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\noindent
{\bf How Incomplete Information Can Solve the Coordination Game Problem in Repeated Games } (April 10, 2007) \\
Consider a ranked coordination game with $n \geq 2$ players indexed by $i$ who simultaneously choose actions $x_1, ..., x_n$ from the interval [0,100]. If $m$ players choose the same action $x_i$, the per-period payoff to player $i$ is $\pi_i(x_i,m)$, with:
\begin{equation}\begin{array}{l}
(a) \frac{ \partial \pi_i(x_i, x_{-i},m)}{\partial x_i} >0\;\;\;\; (b) \frac{ \Delta \pi_i(x_i, x_{-i},m)}{\Delta m} >0,\;\;\;\;
(c) \frac{ \partial^2 \pi_i(x_i, x_{-i},m)}{\partial x_i \partial x_j} =0\\
\\
(d) \pi_i(0, x_{-i},n) > \pi_i(1, x_{-i},n-1),\;\;\;\; (e) \pi_i(w, x_{-i},l) > \pi_i(w', x_{-i},l-1) \forall l, w, w'\neq x\\
\end{array}
\end{equation}
(e) is a more general form of (d). I think I need either (c) or (e), but I've forgotten why.
We will normalize $\pi_i(0, x_{-i},n) =0$, which is to say that the payoff when all the players choose $x=0$ is $0$. The assumptions then imply that full coordination on $x>0$ yields positive payoffs and incomplete coordination yields negative payoffs.
The game is repeated a finite number $T$ times, with the players observing each other's choices after each round and with no discounting (an easily relaxed assumption).
If $T=1$, the game has a continuum of pure strategy equilibria, with $x$ on the continuum from 0 to 1, as well as mixed strategy equilibria.
All players prefer the equilibrium in which $x=1$, the pareto-optimal outcome.
In a {\bf time-dependent equilibrium}, some player's strategy in a round depends on which round it is. If the strategies are the same each round, the equilibrium is {\bf time-independent}.
In a {\bf history-dependent equilibrium}, some player's strategy in a round depends on the history of play up to that point. If the strategies do not depend on past play, the equilibrium is {\bf history-independent}.
\noindent
{\it Players are of two types. With some small probability $p>0$, a player $i$ is ``constrained'' to always play $x_i = z$, in every round of the game, where $z$ is chosen from [0, 100] using the atomless density $f(z)$.
If $T$ is large enough, the game now has a much smaller interval of equilibria, and the average payoff becomes arbitrarily close to 100. Formally:
\noindent
Proposition 1: For any $\epsilon$, there exists $T$ large enough that in all equilibria $\frac{ \sum_{t=1}^T \pi_{it} }{T}> 100-\epsilon$.
\noindent
{\it AN ALTERNATIVE ASSUMPTION: With small probability $p$, player 1 is ``constrained'' to play tit-for-tat: in equilibrium he begins with whatever $x_1=z$ maximizes his equilibrium payoff, but thereafter he plays the action $y$ that is played by the most other players in $(t-1)$, randomizing among the possibilities if there is a tie for action $y$. }
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