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\noindent 
February 24, 2014,  Eric Rasmusen, Erasmuse@indiana.edu.
Http://www.rasmusen.org.



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

   \begin{LARGE}
{\bf 

 \includegraphics[width=6in]{fig06-01.jpg}
 
 

\noindent
 {\bf Entry Deterrence II: Fighting Is Never Profitable: X=1}

 
Subgame perfectness does not rule out any Nash
equilibria. The only subgame is the entire game. 

\newpage

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

\noindent
  Trembling-hand perfectness --- Selten
(1975) says     a  strategy that is to   be part of an equilibrium
must   be optimal for the player even if there is a   small chance
  that the
other player's hand will ``tremble'' :

 .

 \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 $s^*$.}

 This is hard to use, and undefined when games have 
  continuous strategy
spaces because it is hard to work with mixtures of a continuum).
 
\newpage

\bigskip \noindent
  {\bf Perfect Bayesian Equilibrium and Sequential Equilibrium (Kreps \&
Wilson (1982))}

 
 The
profile of beliefs and strategies is called an {\bf assessment}.

   On the equilibrium path, all that the players need to update their beliefs
are their priors and Bayes' s Rule. Off the equilibrium path,  this is not
enough.  Suppose that in equilibrium, the entrant always enters. If   the entrant stays out, what is the
incumbent to think about the probability   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.   
  \begin{equation}\label{e6.1}\hspace{-1in} 
  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  this 
requires dividing by zero.

 \newpage

 \noindent {  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 {  (1) The strategies for the remainder of the game are Nash given
the beliefs and strategies of the other players.}\\
 { (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  to restrict beliefs   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:}
$$\hspace{-1in} 
 (\mu^*,s^*) = Lim_{n \rightarrow  \infty} (\mu^n,s^n) \;for\; some\;sequence \;
(\mu^n,s^n) \; in\; \; \{\mu, s \}. $$

  
 \newpage

 
 \noindent
 {\bf Back to  Entry Deterrence II}

A PBE  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}

 
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.

  
\newpage

  


 \includegraphics[width=6in]{fig06-01.jpg}


\noindent
 In   {   Entry Deterrence III},  assume  $X=60$, not $X=1$. Fighting is  now more profitable for the incumbent   than collusion  if
the entrant is $Weak$. 

 The first  equilibrium  we'll examine uses 

{\it passive conjectures}--- ``posterior equals prior'' 

  for out-of-equilibrium beliefs, but could use ANY beliefs---  it is completely robust. 

 \newpage 
 
\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.

 

\newpage

\includegraphics[width=6in]{fig06-01.jpg}


\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}

  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.

   
 
\newpage
\includegraphics[width=6in]{fig06-01.jpg}


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

 This turns out not to be an equilibrium. 

\newpage
\begin{Large}
  \noindent
 {\bf  A Mixed-Strategy  Equilibrium for Entry Deterrence III}

The prior for the probability that the entrant is strong is .5. 
 In this game, the weak and the strong entrant both get the same payoff from entering. The strong entrant is strong only in the sense that the incumbent doesn't want to fight him. 

 Let the probability that the incumbent colludes be $\alpha$. 
 $$
\pi_(enter) = \alpha (40) + (1-\alpha) (-10) = \pi_(stay ;out) =0
$$
 Thus, $\alpha = .2$.   

Let $\theta$ be the posterior  probability that  an    entrant who enters  is Strong. 
$$
\pi_(fight) = \theta (0) + (1-\theta) (60) = \pi_(collude) =50
$$
Thus,  $\theta = 1/6$. 

 Let $\beta_s$ and $\beta_w$ be the probabilities with which the  strong and weak entrants enter. 
We need  
$$
\theta = \frac{1}{6} = \frac{.5 \cdot \beta_s}{.5 \cdot \beta_w}
$$
 There are lots of values which satisfy this condition, e.g.  $\beta_s= 1/6, \beta _w=1$ or $\beta_s= 1/12, \beta _w=1/2$ or $\beta_s= 1/10, \beta _w=6/10$.

 The weak entrant is more likely to enter!   The reason is that if the strong entrant were to enter with greater probability, the incumbent would want to $Collude$. 

\end{Large}
 \newpage

  \noindent
 {\bf  The PhD Admissions Game: A Separating    Equilibrium}
 

 \includegraphics[width=6in]{fig06-02.jpg}


 
  
\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}

 
 
 \newpage



 \includegraphics[width=6in]{fig06-02.jpg}

 

 \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}



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

  This supports the pooling equilibrium.

\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. 

\newpage
 
\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.

 

\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.



\newpage
  
\noindent
  {\bf  The Beer-Quiche
Game}  of Cho \& Kreps (1987).   Player I is
weak or strong and doesn't want to duel.  Player II wants to duel only if player I is weak.
  Player II  does not know  player I's type,  but
he observes what  player I has for breakfast. Weak players prefer
quiche for breakast,   strong players prefer beer.   

 

 \includegraphics[width=6in]{fig06-05.jpg}
 
\noindent
  $E_1$:  Player I has beer.   
Player II doesn't duel if beer, does duel if quiche.  Out-of-equilibrium
belief:    a quiche-eating player I is weak with probability over 0.5.

\noindent
 $E_2$: Player
I has quiche.  Player II  
duel if beer  doesn't duel if quiche.  Out-of-equilibrium
belief:  a beer-drinking
player I is weak with probability over 0.5.

\newpage

 \includegraphics[width=6in]{fig06-05.jpg}
  
\noindent
 $E_2$: Player
I has quiche.  Player II  
duel if beer  doesn't duel if quiche.  Out-of-equilibrium
belief:  a beer-drinking
player I is weak with probability over 0.5.

Intuitive Criterion:  player I  could
deviate to BEER  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}

  
 
 \newpage
 \includegraphics[width=6in]{fig06-01.jpg}
 
\noindent
 {\bf Entry Deterrence IV:  The Incumbent Benefits from His Own  Ignorance}

\noindent
 Let $X = 300$.  The entrant knows his type, but the incumbent does not. 

 \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}

\newpage
  \includegraphics[width=4in]{fig06-01.jpg}
 
    There is no pure-strategy 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.   Nor is there a pure-strategy separating equilibrium. 

 There exists a   mixed-strategy equilibrium too.

\newpage

 \noindent
{\bf Entry Deterrence V: Lack of Common Knowledge of Ignorance}:
Both the entrant and the
incumbent know the payoff from ({\it Enter, Fight}), but the entrant does not
know whether the incumbent knows.  
 

 \includegraphics[width=6in]{fig06-03.jpg}


 
 \begin{quotation}
 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}

\newpage 
 \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}

  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 $-5$ ($= [0.9][-10] + 0.1[40]$).

 
\newpage

 \noindent
{\bf    Kreps, Milgrom, Roberts, Wilson (1982) : The Gang
of Four Model}
 
One way to incorporate incomplete information    would be  to  assume that with 30\% 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  {\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. 

 But it turns out that even a small probability of a Tit-for-Tat player  can make a big difference. 
 

\newpage
  
 


 
 \noindent
 {\bf Theorem 6.1: The Gang of Four Theorem}\\
{  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.}

 In equilibrium, $Blame$ is played in the periods near T.  Before that there is a period of mixing, and before that they play $Silence$. 

 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  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

\noindent
 {\bf Gang of Four Intuition}

 \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}

Consider what would happen
in a  10,001-period  PD  with a   probability of 0.01 that Row is playing the
Grim Strategy.   \\
 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 $Blame$). \\
   Column's payoff will be     50,010 (= (10,000)(5) + 10).  {\it Blame Always} would just yield 10 as a payoff. 

Suppose instead that if Row is not Grim, he  
will choose $Blame$   every period.  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   is 495.15. 

If instead  Row  follows a
strategy of ($Blame$  every period), his expected payoff is just 0.1 ($=0.01
(10) + 0.99 (0)$). 
 

\newpage 

 \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}

The aggressive strategy  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   $Blame$. Row's payoff would rise from 10 to (9,999)(5)   + 10. 

 Given that Column is cooperating in the
early periods, Row will cooperate also. Still not Nash, but we see why
  Column  chooses $Silence$
in the first period. 



 \newpage
\noindent
{\bf Theorem 6.2: The Incomplete Information Folk Theorem}(Fudenberg \& Maskin
[1986] p. 547) 

 { 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.}

 
\newpage

 THE AXELROD TOURNAMENT: 
 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.  

Strategies were submitted in the form of computer programs.  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.

   What strategy could have beat Rapoport and all the others? 

     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 with Tit-for-Tat.

\newpage

Before choosing his tournament strategy,  Rapoport had written an entire book on
The   {  Prisoner's Dilemma}  in analysis, experiment, and simulation.

  Why did he choose such a simple strategy as 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}

 
 Tit-for-Tat never beats any other strategy in a one-on-one contest.   In an elimination tournament, Tit-for-
Tat would be eliminated  early, because it scores {\it high} payoffs but
never the {\it highest} payoff.

In  a game in which players occasionally blamed because of  trembles,
two Tit-for-Tat players facing each other would do very badly.  

   
 \newpage

 \noindent
 {\bf  ``Reputation Acquisition in Debt Markets'' JPE, 1989 }

 

   Douglas Diamond (1989)   explains  why old firms are less likely than young
firms to default on debt.  The three
types of risk-neutral firms, R, S, and RS, are ``born'' at time zero and borrow to finance
projects at the start of each of $T$ periods.  

 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
see which project is chosen or   a firm's current profits, 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.


\newpage

  \includegraphics[width=6in]{fig06-04.jpg}

  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   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. 


\newpage

 \includegraphics[width=6in]{fig06-04.jpg}

 After the interest
rate has fallen enough, the second phase of the game begins when the RS firms
switch to safe projects, at   $t_1$. 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   falls  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.   



 
\newpage
 \includegraphics[width=4in]{fig06-04.jpg}

 

   Towards   $T$, the value of future profits from safe projects declines
and  the RS's are again tempted to choose risky
projects.


 Between $t_2$ and $t_3$, the RS's follow a mixed
strategy, an increasing number  choosing risky projects. The interest rate rises as a result. 

  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 then falls as the number of RS's diminishes
because of failed risky projects.

\newpage

Why  
three types of firms, not 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 a repeated game.

 }
 \end{LARGE}
\end{document}



