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\begin{LARGE}
\parindent 24pt \parskip 10pt
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}
\end{document}