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\noindent    22  November 2005.  Eric Rasmusen, Erasmuse@indiana.edu.
Http://www.rasmusen.org.

\bigskip

\begin{LARGE}
 \begin{center} {  \bf 6 Dynamic Games with Incomplete Information }
\end{center}
\end{LARGE}

 \vspace{1in}
\noindent
 { \bf 6.1 Perfect Bayesian Equilibrium: Entry Deterrence  II  and III }

 \noindent
 Asymmetric information, and, in particular, incomplete information, is
enormously important in   game theory. This is particularly true for dynamic
games, since when the players  have several moves in sequence, their earlier
moves may convey private information  that is relevant to the decisions of
players moving later on. Revealing and concealing information are the basis of
much of   strategic behavior  and are especially useful as ways of explaining
actions that would be irrational in a  nonstrategic  world.

   Chapter 4 showed that  even if there is symmetric information in a  dynamic
game, Nash equilibrium may need to be refined using subgame perfectness if the
modeller is  to make sensible predictions. Asymmetric information requires a
somewhat different refinement to capture the idea of sunk costs and credible
threats, and Section 6.1  sets out the standard refinement of perfect bayesian
equilibrium. Section 6.2 shows that even  this may not be enough refinement to
guarantee uniqueness and discusses further refinements based on out-of-
equilibrium beliefs. Section 6.3 uses the idea to show that a player's
ignorance may work to his advantage, and to explain how   even when   all
players know something, lack of common knowledge still affects the game.
Section 6.4    introduces incomplete information into the repeated   {
prisoner's dilemma}   and shows the Gang of Four  solution to the Chainstore
Paradox of Chapter 5. Section 6.5 describes the celebrated Axelrod tournament,
an experimental approach to the same paradox. Section 6.6 applies the idea of a
dynamic game of incomplete information to the evolution of creditworthiness
using the model of Diamond (1989).




\bigskip \noindent
{\bf  Subgame Perfectness Is Not Enough}

 \noindent In games of asymmetric information, we will still require   that an
equilibrium be subgame perfect,  but the mere forking of the game tree might not
be relevant to a player's decision, because with asymmetric information he does
not know which fork the game has taken.   Smith might know he is at one of two
different nodes depending on whether Jones has high or low production costs, but
if he  does not know the exact node,  the ``subgames'' starting at each node are
irrelevant to his decisions. In fact, they are not even  subgames as we have
defined them, because they cut across Smith's information sets.  This  can be
seen in    an asymmetric information version of     {   Entry Deterrence I}
(Section 4.2).   In    {   Entry Deterrence I}, the incumbent colluded with the
entrant because fighting him was more costly than colluding   once the entrant
had entered. Now, let us set up the game to allow some entrants to be $Strong$
and some $Weak$  in the sense that it is more costly for the incumbent to choose
$Fight$ against a $Strong$ entrant than a $Weak$ one. The incumbent's payoff
from $Fight|Strong$ will be 0, as before, but his payoff from $Fight|Weak$ will
be $X$, where  $X$ will take  values ranging from 0 (  {   Entry Deterrence I} )
to 300 (Entry Deterrence  IV and V) in different versions of the game.

  Entry Deterrence  II, III, and IV  will all  have the extensive form shown in
Figure  1.  With 50 percent probability,  the incumbent's payoff from {\it
Fight} is $X$ rather than the 0 in   {   Entry Deterrence I}, but the incumbent
does not know which  payoff is the correct one in the particular realization of
the game.     This  is modelled as an initial move by Nature, who chooses
between the entrant being  $Weak$  or $Strong$, unobserved by the incumbent.

 \includegraphics[width=150mm]{fig06-01.jpg}




\begin{center}
  {\bf Figure 1:   Entry Deterrence   II, III,  and IV} \end{center}

\noindent
 {\bf Entry Deterrence II: Fighting Is Never Profitable}

In    {   Entry Deterrence II},   $X=1$, so information is not very asymmetric.
It is common knowledge that the incumbent never benefits from {\it Fight}, even
though his exact payoff might be zero or might be one.  Unlike in the case of
Entry Deterrence I, however,  subgame perfectness does not rule out any Nash
equilibria, because the only subgame is the subgame starting at node $N$, which
is the entire game.  A subgame cannot start at nodes $E_1$ or $E_2$, because
neither of those nodes are singletons in the information partitions. Thus,  the
implausible Nash equilibrium, ({\it Stay Out}, {\it Fight}), escapes elimination
by a technicality.

 The equilibrium concept needs to be  refined in order to eliminate the
implausible equilibrium.    Two general approaches can be taken:  either
introduce small ``trembles'' into the game, or require that strategies be best
responses given rational beliefs.  The first approach takes us to the
``trembling hand-perfect'' equilibrium, while the second takes us to the
``perfect bayesian'' and ``sequential'' equilibrium.  The results are similar
whichever approach is taken.

\bigskip \noindent
  {\bf Trembling-Hand Perfectness}

\noindent
  Trembling-hand perfectness is an equilibrium concept introduced by Selten
(1975)  according to  which   a strategy that is to   be part of an equilibrium
must continue to be optimal for the player even if there is a   small chance
that the other player will pick an out-of-equilibrium action (i.e., that the
other player's hand will ``tremble'').

 Trembling-hand perfectness is defined for games with finite action sets as
follows.

 \noindent {\it The strategy profile $s^*$ is a {\bf trembling-hand perfect}
equilibrium if for any $\epsilon$ there is a vector of positive numbers
$\delta_1,\ldots,\delta_n\in [0,1]$ and a vector of completely mixed strategies
$\sigma_1,\ldots \sigma_n$ such that the perturbed game where every strategy is
replaced by $(1-\delta_i)s_i + \delta_i \sigma_i$ has a Nash equilibrium in
which every strategy is within distance $\epsilon$ of $s^*$.}

   Every trembling-hand perfect equilibrium is subgame perfect; indeed,
Section 4.1   justified subgame perfectness using a tremble argument.
Unfortunately, it is often hard to tell whether a strategy profile is trembling-
hand perfect, and the concept is undefined for games with continuous strategy
spaces because it is hard to work with mixtures of a continuum (see note N3.1).
Moreover, the equilibrium depends on which trembles are chosen, and deciding why
one tremble should be more common than another  may be difficult.

\bigskip \noindent
  {\bf Perfect Bayesian Equilibrium and Sequential Equilibrium}

\noindent The second approach to asymmetric information, introduced by Kreps \&
Wilson (1982b) in the spirit of Harsanyi (1967), is to start with prior beliefs,
common to all players, that specify the probabilities with which Nature chooses
the types of the players at the beginning of the game. Some of the players
observe Nature's move and update their beliefs, while other players can update
their beliefs only by deductions they make from observing the actions of the
informed players.

 The deductions used to update beliefs are based on the actions specified by the
equilibrium.  When players update their beliefs, they assume that the other
players are following the equilibrium strategies, but since the strategies
themselves depend on the beliefs, an equilibrium can no longer be defined based
on strategies alone.  Under asymmetric information, an equilibrium is a strategy
profile and a set of beliefs such that the strategies are best responses. The
profile of beliefs and strategies is called an {\bf assessment}  by Kreps and
Wilson.

   On the equilibrium path, all that the players need to update their beliefs
are their priors and Bayes' s Rule, but off the equilibrium path this is not
enough.  Suppose that in equilibrium, the entrant always enters. If  for
whatever reason  the impossible happens   and the entrant stays out, what is the
incumbent to think about the probability that the entrant is weak?  Bayes' s
Rule does not help, because when $Prob(data)=0$, which is the case  for data
such as $Stay\; Out$ which is never observed in equilibrium, the posterior
belief cannot be calculated using Bayes' s Rule.  From section 2.4,
  \begin{equation}\label{e6.1}
  Prob(Weak|Stay\; Out) = \frac{ Prob(Stay \;Out|Weak)Prob( Weak)} {Prob(Stay \;
Out)}. \end{equation}
 The posterior $Prob(Weak|Stay \; Out)$ is undefined, because (\ref{e6.1})
requires dividing by zero.

 A natural way to define equilibrium is as a strategy profile consisting of best
responses given that equilibrium beliefs follow Bayes' s Rule and out-of-
equilibrium beliefs follow a specified pattern that does not contradict Bayes' s
Rule.

 \noindent {\it A {\bf perfect bayesian equilibrium} is a strategy profile s and
a set of beliefs $\mu$ such that at each node of the game: }

\noindent {\it (1) The strategies for the remainder of the game are Nash given
the beliefs and strategies of the other players.}\\
 {\it (2) The beliefs at each information set are rational given the evidence
appearing thus far in the game (meaning that they are based, if possible, on
priors updated by Bayes' s Rule, given the observed actions of the other players
under the hypothesis that they are in equilibrium). }

     Kreps \& Wilson (1982b) use  this idea to  form  their equilibrium concept
of  sequential equilibrium,  but they impose a third condition, defined only for
games with discrete strategies, to restrict beliefs a little further:

{\it (3) The beliefs are the limit of a sequence of rational beliefs, i.e., if
$(\mu^*,s^*)$ is the equilibrium assessment, then some sequence of rational
beliefs and completely mixed strategies converges to it:} $$
 (\mu^*,s^*) = Lim_{n \rightarrow  \infty} (\mu^n,s^n) \;for\; some\;sequence \;
(\mu^n,s^n) \; in\; \; \{\mu, s \}. $$

 Condition (3) is quite reasonable and  makes sequential equilibrium close to
trembling-hand perfect equilibrium, but it adds more to the concept's difficulty
than to its usefulness. If players are using the sequence of completely mixed
strategies $s^n$, then every action is taken with some positive probability, so
Bayes'Rule can be applied to form the beliefs $\mu^n$ after any action is
observed. Condition (3) says that the equilibrium assessment has to be the limit
of some such sequence (though not of every such sequence).  For the rest of the
book we will   use perfect bayesian equilibrium  and dispense with condition
(3), although it usually can be satisfied.

     Sequential equilibria are always subgame perfect (condition (1) takes care
of that).  Every trembling-hand perfect equilibrium is a sequential equilibrium,
and  ``almost every'' sequential equilibrium is trembling hand perfect.  Every
sequential equilibrium is perfect bayesian, but not every  perfect bayesian
equilibrium is sequential.

 \bigskip
 \noindent
 {\bf Back to  Entry Deterrence II}

\noindent Armed with the concept of the perfect bayesian equilibrium, we can
find a sensible equilibrium   for   {   Entry Deterrence II} .

  \begin{quotation} \noindent Entrant: $ Enter|Weak$,  $Enter|Strong$\\
Incumbent:  {\it Collude}  \\ Beliefs:  $Prob$( $Strong|$ {\it Stay Out}) = 0.4
\end{quotation}

\noindent In this  equilibrium  the entrant enters whether he is $Weak$ or
$Strong$. The incumbent's strategy is $Collude$, which is not conditioned on
Nature's move, since he does not observe it. Because the entrant enters
regardless of Nature's move, an out-of-equilibrium belief for the incumbent if
he should observe $Stay\; Out$ must be specified, and this belief is arbitrarily
chosen to be that the incumbent's subjective probability  that the entrant is
$Strong$ is 0.4  given his observation that   the entrant deviated by choosing
{\it Stay Out}. Given this  strategy profile and out-of-equilibrium belief,
neither player has incentive to change his strategy.

There is no perfect bayesian equilibrium in which the entrant chooses {\it Stay
Out}.  $Fight$ is a bad response even under the most optimistic possible belief,
that  the entrant is $Weak$ with probability 1. Notice that  perfect bayesian
equilibrium is not defined structurally, like subgame perfectness, but rather in
terms of optimal responses.  This enables it to come closer to the   economic
intuition which we wish to capture by an equilibrium refinement.

  Finding the perfect bayesian equilibrium  of a game, like finding the Nash
equilibrium, requires intelligence. Algorithms are not useful. To find a Nash
equilibrium, the modeller thinks about his game, picks a plausible strategy
profile, and tests whether the strategies are best responses to each other. To
make it a perfect bayesian equilibrium, he notes which actions are never taken
in equilibrium and specifies the beliefs that players use to interpret those
actions. He then tests whether each player's strategies are best responses given
his beliefs at each node, checking in particular whether any player would like
to take an out-of-equilibrium action in order to set in motion the other
players' out-of-equilibrium beliefs and strategies. This process does not
involve testing whether a player's beliefs are beneficial to the player, because
players do not choose their own beliefs; the priors and out-of-equilibrium
beliefs are exogenously specified by the modeller.

  One might wonder why the beliefs have to be specified in Entry Deterrence II.
Does not the game tree specify the probability  that the entrant is $Weak$?
What difference does it make if the entrant stays out? Admittedly, Nature does
choose each type with probability 0.5, so if the incumbent had no other
information than this prior, that would be his belief. But the entrant's action
might convey additional information.  The concept of perfect bayesian
equilibrium leaves the modeller free to specify how the players form beliefs
from that additional information, so long as the beliefs do not violate Bayes'
Rule.  (A technically valid choice of beliefs by the modeller might still be met
with scorn, though, as with any silly assumption. ) Here, the equilibrium says
that if the entrant stays out, the incumbent believes   he is $Strong$ with
probability 0.4 and $Weak$ with probability 0.6, beliefs that are arbitrary  but
do not contradict Bayes' s Rule.

 In   {   Entry Deterrence II}    the out-of-equilibrium beliefs do not and
should not matter. If the entrant chooses {\it Stay Out}, the game ends,  so the
incumbent's beliefs are irrelevant. Perfect bayesian equilibrium was  only
introduced   as a way out of a technical problem. In the next section, however,
the precise out-of- equilibrium beliefs will be crucial to which strategy
profiles are equilibria.

   \bigskip

 \noindent
  {\bf  6.2 Refining Perfect Bayesian Equilibrium: The PhD Admissions Game}

\bigskip \noindent {\bf    Entry Deterrence III: Fighting Is Sometimes
Profitable }

\noindent
 In   {   Entry Deterrence III},  assume  that $X=60$, not $X=1$. This
means that  fighting is more profitable for the incumbent   than collusion  if
the entrant is $Weak$. As before,  the entrant knows  if he is $Weak$, but the
incumbent does not. Retaining the prior after observing out- of-equilibrium
actions, which in this game is $Prob( Strong) = 0.5$, is a convenient way to
form beliefs that is called {\bf passive conjectures.}  The following is a
perfect bayesian equilibrium which uses passive conjectures.

\begin{quotation}
\noindent {\bf A plausible  pooling  equilibrium for Entry Deterrence III} \\
Entrant:  $Enter|Weak$, $Enter|Strong$\\
 Incumbent:   {\it Collude},   
  Out-of-equilibrium beliefs:  $Prob ( Strong|$ {\it Stay Out}) = 0.5
\end{quotation}

    In choosing whether to enter, the entrant must predict the incumbent's
behavior.  If the probability  that the entrant is $Weak$ is 0.5, the expected
payoff to the incumbent from choosing {\it Fight} is 30 ($=0.5[0] +0.5[60]$),
which is less than the   payoff of 50 from $Collude$.  The incumbent will
collude, so the entrant enters. The entrant may know that the incumbent's payoff
is actually 60, but that is irrelevant to the incumbent's behavior.

  The out-of-equilibrium belief does not matter to this first equilibrium,
although it will  in other equilibria of the same game.  Although beliefs in a
perfect bayesian equilibrium must follow Bayes' s Rule, that puts very little
restriction on how players interpret out-of-equilibrium behavior.  Out-of-
equilibrium behavior is ``impossible,'' so when it does occur there is no
obvious way the player should react. Some beliefs may seem more reasonable than
others, however, and      { Entry Deterrence III}  has another equilibrium  that
requires less plausible beliefs off the equilibrium path.

\begin{quotation}
 \noindent
 {\bf An  implausible  equilibrium for Entry Deterrence III}\\
  Entrant:  {\it
Stay Out}$| Weak$,   {\it Stay Out}$| Strong$\\
 Incumbent:    {\it Fight}, 
Out-of-equilibrium beliefs: $Prob(Strong | Enter) = 0.1$ 
 \end{quotation}

  This is an equilibrium because if the entrant were to deviate and enter, the
incumbent would calculate his payoff from fighting to be 54 ($=0.1[0] +0.9[60]
$), which is greater than the {\it Collude} payoff of 50. The entrant would
therefore stay out.

 The beliefs in the implausible equilibrium are different and less reasonable
than  in the  plausible equilibrium.  Why should the incumbent believe that weak
entrants would enter mistakenly nine times as often as strong entrants?  The
beliefs do not violate Bayes' s Rule, but they have no justification.

  The reasonableness of the beliefs is important   because if the incumbent uses
passive conjectures, the implausible equilibrium breaks down. With passive
conjectures, the incumbent would want to change his strategy to {\it Collude},
because the expected payoff from {\it Fight} would be less than 50.  The
implausible equilibrium is less robust with respect to beliefs than the
plausible equilibrium, and it requires beliefs that are harder to justify.

  Even though dubious outcomes may be perfect bayesian equilibria, the concept
does have some bite, ruling out other dubious outcomes. There does not, for
example,  exist an equilibrium in which the entrant enters only if he is
$Strong$  and stays out  if he is $Weak$ (called a ``separating  equilibrium''
because it separates out different types of players). Such an equilibrium would
have to look like this:

\begin{quotation}
 \noindent
 {\bf A conjectured separating  equilibrium for Entry Deterrence III}
\\ 
Entrant:  {\it Stay Out}$| Weak$,   {\it Enter}$| Strong$\\
 Incumbent:     $
Collude $
 \end{quotation}

No out-of-equilibrium beliefs are specified for the conjectures in the
separating equilibrium   because there is no    out-of-equilibrium behavior
about which to specify them. Since the incumbent might observe either {\it Stay
out} or $Enter$ in equilibrium, the incumbent will always use Bayes' s Rule to
form his beliefs.  He will believe that an entrant who stays out  must be  weak
and an entrant who enters must be strong. This conforms to the idea behind Nash
equilibrium that each player assumes that the other follows the equilibrium
strategy, and then decides how to reply. Here, the incumbent's best response,
given his beliefs, is  $ Collude |Enter$, so that is the second part of the
proposed equilibrium. But this cannot be an equilibrium, because the entrant
would want to deviate. Knowing that entry would be followed by collusion, even
the weak entrant would enter. So there cannot be an equilibrium in which the
entrant enters only when strong.

\bigskip \noindent
 {\bf  The PhD Admissions Game}

\noindent
 Passive conjectures may not always be the most satisfactory belief, as the next
example shows.  Suppose that a university knows that 90 percent of the
population hate  economics  and   would be unhappy in its PhD program,  and 10
percent love economics  and would do well.  In addition, it   cannot observe the
applicant's type. If the university rejects an application, its payoff is 0 and
the applicant's is $-1$  because of  the trouble needed to apply.  If the
university accepts the application of someone who hates economics, the  payoffs
of both university and student are  $-10$, but if the applicant loves economics,
the payoffs are   +20 for each player.  Figure 2 shows this game   in extensive
form. The population proportions are represented by a node at which Nature
chooses the student to be a {\it Lover} or   {\it Hater} of economics.



 \includegraphics[width=150mm]{fig06-02.jpg}




\begin{center}
{\bf Figure 2: The    PhD Admissions Game} \end{center}

  The   {  PhD Admissions Game } is a signalling game of the kind we will look
at in Chapter 10. It has various perfect bayesian equilibria  that differ in
their  out-of-equilibrium beliefs, but   the equilibria   can be divided into
two distinct categories, depending on the outcome:     the {\bf separating
equilibrium}, in which the lovers of economics apply and the haters do not,
and the  {\bf pooling equilibrium},  in which neither type of student applies.

\begin{quotation}
 \noindent
 {\bf A separating equilibrium for the PhD Admissions Game}\\
 Student:  {\it Apply} $|${\it Lover}, {\it Do Not Apply} $|$ {\it Hater} \\
University: $Admit$ 
\end{quotation}

\noindent
 The separating equilibrium does not need to specify out-of-equilibrium beliefs,
because Bayes' s Rule can always be applied whenever both of the two possible
actions {\it Apply} and {\it Do Not Apply} can occur in equilibrium.

 \begin{quotation}
 \noindent
 {\bf A pooling equilibrium for the PhD Admissions Game}\\
Student:  {\it Do Not Apply} $|${\it Lover},  {\it Do Not Apply} $|${\it Hater}
\\
 University:  $Reject$,     Out-of-equilibrium beliefs:  $Prob$({\it Hater}
$|${\it Apply}) = 0.9
(passive conjectures)
 \end{quotation}

\noindent
 The pooling equilibrium is supported by passive conjectures.   Both types of
students   refrain from applying because they  believe correctly that they would
be rejected and receive a payoff of $-1$;   and the university is willing to
reject any student who foolishly applied, believing that he is a   {\it Hater}
with 90 percent probability.

  Because the perfect bayesian equilibrium concept imposes no restrictions on
out-of-equilibrium beliefs,  economists have come
up with a variety of  exotic refinements  of the equilibrium concept.   Let us
consider whether various alternatives to passive conjectures would support  the
pooling equilibrium in PhD Admissions.

\bigskip

 \noindent
 {\it Passive Conjectures.} $Prob$({\it Hater}$|${\it Apply}) = 0.9

\noindent This is the belief specified above, under which out-of-equilibrium
behavior leaves beliefs unchanged from the  prior. The argument for  passive
conjectures is that the student's application is a mistake, and that both types
are equally likely to make mistakes, although {\it Haters} are more common in
the population.  This supports the pooling equilibrium.

\bigskip

\noindent
 {\it The Intuitive Criterion.} $Prob$({\it Hater}$|${\it Apply}) = 0

  Under the Intuitive Criterion of Cho \& Kreps (1987), if there is a type of
informed player who could not benefit from the out-of-equilibrium action no
matter what beliefs were held by the uninformed player, the uninformed player's
belief must put zero probability on that type. Here, the {\it Hater} could not
benefit from applying under any possible beliefs of the university, so the
university puts zero probability on an applicant being a {\it Hater}. This
argument will not support the pooling equilibrium, because if the university
holds this belief, it will want to admit anyone who applies.

\noindent
 {\it Complete Robustness.} $Prob$({\it Hater}$|${\it Apply}) $=m$, $0 \leq m
\leq 1$

    Under this approach, the equilibrium strategy profile must consist of
responses that are best, given any and all out-of-equilibrium beliefs. Our
equilibrium for   {   Entry Deterrence II} satisfied this requirement. Complete
robustness rules out a pooling equilibrium in the PhD Admissions Game, because a
belief like $m=0$ makes accepting applicants a best response, in which case only
the {\it Lover} will apply. A useful first step in analyzing conjectured
pooling equilibria is to test whether they can be supported by   extreme beliefs
such as $m=0$ and $m=1$.

\bigskip

\noindent
 {\bf An Ad Hoc Specification.} $Prob$({\it Hater}$|${\it Apply}) = 1

    Sometimes the modeller can justify beliefs by the circumstances of the
particular game. Here, one could argue that anyone so foolish as to apply
knowing that the university would  reject them could not possibly have the good
taste to love economics. This supports the pooling equilibrium also.

   An alternative approach to the problem of out-of-equilibrium beliefs is to
remove its origin by building a model in which every outcome is possible in
equilibrium  because different types of players take different equilibrium
actions. In the PhD Admissions Game, we could assume that there are a  few
students who both love economics and actually enjoy writing applications.  Those
students would always apply in equilibrium, so there would never be a pure
pooling equilibrium in which nobody applied, and Bayes' s Rule could always be
used. In equilibrium, the university would always accept someone who applied,
because applying is never out-of-equilibrium behavior and it always indicates
that the applicant is a {\it Lover}. This approach is especially attractive if
the modeller   takes the possibility of trembles literally, instead of just
using it as a technical tool.

    The  arguments for different kinds of beliefs can also be applied to Entry
Deterrence III, which had two different pooling equilibria and no separating
equilibrium.  We used passive conjectures in the ``plausible'' equilibrium. The
intuitive criterion would not restrict beliefs at all, because   both types
would enter if the incumbent's beliefs were such as to make him collude, and
both would stay out if they made him fight.  Complete robustness would rule out
as an equilibrium the strategy profile  in which the entrant  stays out
regardless of type, because the optimality of staying out depends on the
beliefs. It would support the strategy profile in which the entrant enters and
out-of-equilibrium beliefs do not matter.

  \bigskip

 \noindent
  {\bf 6.3  The Importance of Common Knowledge: Entry Deterrence   IV and  V }

 \noindent
 To demonstrate the importance of common knowledge, let us consider two more
versions of Entry Deterrence.  We will use passive conjectures in both.  In   {
Entry Deterrence III}, the incumbent was hurt by his ignorance.   {   Entry
Deterrence IV}  will show how he can benefit from it, and   {   Entry Deterrence
V}  will show what can happen when the incumbent has the same information as the
entrant but the information is not common knowledge.

\noindent
 {\bf Entry Deterrence IV:  The Incumbent Benefits from Ignorance}

\noindent
 To construct   {   Entry Deterrence IV}, let $X = 300$ in Figure 1, so fighting
is even more profitable than in   {   Entry Deterrence III} but the game is
otherwise the same: the entrant knows his type, but the incumbent does not.  The
following is the unique perfect bayesian equilibrium in pure
strategies.\footnote{ There exists a plausible mixed-strategy equilibrium too:
{\it Entrant:  Enter  if Strong, Enter with probability $m = 0.2$ if Weak;
Incumbent: Collude with probability $n = 0.2$.}  The payoff from this is only
150, so if the equilibrium were one in mixed strategies, ignorance would $not$
help.}

  \begin{quotation}
 \noindent {\bf    Equilibrium for Entry Deterrence IV}\\
Entrant: {\it Stay Out} $|Weak$, {\it Stay Out} $|Strong$  \\
 Incumbent:  {\it Fight}, Out-of-equilibrium beliefs: 
   $Prob$({\it Strong}$|${\it Enter}) = 0.5  (passive conjectures)
\end{quotation}

 This equilibrium can be supported by other out-of-equilibrium beliefs, but no
equilibrium is possible in which the entrant enters.  There is no pooling
equilibrium in which both types of entrant enter, because then the incumbent's
expected payoff from {\it Fight} would be  150 ($=0.5[0] +0.5[300]$), which is
greater than the {\it Collude} payoff of 50. There is no separating equilibrium,
because if only the strong entrant entered and the incumbent always colluded,
the weak entrant would be tempted to imitate him and enter as well.

 In    {   Entry Deterrence IV}, unlike   {   Entry Deterrence III}, the
incumbent benefits from his own ignorance, because he would always fight entry,
even if the payoff were (unknown to himself) just zero. The entrant would very
much like to communicate the costliness of fighting, but the incumbent would not
believe him, so entry never occurs.

\bigskip

 \noindent
{\bf Entry Deterrence V: Lack of Common Knowledge of Ignorance}

\noindent
 In   {   Entry Deterrence V}, it  may happen that both the entrant and the
incumbent know the payoff from ({\it Enter, Fight}), but the entrant does not
know whether the incumbent knows. The  information is  known to both players,
but is not common knowledge.

 Figure 3 depicts this somewhat complicated situation. The game begins with
Nature assigning the entrant a type, {\it Strong} or {\it Weak} as before. This
is observed by the entrant but not by the incumbent. Next, Nature moves again
and either tells the incumbent the entrant's type or remains silent. This is
observed by the incumbent, but not by the entrant.   The four games starting at
nodes $G_1$ to $G_4$ represent different profiles of payoffs from ({\it Enter,
Fight}) and knowledge of the incumbent. The entrant does not know how well
informed the incumbent is, so the entrant's information partition is
$(\{G_1,G_2\}, \{G_3, G_4\})$.

 \includegraphics[width=150mm]{fig06-03.jpg}




 \begin{center}
 {\bf Figure 3   Entry Deterrence V  }
 \end{center}

 \begin{quotation}
\noindent
 {\bf Equilibrium for Entry Deterrence  V}\\
 Entrant:  {\it Stay Out}$|Weak$, {\it Stay Out}$|Strong$ \\
 Incumbent: {\it Fight}$|${\it  Nature said ``Weak''},  {\it Collude} $|${\it
Nature said ``Strong''},  {\it Fight} $|${\it  Nature said nothing}, Out-of-
equilibrium
beliefs:  $Prob$(  $Strong|${\it Enter, Nature said nothing)} = 0.5 (passive
conjectures)
  \end{quotation}

   Since the entrant puts a high probability on the incumbent not knowing, the
entrant should stay out, because the incumbent will fight for either of two
reasons. With probability 0.9, Nature has said nothing and the incumbent
calculates his expected payoff from {\it Fight} to be 150, and with probability
0.05 $(=0.1[0.5]$) Nature has told the incumbent that the entrant is weak and
the payoff from {\it Fight} is 300.  Even if the entrant is strong  and  Nature
tells this to the incumbent,   the entrant would choose {\it Stay Out}, because
he does not know that the incumbent knows, and his expected payoff from {\it
Enter} would be $-5$ ($= [0.9][-10] + 0.1[40]$).

If it were common knowledge that the entrant was strong, the entrant would enter
and the incumbent would collude. If it is known by both players, but not common
knowledge, the entrant stays out, even though the incumbent would collude if he
entered. Such is the importance of common knowledge.

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

\noindent
Chapter 5 explored various ways to steer between the Scylla of the Chainstore
Paradox and the Charybdis of the Folk Theorem to find a resolution to the
problem of repeated games.   In the end,  uncertainty turned out to make little
difference to the problem, but incomplete information was   left unexamined in
Chapter 5.   One might imagine that if the players did not know each others'
types,  the resulting confusion might allow cooperation.  Let us investigate
this by adding incomplete information  to the  finitely repeated    { Prisoner's
Dilemma} (whose payoffs are repeated in Table 1)   and  finding   the perfect
bayesian equilibria.

\begin{center} {\bf Table  1:   The  Prisoner's Dilemma   }
  \end{center}


 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Column}\\
&       &             &   {\it Silence}  &   & {\it Blame }     \\ &   &  {\it
Silence }      &     5,5 &    & -5,10 \\ & {\bf Row:} && & &   \\ &  &
{\it   Blame    }     &     10,-5  &    & {\bf 0,0} \\ \multicolumn{6}{l}{\it
Payoffs to: (Row,Column) }
 \end{tabular}

One way to incorporate incomplete information    would be  to assume that a
large number of players are irrational, but  that a given player does not know
whether any other player is of the irrational type or not.  In this vein, one
might assume that with high probability Row is a player who blindly follows the
strategy of Tit-for-Tat.  If Column thinks he is playing against a Tit-for-Tat
player, his optimal strategy is to {\it Silence} until near the last period (how
near depending on the parameters), and then {\it Blame}.  If he  were not
certain of this, but the probability were high that he faced a Tit-for-Tat
player, Row would choose that same  strategy. Such a model begs the question,
because it is not the incompleteness of the information that drives the model,
but the high probability that one player  blindly uses  Tit-for- Tat. Tit-for-
Tat is not a rational strategy, and to assume   that many players use it   is to
assume away the problem. A more  surprising result is  that a small  amount of
incomplete information can make a big difference to the
outcome.\footnote{Begging the question  is not as illegitimate in modelling as
in rhetoric, however, because it may  indicate that the question is a vacuous
one in the first place.  If the payoffs of the   { Prisoner's Dilemma}  are not
those of most of the people one is trying to model, the Chainstore Paradox
becomes irrelevant.}

\bigskip



 \noindent
 {\bf The Gang of Four Model}

\noindent
 One of the most important explanations of reputation is that of Kreps, Milgrom,
Roberts  \& Wilson (1982), hereafter referred to as the Gang of Four. In their
model, a few players are genuinely unable to play any strategy but Tit-for-Tat,
and many players pretend to be of that type.  The beauty of the model is that it
requires only a small amount of incomplete information, and a low probability
$\gamma$ that player Row is a Tit-for-Tat player.  It is not unreasonable to
suppose that the world  contains a few mildly irrational tit-for-tat players,
and such behavior is especially plausible among consumers, who are subject to
less evolutionary pressure than firms.



It  may even be   misleading to call  Tit-for-Tat ``irrational'', because they
may just have unusual payoffs, particularly  since we will assume that they are
rare. The unusual players have a small direct influence, but they matter because
other players imitate them.  Even if Column knows that with high probability Row
is just pretending to be a Tit-for-Tat player, Column does not care what the
truth is so long as Row keeps on pretending. Hypocrisy is not only the tribute
vice pays to virtue;  it can be just as good for  deterring misbehavior.

\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.

 The Gang of Four Theorem characterizes the equilibrium outcome rather than the
equilibrium. Finding perfect bayesian equilibria is  difficult and tedious,
since the modeller must check  all the out-of-equilibrium subgames, as well as
the equilibrium path. Modellers usually content themselves with describing
important characteristics of the equilibrium strategies and payoffs.

  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. Using Table 1's payoffs, 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).
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 ($=0.01
(10) + 0.99 (0)$). It is clearly in Column's advantage to take a chance by
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. This argument has  not described what
the Nash equilibrium actually  is, since the iteration back and forth between
Row and Column can be continued, but it does show why Column  chooses $Silence$
in the first period, which is the leverage the argument needs: the payoff is so
great if Row is actually the grim player  that it is worthwhile for Column to
risk a low payoff for one period.

 The Gang of Four Theorem provides a way out of the Chainstore Paradox, but it
creates a problem of multiple equilibria in much the same way as the infinitely
repeated game.  For one thing, if the asymmetry is two-sided, so   both players
might be  unusual types, it becomes much less clear what happens in threat games
such as Entry Deterrence.    Also, what happens depends on which unusual
behaviors have positive, if small, probability.  Theorem 6.2 says that    the
modeller can make the average payoffs take any particular values by making the
game last long enough and choosing the form of the irrationality carefully.

\noindent
{\bf Theorem 6.2: The Incomplete Information Folk Theorem}(Fudenberg \& Maskin
[1986] p. 547)\\
 {\it For any two-person repeated game without discounting, the modeller can
choose a form of irrationality so that for any probability $\epsilon>0$ there is
some finite number of repetitions such that with probability $(1-\epsilon)$ a
player is rational and the average payoffs in some sequential equilibrium are
closer than $\epsilon$ to any desired payoffs greater than the minimax payoffs.}

\vspace{1in}

 \noindent
{\bf 6.5  The Axelrod Tournament  }

\noindent
 Another way to approach the repeated   {  Prisoner's Dilemma}   is through
experiments, such as the round robin tournament described by  political
scientist Robert Axelrod   in his 1984 book. Contestants submitted strategies
for a 200-repetition    {  Prisoner's Dilemma} . Since the strategies could not
be updated during play, players could precommit, but the strategies could be as
complicated as they wished.  If a player wanted to specify a strategy which
simulated subgame perfectness by adapting to past history just as a noncommitted
player would, he was free to do so, but he could also submit a non-perfect
strategy such as Tit-for-Tat or the slightly more forgiving Tit-for-Two-Tats.
Strategies were submitted in the form of computer programs that were matched
with each other and played automatically.  In Axelrod's first tournament, 14
programs were submitted as entries.  Every program played every other program,
and the winner was the one with the greatest sum of payoffs over all the plays.
The winner   was Anatol Rapoport, whose strategy was Tit-for-Tat.

   The tournament helps to show which strategies are robust against a variety of
other strategies in a game with given parameters. It is quite different from
trying to find a Nash equilibrium, because it is not common knowledge what the
equilibrium is in such a tournament. The situation could be viewed as a game of
incomplete information in which Nature chooses the number and cognitive
abilities of the players  and their priors regarding each other.

     After the results of the first tournament were announced, Axelrod ran a
second tournament, adding a probability $\theta = 0.00346$ that the game would
end each round so as to avoid the Chainstore Paradox. The winner among the 62
entrants was again Anatol Rapoport, and again he used Tit-for-Tat.

Before choosing his tournament strategy,  Rapoport had written an entire book on
The   {  Prisoner's Dilemma}  in analysis, experiment, and simulation (Rapoport
\& Chammah [1965]). Why did he choose such a simple strategy as Tit-for-Tat?
Axelrod points out that 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}

Despite these advantages, care must be taken in interpreting the results of the
tournament. It does not follow that   Tit-for-Tat is the best strategy, or  that
cooperative behavior should always be expected in repeated games.

First,  Tit-for-Tat never beats any other strategy in a one-on-one contest. It
won the tournament by piling up points through cooperation, having lots of high
score plays and very few low score plays. In an elimination tournament, Tit-for-
Tat would be eliminated very early, because it scores {\it high} payoffs but
never the {\it highest} payoff.

 Second, the other players' strategies matter  to the success of Tit-for-Tat.
In neither tournament were the strategies submitted a Nash equilibrium. If a
player knew what strategies he was facing, he would want to revise his own. Some
of the strategies submitted in the second tournament would have won the first,
but they did poorly because the environment had changed. Other programs,
designed to try to probe the strategies of their opposition, wasted too many
$(Blame, Blame)$ episodes on the learning process, but if the games had lasted a
thousand repetitions they  would have done better.

  Third, in  a game in which players occasionally blamed because of  trembles,
two Tit-for-Tat players facing each other would do very badly.  The strategy
instantly punishes a blaming player, and it has no  provision for ending the
punishment phase.

   Optimality depends on the environment.  When information is complete and the
payoffs are all common knowledge,    blaming is the only equilibrium outcome. 
 In practically any real-world setting, however,    information is slightly
incomplete,
so cooperation becomes more plausible.  Tit-for-Tat is suboptimal for any given
environment, but it is robust across environments, and that is its advantage.

\vspace{1in}

 \noindent
 {\bf 6.6 Credit and the Age of the Firm: The Diamond Model}

\noindent An example of another way to look at reputation is Diamond's model of
credit terms, which seeks to explain why older firms get cheaper credit using a
game similar to the Gang of Four model. Telser (1966)   suggested that predatory
pricing would be a credible threat if the incumbent had access to cheaper credit
than the entrant, and so could hold out for more periods of losses before going
bankrupt.  While one might wonder whether this is effective protection against
entry--- what if the entrant is a large old firm from another industry?--- we
shall focus on how better-established firms might get cheaper credit.

   D. Diamond (1989) aims to explain why old firms are less likely than young
firms to default on debt. His model has both adverse selection, because firms
differ in type, and moral hazard, because they take hidden actions. The three
types of firms, R, S, and RS, are ``born'' at time zero and borrow to finance
projects at the start of each of $T$ periods.  We must imagine that there are
overlapping generations of firms, so that at any point in time a variety of ages
are coexisting, but the model looks at the lifecycle of only one generation.
All the players are risk neutral.  Type RS firms can choose independently risky
projects with negative expected values or safe projects with low but positive
expected values.  Although the risky projects are worse in expectation, if they
are successful the return is much higher than from safe projects. Type R firms
can only choose risky projects, and type S firms only safe projects.  At the end
of each period the projects bring in their profits and loans are repaid, after
which new loans and projects are chosen for the next period.  Lenders cannot
tell which project is chosen or what a firm's current profits are, but they can
seize the firm's assets if a loan is not repaid, which always happens if the
risky project was chosen and turned out unsuccessfully.

 This game foreshadows  two other models of credit that  will be described in
this  book,  the {   Repossession Game} of section 8.4 and the Stiglitz-Weiss
model of section 9.6.  Both will be one-shot games in which the bank worried
about not being repaid; in  the  { Repossession Game} because the borrower did
not exert enough effort, and in the Stiglitz-Weiss model because he was of an
undesirable type that could not repay. The Diamond model is a mixture of adverse
selection and moral hazard: the borrowers differ in type, but some borrowers
have a choice of action.

  The equilibrium path has three parts.  The RS firms start by choosing risky
projects. Their downside risk is limited by bankruptcy, but if the project is
successful the firm keeps large residual profits after repaying the loan. Over
time, the number of firms with access to the risky project (the RS's and R's)
diminishes through bankruptcy, while the number of S's remains unchanged.
Lenders can therefore maintain zero profits while lowering their interest rates.
When the interest rate falls, the value of a stream of safe investment profits
minus interest payments rises relative to the expected value of the few periods
of risky returns minus interest payments before bankruptcy.  After the interest
rate has fallen enough, the second phase of the game begins when the RS firms
switch to safe projects at a period we will call $t_1$. Only the tiny and
diminishing group of type R firms continue to choose risky projects. Since the
lenders know that the RS firms switch, the interest rate can fall sharply at
$t_1$. A firm that is older is less likely to be a type R, so it is charged a
lower interest rate.  Figure 4 shows the path of the interest rate over time.


 \includegraphics[width=150mm]{fig06-04.jpg}


\begin{center}
 {\bf Figure 4   The Interest Rate over Time }
  \end{center}


   Towards period $T$, the value of future profits from safe projects declines
and even with a low interest rate the RS's are again tempted to choose risky
projects. They do not all switch at once, however, unlike in period $t_1$. In
period $t_1$, if a few RS's had decided to switch to safe projects, the lenders
would have been willing to lower the interest rate, which would have made
switching even more attractive.  If a few firms switch to risky projects at some
time $t_2$, on the other hand, the interest rate rises and switching to risky
projects becomes more attractive--- a result that will also be seen in  the
Lemons model in Chapter 9. Between $t_2$ and $t_3$, the RS's follow a mixed
strategy, an increasing number of them choosing risky projects as time passes.
The increasing proportion of risky projects causes the interest rate to rise. At
$t_3$, the interest rate is high enough and the end of the game is close enough
that the RS's revert to the pure strategy of choosing risky projects. The
interest rate declines during this last phase as the number of RS's diminishes
because of failed risky projects.

 One might ask, in the spirit of  modelling by example,  why the  model contains
three types of firms rather than two. Types S and RS  are clearly needed, but
why type R? The little extra detail in the game description allows
simplification of the equilibrium, because with three types  bankruptcy is never
out-of-equilibrium behaviour,  since the failing firm might be a type R. Bayes's
Rule can therefore always be applied, elminating the  problem of ruling out
peculiar   beliefs and  absurd perfect bayesian equilibria.

This is a Gang of Four model  but   differs from previous examples in an
important respect:  the Diamond model  is not stationary, and as time
progresses, some firms of types R and RS go bankrupt, which changes the lenders'
payoff functions.  Thus, it is not, strictly speaking, a repeated game.

\newpage

\begin{small}

\bigskip \noindent
 {\bf Notes}

\noindent
{\bf N6.1} {\bf  Perfect Bayesian Equilibrium: Entry Deterrence  I and II}
\begin{itemize}

 \item Section 4.1 showed that even in games of  perfect information, not every
subgame perfect equilibrium is trembling-hand perfect. In games of perfect
information, however, every subgame perfect equilibrium is a perfect bayesian
equilibrium, since no out-of-equilibrium beliefs need to be specified.

 \end{itemize}

\noindent
 {\bf N6.2} {\bf Refining Perfect Bayesian Equilibrium: The PhD Admissions Game}
 \begin{itemize}
\item Fudenberg \& Tirole (1991b) is a careful analysis of the issues involved
in defining perfect bayesian equilibrium.

\item
 Section 6.2  is about debatable ways of restricting beliefs such as passive
conjectures or equilibrium dominance, but   less controversial restrictions
are sometimes useful. In a three-player game, consider what happens when   Smith
and Jones have incomplete information about Brown,  and then Jones  deviates. If
it was Brown himself who had deviated, one might think that the other players
might deduce something about Brown's type. But should they update their priors
on Brown because Jones has deviated? Especially, should Jones updated  his
beliefs, just because he himself deviated?  Passive conjectures seems   much
more reasonable.

    If, to take a second possibility,  Brown himself does deviate, is it
reasonable for the out-of-equilibrium beliefs to specify that    Smith and Jones
update their beliefs about Brown in different ways?  This seems dubious in light
of the Harsanyi doctrine that everyone begins with the same priors.

On the other hand, consider a tremble interpretation of out-of- equilibrium
moves. Maybe if Jones trembles and picks the wrong strategy, that really does
say something about Brown's type. Jones might tremble more often, for example,
if Brown's type is strong than if it is weak. Jones  himself might learn from
his own trembles. Once we are in the realm of non-bayesian beliefs, it is hard
to know what to do without a real-world context.


\item[]
  Dominance and tremble arguments  used to rule out Nash  equilibria apply to
past,
present (in simultaneous move games), and future actions of
the other player. Belief arguments only   depend on  past
actions, because they rely on the uninformed player
observing behavior and interpreting it. Thus, for example, a tremble or weak
dominance argument might say a player should take action 1 instead of 2 because
although their payoffs are equal, action 2   would  lead to  a very low payoff
if the other player later trembled and chose an unintended action that hurt both
of them. An argument based on beliefs  would not work in such a game.



 \item For discussions of the appropriateness of different equilibrium concepts
in actual economic models see Rubinstein (1985b) on bargaining, Shleifer \&
Vishny (1986) on greenmail and D. Hirshleifer \& Titman (1990) on tender offers.

\item
{\bf Exotic refinements.} Binmore (1990) and Kreps (1990b) are booklength
treatments of rationality and equilibrium concepts. Van Damme (1989) introduces the curious ``money burning'' idea of ``forward induction.''


 \item {\bf  The Beer-Quiche
Game}  of Cho \& Kreps (1987). To illustrate their ``intuitive criterion'', Cho
and Kreps use  the  Beer-Quiche Game. In this game, Player I might be either
weak or strong in his duelling ability, but he wishes   to avoid a duel even if
he thinks he can win. Player II wishes to fight a duel only if player I is weak,
which has a probability of 0.1. Player II  does not know  player I's type,  but
he observes what  player I has for breakfast. He  knows that weak players prefer
quiche for breakast, while strong players prefer beer.   The payoffs are   shown
in Figure 5.

Figure 5 illustrates a few twists on how to draw an extensive form. It begins
with Nature's choice of $Strong$ or $Weak$ in the middle of the diagram. Player
I then chooses whether to breakfast on $beer$ or $quiche$. Player II's nodes are
connected by a dotted line if they are in the same information set. Player II
chooses $Duel$ or $Don't$, and payoffs are then received.



 \includegraphics[width=150mm]{fig06-05.jpg}

\begin{center}
{\bf Figure 5   The Beer-Quiche Game}
 \end{center}

This game has two perfect bayesian equilibrium outcomes, both of which are
pooling. In $E_1$, player I has beer for breakfast regardless of type, and
Player II chooses not to duel.  This is supported by the out-of-equilibrium
belief  that a quiche-eating player I is weak with probability over 0.5, in
which case player II would choose to duel on observing quiche. In $E_2$, player
I has quiche for breakfast regardless of type, and player II chooses not to
duel. This is supported by the out-of-equilibrium belief  that a beer-drinking
player I is weak with probability greater than 0.5, in which case player II
would choose to duel on observing  beer.

 Passive conjectures and the intuitive criterion both   rule out equilibrium
$E_2$. According to  the  reasoning of the intuitive criterion, player I  could
deviate without fear of a duel by  giving the following convincing speech,
\begin{quotation}

 I am having beer for breakfast, which ought to convince you I am strong.  The
only conceivable benefit to me of breakfasting on beer  comes if I am strong.
I would never wish to have beer for breakfast if I were weak, but  if I am
strong  and  this message  is convincing, then  I benefit from having beer for
breakfast.

 \end{quotation}

  \end{itemize}


\bigskip \noindent
{\bf N6.5} {\bf The Axelrod tournament}
 \begin{itemize}
  \item Hofstadter (1983) is a nice discussion of  the    {  Prisoner's Dilemma}
and the Axelrod tournament by an intelligent computer scientist who came to the
subject untouched by the preconceptions or training of economics. It is useful
for elementary economics classes. Axelrod's 1984 book provides a fuller
treatment. \end{itemize}



\newpage

\noindent {\bf Problems}

\noindent {\bf 6.1.  Cournot Duopoly under Incomplete Information about Costs }
(hard)
\\
  This problem introduces incomplete information into the  Cournot model of
Chapter 3 and allows for a continuum of player types.
  \begin{enumerate}
  \item[(a)] Modify  the   Cournot Game  of Chapter 3 by specifying that Apex's
average cost of production be $c$ per unit, while Brydox's remains  zero. What
are the outputs of each firm if the costs are common knowledge?  What are the
numerical values if  $c=10$?

  \item[(b)]
 Let  Apex's  cost $c$    be  $c_{max}$ with probability $\theta$ and  0 with
probability $1-\theta$, so Apex is one of two types. Brydox does not know Apex's
type.      What are the outputs of each firm?

  \item[(c)]
  Let  Apex's  cost $c$    be  drawn from the  interval $[0,c_{max}]$  using the
uniform distribution, so there is a continuum of types. Brydox does not know
Apex's type.     What are the outputs of each firm?

  \item[(d)]
  Outputs were  40 for each firm in the zero-cost game in chapter 3.  Check your
answers in parts (b) and  (c) by seeing what happens if $c_{max}=0$.

  \item[(e)]
  Let   $c_{max}=20$ and $\theta =0.5$, so the expectation of  Apex's  average
cost is 10 in parts (a), (b), and (c). What are the average outputs for Apex in
each case?

  \item[(f)]
  Modify  the model of part (b) so that   $c_{max}=20$ and $\theta=0.5$, but
somehow $c= 30$. What outputs do your formulas from part (b) generate?  Is there
anything this could sensibly model? \end{enumerate}

\bigskip
 \noindent
 \textbf{Problem 6.2. Limit Pricing}  (medium)  (see Milgrom and Roberts
[1982a])\\
An incumbent firm operates in the local computer market, which is a natural
monopoly in which only one firm can survive. The incumbent knows  his own
operating cost $c$, which is 20 with probability 0.2 and 30 with probability
0.8.

In the first period,  the incumbent can price $Low$, losing 40 in profits, or
$High$, losing nothing if  his cost is $c=20$.  If  his  cost is $c=30$,
however, then pricing $Low$  he loses   180  in profits. (You might imagine that
all  consumers have a reservation price that is $High$, so a static monopolist
would choose that price whether marginal cost was 20 or 30.)

  A potential entrant knows those probabilities, but not the incumbent's exact
cost. In the second period, the entrant can enter at a cost of 70, and  his
operating cost of 25 is common knowledge. If there are two firms in the market,
each incurs an immediate loss of 50, but one then drops out and the survivor
earns the monopoly revenue of 200 and pays  his operating cost. There is no
discounting: $r=0$.

\begin{enumerate}

\item[(a)]
In a perfect bayesian equilibrium in which the incumbent prices $High$
regardless of its costs (a pooling equilibrium), about what do out-of-
equilibrium beliefs have to be specified?

\item[(b)]
Find  a  pooling perfect bayesian  equilibrium,  in which the incumbent  always
chooses the same price no matter what his costs may be.

\item[(c)]
What is a set of out-of-equilibrium beliefs that do not support a pooling
equilibrium at a $High$ price?

\item[(d)]
 What is a separating equilibrium for this game?

 \end{enumerate}

\bigskip
\noindent
{\bf 6.3. Symmetric Information and Prior Beliefs  } (medium) \\
  In   the  Expensive-Talk Game of Table   2, the   Battle of the Sexes  is
preceded by by a communication move in which the man chooses $Silence$ or
$Talk$. $Talk$ costs 1 payoff unit, and consists of a declaration by the man
that he is going to the prize fight.  This declaration is just talk; it is not
binding on him.
 \begin{center}
 {\bf Table 2:  Subgame Payoffs in the Expensive-Talk Game }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Woman}\\
&       &             &  {\it  Fight}   &   &  $ Ballet$      \\ &   &  $ Fight$
&     3,1 &    &  $0,0$ \\ & {\bf Man:} && & & \\ & & {\it Ballet } & $0,0$ & &
1,3 \\ \multicolumn{6}{l}{\it Payoffs to: (Man, Woman) } \end{tabular}

\end{center}
 \begin{enumerate}
 \item[(a)]
Draw the extensive form for this game, putting the man's move first in the
simultaneous-move subgame.

       \item[(b)]
What are the strategy sets for the game? (Start with the woman's.)

   \item[(c)]
What are the three perfect pure-strategy equilibrium outcomes in terms of
observed actions? (Remember: strategies are not the same thing as outcomes.)

   \item[(d)] Describe the equilibrium strategies for a perfect equilibrium in
which the man chooses to talk.

     \item[(e)]The idea of ``forward induction'' says that an equilibrium should
remain an equilibrium even if strategies dominated in that equilibrium are
removed from the game and the procedure is iterated.  Show that this procedure
rules out SBB as an equilibrium outcome.(See Van Damme [1989]. In fact, this
procedure rules out TFF {\it (Talk, Fight, Fight)} also.)

  \end{enumerate}

 %---------------------------------------------------------------

        \bigskip
 \noindent
 {\bf  6.4. Lack of Common Knowledge } (medium)\\
 This problem looks at what happens if the parameter values in Entry Deterrence
V   are changed.
 \begin{enumerate}
 \item[(a)]
  Why does $Pr(Strong|Enter, Nature \;said\; nothing) =0.95$ not support the
equilibrium in Section 6.3?


   \item[(b)]
 Why is the equilibrium in Section 6.3 not an equilibrium if 0.7 is the
probability that Nature tells the incumbent?

   \item[(c)]
  Describe the equilibrium if 0.7 is the probability that Nature tells the
incumbent. For what out-of-equilibrium beliefs does this remain the equilibrium?

  \end{enumerate}

   %---------------------------------------------------------------

 \newpage

 \begin{center}
 {\bf  The Repeated Prisoner's Dilemma under Incomplete Information: A Classroom
Game for  Chapter 6}
 \end{center}



  
 Consider the  Prisoner's Dilemma in Table 3, obtained by adding 8 to each
payoff in   Chapter 1's Table 2, and identical to Chapter 5's Table 10:
 
\begin{center} 
 {\bf Table  3:  The Prisoner's Dilemma}

 \begin{tabular}{lllccc}
   &       &             &\multicolumn{3}{c}{\bf Column}\\ 
&       &             &   {\it Silence}  &   & {\it Blame}     \\
 &   &  {\it Silence}      &     7,7 & $\rightarrow$  & -2, 8 \\ 
& {\bf Row} &&$\downarrow$& & $\downarrow$ \\
 &  &       {\it   Blame}     &      8,-2  & $\rightarrow$  & {\bf 0,0} \\ 
\multicolumn{6}{l}{\it Payoffs to: (Row,Column) } 
\end{tabular}
 \end{center}

This game will be repeated five times, and your
  objective is to get as high a summed, undiscounted, payoff as possible ({\it
not } just to get a higher summed payoff than anybody else).  Remember, too,
that  there are lots of pairing of Row and Column in the class, so to just beat
your immediate opponent would not even be the right tournament strategy.


  The instructor will form groups of three students each to represent $Row$, and
groups of one student each to represent $Column$. Each $Row$ group will play
against multiple $Columns$.

 The five-repetition games  will be  different in how $Column$ behaves. 

\noindent
Game (i) Complete Information:  Column will   seek to maximize his payoff
according to Table 3.

\noindent
Game (ii) 80\% Tit-for-Tat: With 20\% probability, Column will  seek to maximize
his payoff according to Table 3.  With 80\% probability, Column is a ``Tit-for-
Tat Player'' and must use the strategy of ``Tit-for-Tat,'' starting with
$Silence$ in Round 1 and after that imitating what Row did in the previous
round.

\noindent
Game (iii) 10\% Tit-for-Tat: With 90\% probability, Column will   seek to
maximize his payoff according to Table 3.  With 10\% probability, Column  is a
``Tit-for-Tat Player'' and must use the strategy of ``Tit-for-Tat,'' starting
with $Silence$ in Round 1 and after that imitating what Row did in the previous
round.  The identities of the Game (ii).

 The probabilities are independent, so although in Game(ii)   the most likely
outcome is that 8 of 10  Column players use tit-for-tat, it is possible that 7
or 9 do, or even (improbably)   0 or 10.






\end{small}

\end{document}



