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\begin{LARGE}

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 31 January 2006  Eric Rasmusen, Erasmuse@indiana.edu.
Http://www.rasmusen.org. Overheads for Parts of  Sections 5.1, 5.2, 5.4, 6.4, 6.5 of {\it Games and Information}

  

   \noindent
{\bf 5.4  Product Quality  in an Infinitely Repeated Game}

  
 {\bf Players}\\
 An infinite number of potential firms and a continuum of consumers.

 \noindent
{\bf The Order of Play}\\
1  An endogenous number $n$ of firms decide to  enter the market at cost $F$.\\
2  A firm that has entered chooses its quality to be $High$ or $Low$, incurring
the constant marginal cost $c$ if it picks $High$ and zero if it picks $Low$.
The choice is unobserved by consumers. The firm also picks a price $p$.\\ 3
Consumers decide which firms (if any) to buy from, choosing firms randomly if
they are indifferent.  The amount bought from firm $i$ is denoted $q_i$.  \\ 4
All consumers observe the quality of all goods purchased in that period.\\ 5
The game returns to (2) and repeats.

\newpage



   \noindent
{\bf    Product Quality  in an Infinitely Repeated Game}



\noindent {\bf Payoffs}\\
 The   consumer benefit from a product of low quality is zero, but consumers are
willing to buy quantity $q(p) = \sum_{i=1}^n q_i$ for a product believed to be
high quality, where $\frac{dq}{dp} < 0$.  


If a firm stays out of the market,
its payoff is zero. 

If firm $i$ enters, it receives $-F$ immediately. Its
current end-of-period payoff is $q_ip$ if it produces $Low$ quality and $q_i(p-
c)$ if it produces $High$ quality. 

 The discount rate is $r \geq 0$.

   

\newpage 

\noindent
AN EQUILIBRIUM


\noindent {\bf Firms.} $\tilde{n}$ firms enter. Each produces high quality and
sells at price $\tilde{p}$. If a firm ever deviates from this, it thereafter
produces low quality (and sells at  the same price $\tilde{p} $).  The values of
$\tilde{p}$ and $\tilde{n}$ are given by equations (\ref{e5.4}) and (\ref{e5.8})
below.

\noindent {\bf Buyers.} Buyers start by choosing randomly among the firms
charging $\tilde{p}$. Thereafter, they remain with their initial firm unless it
changes its price or  quality, in which case they switch randomly to a firm that
has not changed its price or quality.

\noindent
 This strategy profile is a perfect equilibrium. 

 Each firm is willing
to produce high quality and refrain from price-cutting because otherwise it
would lose all its customers. 

 If it has deviated, it is willing to produce low
quality because the quality is unimportant, given the absence of customers.


Buyers stay away from a firm that has produced low quality because they know it
will continue to do so, and they stay away from a firm that has cut the price
because they know it  will produce low quality. 


\newpage

The equilibrium must satisfy three constraints: incentive compatibility, competition, and market
clearing.

 The {\bf incentive compatibility} constraint says that the individual firm must
be willing to produce high quality.  Given the buyers' strategy, if the firm
ever produces low quality it receives a one-time windfall profit, but loses its
future profits.  The tradeoff is represented by constraint (\ref{e5.3}), which
is satisfied if the discount rate is low enough. \begin{equation}\label{e5.3}
\frac{q_i p}{1+r} \leq \frac{q_i(p-c)}{r}  \;\;\;\;\;\;\;\;\; (incentive \;
compatibility). \end{equation} Inequality (\ref{e5.3}) determines a lower bound
for the price, which must satisfy \begin{equation}\label{e5.4} \tilde{p} \geq
(1+r)c. \end{equation}  

\newpage

   The second constraint is that competition drives profits to zero, so firms
are indifferent between entering and staying out of the market.
 \begin{equation}
\label{e5.5} \frac{q_i(p-c)}{r} = F  \;\;\;\;\;\;\;\;\; (competition)
\end{equation}
  Treating (\ref{e5.3}) as an equation and using it to replace $p$
in equation (\ref{e5.5}) gives \begin{equation}\label{e5.6} q_i =  \frac{F }{c}.
\end{equation}

 We have now determined $p$ and $q_i$, and only $n$ remains, which
is determined by the equality of supply and demand.  
  \begin{equation}\label{e5.7} nq_i = q(p). \;\;\;\;\;\;\;\;\; (market
\;\; clearing) \end{equation} Combining equations (\ref{e5.3}), (\ref{e5.6}),
and (\ref{e5.7}) yields \begin{equation}\label{e5.8}
  \tilde{ n} = \frac{cq([1+r]c)}{F }. \end{equation}  


\newpage



  \begin{center} {\bf Table  2:    Prisoner's Dilemmas  }
  \end{center}

\noindent (a) Two-Sided (conventional)

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Column}\\
&       &             &   {\it Silence}  &   & {\it Blame }     \\ &   &  {\it
Silence }      &     5,5 & $\rightarrow$  & -5,10 \\ & {\bf Row:}
&&$\downarrow$& & $\downarrow$ \\ &  &       {\it   Blame    }     &     10,-5
& $\rightarrow$  & {\bf 0,0} \\
 \end{tabular}

 {\it Payoffs  to:  (Row, Column). Arrows show how a player can increase his
payoff.   }

\newpage


\noindent
  REPEATED GAME IDEAS  

 {\bf The Prisoner's Dilemma}: Each prisoner can Blame or be Silent, and in equilibrium, they choose (Blame, Blame). (Tucker) 

 {\bf The Prisoner's Dilemma and Duopoly Collusion:} Each firm can choose Low or High price, and in equilibrium they choose (Low, Low).   (Bertrand, Cournot both are like that)

 {\bf The Chainstore Paradox:} With finite repetitions, the only perfect equilibrium is (Blame, Blame) in each period. (Selten) 

 {\bf The Folk Theorem: }With infinite repetitions, the number of equilibria is infinite, and in any set of $T$ periods, any behavior could be equilibrium behavior. (Aumann)

 \noindent
{\bf The Grim Strategy}\\
{\it  1  Start by choosing {\it Silence}.\\
 2  Continue to choose {\it Silence} unless some player  has chosen $Blame$, in
which case choose $Blame$ forever.}

  

 \noindent
{\bf Tit-for-Tat}\\
 {\it 1  Start by choosing {\it Silence}.\\
2  Thereafter, in period $n$ choose the action that the other player chose in
period $(n-1)$.}


\newpage
  
 
 \noindent
{\bf  6.4 Incomplete Information in the  Repeated  Prisoner's Dilemma: The Gang
of Four Model}

 
 

\noindent
 One of the most important explanations of reputation is that of Kreps, Milgrom,
Roberts  \& Wilson (1982).

  

 
 
\bigskip
 \noindent
 {\bf Theorem 6.1: The Gang of Four Theorem}\\
{\it Consider a    T-stage, repeated Prisoner's Dilemma,  without discounting
but with a probability $\gamma$ of a Tit-for-Tat player. In   any perfect
bayesian equilibrium,  the number of stages in which either player chooses {\rm
Blame} is less than some number M that depends on $\gamma$ but not on T.}

   The significance of the Gang of Four theorem is that while the players do
resort to $Blame$ as the last period approaches, the number of periods during
which they $Blame$ is independent of the total number of periods. Suppose $M=
2,500$. If $T=2,500$, there might be   $Blame$ every period.  But if $T=10,000$,
there are 7,500 periods without a $Blame$ move. For reasonable probabilities  of
the unusual type, the number of periods of cooperation can be much larger.
Wilson (unpublished) has set up an entry deterrence model in which the incumbent
fights entry (the equivalent of {\it Silence} above) up to seven periods from
the end, although the probability the entrant is  of the unusual type is only
0.008.

 \newpage

  To get a feeling for why  Theorem 6.1 is correct, consider what would happen
in a  10,001  period game with a   probability of 0.01 that Row is playing the
Grim Strategy of $Silence$ until the first $Blame$,  and $Blame$   every period
thereafter.



A best response for Column to  a known Grim
player is  ($Blame$ only in the last period, unless Row chooses $Blame$ first,
in which case respond with $Blame$). 


 Both players will choose $Silence$ until
the last period, and Column's payoff will be     50,010 (= (10,000)(5) + 10).


The Grim Strategy is NOT an equilibrium strategy in the complete-information game, though. Row would deviate to $Blame$ in the second-to-last period. 

\newpage


Suppose for  the moment that if Row is not Grim, he is highly aggressive, and
will choose $Blame$   every period.  If Column follows the strategy just
described, the outcome will be ($Blame, Silence$) in the first period and {\it
(Blame, Blame}) thereafter, for a payoff to Column of   $ -5 (= -5 +  (10,000)
(0) )$.


 If   the probabilities of  the two outcomes are 0.01 and 0.99, Column's
expected payoff from the strategy described is 495.15. If instead he follows a
strategy of ($Blame$  every period), his expected payoff is just 0.1 ($=0.01
(10) + 0.99 (0)$).

 Column  should risk
cooperating with Row  even if Row has a 0.99 probability of following a very
aggressive strategy.

The aggressive strategy, however, is not Row's best response to Column's
strategy. A better response is for Row to choose $Silence$ until the second-to-
last period, and then to choose $Blame$. Given that Column is cooperating in the
early periods, Row will cooperate also. 


 \newpage

\noindent
 TIT-FOR-TAT

  Tit-for-Tat has three strong points.
  \begin{enumerate} \item
It never initiates blaming ({\bf niceness});

\item
 It retaliates instantly against blaming ({\bf provocability});

\item
It forgives someone who plays $Blame$ but then   goes back to cooperating (it is
{\bf forgiving}).
  \end{enumerate}

 



\end{LARGE}
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