\documentclass[12pt,epsf]{article}

 \usepackage{mathpazo}

\usepackage{ccaption} \captiondelim{}\renewcommand{\thefigure}{}
\renewcommand{\figurename}{}

\renewcommand\floatpagefraction{.9}
\renewcommand\topfraction{.9}
\renewcommand\bottomfraction{.9}
\renewcommand\textfraction{.1}
\setcounter{totalnumber}{50}
\setcounter{topnumber}{50}
\setcounter{bottomnumber}{50}

\usepackage{hyperref} \hypersetup{breaklinks=true, pagecolor=white, colorlinks=
true, linkcolor=black, hyperfootnotes= false, urlcolor=blue } \urlstyle{rm}

\usepackage{graphicx}

\usepackage{amsmath}

 \usepackage{amssymb}


  \reversemarginpar
  \topmargin  -.3in
  \oddsidemargin -.1in
  \textheight  10in
  \textwidth 7in
  \baselineskip 16pt



         \begin{document}

\titlepage

 

 \parindent 24pt

 \parskip 10pt
 
 %\begin{small}

 
 
\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 $y$.  }

 



 

%\end{small}

 

\end{document}


