct $(x,x')\in (X_A\cap X_B)^2$.

Proof

Recall that $r(z_x, \mathcal {M})$ equals (18).

Now consider $r(z_x,z_{x'})$. On the minimum cost path from $z_x$ to $z_{x'}$ one of two things must occur. Starting from $z_x$, through a series of errors, either (i) a mutually impermissible action becomes a best response, in which case $r(z_x,z_{x'})\ge r(z_x, \mathcal {M})$, or (ii) an action $x''\in X_A \cap X_B$ (possibly $x'$) becomes a best response. The path with minimum cost that conforms to the case (ii) involves $\beta$ erroneous plays of $x''$ by $k'$ members, where $\beta$ equals (19).

We established in the proof of Lemma 4 that (19) is greater than (18). This establishes the Lemma. $\square$

Proof of Proposition 4 To prove that (9) is sufficient for $\mathcal {M}$ to be stochastically stable we use the same tree-surgery argument as in Proposition 2(i). Starting with an arbitrary x-tree, we transform it through steps 1–2 into an $\mathcal {M}$-tree. Again the difference in resistance between the original x-tree and the resultant $\mathcal {M}$-tree is given by (13). If (9) holds, then the first term is positive by Lemma 4 and the second term, if it exists, is positive by Corollary 2. Hence (9) is sufficient.

To establish that (9) is also necessary, note that Lemma 5 implies that the minimum cost $\mathcal {M}$-tree is composed entirely of direct links, i.e., edges $(z_x, \mathcal {M})$ for each $x\in X_A\cap X_B$.

Suppose that (9) were violated, so that $2\min _{k\in \{A,B\}}D_{kx} + \max _{k\in \{A,B\}}D_{kx} \le 1$ for some $x\in X_A\cap X_B$. Recall that the minimum cost $\mathcal {M}$-tree contains edge $(z_x, \mathcal {M})$. Replace this edge with $(\mathcal {M},z_x)$. This operation transforms the minimum cost $\mathcal {M}$-tree into an x-tree. The difference in resistance between the original $\mathcal {M}$-tree and the resultant x-tree is $r(z_x, \mathcal {M})-r(\mathcal {M},z_x)$ which is nonnegative as (9) is violated [see (20)]. Hence $\mathcal {M}$ is not the unique stochastically stable class. $\square$