Cooperation under incomplete information on the discount factors

Abstract

In repeated games, cooperation is possible in equilibrium only if players are sufficiently patient, and long-term gains from cooperation outweigh short-term gains from deviation. What happens if the players have incomplete information regarding each other’s discount factors? In this paper we look at repeated games in which each player has incomplete information regarding the other player’s discount factor, and ask when full cooperation can arise in equilibrium. We provide necessary and sufficient conditions that allow full cooperation in equilibrium that is composed of grim trigger strategies, and characterize the states of the world in which full cooperation occurs. We then ask whether these “cooperation events” are close to those in the complete information case, when the information on the other player’s discount factor is “almost” complete.

Appendix: Proof of Theorem 3.2

Because a strategy of player $i$ is a $\Sigma _i$-measurable function, a necessary condition for $\eta ^*_i(K_i)$ to be a strategy is that $K_i$ is $\Sigma _i$-measurable.

For each $i \in \{1,2\}$, let $K_i$ be a $\Sigma _i$-measurable event. We will check when a deviation from $\eta ^*(K_1,K_2)$ can be profitable.

Assume first that $\omega \in K_i$. In this case player $i$ is supposed to follow the grim trigger course of action $GT^*$.

If $\omega \not \in K_j$, player $j$ will play $D$ all along the game, regardless of the play of player $i$. If $\omega \in K_j$, player $j$ will follow the grim trigger course of action $GT^*$. It then follows that if player $i$ deviates and plays $D$ at stage $k$, then he receives the highest payoff if he continues to play $D$ after stage $k$. Because payoffs are discounted, if a deviation to $D$ at stage $k$ is profitable to player $i$, then a deviation at stage 1 provides a higher profit. Therefore, the best possible deviation of player $i$ is to the course of action $D^*$. This deviation is not profitable if and only if

$$\begin{aligned} \gamma _i(\eta ^*(K_i,K_j)\mid \omega ) \ge \gamma _i(D^*_i,\eta ^*_j(K_i,K_j)\mid \omega ), \end{aligned}$$

(8)

where $D^*_i$ is player $i$’s strategy in which he always plays $D$. One can verify that

$$\begin{aligned} \gamma _i(\eta ^*(K_i,K_j)\mid \omega )\!=\! P_i(K_j\mid \omega )\left( \frac{3}{1\!-\!\lambda _i(\omega )}\right) \!+\!(1-P_i(K_j\mid \omega ))\frac{\lambda _i(\omega )}{1\!-\!\lambda _i(\omega )}.\quad \quad \end{aligned}$$

(9)

Indeed, according to player $i$’s belief, with probability $P_i(K_j\mid \omega )$ player $j$ plays $GT^*$, so they will play $(C,C)$ at every stage and his payoff will be $\frac{3}{1-\lambda _i(\omega )}$, and with probability $(1-P_i(K_j\mid \omega ))$ player $j$ plays $D^*$, so in the first stage he will get 0, and afterwards the players will play $(D,D)$; therefore in this case player $i$’s payoff is $\frac{\lambda _i(\omega )}{1-\lambda _i(\omega )}$. Similarly,

$$\begin{aligned} \gamma _i(D_i^*,\eta _j^*(K_j)\mid \omega )= P_i(K_j\mid \omega )\left( 4+\frac{\lambda _i(\omega )}{1-\lambda _i(\omega )}\right) +(1-P_i(K_j\mid \omega ))\left( \frac{1}{1-\lambda _i(\omega )}\right) .\!\!\!\nonumber \\ \end{aligned}$$

(10)

Plugging in (9) and (10) in (8) yields that $D^*_i$ is not a profitable deviation for player $i$ if and only if $P_i(K_j\mid \omega )\ge f_i(\omega )$, which is condition (b). Since $f_i(\omega )>1$ for $\omega \notin \Lambda _i$, this also implies that $K_i \subseteq \Lambda _i$, which is condition (a).

Now assume that $\omega \notin K_i$. In this case under $\eta ^*_i(K_1,K_2)$ player $i$ follows the course of action $D^*$. If player $i$ follows $\eta ^*_i(K_1,K_2)$ at the first stage, then from the second stage on player $j$ will play $D$, and then the best reply of player $i$ is to play $D$.

If $\omega \not \in K_j$, player $j$ follows $D^*$, and $\eta ^*_i(K_1,K_2)$ is player $i$’s best response. If $\omega \in K_j$, player $j$ follows $GT^*$, and therefore if there is a profitable deviation to player $i$, he must deviate at the first stage. Thus, to check whether $\eta ^*(K_1,K_2)$ is a Bayesian equilibrium we need to check whether $\gamma _i(\eta ^*(K_1,K_2)) \ge \gamma _i(GT^*_i,\eta ^*_j(K_1,K_2))$, where $GT^*_i$ is the grim trigger course of action of player $i$.

One can verify that

$$\begin{aligned} \gamma _i(\eta ^*(K_1,K_2)\mid \omega )=(1-P_i(K_j\mid \omega ))\frac{1}{1-\lambda _i(\omega )}+ P_i(K_j\mid \omega )\left( 4+\frac{\lambda _i}{1-\lambda _i}\mid \omega \right) \end{aligned}$$

and

$$\begin{aligned} \gamma _i(C_i^*,\eta _j^*(K_j)\mid \omega )=P_i(K_j\mid \omega )\frac{3}{1-\lambda _i(\omega )}+ (1-P_i(K_j\mid \omega ))\left( \frac{\lambda _i(\omega )}{1-\lambda _i(\omega )}\right) . \end{aligned}$$

The inequality $\gamma _i(\eta ^*(K_1,K_2)) \ge \gamma _i(GT^*_i,\eta ^*_j(K_1,K_2))$ then reduces to $P_i(K_j\mid \omega )\le f_i(\omega )$, and condition (c) is obtained.

Finally we need to verify that $f_i$ is $\Sigma _i$-measurable. This function is a rational function of $\lambda _i$, and therefore it is a Borel function of $\lambda _i$. Since $\Sigma _i$ contains the sets $\{\omega \mid \lambda _i(\omega )\!\in \! B\}$, for every open set $B\!\subseteq \![0,1)$, $f_i$, as a function of $\omega$, is indeed $\Sigma _i$-measurable.