The emergence of cooperation through leadership

Abstract

We study the long-run outcomes of noisy asynchronous repeated games with players that are heterogeneous in their patience. The players repeatedly play a $2\times 2$ coordination game with random pair-wise matching. The games are noisy because the players may make mistakes when choosing their actions and are asynchronous because only one player can move in each period. We characterize the long-run outcomes of Markov perfect equilibrium that are robust to the mistakes and show that if one player is sufficiently patient whereas the other players are not so patient, the efficient state can be the unique robust outcome even if it is risk-dominated. Because we need heterogeneity for the result, we argue that it enables the most patient player in effect to be the leader.

Appendix

In this appendix, proofs omitted from the main text are provided.

Proof of Lemma 1

Since $V_i(x) \le \frac{d}{1-\delta _i} <\infty$,

$$\begin{aligned} V_i(0, x_j) - V_i(1, x_j)&= u_\beta (x_j) - u_\alpha (x_j+1) \\&\quad + \frac{\delta _i}{2} \bigg [ (1-\varepsilon ) \big ( V_i (0, \psi ^*_j (0))- V_i (1, \psi ^*_j (1)) \big ) \\&\qquad \qquad + \varepsilon \big ( V_i (0, 1- \psi ^*_j (0))- V_i (1, 1- \psi ^*_j (1)) \big ) \bigg ] \\&\rightarrow u_\beta (x_j) - u_\alpha (x_j+1) + \frac{\delta _i}{2}\big ( V_i (0, \psi ^*_j (0))- V_i (1, \psi ^*_j (1)) \big )\\&\quad \equiv \Delta V_i (x_j) \end{aligned}$$

as $\varepsilon \rightarrow 0$. Thus, we focus on $\Delta V_i (x_j)$’s and invoke their continuity in $\varepsilon$. Note that $\Delta V_i(0) -\Delta V_i(1) = u_\beta (0) -u_\alpha (1) - u_\beta (1) +u_\alpha (2) = a-c +d-b >0$. Suppose a MPE $\psi ^*$ exists. Let $\psi ^*_i = (\psi _i^*(0), \psi ^*_i (1) )$. We first show that $\Delta V_2(0) >0$ and $\Delta V_2(1) <0$ at equilibrium in the following two claims.

Claim 1

$\Delta V_2(0) >0$.

Proof

If $\psi ^*_1= (0, 0)$, $\Delta V_2 (0) = (1- \delta _2/2) ^{-1} (u_\beta (0)-u_\alpha (1) ) >0$. If $\psi ^*_1 = (1, 1)$,

$$\begin{aligned} \Delta V_2 (0)&= u_\beta (0) -u_\alpha (1) + \frac{\delta _2}{2} \Delta V_2(1) \\&= u_\beta (0) -u_\alpha (1) + \frac{\delta _2}{2} \left( \Delta V_2(0) - u_\beta (0) + u_\alpha (1) + u_\beta (1) -u_\alpha (2) \right) . \end{aligned}$$

Therefore, $\Delta V_2(0) = u_\beta (0) -u_\alpha (1) + \frac{\delta _2}{2-\delta _2} ( u_\beta (1)- u_\alpha (2) ) >0$ where the last inequality follows because $\delta _2 < \frac{2(d-b)}{a-c+d-b}$. Finally, suppose $\psi ^*_1= (0, 1)$ but $\Delta V_2(0) \le 0$. (Note that $\psi _1^*\ne (1, 0)$ because, if so, $\Delta V_1(0) \le \Delta V_1(1)$.) Then, $\Delta V_2 (0) = u_\beta (0) -u_\alpha (1) + \frac{\delta _2}{2} \left( V_2(0, 0)- V_2(1, 1) \right)$. Because $V_2(0, 0) = u_\beta (0) + \frac{\delta _2}{2} V_2(0, \psi _2^*(0) ) +\frac{\delta _2}{2} V_2( 0, 0)$ and $V_2(0, \psi _2^*(0) ) \ge V_2(0, 0)$, we have $V_2(0, 0) \ge \frac{u_\beta (0)}{ 1-\delta _2}$. Similarly, $V_2(1, 1) = u_\alpha (2) + \frac{\delta _2}{2} V_2(1, \psi _2^*(1) ) +\frac{\delta _2}{2} V_2( 1, 1)$. But, because $\Delta V_2 (1) < \Delta V_2 (0) \le 0$, $\psi _2^*(1)=1$. Thus, $V_2(1, 1)=\frac{u_\alpha (2)}{1-\delta _2}$. Therefore, $\Delta V_2(0) \ge u_\beta (0)- u_\alpha (1) + \frac{\delta _2}{2(1-\delta _2)} ( u_\beta (0) - u_\alpha (2) ) >0$. $\square$

Claim 2

$\Delta V_2(1) <0$.

Proof

If $\psi ^*_1= (1, 1)$, $\Delta V_2 (1) = (1-\delta _2/2)^{-1} ( u_\beta (1) -u_\alpha (2) )<0$. If $\psi ^*_1 = (0, 0)$,

$$\begin{aligned} \Delta V_2 (1)&= u_\beta (1) -u_\alpha (2) + \frac{\delta _2}{2} \Delta V_2(0) \\&= u_\beta (1) -u_\alpha (2) + \frac{\delta _2}{2} \left( \Delta V_2(1) + u_\beta (0) - u_\alpha (1) - u_\beta (1) +u_\alpha (2) \right) . \end{aligned}$$

Thus, $\Delta V_2 (1) = u_\beta (1)- u_\alpha (2) + \frac{\delta _2}{2-\delta _2} (u_\beta (0)-u_\alpha (1) ) < 0$. Finally, if $\psi ^*_1 = (0, 1)$, $\Delta V_2(1) \le u_\beta (1)- u_\alpha (2) + \frac{\delta _2}{2(1-\delta _2)} ( u_\beta (0) - u_\alpha (2) ) <0$ because $V_2(0, 0) = \frac{u_\beta (0)}{ 1-\delta _2}$ (by Claim 1), $V_2(1, 1)\ge \frac{u_\alpha (2)}{1-\delta _2}$, and $\delta _2 < \frac{2(a-c)}{a+d-2c}$. $\square$

Now, suppose $\Delta V_1(1) \le 0$. By Claims 1 and 2, $\psi ^*_2 =(0, 1)$. Then, since $V_1(0, 0)\ge \frac{u_\beta (0)}{1-\delta _1}, V_1(1, 1) = \frac{u_\alpha (2)}{1-\delta _1}$, and $\delta _1 > \frac{2(a-c)}{a+d-2c}$,

$$\begin{aligned} \Delta V_1(1) \ge u_\beta (1)- u_\alpha (2) + \frac{\delta _1}{2(1-\delta _1)} ( u_\beta (0) - u_\alpha (2) ) >0. \end{aligned}$$

(17)

Hence, we must have $\Delta V_1(1) >0$ and therefore $\Delta V_1 (0) > \Delta V_1(1) >0$. Thus, if a MPE $\psi ^*$ exists, it is uniquely given by $\psi ^*_1 = (0, 0)$ and $\psi ^*_2 = (0, 1)$. But, because our state and action spaces are finite, the existence of MPE follows by a standard argument. $\square$

Proposition A1 Suppose $N\!=\!2$ and $\delta _1 \!=\! \delta _2$. If $\frac{2(a-c)}{a+d-2c} \!\ge \! \frac{2(d-b)}{a-c+d-b}$, $\lim _{\varepsilon \rightarrow 0} \pi ^\varepsilon ((0, 0)) \ne 1$ whenever a MPE uniquely exists.

Proof

Let $\delta _1 =\delta _2= \delta$ and $\varepsilon =0$. Suppose $\psi _j^*= (0, 1)$. We first show $\psi _i^*(1) =0 \Leftrightarrow \delta > \frac{2(a-c)}{a+d-2c}$. If $1\in \psi _i^*(1)$, $V_i (1, 1) = \frac{u_\alpha (2)}{1-\delta }$. Then, because $V_i( 0, 0) = \frac{u_\beta (0)}{1-\delta }$, $\Delta V_i(1) = u_\beta (1)- u_\alpha (2) + \frac{\delta }{2-\delta } (u_\beta (0)-u_\alpha (1) ) \le 0 \Rightarrow \delta \le \frac{2(a-c)}{a+d-2c}$. If $\psi _i^*(1)= 0$, we have

$$\begin{aligned} V_i (1, 0)&= u_\beta (1) + \frac{\delta }{2} V_i (1, 0) + \frac{\delta }{2} V_i (0, 0), \\ V_i (1, 1)&= u_\alpha (2) + \frac{\delta }{2} V_i (1, 0) + \frac{\delta }{2} V_i (1, 1). \end{aligned}$$

Because $V_i( 0, 0) = \frac{u_\beta (0)}{1-\delta }$, we can solve the above equations for $V_i(1, 0)$ and $V_i(1, 1)$. Then, it follows that $\Delta V_i(1) >0 \Rightarrow \delta > \frac{2(a-c)}{a+d-2c}$.

Therefore, by the previous argument (see the proofs of Claims 1 and 2),

$$\begin{aligned} \psi _j^*= (0, 0)&\Rightarrow \psi _i^*= (0, 1), \\ \psi _j^*= (0, 1)&\Rightarrow \psi _i^*= \left\{ \begin{array}{l@{\quad }l} (0, 1) &{} \text {if}\quad \delta < \frac{2(a-c)}{a+d-2c}, \\ (0, 0) &{} \text {if}\quad \delta > \frac{2(a-c)}{a+d-2c}, \end{array}\right. \\ \psi _j^*= (1, 1)&\Rightarrow \psi _i^*= \left\{ \begin{array}{l@{\quad }l} (0, 1) &{} \text {if}\quad \delta < \frac{2(d-b)}{a-c+d-b}, \\ (1, 1) &{} \text {if}\quad \delta > \frac{2(d-b)}{a-c+d-b}. \end{array}\right. \end{aligned}$$

Note that $\psi _i= (1, 0)$ is strictly dominated. Thus, there can exist three types of (pure) Markov perfect equilibria: (i) $\psi _i^*= \psi _j^*= (0, 1)$; (ii) $\psi _i^*= \psi _j^*= (1, 1)$; and (iii) $\psi _i^*= (0, 1), \psi _j^*= (0, 0)$. In the first MPE, the limit invariant distribution puts half of the mass on each of the static equilibria because both of the players behave as if they are myopic. In the second MPE, the limit invariant distribution puts all the mass on the risk dominant state because both of the players always take action $\alpha$. The third MPE is the same as the one we attained in Lemma 1, and therefore the limit invariant distribution puts all the mass on the efficient state.

Now, suppose $\delta \le \frac{2(d-b)}{a-c+d-b}$. Then, because $\delta < \frac{2(a-c)}{a+d-2c}$ by assumption, the first MPE attains. But, if $\delta > \frac{2(d-b)}{a-c+d-b}$, the second MPE attains. Therefore, there always exists a MPE under which $\lim _{\varepsilon \rightarrow 0} \pi ^\varepsilon ((0, 0)) \ne 1$.²⁴ $\square$

Proof of Lemma 2

As in Lemma 1, we let $\varepsilon =0$ and invoke continuity in $\varepsilon$. Suppose $S_N>1$ so that there exists $m \in \{0, 1, \ldots , N-1\}$ such that $m < S_N-1$. Then, by (14),

$$\begin{aligned} V_1( 0, m)&= u_\beta (m) \!+\! \frac{\delta }{N} \left( V_1 (\psi _1^*(m), m) \!+\! m V_1 (0, m-1) \!+\! (N-m-1) V_1(0, m) \right) , \end{aligned}$$

(18)

$$\begin{aligned} V_1( 1, m)&= u_\alpha (1+m) + \frac{\delta }{N} \left( V_1 (\psi _1^*(m), m) \!+\! m V_1 (1, m-1)\right. \nonumber \\&\qquad \qquad \qquad \qquad \qquad \quad \left. + (N-m-1) V_1(1, m) \right) \end{aligned}$$

(19)

for $0\le m < S_N -1$ because $\psi _\alpha ^*(x) = \psi _\beta ^*(x) =0$. Therefore,

$$\begin{aligned} \left( 1 - \delta \frac{N-m-1}{N} \right) \Delta V_1(m) = u_\beta (m) - u_\alpha (1+m) +\delta \frac{m}{N} \Delta V_1(m-1) \end{aligned}$$

(20)

where $\Delta V_1(m) = V_1(0, m)- V_1(1, m)$. On the other hand, because $S_N < N-2$ when $N\ge 4$, there exists $m \in \{0, 1, \ldots , N-1\}$ such that $m > S_N+1$. Then, we get

$$\begin{aligned} \left( 1- \delta \frac{m}{N} \right) \Delta V_1(m) = u_\beta (m) - u_\alpha (1+m) + \delta \frac{N-m-1}{N} \Delta V_1(m+1) \end{aligned}$$

(21)

for $S_N +1 < m \le N-1$. But, by (20),

$$\begin{aligned} \Delta V_1 (0) = \left( 1- \delta \frac{N-1}{N} \right) ^{-1} \big ( u_\beta (0) - u_\alpha (1) \big ) >0 . \end{aligned}$$

(22)

This, in turn, implies that

$$\begin{aligned} \Delta V_1(1) = \left( 1- \delta \frac{N-2}{N} \right) ^{-1} \left( u_\beta (1) - u_\alpha (2) + \delta \frac{1}{N} \Delta V_1(0) \right) >0 \end{aligned}$$

(23)

by (20). Continuing in this way, we induce $\Delta V_1(m) >0$ for $0\le m < S_N-1$. The same argument is applied to (21) to show that $\Delta V_1(m) <0$ for $S_N+1 < m \le N-1$. $\square$

Proof of Lemma 4

As in the previous lemmas, we let $\varepsilon =0$ and invoke continuity in $\varepsilon$. Then, we have

$$\begin{aligned} \Delta V_1 (m^*+1)&= u_\beta (m^*+1)- u_\alpha ( m^*+2) + \delta \frac{N-m^*-2}{N} \Delta V_1 (m^*+2) \nonumber \\&+\, \delta \frac{m^*+1}{N} \big ( V_1(0, m^*) - V_1(1, m^*+1) \big ). \end{aligned}$$

(24)

Suppose $\Delta V_1(m^*+1) \le 0$. Then,

$$\begin{aligned} \left( 1- \delta \frac{1}{N}- \delta \frac{N-m^*-1}{N} \right) V_1(0, m^*)&\ge u_\beta (m^*) + \delta \frac{m^*}{N} V_1(0, m^*-1), \end{aligned}$$

(25)

$$\begin{aligned} \left( 1- \delta \frac{1}{N} - \delta \frac{m^*+1}{N} \right) V_1(1, m^*+1)&= u_\alpha (m^*\!+\!2) +\delta \frac{N-m^*-2}{N} V(1, m^*\!+\!2). \end{aligned}$$

(26)

(25) holds with equality iff $\Delta V_1(m^*) \ge 0$. If $4\le N < \frac{2(a-c)}{a-c-(d-b)}$, it follows that $m^*= \frac{N-2}{2}$. Therefore, (25) and (26) are combined and simplified to

$$\begin{aligned} V_1(0, m^*) \!-\! V_1(1, m^*+1)&\ge \big ( u_\beta (m^*)-u_\alpha (m^*+2) \big ) Q(1) \nonumber \\&+\, \delta \frac{m^*}{N} \big ( V_1(0, m^*-1)- V_1(1, m^*+2 ) \big ) Q(1) \qquad \quad \end{aligned}$$

(27)

where $Q(n) = \left( 1 - \delta \frac{1}{N} - \delta \frac{m^*+n}{N} \right) ^{-1}$. Then, solving (27) iteratively, we get

$$\begin{aligned} V_1(0, m^*)- V_1(1, m^*+1)&\ge \sum _{k=1}^{m^*}\delta ^{k -1} \frac{f(k)}{N^{k-1}} \big ( u_\beta (m^*- k +1 )- u_\alpha (m^*+k+1 ) \big ) \prod _{n=1}^k Q(n) \nonumber \\&+\, \delta ^{m^*}\frac{m^*!}{N^{m^*}} \big ( V_1(0, 0) - V_1(1, N-1) \big ) \prod _{n=1}^{m^*} Q(n) \end{aligned}$$

(28)

where $f(1)=1, f(2)=m^*, f(3)= m^*(m^*-1)$, and $f(k)= m^*(m^*-1)\cdots ( m^*-k+2)$ for $k \ge 4$. Then, since $V_1(0, 0) \!-\! V_1(1, N-1) \!=\! \frac{d-a}{1-\delta }$, $V_1(0, m^*)- V_1(1, m^*+1) \!\rightarrow \! \infty$ as $\delta \rightarrow 1$. Hence, by (24), there exists $\delta ^*\in (0, 1)$ such that $\Delta V_1(m^*+1) >0$ for all $\delta \in (\delta ^*, 1)$. Therefore, we must have $\delta \in (0, \delta ^*]$. Thus, $\Delta V_1(m^*+1)>0$ if $\delta \in (\delta ^*, 1)$. Moreover, for such $\delta$’s,

$$\begin{aligned} \Delta V_1(m^*)&= u_\beta (m^*)- u_\alpha ( m^*+1) + \delta \frac{m^*}{N} \Delta V_1 (m^*-1) \nonumber \\&+\, \delta \frac{N- m^*-1}{N} \big ( V_1(0, m^*) - V_1(1, m^*+1) \big ) >0 \end{aligned}$$

(29)

since $u_\beta (m^*) > u_\alpha ( m^*+1)$ and $\Delta V_1 (m^*-1) >0$. $\square$