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

\begin{LARGE} \begin{center}
{\bf  3 Mixed and Continuous Strategies}
 \end{center}
\end{LARGE}

  \bigskip

 \noindent
 {\bf  3.1 Mixed Strategies: The { Welfare Game} }

\noindent
The games we have looked at have so far  been simple in at least one respect:
the number of moves in the action set has been finite. In this chapter we allow
a continuum of moves, such as when a player chooses a price between 10 and 20 or
a purchase probability between 0 and 1.   Chapter 3 begins by showing how to
find mixed-strategy equilibria  for a game with no pure-strategy equilibria. In
Section 3.2 the mixed-strategy equilibria are found by the payoff-equating
method, and mixed strategies are applied to  two  dynamic games, the War of
Attrition and Patent Race for a New Market.   Section 3.3   takes a more general
look at mixed strategy equilibria and extends the analysis to three or more
players. Section 3.4  distinguishes   between mixed strategies and random
actions in  the importance class of ``auditing games''. Section 3.5  switches
from the continuous strategy spaces of mixed strategies to  strategy spaces that
are continuous  even with pure strategies, using the   Cournot duopoly model, in
which   two firms choose output on the continuum between zero and infinity.
Section 3.6 looks at the Bertrand model and strategic substitutes. Section 3.7
switches gears a bit and talks about four reasons why a Nash equilibrium might
not exist.
These last sections introduce   a number of   ideas besides  simply how to find
equilibria, ideas   that will be  built upon in later chapters--- dynamic games
in Chapter 4, auditing and agency in Chapters 7 and 8, and Cournot oligopoly in
Chapter 14.

\bigskip

         We invoked the concept of Nash equilibrium to provide predictions of
outcomes without  dominant strategies, but some games lack even a Nash
equilibrium.  It is often useful and realistic to expand the strategy space to
include random strategies, in which case a Nash equilibrium  almost always
exists. These random strategies are called ``mixed strategies.''

\noindent
{\it A {\bf pure strategy} maps each of a player's possible information sets to
one action. $s_i: \omega_{i} \rightarrow a_i$.}

\noindent
{\it A {\bf mixed strategy} maps each of a player's possible information sets to
a probability distribution over actions.} $$ s_i: \omega_i \rightarrow m (a_i),
\;\; where \;\; m \geq 0  \;\;{\rm and} \;\; \int_{A_i} m(a_i) da_i = 1. $$

\noindent
 {\it A {\bf completely mixed} strategy puts positive probability on every
action, so $m > 0$.}

\noindent
{\it The version of a game expanded to allow mixed strategies is called the {\bf
mixed extension} of the game.}

  A pure strategy constitutes a rule that tells the player what action to
choose, while a mixed strategy constitutes a rule that tells him what dice to
throw in order to choose an action.  If a player pursues a mixed strategy, he
might choose any of several different actions in a given situation, an
unpredictability which can be helpful to him.  Mixed strategies occur frequently
in the real world. In American football games, for example, the offensive team
has to decide whether to pass or to run. Passing generally gains more yards, but
what is most important is to choose an action not expected by the other team.
Teams decide to run part of the time and pass part of the time in a way that
seems random to observers but rational to game theorists.

\bigskip

\noindent
 {\bf The { Welfare Game} }

\noindent
The { Welfare Game}  models a government that wishes to aid a pauper if he
searches for work but not otherwise, and a pauper who searches for work only if
he cannot depend on government aid.

 Table 1 shows payoffs which represent the situation.``Work'' represents trying
to find work, and ``Loaf'' represents not trying.  The government wishes to help
a pauper who is trying to find work, but not one who does not try. Neither
player has a dominant strategy, and with a little thought we can see that no
Nash equilibrium exists in pure strategies either.

 \begin{center} {\bf Table  1:  The { Welfare Game}  }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Pauper}\\
&       &             &  {\it  Work} ($\gamma_w$)  &   &  $ Loaf$ ($1-\gamma_w$)
\\ &   &  $ Aid$   ($\theta_a$)    &     3,2 & $\rightarrow$  &  $-1,3$ \\ &
{\bf Government } &&$\uparrow$& & $\downarrow$ \\ &  &    {\it No Aid }
($1-\theta_a$)    &      $-1,1$  & $\leftarrow$ & 0,0 \\
   \end{tabular} \end{center}
\vspace{-24pt}
 {\it Payoffs  to:  (Government, Pauper). Arrows show how a player can increase
his payoff.   }

\bigskip

 Each strategy profile must be examined in turn to check for Nash equilibria.

\noindent
 \hspace*{16pt}  1  The strategy profile ({\it Aid, Work}) is not a Nash
equilibrium, because the pauper would respond with {\it Loaf} if the government
picked {\it Aid.}\\
 \hspace*{16pt}  2 ({\it Aid, Loaf}) is not Nash, because the government would
switch to {\it No Aid}.\\ \hspace*{16pt}  3   ({\it No Aid, Loaf}) is not Nash,
because the pauper would switch to {\it Work}.\\
 \hspace*{16pt}  4  ({\it No Aid, Work}) is not Nash, because the government
would switch to {\it Aid}, which brings us back to  (1).

         The   Welfare Game  does have a mixed-strategy Nash equilibrium, which
we can calculate. The players' payoffs are the expected values of the payments
from Table   1.  If the government plays {\it Aid} with probability $\theta_a$
and the pauper plays {\it Work} with probability $\gamma_w$, the government's
expected payoff is \begin{equation} \label{e3.1} \begin{array}{ll}
\pi_{Government} &= \theta_a[3\gamma_w + (-1)(1-\gamma_w)]+[1- \theta_a]
[-1\gamma_w+0(1-\gamma_w)] \\ & \\ &= \theta_a[3\gamma_w - 1 + \gamma_w] -
\gamma_w + \theta_a\gamma_w \\ & \\ &= \theta_a[5\gamma_w-1] - \gamma_w.\\
\end{array} \end{equation}

  If only pure strategies are allowed, $\theta_a$ equals zero or one, but in the
mixed extension of the game, the government's action of $\theta_a$ lies on the
continuum from zero to one, the pure strategies being the extreme values.   If
we followed  the usual procedure for solving a maximization problem, we would
differentiate the payoff function with respect to the choice variable to obtain
the first-order condition. That procedure is actually not the best way to find
mixed- strategy equilibria, which is the ``payoff-equating method'' I will
describe  in the next section. Let us use the maximization approach here,
though, because it is good for helping you understand how mixed strategies work.
The first-order condition for the government would be
   \begin{equation} \label{e3.2}
 \begin{array}{ll}
 &0 = \frac{d
\pi_{Government} }{d\theta_a\;\;\;}  = 5\gamma_w - 1 \\ & \\ \Rightarrow &
\gamma_w = 0.2.
 \end{array} 
\end{equation}

 In the mixed-strategy equilibrium, the pauper selects {\it Work} 20 percent of
the time. This is a bit strange, though: we obtained the pauper's strategy by
differentiating  the government's payoff! That is because we have not used
maximization in the standard way. The problem has a corner solution, because
depending on the pauper's strategy, one of three strategies maximizes the
government's payoff: (i) Do not aid ($\theta_a =0$) if the pauper is unlikely
enough to try to work; (ii) Definitely aid ($\theta_a =1$) if the pauper is
likely enough to try to work;(iii) any probability of aid, if the government is
indifferent because the pauper's probability of work is right on the border line
of  $\gamma_w = 0.2$.

 It is possibility (iii) which allows a mixed strategy equilibrium to exist.  To
see this, go through the following 4 steps:

\begin{quotation} 
\noindent
 \hspace*{16pt} 1  I assert that an optimal mixed
strategy exists for the government.

\noindent
\hspace*{16pt}  2   If the pauper selects {\it Work} more than 20 percent of the
time, the government always selects {\it Aid}. If the pauper selects {\it Work}
less than 20 percent of the time, the government never selects {\it Aid}.

\noindent
 \hspace*{16pt}  3 If a mixed strategy is to be optimal for the government, the
pauper must therefore select {\it Work} with probability exactly 20 percent.
\end{quotation}

  To obtain the probability of the government choosing $Aid$, we must turn to
the pauper's payoff function, which is
  \begin{equation} \label{e3.3}
 \begin{array}{ll}
 \pi_{Pauper} &= \gamma_w(2 \theta_a +1 [1-\theta_a]) + (1-\gamma_w) (3\theta_a+
[0][1-\theta_a])\\
 & \\
 &= 2\gamma_w\theta_a + \gamma_w - \gamma_w\theta_a +3\theta_a -3\gamma_w
\theta_a \\
 & \\
 &= -\gamma_w(2\theta_a -1) + 3\theta_a.\\
\end{array}
 \end{equation}
 The first-order condition is 
\begin{equation} \label{e3.4}
 \begin{array}{l}
\frac{d \pi_{Pauper}}{d\gamma_w\;\;\;\;\;\;\;\;\;\;\; } = -(2\theta_a - 1) =
0,\\ 
 \\
  \Rightarrow \theta_a = 1/2.\\
 \end{array} 
\end{equation}

   If the pauper selects {\it Work} with probability 0.2, the government is
indifferent among selecting {\it Aid} with probability 100 percent, 0 percent,
or anything in between. If the strategies are to form a Nash equilibrium,
however, the government must choose $\theta_a =0.5$.  In the mixed-strategy Nash
equilibrium, the government selects {\it Aid} with probability 0.5 and the
pauper selects {\it Work} with probability 0.2.  The equilibrium outcome could
be any of the four entries in the outcome matrix.  The entries having the
highest probability of occurence are {\it (No Aid, Loaf)} and {\it (Aid, Loaf)},
each with probability 0.4 ($=0.5[1-0.2]$).

  \bigskip \noindent
  {\bf Interpreting  Mixed Strategies }

\noindent
  Mixed strategies are not as intuitive as pure strategies, and many modellers
prefer to restrict themselves to pure-strategy equilibria in games which have
them. One objection to mixed strategies is that people in the real world do not
take random actions. That is not a compelling objection, because  all that a
model with mixed strategies requires to be a good description of the world is
that the  actions appear random to observers, even if the player himself has
always been sure what action he would take.  Even explicitly random actions are
not uncommon, however---   the Internal Revenue Service randomly selects which
tax returns to audit, and telephone companies randomly monitor their operators'
conversations to discover whether they are being polite.

         A more troubling objection is that a player who selects a mixed
strategy is always indifferent between two pure strategies. In the { Welfare
Game}, the pauper is indifferent between his two pure strategies and a whole
continuum of mixed strategies, given the government's mixed strategy. If the
pauper were to decide not to follow the particular mixed strategy $\gamma_w =
0.2$, the equilibrium would collapse because the government would change its
strategy in response. Even a small deviation in the probability selected by the
pauper, a deviation that does not change his payoff if the government does not
respond, destroys the equilibrium completely because the government does
respond. A mixed-strategy Nash equilibrium is weak in the same sense as the
({\it North}, {\it North}) equilibrium in   the Battle of the Bismarck Sea: to
maintain the equilibrium  a player who is indifferent between strategies must
pick a particular strategy from out of the set of strategies.

    One way to reinterpret the { Welfare Game}  is to imagine that instead of a
single pauper there are many, with identical tastes and payoff functions, all of
whom must be treated alike by the government. In the mixed-strategy equilibrium,
each of the paupers chooses {\it Work} with probability 0.2, just as in the one-
pauper game. But the many-pauper game has a pure-strategy equilibrium: 20
percent of the paupers choose the pure strategy {\it Work} and 80 percent choose
the pure strategy {\it Loaf}. The problem persists of how an individual pauper,
indifferent between the pure strategies, chooses one or the other, but it is
easy to imagine that individual characteristics outside the model could
determine which actions are chosen by which paupers.

         The number of players needed so that mixed strategies can be
interpreted as pure strategies in this way depends on the equilibrium
probability $\gamma_w$, since we cannot speak of a fraction of a player. The
number of paupers must be a multiple of five in the { Welfare Game} in order to
use this interpretation, since the equilibrium mixing probability is a multiple
of $1/5$.  For the interpretation to apply no matter how we vary the parameters
of a model we would need a {\it continuum} of players.

 Another interpretation of mixed strategies, which works even in the single-
pauper game, assumes that the pauper is drawn from a population of paupers, and
the government does not know his characteristics.  The government only knows
that there are two types of paupers, in the proportions (0.2, 0.8): those who
pick {\it Work} if the government picks $\theta_a =0.5$, and those who pick {\it
Loaf}. A pauper drawn randomly from the population might be of either type.
Harsanyi (1973) gives  a      careful interpretation of this situation.



\bigskip
\noindent
 {\bf   Mixed Strategies Can Dominate Otherwise Undominated Pure Strategies }

 Before we continue with methods of calculating mixed strategies, it is worth
taking a moment to show how they can be used to simplify the set of rational
strategies players might use in a game.    Chapter 1 talked about using the
ideas of  dominated strategies and   iterated dominance as an alternative to
Nash equilibrium, but it  ignored the possibility of mixed strategies.  That is
a meaningful omission, because some pure
  strategy  in a game may be strictly dominated by a mixed strategy, even if it
is not dominated by any of the other pure strategies.    The example in Table 2
illustrates this.

\begin{center} {\bf Table 2: Pure Strategies Dominated by a Mixed Strategy    }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Column}\\
&       &             &    $North$  & & $South$  \\
 &   &  $North$   &          0,0 &   & 4,-4 \\
 &  &     &    & &   \\
 & {\bf Row} &   $South$  &   4,-4 &    & 0,0 \\
 & & & &\\
 &  &       $Defense$     &  1,-1 &    & 1,-1 \\
 & & & &\\
\multicolumn{6}{l}{\it Payoffs to: (Row, Column)} 
\end{tabular} 
\end{center}

In the zero-sum game of Table 2, Row's army can  attack in the  North, attack in
the South, or remain on the defense. An unexpected attack gives Row a payoff of
4, an expected attack a payoff of 0, and defense a payoff of 1. Column can
respond by preparing to defend in the North or in the South.

 Row could guarantee himself a payoff of 1 if he chose {\it  Defense}. But
suppose he plays {\it  North} with probability 0.5 and {\it  South} with
probability 0.5.  His expected payoff from this mixed strategy if Column plays
{\it  North} with probability $N$ is
\begin{equation} \label{e100}
 0.5(N)(0) + 0.5(1-N)(4) + 0.5(N)(4) + 0.5 (1-N)(0) = 2, \end{equation} 
so whatever response Column picks, Row's expected payoff is higher from the
mixed strategy than his payoff of 1 from {\it  Defense}. For Row,  {\it  Defense
} is strictly dominated  by (0.5 {\it  North}, 0.5 {\it  South}).

   ``What if Row  is risk averse?'',  you may ask.   ``Might he not prefer  the
sure payoff of 1 from playing $Defense$?'' No.  Payoffs are   specified in units
of utility, not of money or some other input into a utility function.   In Table
2, it might be that Row's payoff of 0 represents gaining   no territory, 1
represents   100 square miles, and 4 represents   800 square miles, so the
marginal payoff of territory acquisition (or loss, for $South$) is declining.
When using mixed strategies it is particularly important to keep track of the
difference between utility and the inputs into utility.


  Thus, regardless of risk aversion,  in  the unique Nash equilibrium of Pure
Strategies Dominated by a Mixed Strategy, Row and Column would  both  choose
{\it  North} with probability $N=0.5$ and {\it   South} with probability 0.5.
This is  a player's unique equilibrium action because any other choice    would
cause  the other player to deviate to whichever direction   was not  being
guarded as often.



   \bigskip \noindent
  {\bf  3.2   The Payoff-Equating Method and Games of Timing   }



\noindent
 The next game illustrates why we might decide that a mixed-strategy equilibrium
is best even if pure-strategy equilibria also exist.  In the game of   {
Chicken}, the players are two Malibu teenagers, Smith and Jones.  Smith drives a
hot rod south down the middle of Route 1, and Jones drives   north. As
collision threatens, each decides whether to $Continue$ in the middle or
$Swerve $ to the side.  If a player is the only one to $Swerve$, he loses face,
but if neither player picks $Swerve$ they are both killed, which has an even
lower payoff.  If a player is the only one to $Continue$, he is covered with
glory, and if both $Swerve$ they are both embarassed.  (We will assume that  to
$Swerve$ means by convention to $Swerve$ right; if one swerved to the left and
the other to the right, the result would be both death and humiliation.) Table
3 assigns numbers to these four outcomes.

 \begin{center} {\bf Table  3:      Chicken   }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Jones}\\
&       &             &    $Continue$ ($\theta$)  &   &  $Swerve$ ($1- \theta$)
\\ &   &  $Continue$ ($\theta$)      &     $-3,-3$ & $\rightarrow$  & {\bf 2, 0}
\\ & {\bf Smith:} &&$\downarrow$& & $\uparrow$ \\ &  &         $Swerve$
($1-\theta$)       &    {\bf 0, 2}  & $\leftarrow$  & 1, 1 \\
 \end{tabular} \end{center}
\vspace{-24pt}
 {\it Payoffs  to:  (Smith, Jones). Arrows show how a player can increase his
payoff.   }

\bigskip

      { Chicken}  has two pure-strategy Nash equilibria, $(Swerve, Continue)$
and $(Continue, Swerve)$, but they have the defect of asymmetry.  How do the
players know which equilibrium is the one that will be played out?  Even if they
talk  before the game started, it is not clear how they could arrive at an
asymmetric result. We encountered the same dilemma in choosing an equilibrium
for   { Battle of the Sexes} . As in that game,  the best prediction in   {
Chicken}  is perhaps the mixed-strategy equilibrium, because its   symmetry
makes it a focal point of sorts, and does not require any differences between
the players.

  The {\bf payoff-equating} method used here to   calculate the mixing
probabilities for   { Chicken}   will be based on the logic followed in Section
3.1, but  it does not use  the calculus of maximization.  The basis of the
payoff-equating method is that
{\bf  when a player uses a mixed strategy in equilibrium, he must be getting the
same payoff from each of the pure strategies used in the mixed strategy}.   If
one of his mixing strategies has a higher payoff, he should deviate to use just
that one instead of mixing. If one has a lower payoff, he should deviate by
dropping it from his mixing.

 In Chicken, therefore, Smith's payoffs from  the pure strategies  of $Swerve$
and $Continue$ must be equal.
Moreover, { Chicken}, unlike  the { Welfare Game}, is a symmetric game, so we
can guess that in   equilibrium   each player will choose the same mixing
probability. If that is the case, then, since the payoffs from each of Jones'
pure strategies must be equal in a mixed-strategy equilibrium, it is true that
\begin{equation} \label{e3.5}
 \begin{array}{lll} \pi_{Jones}(Swerve) & = (\theta_{Smith})\cdot (0) +(1-
\theta_{Smith}) \cdot (1) & \\ &\\ & = (\theta_{Smith}) \cdot (-3) +
(1-\theta_{Smith}) \cdot (2) = \pi_{Jones}(Continue).
 \end{array} 
\end{equation}
From equation (\ref{e3.5}) we can conclude that $1-\theta_{Smith}=2 -
5\theta_{Smith}$, so $\theta_{Smith} = 0.25.$  In the symmetric equilibrium,
both players choose the same probability, so we can replace $\theta_{Smith}$
with simply $\theta$. As for the question which represents the greatest interest
to their mothers, the two teenagers will survive with probability $ 1- (\theta
\cdot \theta) = 0.9375.$

The  payoff-equating  method is easier to use than the  calculus method if the
modeller is sure which strategies will be mixed, and it can also be used in
asymmetric games. In the { Welfare Game}, it would start with $V_g(Aid) =
V_g(No\;Aid) $ and $V_p(Loaf ) = V_p(Work ) $, yielding two equations for the
two unknowns, $\theta_a$ and $\gamma_w$,  which when solved  give the same
mixing probabilities as were found earlier for that game.   The reason why the
payoff-equating and   calculus maximization methods reach the same result is
that the expected payoff is linear in the possible payoffs, so differentiating
the expected payoff equalizes the possible payoffs.  The only difference from
the symmetric-game case is that two equations are solved for two different
mixing probabilities instead of a single equation for  the one mixing
probability that both players use.

It is interesting to see what happens if the payoff of $-3$ in the northwest
corner of Table   3  is generalized to $x$. Solving the analog of  equation
(\ref{e3.5})  then  yields
 \begin{equation} \label{e3.6} \theta =  \frac{1}{1-x}. \end{equation}
 If
$x=-3$, this yields $\theta = 0.25$, as was just calculated, and if $x= -9$,  it
yields $\theta = 0.10 $. This makes sense; increasing the loss from crashes
reduces the equilibrium probability of continuing down the middle of the road.
But what if $x= 0.5$? Then the equilibrium probability of continuing appears to
be $\theta =  2 $, which is impossible; probabilities are bounded by zero and
one.

 When a mixing probability is calculated to be greater than one  or less than
zero, the implication is either that the modeller has made an arithmetic mistake
or, as in this case, that he is wrong in thinking that the game has a mixed-
strategy equilibrium. If $x=0.5$, one can still try to solve for the mixing
probabilities, but, in fact, the only equilibrium is in pure strategies--- {\it
(Continue, Continue) } (the game has become a   { prisoner's dilemma} ). The
absurdity of probabilities greater than one or less than zero is a valuable aid
to the fallible modeller  because such results show that he is wrong about the
qualitative nature of the equilibrium--- it is pure, not mixed. Or, if the
modeller is not sure whether the equilibrium is mixed or not, he can use this
approach to prove that the equilibrium is not in mixed strategies.



\bigskip
\noindent
 {\bf The  War of Attrition}

\noindent
    After the start of the book with the Dry Cleaners Game, we have been looking
at games that are either simultaneous or have the players move in sequence. Some
situations, however, are naturally modelled as   flows of time during which
players  repeatedly choose their moves.
 The  War of Attrition is  one of these. It is a game something like   {
Chicken} stretched out over time, where both players start with $Continue$, and
the game ends when the first one picks $Swerve$. Until the game ends, both earn
a negative amount per period, and when one exits, he earns zero and the other
player earns a reward for outlasting him.

   We will look at a war of attrition in discrete time. We will continue with
Smith and Jones, who have both survived to maturity  and now play games with
more expensive toys: they control two  firms in an industry which is a natural
monopoly, with   demand  strong enough for one firm to operate profitably, but
not two. The possible  actions are to $Exit$ or to $Continue$. In each period
that both $Continue$,   each earns $- 1$. If a firm exits, its losses cease and
the remaining firm obtains the value of the market's monopoly profit, which we
set equal to 3.  We will set the discount rate equal to $r > 0$, although that
is inessential to the model, even if the possible length of the game is infinite
(discount rates will be discussed in detail in Section 4.3).

  The War of Attrition has a continuum of Nash equilibria.  One simple
equilibrium is for Smith to choose ($Continue$ regardless of what Jones does)
and for Jones to choose ($Exit$ immediately), which are best responses to each
other. But we will solve for a symmetric equilibrium in which each player
chooses the same mixed strategy: a constant probability $\theta$ that the player
picks $Exit$ given that the other player has not yet exited.

  We can calculate $\theta$ as follows, adopting the perspective of Smith.
Denote the expected discounted value of Smith's payoffs by $V_{stay}$ if he
stays and $V_{exit}$ if he exits immediately.  These two pure strategy payoffs
must be equal in a mixed strategy equilibrium (which was the basis for the
payoff-equating method). If Smith exits, he obtains $V_{exit} =0$.  If Smith
stays in, his payoff depends on what Jones does.  If Jones stays in too, which
has probability $(1-\theta) $, Smith gets $-1$ currently and his expected value
for the following period, which is discounted using $r$, is unchanged. If Jones
exits immediately, which has probability $\theta$, then Smith receives a payment
of 3. In symbols,
\begin{equation}\label{e3.7}
 V_{stay} = \theta \cdot (3) + \left( 1-\theta \right) \left(-1 + \left[
\frac{V_{stay}}{1+r}\right] \right), \end{equation} which,  after a little
manipulation, becomes \begin{equation}\label{e3.8} V_{stay} = \left( \frac{1+r}
{r+\theta} \right) \left( 4\theta - 1 \right). \end{equation}
Once we equate $V_{stay}$ to $V_{exit}$, which equals zero, equation
(\ref{e3.8}) tells us that $\theta = 0.25$ in equilibrium, and that this is
independent of the discount rate $r$.

   Returning from arithmetic to ideas, why does Smith $Exit$ immediately with
positive probability, given that  Jones will exit first if Smith waits long
enough?  The reason is that Jones might choose to continue for a long time   and
both players would earn $-1$ each period until Jones exited. The equilibrium
mixing probability is calculated so that both of them are likely to stay in long
enough so that their losses soak up the gain from being the survivor. Papers on
the War of Attrition include Fudenberg \& Tirole (1986b), Ghemawat \& Nalebuff
(1985), Maynard Smith (1974), Nalebuff \& Riley (1985), and Riley (1980).  All
are examples of  ``rent-seeking'' welfare losses. As Posner (1975) and Tullock
(1967) have pointed out, the real costs of acquiring rents can be much bigger
than the  second-order triangle losses from allocative distortions, and the war
of attrition shows that the big loss from a natural monopoly might be not the
reduced trade that results from higher prices, but the cost of the battle to
gain the monopoly.

    We are likely to see wars of attrition in business when new markets open up,
either new geographic markets for old goods or new goods, especially when it
appears the market may be a natural monopoly, as in situations of network
externalities. McAfee (2002, p. 76, 364) cites as examples  the fight between
Sky Television and British Satellite Broadcasting for the British satellite TV
market; Amazon versus Barnes and Noble for the Internet book market; and Windows
CE versus Palm in the market for handheld computers. Wars of attrition can also
arise in declining industries, as a contest for which firm can exit the latest.
In the United States, for example, the number of firms making rockets declined
from six in 1990 to two by 2002 (McAfee [2002, p. 104]).


   In  the War  of Attrition, the reward goes to the player who does not choose
the     move which ends the game, and a cost is paid each period that both
players refuse to end it. Various other {\bf timing games} also exist.  The
opposite of a war of attrition is a {\bf pre-emption game}, in which the
reward goes to the player who  chooses the move which ends the game,  and a cost
is paid if both players choose that move, but no cost is incurred in a period
when neither player chooses it. The game of {\bf Grab the Dollar} is an example.
A dollar is placed on the table  between Smith and Jones, who each must decide
whether to grab for it or not. If both grab, both are fined one dollar.  This
could be set up as a one-period game, a $T$ period game, or an infinite- period
game, but the game definitely ends when  someone grabs the dollar. Table 4 shows
the payoffs.

  \begin{center} {\bf Table  4:  Grab the Dollar }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Jones}\\
&       &             &  {\it  Grab}    &   &  {\it Don't Grab } \\ &   &
{\it Grab} &      $-1, -1$ & $\rightarrow$  &   {\bf 1,0} \\ & {\bf Smith:}
&&$\downarrow$& & $\uparrow$ \\ &  &    {\it Don't Grab }       &       {\bf
0,1}  & $\leftarrow$  & $0, 0$ \\
  \end{tabular}
 \end{center}
\vspace{-24pt}
 {\it Payoffs  to:  (Smith, Jones). Arrows show how a player can increase his
payoff.   }

\bigskip


 Like the War of Attrition, Grab the Dollar has asymmetric equilibria in pure
strategies, and a symmetric equilibrium in mixed strategies.  In the infinite-
period version, the equilibrium probability of grabbing is 0.5 per period in the
symmetric equilibrium.

 Still another class of  timing games are    duels, in which  the actions are
discrete occurrences which the players locate at particular points in continuous
time. Two players with guns approach each other and must decide when to shoot.
In a {\bf noisy duel}, if a player shoots and misses, the other player observes
the miss and can kill the first player at his leisure. An equilibrium   exists
in pure strategies for the noisy duel.  In a {\bf silent duel}, a player does
not know when the other player has fired, and the equilibrium is in mixed
strategies. Karlin (1959) has details  on duelling games, and  Chapter 4 of
Fudenberg \& Tirole (1991a) has an excellent discussion of games of timing in
general. See also Shubik (1954) on the rather different problem of who to shoot
first in a battle with three or more sides.

  We will go through one more game of timing to see how to  derive a continuous
mixed strategies probability distribution, instead of just the single number
derived earlier. In presenting this game, a  new presentation scheme  will be
useful. If a game has a continuous strategy set, it is harder or impossible  to
depict the payoffs using   tables or the extensive form using  a tree. Tables of
the sort we have been using so far would require a continuum of rows and
columns, and trees a continuum of branches.  A new format for game descriptions
of the players, actions, and payoffs will be used for the rest of the book. The
new format will be similar to the way the rules of  the  {Dry Cleaners Game}
were presented in Section 1.1.

\begin{center}
{\bf    Patent Race for a New Market }
 \end{center}
{\bf Players}\\
 Three identical firms, Apex, Brydox, and Central.

\noindent {\bf The Order of Play }\\ 
Each firm simultaneously chooses research
spending $x_i \geq 0$, $ (i = a,b,c)$.

 \noindent {\bf Payoffs}\\
Firms are risk neutral and the discount rate is zero. Innovation occurs at time
$T(x_i)$ where $T' <0$.  The value of the patent is $V$, and if several players
innovate simultaneously they share its value. Let us look at the payoff of firm
$i= a, b,c,$ with  $j$ and $k$ indexing the other two firms:

 \begin{tabular}{ll}
 $ \pi_i =$& $\left\{ \begin{tabular}{lll}
  $V-x_i$ & if
$T(x_i) < Min\{ T(x_j, T(x_k) \} $  & (Firm $i$ gets the patent)\\
 & & \\
$\frac{V}{2} - x_i$ & if $T(x_i) = Min \{T(x_j),T(x_k)\}    $ & (Firm $i$ shares
the patent
with\\
 & $\;\;\;\;\;\;\;\;\;\;\; <Max \{T(x_j),T(x_k)\}$ & 1 other firm)\\ 
 & & \\
$\frac{V}{3} - x_i$ & if $T(x_i) = T(x_j=T(x_k)  $ & (Firm $i$ shares the patent
with\\
 & & 2 other firms)\\ 
& & \\ 
$- x_i$ & if $T(x_i) > Min\{ T(x_j, T(x_k) \}$  & (Firm $i$ does not get the
patent) \\
\end{tabular} \right.$
\end{tabular}\\





 The format first assigns the game a title, after which it lists the players,
the order of play  (together with who observes what), and the payoff functions.
Listing the players is redundant, strictly speaking, since they can be deduced
from the  order of play,   but it is useful for letting the reader know what
kind of model to expect.  The format includes very little explanation; that is
postponed, lest it obscure the description. This exact format is not standard in
the literature, but every good article begins its technical section by
specifying the same information, if   in a less structured way, and the novice
is strongly advised to use all the structure he can.

  The game Patent Race for  a New Market does not have any pure strategy Nash
equilibria, because the payoff functions are discontinuous. A slight difference
in research by one player can make a big difference in the payoffs, as shown in
Figure 1   for fixed values of $x_b$ and $x_c$. The research levels shown in
Figure 1 are not equilibrium values.  If Apex chose any research level $x_a$
less than $V$, Brydox would respond with $x_a + \varepsilon$ and win the patent.
If Apex chose $x_a = V$, then Brydox and Central would respond with $x_b = 0$
and $x_c = 0$, which would make Apex want to switch to $x_a = \varepsilon.$

\includegraphics[width=150mm]{fig03-01.jpg}



\begin{center} 
{\bf Figure 1:} The Payoffs in Patent Race for a New Market
\end{center}

   There does exist a symmetric mixed strategy equilibrium. Denote the
probability that firm $i$ chooses a research level less than or equal to $x$ as
$M_i(x)$. This function describes the firm's mixed strategy.  In a mixed-
strategy equilibrium a player is indifferent between any of the pure strategies
among which he is mixing (the basis of  Section 3.2's  payoff-equating method).
Since we know that the pure strategies $x_a=0$ and
$x_a= V$ yield zero payoffs, if Apex mixes over the support $[0,V]$ then the
expected payoff for every strategy mixed between must also equal zero.  The
expected payoff from the pure strategy $x_a$ is the expected value of winning
minus the cost of research. Letting $x$ stand for nonrandom and $X$ for random
variables, this is
\begin{equation} \label{e9}
 \pi_a(x_a)= V \cdot Pr (x_a \geq X_b, x_a \geq X_c) - x_a = 0= \pi_a(x_a=0),
\end{equation}
 which can be rewritten as
\begin{equation} \label{e10}
 V \cdot Pr (X_b \leq x_a) Pr(X_c \leq x_a) - x_a =
0, \end{equation}
 or
 \begin{equation} \label{e14.11}
 V \cdot M_b(x_a)M_c(x_a) -
x_a = 0. \end{equation}
 We can rearrange equation (\ref{e14.11}) to obtain
\begin{equation} \label{e12} M_b(x_a)M_c(x_a) =\frac{ x_a}{V}. \end{equation} If
all three firms choose the same mixing distribution $M$, then
  \begin{equation} \label{e13} M(x) = \left( \frac{x}{V} \right)^{1/2} \;{\rm
for}\; 0 \leq x \leq V.
 \end{equation}

  What is noteworthy about a patent race is not the nonexistence of a pure
strategy equilibrium  but the overexpenditure on research. All three players
have expected payoffs of zero, because the patent value $V$ is completely
dissipated in the race.  As in Brecht's {\it Threepenny Opera} (Act III, Scene
7), ``When all race after happiness/Happiness comes in last.''   To be sure, the
innovation is made earlier than it would have been by a monopolist, but hurrying
the innovation is not worth the cost, from society's point of view, a result
that would persist even if the discount rate were positive.   Rogerson (1982)
uses a game very similar to  Patent Race for a New Market  to analyze
competition for a government monopoly franchise. Also, we will see in Chapter 13
that this is an example of an ``all-pay auction'', and the techniques and
findings of auction theory can be quite useful when modelling this kind of
conflict.

\bigskip \noindent {\bf Correlated Strategies}

\noindent One example of a war of attrition is setting up a market for a new
security, which may be a natural monopoly for reasons to be  explained in
Section 8.5. Certain stock exchanges have avoided the destructive symmetric
equilibrium by using lotteries to determine which of them would trade newly
listed stock options  under a system similar to the football
draft.\footnote{``Big Board will Begin Trading of Options on 4 Stocks it
Lists,'' {\it Wall Street Journal},p. 15 (4 October 1985).} Rather than waste
resources fighting, these exchanges use the lottery as a coordinating device,
even though it might not be a binding agreement.

  Aumann (1974, 1987) has pointed out that it is often important whether players
can
use the same randomizing device for their mixed strategies. If they can, we
refer to the resulting strategies as {\bf correlated strategies}. Consider the
game of   { Chicken}.  The only mixed-strategy equilibrium is the symmetric one
in which each player chooses $Continue$ with probability 0.25 and the expected
payoff is 0.75. A correlated equilibrium would be for the two players to flip a
coin and for Smith to choose $Continue$ if it comes up heads and for Jones to
choose $Continue$ otherwise. Each player's strategy is a best response to the
other's, the probability of each choosing $Continue$ is 0.5, and the expected
payoff for each is 1.0, which is better than the 0.75 achieved without
correlated strategies.

 Usually the randomizing device is not modelled explicitly when a model refers
to correlated equilibrium. If it is, uncertainty over variables that do not
affect preferences, endowments, or production is called {\bf extrinsic
uncertainty.} Extrinsic uncertainty is the driving force behind {\bf sunspot
models}, so called because the random appearance of sunspots might cause
macroeconomic changes via correlated equilibria (Maskin \& Tirole [1987]) or
bets made between players (Cass \& Shell [1983]).

  One way to model correlated strategies   is to specify a move in which Nature
gives each player  the ability to commit first to an action such as $Continue$
with equal probability. This is often realistic  because it amounts to a zero
probability of both players entering the industry at exactly the same time
without anyone knowing in advance who will be the lucky starter.  Neither firm
has an  a priori advantage, but the outcome is efficient.

 The population interpretation of mixed strategies cannot be used for correlated
strategies.  In ordinary mixed strategies, the mixing probabilities are
statistically independent, whereas in correlated strategies they are not. In   {
Chicken}, the usual mixed strategy can be interpreted as populations of Smiths
and Joneses, each population consisting of a certain proportion of pure swervers
and pure stayers. The correlated equilibrium has no such interpretation.

   Another coordinating device, useful in games that, like {Battle of the
Sexes}, have a coordination problem, is {\bf cheap talk} (Crawford \& Sobel
[1982], Farrell [1987]). Cheap talk refers to costless communication before the
game proper begins. In { Ranked Coordination}, cheap talk instantly allows the
players to make the desirable outcome a focal point. In   { Chicken}, cheap talk
is useless, because it is dominant for each player to announce that he will
choose $Continue$. But in  The    Battle of the Sexes, coordination and conflict
are combined. Without communication, the only symmetric equilibrium is in mixed
strategies. If both players know that making inconsistent announcements will
lead to the wasteful mixed-strategy outcome, then they are willing to mix
announcing whether they will go to the ballet or the prize fight. With many
periods of announcements before the final decision, their chances of coming to
an agreement are high. Thus communication can help reduce inefficiency even if
the two players are in conflict.

 \bigskip
 \noindent
 {\bf *3.3  Mixed Strategies  with General  Parameters and $N$ Players: The
Civic Duty Game  }

  Having looked at a number of specific games with mixed-strategy equilibria,
let us now apply the method to    the    general game  of Table   5.

\begin{center}
 {\bf Table  5:   The  General 2-by-2 Game}

  \begin{tabular}{lllccc}
 &       &             &\multicolumn{3}{c}{\bf Column}\\
 &       &   &     {\it Left} ($\theta$) &   &   $Right$  ($1- \theta$) \\ &   &
$Up$ ($\gamma$)      &    $a,w$ &    & $b,x$ \\
& {\bf Row:} && & &   \\
 &  &     $Down$ ($1-\gamma$)     &  $c,y$  &    &  $d,z$ \\ \multicolumn{6}{l}
{\it Payoffs to: (Row, Column) } \end{tabular} \end{center}

 To find the game's equilibrium, equate the payoffs from the pure strategies.
For Row, this yields \begin{equation}\label{e3.9}
 \pi_{Row} (Up) =   \theta a + (1-\theta) b \end{equation} and \begin{equation}
\label{e3.10} \pi_{Row} (Down) =   \theta c + (1-\theta) d. \end{equation}
Equating (\ref{e3.9}) and (\ref{e3.10}) gives us
  \begin{equation}\label{e3.11}
\theta  (a + d -b-c) + b -d = 0, \end{equation} which yields \begin{equation}
\label{e3.12} \theta^* = \frac{d-b}{(d-b) + (a-c)}. \end{equation}
 Similarly,
equating the payoffs for Column gives \begin{equation}\label{e3.13}
 \pi_{Column}
(Left) =   \gamma w + (1-\gamma) y =  \pi_{Column} (Right) =   \gamma x +
(1-\gamma) z, \end{equation}
 which yields 
\begin{equation}\label{e3.14} \gamma^*
= \frac{z-y}{(z-y) + (w-x)}. \end{equation} The equilibrium represented by
(\ref{e3.12}) and (\ref{e3.14}) illustrates a number of features of mixed
strategies.

  First, it is possible, but wrong,  to follow the payoff-equating method for
finding a mixed strategy even if no mixed strategy equilibrium actually exists.
Suppose, for example, that $Down$  is a strictly dominant strategy for Row, so
$c>a$ and $d>b$. Row is unwilling to   mix, so  the equilibrium is not in mixed
strategies. Equation (\ref{e3.12}) would be misleading, though some idiocy would
be required to stay misled for very long  since the equation implies that
$\theta^* >1$  or $\theta^* \leq 0$  in cases like that.

Second,  the exact features of the  equilibrium in mixed strategies depend
heavily on the cardinal values of the payoffs, not just on their ordinal values
like the pure strategy equilibria in other 2-by-2 games. Ordinal rankings are
all that is needed to know that  an  equilibrium exists in mixed strategies, but
cardinal values are needed to know the exact mixing probabilities. If the payoff
to   Column   from {\it (Confess, Confess}) is changed slightly in   the   {
Prisoner's Dilemma} it makes no difference at all to  the equilibrium. If the
payoff  of $z$  to  Column  from {\it (Down, Right)}  is increased slightly in
the General 2-by-2 Game, equation (\ref{e3.14}) says that the mixing probability
$\gamma^*$ will change also.

Third, the payoffs can be changed by affine transformations without changing the
game   substantively, even though cardinal payoffs do matter (which is to say
that   monotonic but non-affine transformations do make a difference).  Let each
payoff $\pi$ in Table 5 become $\alpha + \beta \pi$. Equation  (\ref{e3.14})
then becomes
 \begin{equation}\label{e3.15}
\begin{array}{ll}
\gamma^*&  = \frac{\alpha + \beta z- \alpha - \beta y}{(\alpha + \beta z-\alpha
- \beta y) + (\alpha + \beta w-\alpha - \beta x)} \\ & \\
&=\frac{z-y}{(z-y) + (w-x)}. \end{array}
 \end{equation}
 The affine transformation has left the equilibrium strategy unchanged.

Fourth, as was mentioned earlier in connection with   the { Welfare Game},
each player's mixing probability depends only on the payoff parameters of the
other player. Row's strategy $\gamma^*$ in equation (\ref{e3.14}) depends on the
parameters $w,x,y$ and $z$, which are the payoff parameters for Column, and have
no direct relevance for Row.

  \bigskip \noindent
 {\bf   Categories of Games with Mixed Strategies}

    Table  6 uses the players and actions of Table   5 to  depict three major
categories of  2-by-2 games in which mixed-strategy equilibria are important.
Some games fall in none of these categories--- those with tied payoffs,  such as
The Swiss Cheese Game in which all eight payoffs equal zero---   but the three
games in Table  6 encompass a wide variety of economic phenomena.

 \begin{center} {\bf Table  6:  2-by-2 Games with Mixed Strategy Equilibria  }

 \begin{tabular}{|lllll lllll lllll|}
 \cline{1-3} \cline{5-7} \cline{9-11} \cline{13-15 } $a,w$ & $\rightarrow$  &
\multicolumn{1}{l|} {$ b,x $}  &  & \multicolumn{1}{|l } {$a,w$ }& $\leftarrow$
&  \multicolumn{1}{l|} {$ b,x$}     &  & \multicolumn{1}{|l } { $\boldmath{a,w}
$}& $\leftarrow$ & \multicolumn{1}{l|} { $ b,x $ } &  & \multicolumn{1}{|l } {
$a,w$ } & $\rightarrow$  &\multicolumn{1}{l|} { $\boldmath{b,x}$}\\
$\uparrow$ &     & \multicolumn{1}{l|} {$\downarrow$   } &  & \multicolumn{1}{|l
} { $\downarrow$ } &     &  \multicolumn{1}{l|} {$\uparrow$ }  &  &
\multicolumn{1}{|l } { $\uparrow$ } &  & \multicolumn{1}{l|} { $\downarrow$  }
&  & \multicolumn{1}{|l } { $\downarrow$ } &   &\multicolumn{1}{l|} {$\uparrow$}
\\ $c,y$ & $\leftarrow$  & \multicolumn{1}{l|} { $ d,z $}  &  & \multicolumn{1}
{|l } { $c,y$} & $\rightarrow$  & \multicolumn{1}{l|} { $ d,z $} &  &
\multicolumn{1}{|l } { $c,y$} &   $\rightarrow$  & \multicolumn{1}{l|} {
$\boldmath{d,z}$ } &  & \multicolumn{1}{|l } { $\boldmath{c,y}$} &
$\leftarrow$  & \multicolumn{1}{l|} { $ d,z $}\\
 \cline{1-3} \cline{5-7} \cline{9-11} \cline{13-15 } \multicolumn{7}{c}{
Discoordination Games}  &  & \multicolumn{3}{c}{ Coordination Games } &  &
\multicolumn{3}{c}{  Contribution Games }\\
\end{tabular}
 \end{center}
\vspace{-24pt}
 {\it Payoffs  to:  (Row, Column). Arrows show how a player can increase his
payoff.   }

\bigskip

{\bf Discoordination games}  have a single equilibrium, in  mixed strategies.
The  payoffs are such that either (a) $a>c$, $d>b$, $x > w$, and $y>z$, or (b)
$c>a$, $b>d$, $w>x$, and $z>y$. The { Welfare Game} is a discoordination game,
as are   { Auditing Game I}  in the next section and Matching Pennies in problem
3.3.

 {\bf Coordination games } have   three equilibria:  two  symmetric equilibria
in pure strategies and one symmetric equilibrium   in mixed strategies. The
payoffs are such that $a>c$, $d>b$, $w>x$, and $z>y$. { Ranked Coordination}
and The   { Battle of the Sexes}  are two varieties of coordination games in
which the players have the same and opposite rankings of the pure-strategy
equilibria.

 {\bf Contribution games }   have three equilibria: two asymmetric equilibria in
pure strategies and one symmetric equilibrium in mixed strategies.  The  payoffs
are such that $c >a$, $b>d$, $x >w$, and $y> z$.  Also,    it must  be true that
$c<b$ and $y>x$, or $c>b$ and $y<x$

 I have used the name  ``contribution game'' because the type of game described
by this term is often used to model a situation in which two  players each have
a choice of taking some action that contributes to the public good, but would
each like the other player to bear the cost. The difference from the   {
Prisoner's Dilemma} is that each player in a contribution game is willing to
bear the cost alone if necessary.

Contribution games  appear to be quite  different from  the  { Battle of the
Sexes}, but they are essentially the same.   Both of them have two pure-strategy
equilibria,  ranked oppositely by    the two players. In mathematical terms, the
fact that  contribution games have the equilibria in the southwest and northeast
corners of the outcome matrix whereas coordination games   have them in the
northwest and southeast, is unimportant; the location of the equilibria could be
changed by just switching the order of Row's strategies.  We do  view real
situations differently, however, depending  on whether  players choose the same
actions or different actions in equilibrium.

 Let us take a look at a particular contribution game to show how to extend two-
player games to games with several players.  A notorious example in social
psychology is the murder of Kitty Genovese, who was killed in New York City in
1964 despite the presence of numerous neighbors. 
\begin{quotation} \begin{small}
``For more than half an hour 38
respectable, law-abiding citizens in Queens watched a killer stalk and stab a
woman in three separate attacks in Kew Gardens....  Twice the sound of their
voices and the sudden glow of their  bedroom lights interrupted him and
frightened him off. Each time he returned, sought her out, and stabbed her
again. Not one person telephoned the police during the assault; one witness
called after the woman was dead.''   (Martin Gansberg, ``38 Who Saw Murder
Didn't Call Police,'' {\it The New York Times}, March 27, 1964, p. 1. )
\end{small}
 \end{quotation}
  Even as
hardened an economist as myself finds it somewhat distasteful to call this a
``game,'' but game theory does explain what happened.

  I will  use a   less appalling story for our model. In  the { Civic Duty Game}
of  Table   7,   Smith and Jones observe a burglary taking place. Each   would
like someone to call the police and stop the burglary  because having  it
stopped adds 10 to his payoff,  but neither wishes to make the call himself
because the effort subtracts 3. If Smith can be assured that Jones will call,
Smith himself will ignore the burglary.  Table 7 shows the payoffs.

   \begin{center} {\bf Table  7:  The   { Civic Duty Game}  }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Jones}\\
&       &             &  {\it  Ignore} ($\gamma$)   &   & $Telephone$
($1-\gamma$)     \\ &   &    {\it Ignore} ($\gamma$) &      $0,0$ &
$\rightarrow$  & {\bf 10,7}  \\ & {\bf Smith:} &&$\downarrow$& & $\uparrow$ \\ &
&    {\it Telephone } ($1-\gamma$)      &       {\bf 7,10}  & $\leftarrow$  &
$7,7$ \\   \end{tabular}
\end{center}
\vspace{-10pt}
 {\it Payoffs  to:  (Row, Column). Arrows show how a player can increase his
payoff.   }

\bigskip
 The   { Civic Duty Game}  has two  asymmetric pure-strategy equilibria and a
symmetric mixed-strategy equilibrium.  In solving for the mixed- strategy
equilibrium, let us   move from two players to $N$ players. In the N-player
version of the game, the payoff to Smith is 0 if nobody calls, 7 if he himself
calls, and 10 if one or more  of the other $N- 1$ players calls. This game also
has asymmetric pure-strategy and a symmetric mixed-strategy equilibrium. If all
players use the same probability $\gamma$ of $Ignore$, the probability that the
other $N-1$ players besides Smith all choose $Ignore$ is $\gamma^{N-1}$, so the
probability that one or more of them chooses $Telephone$  is $1- \gamma^{N-1}$.
Thus,   equating Smith's pure-strategy payoffs using the payoff-equating method
of equilibrium calculation yields
 \begin{equation}\label{e3.16}
  \pi_{Smith} (Telephone) = 7 =    \pi_{Smith} (Ignore) =  \gamma^{N- 1} (0) +
(1-\gamma^{N-1})  (10). \end{equation} Equation (\ref{e3.16})  tells us that
\begin{equation}\label{e3.17} \gamma^{N-1}  = 0.3 \end{equation} and
\begin{equation}\label{e3.18} \gamma^*   = 0.3^{\frac{1}{N-1}}. \end{equation}
  If $N=2$,  Smith chooses $Ignore$ with a probability of 0.30.  As $N$
increases,   Smith's expected payoff  remains equal to 7 whether $N=2$ or $N=
38$, since his expected payoff   equals his payoff from the pure strategy of
$Telephone$. The probability  of $Ignore$, $\gamma^*$, however,   increases with
$N$. If $N=38$, the value of $\gamma^*$ is about 0.97.    When there are  more
players, each player relies more on somebody else calling.

 The probability that nobody calls is $\gamma^{*N} $. Equation (\ref{e3.17})
shows that $\gamma^{*N-1}= 0.3$, so  $\gamma^{*N} = 0.3\gamma^*$, which is
increasing in   $N$ because $\gamma^*$ is increasing in $N$. If $N=2$, the
probability that neither player phones the police is  $ \gamma^{*2} = 0.09$.
When there are 38 players, the probability rises to $ \gamma^{* 38}$, about
0.29. The  more people that watch a crime,  the less likely  it is to be
reported.

  As in  the   { Prisoner's Dilemma}, the  disappointing result in the { Civic
Duty Game}   suggests a role for active policy.  The  mixed- strategy outcome is
clearly bad. The expected payoff per player remains equal to 7 whether there  is
1 player  or 38, whereas  if the equilibrium played out was the equilibrium in
which one and only one player called the police, the average payoff would rise
from 7 with 1 player to about 9.9 with 38 (=[1(7) + 37(10)]/38). A situation
like this requires  something to make one of the pure-strategy equilibria a
focal point. The problem is divided responsibility. One person must be made
responsible for calling the police, whether by tradition (e.g., the oldest
person on the block always calls the police) or direction (e.g., Smith shouts to
Jones: ``Call the police!'').

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

 \bigskip
  \noindent
 {\bf *3.4    Randomizing Is Not Always Mixing:     The Auditing Game }



 The next three games will illustrate the difference between mixed strategies
and random actions, a subtle but important distinction. In all three games, the
Internal Revenue Service must decide whether to audit a certain class of suspect
tax returns to discover whether they are accurate or not. The  goal of the IRS
is to  either prevent or catch cheating at minimum cost. The  suspects want to
cheat only if they will not  be caught. Let us assume that the benefit of
preventing or catching cheating is 4, the cost of auditing     is $C$, where
$C<4$, the cost to the suspects of obeying the law is 1, and the cost of being
caught is the fine $F >1$.

 Even with all of this information, there are several ways to model the
situation. Table  8 shows one way:  a 2-by-2 simultaneous-move game.

 \begin{center} {\bf Table  8:   { Auditing Game I}  }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Suspects}
\\ &       &  &  {\it Cheat } ($\theta$)  &   &  $ Obey$  ($1-\theta$) \\ &   &
$Audit$   ($ \gamma$) &   $4-C,-F$  & $\rightarrow$  &  $4-C, - 1  $ \\ & {\bf
IRS:} &&$\uparrow$& & $\downarrow$ \\ &  & {\it  Trust }  ($1-\gamma$)    &
0,0    & $\leftarrow$  & $4,- 1$ \\
 \end{tabular} \end{center}
\vspace{-24pt}
 {\it Payoffs  to:  (IRS, Suspects). Arrows show how a player can increase his
payoff.   }

\bigskip

  { Auditing Game I}  is a discoordination game, with only a mixed strategy
equilibrium.       Equations (\ref{e3.12}) and (\ref{e3.14}) or the payoff-
equating method  tell  us that
\begin{equation}\label{e3.19}
\begin{array}{ll} Probability (Cheat)= \theta^*  &= \frac{4 - (4-C)}{(4 - (4-C))
+ ( (4- C)-0)} \\ & \\ & = \frac{C}{4}
\end{array}
 \end{equation}
 and
\begin{equation}\label{e3.20} \begin{array}{ll} Probability (Audit)  =\gamma^*
& = \frac{ -1-0}{(-1-0) + (-F - -1)}\\ & \\ &  = \frac{1}{F}. \end{array}
\end{equation} Using (\ref{e3.19}) and  (\ref{e3.20}), the payoffs are
\begin{equation}\label{e3.21} \begin{array}{ll} \pi_{IRS}  (Audit)  =\pi_{IRS}
(Trust)  & =   \theta^* (0) +(1- \theta^* )(4)\\ & \\ & = 4-C. \end{array}
\end{equation} and \begin{equation}\label{e3.22} \begin{array}{ll} \pi_{Suspect}
(Obey)  =\pi_{Suspect} (Cheat) &  =  \gamma^*(-F) + (1- \gamma^*)(0)\\ & \\ &
=-1.
\end{array}
 \end{equation}

 A second way to model the situation is as a sequential game. Let us call this
{ Auditing Game II}.    The simultaneous game implicitly assumes that both
players  choose their actions without knowing what the other player has decided.
In the sequential game, the IRS chooses government policy first, and the
suspects react to it. The equilibrium in   { Auditing Game II}  is in pure
strategies,   a general feature of sequential games of perfect information.   In
equilibrium,   the IRS chooses $Audit$, anticipating that the suspect will then
choose $Obey$. The    payoffs  are $(4-C)$ for the IRS and     $-1$ for the
suspects, the same for   both players as in Auditing Game  I, although now there
is more auditing and less cheating and fine-paying.

We can go a step further.   Suppose the IRS  does not have to adopt a policy of
auditing or trusting every   suspect, but instead can audit a random sample.
This is not necessarily a mixed strategy. In Auditing Game  I, the equilibrium
strategy was to audit all suspects   with probability $1/F$ and  none of them
otherwise. That is different from announcing in advance that the IRS will audit
a random sample of $1/F$ of the  suspects. For   { Auditing Game III}, suppose
the IRS move first, but let its move consist of the choice of the proportion
$\alpha$ of tax returns to be audited.

 We know that the IRS is willing to deter the suspects from cheating, since it
would be willing to  choose $\alpha = 1$ and replicate the result in   {
Auditing Game II}    if it had to.  It  chooses $\alpha$ so that
\begin{equation}\label{e3.23}
 \pi_{suspect} (Obey)  \geq   \pi_{suspect} (Cheat),
 \end{equation}
 i.e.,
 \begin{equation}\label{e3.24}
 -1  \geq    \alpha (-F) + (1-\alpha ) (0). \end{equation}
 In equilibrium, therefore,   the IRS   chooses $\alpha =  1/F$ and the suspects
respond with $Obey$. The IRS  payoff is $(4 -  \alpha C)$, which is better than
the $(4-C)$   in the other two games, and the suspect's payoff is $-1$, exactly
the same as before.

    The equilibrium of   { Auditing Game III}  is in pure strategies, even
though the IRS's action is random. It is different from {Auditing Game  I  }
because  the IRS must go ahead with the costly audit even if the suspect chooses
$Obey$.     { Auditing Game III}  is different in another way also: its action
set is continuous. In  {Auditing Games I} and   { Auditing Game II}  the action
set is {\it \{Audit, Trust\},} although the strategy set becomes $\gamma \in
[0,1]$ once mixed strategies are allowed. In   { Auditing Game III}, the action
set is $\alpha  \in [0,1]$, and the strategy set would allow mixing of any of
the elements in the action set, although mixed strategies are pointless for the
IRS because the game is sequential.

  Games with mixed strategies are  like games with   continuous strategies
since a      probability is drawn from the continuum between zero and one.  {
Auditing Game III}   also has a   strategy  drawn from the  interval between
zero and one, but it is not a mixed strategy to pick an audit probability of,
say, 70 percent.  An example of a  mixed strategy   would be   the choice
of  a   probability 0.5 of an audit probability of 60 percent  and 0.5 of 80
percent.   The big difference between the pure strategy choice of an audit
probability of 0.70 and the  mixed strategy choice  of (0.5--- 60\%audit, 0.5---
80\% audit), both of which yield an audit probability of 70\%,  is
that the pure strategy is an irreversible choice that might be used even when
the player is not indifferent between pure strategies, but the  mixed strategy
is the result of a player who in equilibrium is indifferent as to what he does.
The next section will show another difference between mixed strategies and
continuous strategies:   the payoffs are linear in the mixed-strategy
probability, as is evident from   payoff equations (\ref{e3.9})   and
(\ref{e3.10}), but  they can be nonlinear in continuous strategies generally.

 
  I have used auditing here mainly to illustrate what mixed strategies
are and are not, but auditing is interesting in itself and  optimal auditing
schemes have many twists to them. An example is the idea of {\bf cross-
checking}. Suppose an auditor is supposed to check  the value of some variable
$x \in   [0, 1]$, but  his employer is worried that he will not report the true
value. This might be because  the auditor  will be lazy and guess rather than go
to the effort of finding $x$, or because some third party will bribe him, or
that certain values of $x$ will trigger punishments or policies the auditor
dislikes (this model applies even if $x$ is the auditor's own performance on
some other task). The idea of   cross-checking  is to hire a second auditor and
ask him to simultaneously report $x$. Then, if both auditors report the same
$x$, they are both rewarded, but if they report different values they are both
punished. There will still be multiple equilibria, because anything in which
they report the same value is an equilibrium. But at least truthful reporting
becomes a possible equilibrium. See Kandori \& Matsushima (1998)  for details
(and compare to the Maskin Matching Scheme of Chapter 10).

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

 \bigskip \noindent
 {\bf 3.5  Continuous Strategies: The { Cournot Game} }

\noindent Most of  the games so far in the book have had  discrete strategy
spaces: {\it Aid} or {\it No Aid}, {\it Confess} or {\it Deny}. Quite often when
strategies are discrete and moves are simultaneous,  no pure- strategy
equilibrium exists. The only sort of compromise possible in the { Welfare Game},
for instance, is  to choose {\it Aid} sometimes and {\it No Aid} sometimes, a
mixed strategy. If ``{\it A Little Aid}'' were a possible action, maybe there
would be a pure-strategy equilibrium. The simultaneous-move game we discuss
next,   the { Cournot Game}, has a continuous strategy space even without
mixing. It models  a duopoly in which two firms choose output levels in
competition with each other.

\begin{center}
 {\bf The { Cournot Game} } \end{center}
  {\bf Players}\\
  Firms Apex and Brydox

 \noindent {\bf The Order of Play}\\
 Apex and Brydox simultaneously choose quantities $q_a$ and $q_b$ from the  set
$[0, \infty)$.

\noindent {\bf Payoffs}\\
 Marginal cost is constant at $c=12$. Demand is a function of the total quantity
sold, $Q= q_a + q_b$,  and we will assume it to be linear (for generalization
see Chapter 14), and, in fact, will use the following specific function:
 \begin{equation} \label{e3.25} p(Q) = 120-q_a - q_b. \end{equation}
 Payoffs are profits, which are given by a firm's price times its quantity minus
its costs, i.e., \begin{equation} \label{e3.26}
\begin{array}{l}
 \pi_{Apex}  =(120-q_a - q_b)q_a - cq_a  =  (120-c)q_a  - q_a^2 - q_a q_b ;\\
 \\
 \pi_{Brydox} = (120-q_a - q_b)q_b - cq_b= (120-c)q_b  - q_a q_b  - q_b^2.
\end{array}
 \end{equation}

\includegraphics[width=150mm]{fig03-02.jpg}
\begin{center}
{\bf  Figure 2:  Reaction Curves in the   Cournot Game }
  \end{center}

  If this game were cooperative (see Section 1.2), firms would end up producing
somewhere on the 45$^\circ$ line in Figure  2, where total output is the
monopoly output and maximizes the sum of the payoffs. The monopoly output
maximizes $pQ-cQ= (120-Q-c)Q$ with respect to the total output of $Q$, resulting
in the first-order condition \begin{equation} \label{e3.27} 120 - c- 2Q = 0,
\end{equation} which implies a total output of $Q= 54$ and a price of  66.
Deciding how much of that output of 54 should be produced by each firm---where
the firm's output should be located on the 45$^\circ$ line---would be a zero-sum
cooperative game, an example of bargaining.  But since the { Cournot Game}  is
noncooperative, the strategy profiles such that $q_a +q_b = 54$ are not
necessarily equilibria despite their Pareto optimality (where Pareto optimality
is defined from the point of view of the two players, not of consumers, and
under the implicit assumption that price discrimination cannot be used).

    Cournot  noted in Chapter 7 of his 1838 book  that this game has a unique
equilibrium when demand curves are linear.  To find that ``Cournot-Nash''
equilibrium, we need to refer to the {\bf best-response functions} for the two
players.  If Brydox produced 0, Apex would produce the monopoly output of 54.
If Brydox produced $q_b = 108$ or greater, the market price would fall to 12 and
Apex would choose to produce zero. The best response function is found by
maximizing Apex's payoff, given in equation (\ref{e3.26}), with respect to his
strategy, $q_a$.  This generates the first-order condition $120 - c-  2q_a - q_b
= 0,$ or
 \begin{equation} \label{e3.28}
 q_a  = 60 - \left( \frac{q_b +c}{2} \right)  =  54 - \left( \frac{1 }{2}\right)
q_b.
\end{equation}

Another name for the best response function, the name usually used in the
context of  the { Cournot Game}, is the {\bf reaction function}. Both names are
somewhat misleading since the players move simultaneously with no chance to
reply or react, but they are useful in imagining what a player would do if the
rules of the game did allow him to move second. The reaction functions of the
two firms are labelled $R_a$ and $R_b$ in Figure  2. Where they cross, point E,
is the {\bf Cournot-Nash equilibrium}, which is simply the Nash equilibrium when
the strategies consist of quantities. Algebraically, it is found by solving the
two reaction functions for $q_a $ and $q_b$, which generates the unique
equilibrium, $q_a =  q_b =    40- c/3 = 36$.  The equilibrium price is then 48
(= 120-36-36).

    In   the { Cournot Game}, the Nash equilibrium has the particularly nice
property of {\bf stability}: we can imagine how starting from some other
strategy profile the players might reach the equilibrium.  If the initial
strategy profile is   point $X $ in Figure  2, for example, Apex's best response
is to decrease $q_a $ and Brydox's is to increase $q_b $, which moves the
profile closer to the equilibrium. But this is special to The { Cournot Game},
and Nash equilibria are not always stable in this way.

\bigskip \noindent
 {\bf Stackelberg Equilibrium}

\noindent There are many ways to model duopoly. The three most prominent are
Cournot, Stackelberg, and Bertrand.   Stackelberg equilibrium differs from
Cournot in that one firm gets to choose its quantity first.   If Apex moved
first, what output would it choose? Apex knows how Brydox will react to its
choice, so it picks the point on Brydox's reaction curve that maximizes Apex's
profit (see Figure  3).

\begin{center}
{\bf The { Stackelberg Game} } \end{center} {\bf Players}\\
  Firms Apex and
Brydox

 \noindent {\bf The Order of Play}\\
 1  Apex   chooses quantity $q_a$   from the  set  $[0, \infty)$.\\
  2 .   Brydox  chooses quantity  $q_b$ from the  set  $[0, \infty)$.

\noindent {\bf Payoffs}\\
 Marginal cost is constant at $c=12$. Demand is a function of the total quantity
sold, $Q= q_a + q_b$:
 \begin{equation} \label{e3.25a}
  p(Q) = 120-q_a - q_b. \end{equation} Payoffs are profits, which are given by a
firm's price times its quantity minus its costs, i.e., \begin{equation}
\label{e3.26b}
 \begin{array}{l}
 \pi_{Apex}  =(120-q_a - q_b)q_a - cq_a  =  (120-c)q_a  - q_a^2 - q_a q_b ;\\
 \\
 \pi_{Brydox} = (120-q_a - q_b)q_b - cq_b = (120-c)q_b  - q_a q_b  - q_b^2.
\end{array} \end{equation}

\includegraphics[width=150mm]{fig03-03.jpg}

\begin{center}
 {\bf Figure  3:  Stackelberg Equilibrium } \end{center}

  Apex, moving first,   is called the {\bf Stackelberg leader} and Brydox is the
{\bf Stackelberg follower.} The distinguishing characteristic of a Stackelberg
equilibrium is that one player gets to commit himself first.  In Figure  3, Apex
moves first intertemporally. If moves were simultaneous  but Apex could commit
himself to a certain strategy, the same equilibrium would be reached as long as
Brydox was not able to commit himself.  Algebraically, since Apex forecasts
Brydox's output to be $q_b  = 60 - \frac{q_a +c}{2}$ from the analog of equation
(\ref{e3.28}), Apex can substitute this into his payoff function in
(\ref{e3.26}) to obtain
 \begin{equation}\label{e3.29}
  \pi_a  =  (120-c)q_a  - q_a ^2 - q_a ( 60 - \frac{q_a +c}{2}).
 \end{equation}
 Maximizing his payoff with respect to $q_a $ yields the first-order condition
  \begin{equation}\label{e3.30}
(120-c) - 2q_a  - 60 + q_a+ \frac{ c}{2} = 0,
  \end{equation}
 which generates  Apex's reaction function, $q_a = 60-   c/2 = 54$ (which only
equals the monopoly output
by coincidence, due to the particular numbers in this example). Once Apex
chooses this output, Brydox chooses his output to be $q_b = 27$. (That Brydox
chooses exactly half the monopoly output is also accidental.)  The market price
is 120-54-27= 39  for both firms, so Apex has benefited from his status as
Stackelberg leader, but industry profits have fallen compared to the Cournot
equilibrium.

\bigskip \noindent
 { \bf  3.6 Continuous Strategies: The Bertrand Game, Strategic Complements, and
Strategic Substitutes  }

\noindent A natural alternative to a duopoly model in which the two firms pick
outputs simultaneously is a model in which they pick prices simultaneously.
This is known as {\bf Bertrand equilibrium}, because the difficulty of choosing
between the two models was stressed in Bertrand (1883), a review discussion of
Cournot's book. We will use the same two-player  linear-demand world as before,
but now the strategy spaces will be the prices, not the quantities.  We will
also use the same demand function, equation (\ref{e3.25}), which implies that if
$p$ is the lowest price, $q = 120 - p$.  In the Cournot model, firms chose
quantities but allowed the market price to vary freely. In  the Bertrand model,
they choose prices and sell as much as they can.

\begin{center}
{\bf The { Bertrand Game} }
\end{center}
{\bf Players}\\
Firms Apex and Brydox

 \noindent {\bf The Order of Play}\\
 Apex  and Brydox simultaneously choose   prices  $p_a$    and $p_b$ from the
set  $[0, \infty)$.

\noindent {\bf Payoffs}\\
 Marginal cost is constant at $c=12$. Demand is a function of the total quantity
sold, $ Q(p) = 120-p.$ The payoff function for Apex (Brydox's would be
analogous) is

  \begin{tabular}{ll}
 $   \;\;\;\;\;\;\;\;  \pi_a =$ & $\left\{ \begin{tabular}{ll}
 $ (120 - p_a)(p_a-c) $ &if $p_a \leq  p_b$  \\
 & \\ $ \frac{(120 - p_a)(p_a-c)}{2}$& if $p_a = p_b $ \\
 & \\ 0        & if $ p_a > p_b$ \\
 \end{tabular} \right.$
\end{tabular}

 The Bertrand Game has a unique Nash equilibrium: $p_a = p_b = c=12$, with $q_a=
q_b=54$.  That this is a weak Nash equilibrium is clear: if either firm deviates
to a higher price, it loses all its customers and so fails to increase its
profits to above  zero.  In fact, this is an example of a Nash equilibrium in
weakly dominated strategies.    That the equilibrium is unique is less clear.
To see why it is, divide the possible  strategy profiles into four groups:
  \begin{enumerate}
 \item[]
\underline{$p_a<c$ or $p_b<c$}.  In either of these cases, the firm with the
lowest price will earn negative profits, and could profitably deviate to a price
high enough to reduce its demand to zero.

 \item[]
\underline{$p_a>p_b>c$ or $p_b>p_a>c$.}  In either of these cases the firm with
the  higher price could deviate to a price below its rival and increase its
profits from zero to some positive value.

 \item[]
 \underline{$p_a=p_b>c$}. In this case,  Apex could deviate to a price
$\epsilon$ less than Brydox and its profit would rise, because it would go from
selling half the market quantity  to selling all of it with an infinitesimal
decline in profit per unit sale.

 \item[] \underline{$p_a>p_b=c$ or $p_b>p_a=c$}. In this case, the firm with the
price of $c$ could move from zero profits to positive profits by increasing its
price  slightly while keeping it below the other firm's price.
 \end{enumerate}

  This proof is a good example of   one common method of  proving uniqueness  of
equilibrium in game theory: partition  the strategy profile space  and show
area by area that deviations would occur. It is such a  good example that I
recommend it to anyone teaching from this book as a good test
question.\footnote{Is it still a good question given that I have just provided a
warning to the students?  Yes. First, it will prove a  filter for discovering
which students have even skimmed the  assigned reading. Second,  questions like
this are  not always easy even if one knows they are on the test. Third,  and
most important, even if in equilibrium every student answers the question
correctly, that very fact shows that the incentive to learn this particular item
has worked -- and that is our main goal, is it not?}

Like the surprising outcome of Prisoner's Dilemma, the Bertrand equilibrium is
less surprising once one thinks about the    model's limitations. What it shows
is that duopoly profits do not arise  just because there are   two firms.
Profits arise from something else, such as multiple periods, incomplete
information, or differentiated products.

   Both the Bertrand and   Cournot models are in common use. The Bertrand model
can be awkward mathematically because of the discontinuous jump from a market
share of 0 to 100 percent after a slight price cut. The Cournot model is useful
as a simple model that avoids this problem  and which predicts that  the price
will  fall gradually as more firms enter the market.   There are also ways to
modify the Bertrand model to obtain   intermediate prices and gradual effects of
entry. Let us  proceed to look at one such modification.

\bigskip \noindent
{\bf The Differentiated Bertrand Game}

\noindent The Bertrand model   generates zero profits because only slight price
discounts are needed to bid away customers.  The assumption behind this is that
the two firms sell identical goods, so   if Apex's price is slightly higher than
Brydox's all the customers go to Brydox.  If customers have brand loyalty or
poor price information, the equilibrium is different.  Let us now move to a
different duopoly market, where the demand curves facing Apex and Brydox are
\begin{equation} \label{e13.21}
 q_a = 24 - 2p_a + p_b
\end{equation}
and
\begin{equation} \label{e13.22}
 q_b = 24 - 2p_b + p_a,
 \end{equation} and they have constant marginal costs of $c=3$.

 The greater the difference in the coefficients on prices in a demand curve like
(\ref{e13.21})or  (\ref{e13.22}), the less substitutable are the products.  As
with standard demand functions  scuh as equation (\ref{e3.25}), we have made
implicit assumptions about the extreme points of equations (\ref{e13.21}) and
(\ref{e13.22}). These equations only apply if the quantities demanded turn out
to be nonnegative, and we might also want to restrict them to prices below some
ceiling, since otherwise the demand facing one firm becomes infinite as the
other's price rises to infinity. A sensible ceiling here is 12, since  if $p_a>
12$ and $p_b = 0$, equation (\ref{e13.21}) would yield a negative quantity
demanded for Apex. Keeping in mind these limitations,  the payoffs are
\begin{equation} \label{e13.22a}
 \pi_a = (24 - 2p_a + p_b) (p_a-c) \end{equation} and \begin{equation}
\label{e13.23}
 \pi_b =   (24 - 2p_b + p_a)(p_b-c).
  \end{equation}

The order of play is the same as in the Bertrand Game (or Undifferentiated
Bertrand Game, as we will call it when that is necessary to avoid confusion):
Apex  and Brydox simultaneously choose prices  $p_a$    and $p_b$ from the  set
$[0, \infty)$.

\noindent
 Maximizing Apex's payoff by choice of $p_a$, we obtain the first- order
condition,
 \begin{equation} \label{e13.24} \frac{ d\pi_a}{d p_a} = 24 - 4p_a + p_b+2c  =
0,
  \end{equation}
and the reaction function,
\begin{equation} \label{e13.25}
 {\displaystyle  p_a = 6 +  \left( \frac{1}{2}  \right) c+ \left(\frac{1}{4}
\right) p_b =  7.5  +\left( \frac{1}{4} \right) p_b. }
 \end{equation}

    Since Brydox has a parallel first-order condition, the equilibrium occurs
where $p_a = p_b =  10.$ The quantity each firm produces is 14, which is below
the 21  each would produce at prices of $p_a=p_b= c= 3$. Figure  4 shows that
the reaction functions intersect. Apex's demand curve has the elasticity
\begin{equation} \label{e13.26}
  \left( \frac{\partial q_a}{\partial p_a} \right) \cdot \left( \frac{p_a}{q_a}
\right) = - 2 \left( \frac{p_a}{q_a} \right), \end{equation} which is  finite
even when $p_a = p_b$, unlike in the undifferentiated-goods Bertrand model.

\includegraphics[width=150mm]{fig03-04.jpg}


\begin{center}
 {\bf Figure  4:  Bertrand Reaction Functions with Differentiated Products  }
\end{center}

  The differentiated-good Bertrand model is   important  because it is often the
most descriptively realistic model of a market. A basic idea in  marketing is
that selling   depends on ``The Four P's'': Product, Place, Promotion, and
Price. Economists have concentrated heavily on price differences between
products, but we realize that differences in product quality and
characteristics, where something is sold, and how  the sellers get  information
about it to the buyers  also matter. Sellers use their prices as control
variables more often than their quantities, but the seller with the lowest price
does not get all the customers.

 Why, then, did I bother to even describe the Cournot and undifferentiated
Bertrand models? Aren't they obsolete? No, because descriptive realism is not
the {\it summum bonum} of modelling. Simplicity matters a lot too. The Cournot
and undifferentiated Bertrand models are simpler, especially when we go to three
or more firms, so they are better models in many applications.

\bigskip \noindent
  {\bf Strategic Substitutes and Strategic Complements}

     You may have noticed an interesting difference between the Cournot and
Differentiated Bertrand reaction curves in Figures 2 and 4: the reaction  curves
have opposite slopes. Figure 5 puts the two together for easier comparison.


\includegraphics[width=150mm]{fig03-05.jpg}

\begin{center} {\bf Figure  5: Cournot vs. Differentiated  Bertrand Reaction
Functions (Strategic Substitutes vs. Strategic Complements) } \end{center}

 In both models, the reaction curves cross once, so there is a unique Nash
equilibrium. Off the equilibrium path, though, there is an interesting
difference. If a Cournot firm increases its output, its rival will do the
opposite and reduce its output. If a Bertrand firm increases its price, its
rival will do the same thing, and increase its price too.

 We can ask of any game: ``If the other players do more of their strategy, will
I do more of my own strategy, or less?'' In some games, the   answer is ``do
more'' and in others it is ``do less''.  Jeremy Bulow, John Geanakoplos  \&
Paul Klemperer (1985) apply the term ``strategic complements''  to  the
strategies in the ``do more''  kind of
game, because when Player 1 does more of his  strategy
that  increases Player 2's marginal  payoff from 2's strategy, just as when I
buy more bread it increases my marginal utility from buying more butter.  If
strategies   are strategic complements,   their reaction curves are upward
sloping, as in the Differentiated Bertrand Game.

 On the other hand, in the ``do less'' kind of game,   when Player 1 does more
of his strategy  that {\it reduces}  Player 2's marginal  payoff from 2's
strategy, just as  my buying potato chips   reduces my marginal utility from
buying more corn chips.  The strategies are therefore ``strategic substitutes''
and  their reaction curves are  downward sloping, as in  the Cournot Game.

  Which way the reaction curves slope also affects whether a player wants to
move first or second.  Esther  Gal-Or (1985) notes that if reaction curves slope
down (as with strategic substitutes and  Cournot) there is a first-mover
advantage, whereas if they slope upwards (as with strategic complements and
Differentiated Bertrand) there is a second-mover advantage.

 We can see that in Figure 5. the Cournot Game in which Player 1 moves first is
simply the Stackelberg Game, which we have already analyzed  using Figure 3. The
equilibrium moves from $E $ to $E^*$ in Figure 5a, Player 1's payoff increases,
and Player 2's payoff falls. Note, too, that the total industry payoff is lower
in Stackelberg than in Cournot-- not only does one player lose, but he loses
more than the other player gains.

   We have not analyzed the Differentiated Bertrand Game when Player 1 moves
first, but since price is a strategic complement, the effect of sequentiality is
very different from in the Cournot Game (and, actually, from the sequential
undifferentiated Bertrand Game-- see the end-of-chapter notes). We cannot tell
what Player 1's optimal strategy is from the diagram alone, but Figure 5
illustrates one possibility. Player 1 chooses  a price  $p^*$ higher than he
would in the simultaneous-move game, predicting that Player 2's response will be
a price somewhat lower than $p^*$, but still greater than the simultaneous
Bertrand price at $E$.  The result is that Player 2's payoff is higher than
Player 1's---a second-mover advantage. Note, however, that    both players are
better off at $E^*$ than at $E$, so both players would favor converting the game
to be sequential.

  Both sequential games could be elaborated further by adding moves beforehand
which would determine which player would choose his price or quantity first, but
I will leave that to you.  The important point for now is that whether a game
has strategic complements or strategic substitutes is hugely important to the
incentives of the players.

     The point is simple enough and important enough that I devote an entire
session of my MBA game theory course to strategic complements and strategic
substitutes.  In the practical game theory that someone with a Master of
Business Administration degree ought to know, the most important thing is to
learn how to describe a situation in terms of players, actions, information, and
payoffs. Often there is not enough data to use a specific functional form, but
it is possible to figure out with a mixture of qualitative and quantitative
information whether the relation between actions and payoffs is one of strategic
substitutes or strategic complements.  The businessman then knows whether, for
example, he should try to be a first mover or a second mover, and whether he
should keep his action secret or proclaim his action to the entire world.

    To understand the usefulness of the idea of strategic complements and
substitutes, think about how you would model   situations like the following
(note that there is no universally right answer for any of them):
  \begin{enumerate} \item Two firms are choosing their  research and development
budgets. Are the budgets strategic complements or strategic substitutes?

\item Smith and Jones are both trying to be elected President of the United
States. Each must decide how much he will spend on advertising in California.
Are the  advertising budgets strategic complements or strategic substitutes?

\item Seven  firms are each deciding whether to make their products more
special, or more suited to the average consumer.  Is the degree of specialness a
strategic complement  or a strategic substitute?

 \item India and Pakistan are each deciding whether to make their armies larger
or smaller. Is  army size a   strategic complement  or a strategic substitute?



\end{enumerate}

   Economists have a growing appreciation of how powerful the ideas of
substitution and complementarity can be in thinking about the deep structure of
economic behavior. The  mathematical idea of supermodularity, to be discussed in
Chapter 14, is all about complementarity. For an inspiring survey, see
Vives  (2005).



\bigskip \noindent
  { \bf  *3.7 Existence of Equilibrium }

\noindent One of the strong points of Nash equilibria is that  they  exist  in
practically every game one is likely to encounter.   There are four common
reasons why an equilibrium might not exist or might only exist in mixed
strategies.

\bigskip

\noindent
{\bf  (1) An unbounded strategy space }

  Suppose in a stock market game that Smith can borrow money and buy as many
shares  $x$ of stock as he likes, so his strategy set, the amount of stock he
can buy, is $[0, \infty)$, a set which is unbounded above. (Note, by the way,
that we thus assume that he can buy fractional shares, e.g. $x=13.4$,  but
cannot sell short, e.g. $x=-100$.)

 If Smith knows that the price  is lower today  than it will be tomorrow, his
payoff function will be $\pi(x) =x$ and  he will want to buy an infinite number
of shares, which is not an equilibrium purchase.  If the amount he buys is
restricted to be   less than or equal to 1,000, however, then the strategy set
is bounded (by 1,000), and an equilibrium exists--- $x=1,000$.

 Sometimes, as in the  { Cournot Game}  discussed earlier  in this chapter, the
unboundedness of the strategy sets does not matter because the optimum is an
interior solution.  In other games, though, it is important, not just to get a
determinate solution but because    the real world is a rather bounded place.
The   solar system is finite in size, as  is the amount of  human  time past and
future.

\bigskip

\noindent
{\bf  (2) An open  strategy space }

 Again consider Smith. Let his strategy be $x \in [0, 1,000)$, which is the same
as saying that $ 0 \leq x <1,000$,  and his payoff function be $\pi(x) =x$.
Smith's strategy set is bounded (by 0 and 1,000), but it is open rather than
closed, because he can choose any number less than 1,000, but not 1, 000 itself.
This means no equilibrium will exist, because he wants to buy 999.999$\ldots$
shares.   This is just a technical problem; we ought to have specified Smith
strategy space to be $ [0,1,000]$, and then an equilibrium would exist, at $x=
1,000$.

\bigskip

\noindent
{\bf (3) A discrete strategy  space (or, more generally, a nonconvex strategy
space)  }

 Suppose we start with an arbitrary pair of strategies $s_1$ and $s_2$ for two
players. If the players' strategies are strategic complements, then if player 1
increases his strategy in response to $s_2$, then player 2 will increase his
strategy in response to that. An equilibrium will occur where the players run
into diminishing returns or increasing costs, or where they hit the upper bounds
of their strategy sets.  If, on the other hand, the strategies are strategic
substitutes, then if player 1 increases his strategy in response to $s_2$,
player 2 will in turn want to  reduce his strategy. If the strategy spaces are
continuous, this can lead to an equilibrium, but if they are discrete, player 2
cannot reduce his strategy just a little bit-- he has to jump down a discrete
level. That could then induce Player 1 to increase his strategy by a discrete
amount. This jumping of responses can be never- ending--there is no equilibrium.

   That is what is happening in the Welfare Game of Table 1 in this chapter. No
compromise is possible between a little aid and no aid, or between working and
not working-- until we introduce mixed strategies. That allows for each player
to choose a continuous  amount of his strategy.

   This problem is not limited to games such as 2-by-2 games  that have
discrete strategy spaces. Rather, it is a problem of ``gaps'' in the strategy
space. Suppose we had a game in which the government was not limited to amount 0
or 100 of aid, but could choose any amount in the space $\{[0, 10], [90, 100]
\}.$ That is a continuous, closed, and bounded strategy space, but it is non-
convex-- there is gap in it. (For a space $\{x\}$ to be  convex, it must be true
that if $x_1$ and $x_2$ are in the space, so is $\theta x_1 + (1-\theta) x_2$
for any $\theta \in [0, 1]$.) Without mixed strategies, an equilibrium to the
game might well not exist.

\bigskip

\noindent
{\bf  (4)  A discontinuous reaction function arising from nonconcave or
discontinuous payoff functions  }

   Even if the strategy spaces are closed, bounded, and convex, a problem
remains. For a Nash equilibrium to exist, we need for the reaction functions of
the players to intersect. If the reaction functions are discontinuous, they
might not intersect.

 Figure 6 shows this for a  two-player game in which each player chooses a
strategy from the interval between 0 and 1.  Player 1's reaction function,
$s_1(s_2)$, must pick one or more value of $s_1$ for each possible value of
$s_2$, so it must cross from the bottom to the top of the diagram.   Player 2's
reaction function, $s_2(s_1)$, must pick one or more value of $s_2$ for each
possible value of $s_1$, so it must cross from the left  to the  right of the
diagram. If the strategy sets were unbounded or open, the reaction functions
might not exist, but that is not a problem here: they do exist. And in Panel (a)
a Nash equilibrium exists, at the point, $E$,  where the two reaction functions
intersect.

 In Panel (b), however, no Nash equilibrium exists. The problem is that  Firm
2's reaction function $s_2(s_1)$ is discontinuous at the point  $ s_1=0.5.$  It
jumps down from $s_2(0.5)=0.6$ to $s_2(0.50001)=0.4$. As a result, the reaction
curves never intersect, and no equilibrium exists.

 If the two players can use mixed strategies, then an equilibrium will exist
even for the game in Panel (b), though I will not prove that here. I would,
however, like to say why it is that the reaction function might be
discontinuous. A player's reaction functions, remember,  is derived by
maximizing his payoff  as a function of his own strategy given the strategies of
the other players.

Thus, a first reason why  Player 1's reaction function might be discontinuous in
the other players' strategies   is that   his payoff function is discontinuous
in either his own or the other players' strategies. This is what happens in
Chapter 14's Hotelling Pricing Game, where if Player 1's price drops enough (or
Player 2's price rises high enough),  all of Player 2's customers suddenly rush
to Player 1.

 A second reason why  Player 1's reaction function might be discontinuous in the
other players' strategies   is that his payoff function is  not concave. The
intuition is that if an objective function is not concave, then there might be a
number of maxima that are local but not global, and as the parameters change,
which maximum is the global one can suddenly change.   This means that the
reaction function will suddenly jump from one maximizing choice to another one
that is far-distant, rather than smoothly changing as it would in a more nicely
behaved problem.

\includegraphics[width=150mm]{fig03-06.jpg}

\begin{center} {\bf Figure  6:   Continuous and Discontinuous Reaction Functions
} \end{center}

\bigskip

   Problems (1) and (2) are really  problems in decision theory, not game
theory, because unboundedness and openness lead to nonexistence of the solution
to even a   one-player maximization problem.  Problems (3) and (4) are special
to game theory. They arise because although each player has a best response to
the other players, no profile of best choices is such that everybody has chosen
his best response to everybody else.  They are  similar to the decision theory
problem of nonexistence of an interior solution, but if only one player were
involved, we would at least have a corner solution.

\bigskip

      In this chapter, I have introduced a number of seemingly disparate ideas--
mixed strategies, auditing, continuous strategy spaces, reaction curves,
complentary substitutes and complements, existence of equilibrium...  What ties
them together? The unifying theme is the possibility of   reaching equilibrium
by small changes in behavior, whether that be by changing the probability in  a
mixed strategy or an auditing game or by changing the level of a continuous
price or quantity. Continuous strategies free us from the need to use $n$-by-$n$
tables to  predict behavior in games, and  with a few technical assumptions they
guarantee we will find equilibria.

\newpage

\begin{small}

\bigskip \noindent {\bf NOTES}

           \noindent {\bf N3.1 } {\bf Mixed Strategies: The { Welfare Game} }
\begin{itemize} \item Waldegrave (1713) is  a very early reference to mixed
strategies.

\item
    Mixed strategies come up constantly in recreational games. The  Scissors-
Paper-Stone choosing game, for example, has a unique mixed-strategy equilibrium,
as shown in   Fisher \& Ryan
(1992). There are usually analogs in other spheres; it turns out that three
types of  Californian side-blotched lizard males play the same game, as reported
in Sinevero \& Lively (1996).  It is interesting to  try to see how closely
players come to the theoretically optimal mixing strategies.  Chiappori, Levitt
\& Groseclose (2002)  conclude that players choosing whether to kick right or
left in soccer penalty kicks are following   optimal mixed strategies, but  that
kickers are heterogeneous in their abilities to kick in each direction.



\item The January 1992 issue of {\it Rationality and Society} is devoted to
attacks on and defenses of the use of game theory in the social sciences, with
considerable discussion of mixed strategies and multiple equilibria.
Contributors include   Harsanyi, Myerson,  Rapaport, Tullock,  and Wildavsky.
The Spring 1989 issue of the {\it RAND Journal of Economics} also has an
exchange on the use of game theory, between Franklin Fisher and Carl Shapiro.  I
also recommend the  Peltzman (1991) attack on the game theory approach to
industrial organization  in the spirit of   the ``Chicago School''.

 \item In this book  it will  always  be assumed that players remember their
previous moves.  Without this assumption of {\bf perfect recall}, the definition
in the text is not that for a mixed strategy, but for a {\bf behavior strategy}.
As historically defined, a player pursues a mixed strategy when he randomly
chooses between pure strategies at the starting node, but he plays a pure
strategy thereafter.  Under that definition, the modeller cannot talk of random
choices at any but the starting node.  Kuhn (1953) showed that the definition of
mixed strategy given in the text is equivalent to the original definition if the
game has perfect recall. Since all important games have perfect recall and the
new definition of mixed strategy is better in keeping with the modern spirit of
sequential rationality, I have abandoned the old definition.

 $\;\;\;$ The classic example of a game without perfect recall is {\bf bridge},
where the four players of the actual game can be cutely modelled as two players
who forget what half their cards look like at any one time in the bidding. A
more useful example is a game that has been simplified by restricting players to
Markov strategies (see Section 5.4), but usually the modeller sets up such a
game with perfect recall and then rules out non-Markov equilibria after showing
that the Markov strategies form an equilibrium for the general game.

 

 \item It is {\it not} true that when two pure-strategy equilibria exist a
player would be just as willing to use a strategy mixing the two  even when the
other player is using a pure strategy. In  {Battle of the Sexes}, for instance,
if the man knows the woman is going to the ballet he is not indifferent between
the ballet and the prize fight.

\item A continuum of players is useful not only because the modeller need not
worry about fractions of players, but because he can use more modelling tools
from calculus--- taking the integral of the quantities demanded by different
consumers, for example, rather than the sum. But using a continuum is also
mathematically more difficult: see Aumann (1964a, 1964b).

 \item There is an entire literature on the econometrics of estimating game
theory models.  Suppose we would like to estimate the payoff numbers in a 2-by-2
game, where we observe the actions taken by each of the two players and various
background variables.  The two actions might be, for example, to enter or not
enter, and the background variables might be such things as the size of the
market or the cost conditions facing one of the players. We will of course need
multiple repetitions of the situation to generate enough data to use
econometrics.  There is an identification problem, because there are eight
payoffs in a 2-by-2 payoff matrix, but only four possible action profiles-- and
if   mixed strategies are  being used, the four mixing probabilities have to add
up to one, so there are really only three independent observed outcomes. How can
we estimate 8 parameters with only 3 possible outcomes? For identification, it
must be that some environmental variables affect only one of the players, as
Bajari, Hong \& Ryan (2004) note. In addition, there is the problem that there
may be multiple equilibria being played out, so that additional identifying
assumptions are needed to help us know which equilibria are being played out in
which observations. The  foundational articles in this literature are Bresnahan
\& Reiss (1990, 1991a),  and it is an active area of research.

  \end{itemize}

\bigskip \noindent {\bf N3.2} {\bf   The Payoff-Equating Method and Games of
Timing} \begin{itemize}

\item The game of   { Chicken}  discussed in the text is simpler than the game
acted out in the movie {\it Rebel Without a Cause,} in which the players race
towards a cliff and the winner is the player who jumps out of his car last. The
pure-strategy space in the movie game is continuous and the payoffs are
discontinuous at the cliff's edge, which makes the game more difficult to
analyze technically. (Recall, too,  the importance in the movie  of a disastrous
mistake-- the kind of ``tremble'' that Section 4.1 will discuss.)

 \item Technical difficulties arise in some models with a continuum of actions
and mixed strategies.  In the { Welfare Game}, the government chose a single
number, a probability, on the continuum from zero to one.  If we allowed the
government to mix over a continuum of aid levels, it would choose a function, a
probability density, over the continuum.  The original game has a finite number
of elements in its strategy set, so its mixed extension still has a strategy
space in ${\bf R}^n$. But with a continuous strategy set extended by a continuum
of mixed strategies for each pure strategy, the mathematics become difficult.  A
finite number of mixed strategies can be allowed without much problem, but
usually that is not satisfactory.

$\;\;\;$ Games in continuous time frequently run into this problem. Sometimes it
can be avoided by clever modelling, as in Fudenberg \& Tirole's (1986b)
continuous-time war of attrition with asymmetric information.  They specify as
strategies the length of time firms would proceed to $Continue$ given their
beliefs about the type of the other player, in which case there is a pure
strategy equilibrium.

\item {\bf Differential games} are played in continuous time. The action is a
function describing the value of a state variable at each instant, so the
strategy maps the game's past history to such a function.  Differential games
are solved using dynamic optimization. A book-length treatment is  Bagchi
(1984).

\item Fudenberg \& Levine (1986) show circumstances under which the equilibria
of games with infinite strategy spaces can be found as the limits of equilibria
of games with finite strategy spaces.

 \end{itemize}

\bigskip \noindent {\bf N3.4} {\bf  Randomizing versus Mixing: The Auditing Game
} \begin{itemize}
 \item { Auditing Game I}  is similar to  a game called     The Police Game.
Care must be taken  in such games that one does not  use a simultaneous- move
game when a  sequential game is appropriate. Also,  discrete strategy spaces
can be misleading.  In general, economic analysis assumes  that costs rise
convexly in the amount of an activity and benefits rise concavely.   Modelling a
situation with a 2-by-2 game uses just two discrete levels of the activity, so
the concavity or convexity is lost in the simplification. If the true functions
are linear, as in auditing costs which rise linearly with the probability of
auditing, this is no great loss.  If the true costs rise convexly, as in the
case where the hours a policeman must stay on the street each day are increased,
then a 2-by-2 model can be misleading. Be especially careful not to press the
idea of a mixed-strategy equilibrium too hard if  a pure-strategy equilibrium
would exist when intermediate strategies are allowed. See   Tsebelis (1989) and
the criticism of it in  Jack Hirshleifer \& Rasmusen (1992).

 \item Douglas Diamond (1984) shows the implications of monitoring costs for the
structure of financial markets.  A fixed cost to monitoring investments
motivates the creation of a financial intermediary to avoid repetititive
monitoring by many investors.

\item Baron \& Besanko (1984) study auditing in the context of a government
agency which can at some cost collect information on the true production costs
of a regulated firm.

\item Mookherjee \& Png (1989) and Border \& Sobel (1987) have examined random
auditing in the context of taxation. They find that if a taxpayer is audited he
ought to be more than compensated for his trouble  if it turns out he was
telling the truth. Under the optimal contract, the truth-telling taxpayer should
be delighted to hear that he is being audited.  The reason is that a reward for
truthfulness widens the differential between the agent's payoff when he tells
the truth and when he lies.

Why is such a scheme not used? It is certainly practical, and one would think it
would be popular with the voters. One reason might be the possibility of
corruption; if being audited leads to a lucrative reward, the government might
purposely choose to audit its friends. The current danger seems even worse,
though, since the government can audit its enemies and  burden them with the
trouble of an audit even if they have paid their taxes properly.

 \item
 Government action strongly affects what information is available as well as
what is contractible. In 1988, for example, the United States passed a law
sharply restricting the use of lie detectors for testing or monitoring. Previous
to the restriction, about two million workers had been tested each year. (``Law
Limiting Use of Lie Detectors is Seen Having Widespread Effect'' {\it Wall
Street Journal}, p. 13, 1 July 1988), ``American Polygraph Association,'' http:
//www.polygraph.org/betasite/menu8.html, Eric Rasmusen, ``Bans on Lie Detector
Tests,'' http://mypage.iu.edu/$\sim$erasmuse/archives1.htm\#august10a.)

\item Section 3.4 shows how random  actions come up in auditing and in mixed
strategies. Another use for randomness is to reduce transactions costs.  In
1983, for example, Chrysler was bargaining over how much to pay Volkswagen for a
Detroit factory. The two negotiators locked themselves into a hotel room and
agreed not to leave till they had an agreement. When they narrowed the price gap
from \$100 million to \$5 million, they agreed to flip a coin. (Chrysler won.)
How would you model that?  ``Chrysler Hits Brakes, Starts Saving Money after
Shopping Spree, '' {\it Wall Street Journal}, p. 1,  12 January 1988. See also
David Friedman's ingenious idea in Chapter 15 of {\it Law's Order} of using a
10\% probability of death to replace a 6-year prison term   (http:
//www.daviddfriedman.com/Academic/Course\_Pages/L$\_$and$\_$E$\_$LS$\_ $9 8
/Why$\_$Is $\_$aw/
Why$\_$Is$\_$Law$\_$Chapter$\_$15/Why$\_$Is$\_$Law$\_$Chapter$\_$15.ht ml)

  \end{itemize}

 \noindent {\bf N3.5} {\bf Continuous Strategies: The { Cournot Game} }
\begin{itemize} \item An interesting class of simple continuous payoff games are
the {\bf Colonel Blotto games} (Tukey [1949], McDonald \& Tukey [1949]). In
these games, two military commanders allocate their forces to $m$ different
battlefields, and a battlefield contributes more to the payoff of the commander
with the greater forces there. A distinguishing characteristic is that player
$i$'s payoff increases with the value of player $i$'s particular action relative
to player $j$'s, and $i$'s actions are subject to a budget constraint. Except
for the budget constraint, this is similar to the  tournaments     of Section
8.2.



 \item ``Stability'' is a word used in many different ways in game theory and
economics.  The natural meaning of a stable equilibrium is that it has dynamics
which cause the system to return to that point after being perturbed slightly,
and the discussion of the stability of Cournot equilibrium was in that spirit.
The uses of the term by von Neumann \& Morgenstern (1944) and Kohlberg \&
Mertens (1986) are entirely different.

  \item The term  ``Stackelberg equilibrium'' is not clearly defined in the
literature.  It is sometimes used to denote equilibria in which players take
actions in a given order, but since that is just the perfect equilibrium (see
Section 4.1) of a well-specified extensive form, I prefer to reserve the term
for the Nash equilibrium of the duopoly quantity game in which one player moves
first, which is the context of Chapter 3 of Stackelberg (1934).

$\;\;\;$ An alternative definition is that a Stackelberg equilibrium is a
strategy profile in which players select strategies in a given order and in
which each player's strategy is a best response to the fixed strategies of the
players preceding him and the yet-to-be-chosen strategies of players succeeding
him, i.e., a situation in which players precommit to strategies in a given
order. Such an equilibrium would not generally be either Nash or perfect.

\item Stackelberg (1934) suggested that sometimes the players are confused about
which of them is the leader and which the follower, resulting in the
disequilibrium outcome called {\bf Stackelberg warfare}.

 \item With linear costs and demand, total output is greater in Stackelberg
equilibrium than in Cournot.  The slope of the reaction curve is less than one,
so Apex's output expands more than Brydox's contracts.  Total output being
greater, the price is less than in the Cournot equilibrium.

\item A useful application of Stackelberg equilibrium is to an industry with a
dominant firm and a {\bf competitive fringe} of smaller firms that sell at
capacity if the price exceeds their marginal cost. These smaller firms act as
Stackelberg leaders (not followers), since each is small enough to ignore its
effect on the behavior of the dominant firm.  The oil market could be modelled
this way with OPEC as the dominant firm and producers such as Britain on the
fringe. \end{itemize}

\bigskip \noindent
{ \bf  N3.6 Continuous Strategies: The Bertrand Game, Strategic Complements, and
Strategic Substitutes    }

  \begin{itemize}

\item The text analyzed the simultaneous undifferentiated Bertrand game but not
the sequential one.  $p_a=p_c=c$ remains an equilibrium outcome, but it is no
longer unique. Suppose Apex moves first, then Brydox, and suppose, for a
technical reason to be apparent shortly, that if $p_a= p_b$ Brydox captures the
entire market. Apex cannot achieve more than a payoff of zero, because either
$p_a=c$ or Brydox will choose $p_b= p_a$ and capture the entire market. Thus,
Apex is indifferent between any $p_a \geq c$.

   The game needs to be set up with this tiebreaking rule because if split the
market between Apex and Brydox when $p_a=p_b$, Brydox's best response to $p_a>c$
would be to choose $p_b$ to be the biggest number less than $p_a$-- but with a
continuous space, no such number exists, so Brydox's best response is ill-
defined. Giving all the demand to Brydox in case of price ties gets around this
problem.

\item
The demand curves  (\ref{e13.21}) and (\ref{e13.22})     can   be generated   by
a quadratic utility function. Dixit (1979) tells us that  with respect to  three
goods 0, 1, and 2, the utility function
\begin{equation} \label{e13.75}
 U = q_0 + \alpha_1 q_1 + \alpha_2 q_2 - \frac{1}{2} \left(\beta_1 q_1^2 + 2
\gamma q_1 q_2 + \beta_2 q_2^2 \right)
\end{equation}
 (where the constants $\alpha_1,\alpha_2, \beta_1$, and $\beta_2$ are positive
and $\gamma^2 \leq \beta_1 \beta_2$) generates the inverse demand functions
\begin{equation} \label{e13.76}
 p_1 = \alpha_1 - \beta_1 q_1 - \gamma q_2
\end{equation}
and
\begin{equation} \label{e13.77}
 p_2 = \alpha_2 - \beta_2 q_2 - \gamma q_1.
 \end{equation}

\item
 We can also work out the Cournot equilibrium for demand functions
(\ref{e13.21}) and (\ref{e13.22}), but product differentiation does not affect
it much. Start by expressing the price in the demand curve in terms of
quantities alone, obtaining
\begin{equation} \label{e13.28}
p_a = 12 -  \left( \frac{1}{2} \right) q_a +\left(\frac{1}{2}\right) p_b
 \end{equation}
and \begin{equation} \label{e13.29}
 p_b = 12 - \left(\frac{1}{2}\right)q_b +\left( \frac{1}{2}\right)p_a.
\end{equation}
 After substituting from (\ref{e13.29}) into (\ref{e13.28}) and solving for
$p_a$, we obtain \begin{equation} \label{e13.30}
  p_a = 24 -\left( \frac{2}{3}\right)q_a -\left( \frac{1}{3}\right)q_b.
 \end{equation}
 The first-order condition for Apex's maximization problem is \begin{equation}
\label{e13.30a}
 \frac{d \pi_a}{dq_a} = 24-3  - \left(\frac{4}{3}\right)q_a - \left(\frac{1}{3}
\right)q_b = 0,
 \end{equation} which gives rise to the reaction function \begin{equation}
\label{e13.31}
  q_a = 15.75 -\left( \frac{1}{4}\right)q_b.
 \end{equation}
 We can guess that $q_a = q_b$. It follows from (\ref{e13.31}) that $q_a = 12.6$
and the market price is  11.4. On checking, you would find this to indeed be a
Nash equilibrium.  But reaction function (\ref{e13.31}) has much the same shape
as if there were no product differentiation, unlike when we moved from
undifferentiated   to differentiated Bertrand competition.

\item For more on the technicalities of strategic complements and strategic
substitutes, see    Bulow, Geanakoplos  \&    Klemperer (1985)  and Milgrom \&
Roberts (1990). If the strategies are strategic complements, Milgrom \& Roberts
(1990) and Vives (1990) show that pure-strategy equilibria exist.  These models
often explain peculiar economic phenomenon nicely, as in Peter Diamond (1982) on
search and business cycles and  Douglas Diamond \&  Dybvig (1983) on bank runs.
If the strategies are strategic substitutes, existence of pure-strategy
equilibria is more troublesome; see   Dubey,     Haimanko \&  Zapechelnyuk
(2005).

 \end{itemize}

\newpage

\noindent {\bf Problems}

  \bigskip \noindent {\bf 3.1. Presidential Primaries } (medium)\\
Smith and Jones are fighting it out for the Democratic nomination for President
of the United States. The more months they keep fighting, the more money they
spend, because a candidate must spend one million dollars a month in order to
stay in the race.  If one of them drops out, the other one wins the nomination,
which is worth 11 million dollars. The discount rate is $r$ per month. To
simplify the problem, you may assume that this battle could go on forever if
neither of them drops out.  Let $\theta$ denote the probability that an
individual player will drop out each month in the mixed-strategy equilibrium.

 \begin{enumerate}

  \item[(a)] In the mixed-strategy equilibrium, what is the probability $\theta$
each month that Smith will drop out?  What happens if $r$ changes from 0.1 to
0.15?

 \item[(b)] What are the two pure-strategy equilibria?

\item[(c)] 
 If the game only lasts one period, and the Republican wins the
general election  if both Democrats refuse to
give up (resulting in  Democrat payoffs of zero), what is the probability
$\gamma$ with which each Democrat drops out   in a
symmetric equilibrium?

   \end{enumerate}

  \bigskip \noindent 
{\bf 3.2. Running from the Police    } (medium)\\ 
 Two risk-neutral
men, Schmidt and Braun, are walking south along a street in Nazi Germany when
they see a single policeman coming to check their papers. Only Braun has his
papers (unknown to the policeman, of course).  The policeman will catch both men
if both or neither of them run north, but if just one runs, he must choose which
one to stop--- the walker or the runner.  The penalty for being without papers
is 24 months in prison.  The penalty for running away from a policeman is 24
months in prison, on top of the sentences for any other charges, but the
conviction rate for this offense is only 25 percent.   The two friends want to
maximize their joint welfare, which the policeman wants to minimize.  Braun
moves first, then Schmidt, then the policeman.
  \begin{enumerate} \item[(a)] What
is the outcome matrix for outcomes that might be observed in equilibrium? (Use
$\theta$ for the probability that the policeman chases the runner and $\gamma$
for the probability that Braun runs.)

\item[(b)] What is the probability  that the policeman chases the runner, (call
it $\theta^*$)?
\item[(c)] What is the probability  that Braun runs, (call it $\gamma^*$)?
 \item[(d)] Since Schmidt and Braun share the same objectives, is this a
cooperative game?

\end{enumerate}

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

 \bigskip

\noindent
{\bf  3.3:   Uniqueness in Matching Pennies    } (easy)\\
In the game Matching Pennies, Smith and Jones each show a penny with either
heads or tails up. If they choose the same side of the penny, Smith gets both
pennies; otherwise, Jones gets them.

 \begin{enumerate} \item [(a)] Draw the outcome matrix for Matching Pennies.
\item[(b)] Show that  there is no Nash equilibrium in pure strategies.
\item[(c)] Find the mixed-strategy equilibrium, denoting  Smith's probability of
$Heads$ by $\gamma$ and  Jones's  by $\theta$. \item[(d)] Prove that there is
only one mixed-strategy equilibrium.

   \end{enumerate}

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

  \bigskip

 \noindent 
{\bf 3.4. Mixed Strategies in   the  { Battle of the Sexes} 
} (medium)\\
 Refer back to  The   { Battle of the Sexes}  and   { Ranked Coordination}.
  Denote the probabilities that the man and woman pick {\it Prize
Fight} by $\gamma$ and $\theta$.
  \begin{enumerate}

 \item[(a)] Find an expression for the man's expected payoff.

\item[(b)] What are the equilibrium values of $\gamma$ and $\theta$, and the
expected payoffs?

 \item[(c)] Find the most likely  outcome and  its probability.

\item[(d)] What is the equilibrium payoff in the mixed-strategy equilibrium for
{Ranked Coordination}?

\item[(e)] Why is the mixed-strategy equilibrium a better focal point in   The
{ Battle of the Sexes} than in {Ranked Coordination?   }

 \end{enumerate}

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

 \bigskip
 \noindent
{\bf  3.5. A Voting Paradox } (medium)\\
 Adam, Karl, and Vladimir are the only three voters in Podunk. Only Adam owns
property. There is a proposition on the ballot to tax property-holders 120
dollars and distribute the proceeds equally among all citizens who do not own
property.  Each citizen dislikes having to  go to the polling place and vote
(despite the short lines), and would pay 20 dollars to avoid voting.  They all
must decide whether to vote before going to work. The proposition fails if the
vote is tied.  Assume that in equilibrium Adam votes with probability $\theta$
and Karl and Vladimir each vote with the same probability $\gamma$, but they
decide to vote independently of each other.

\begin{enumerate}

 \item[(a)] What is the probability that the proposition will pass, as a
function of $\theta$ and $\gamma$?

 \item[(b)] What are the two possible equilibrium probabilities $\gamma_1 $ and
$\gamma_2$ with which Karl might vote? Why, intuitively, are there two symmetric
equilibria?

\item[(c)] What is the probability $\theta$ that Adam will vote in each of the
two symmetric equilibria?

\item[(d)] What is the probability that the proposition will pass?

 \end{enumerate}



\bigskip 

\noindent
 {\bf 3.6. Rent Seeking  }(hard) \\
 I mentioned that  Rogerson (1982) uses a game very similar to ``Patent Race for
a New Market''   to analyze competition for a government monopoly franchise.
See if you can do this too. What can you predict about the welfare results of
such competition?




\bigskip 
\noindent
 {\bf 3.7.  Nash Equilibrium} (easy) \\
Find the  unique Nash equilibrium of the   game in Table 9.

   \begin{center} {\bf   Table 9: A Meaningless Game}

 \begin{tabular}{lllccccc} &       &             &\multicolumn{5}{c}{\bf Column}
\\ &       &   &  {\it  Left }   &   &  $Middle$   & &$Right$   \\
 &   && & &  &
\\ &   &  $ Up$       &  1,0 &    &  $10,-1 $&  & $0, 1$\\
 &  && & &  & \\ &
{\bf Row:} &    {\it Sideways }    &    $ -1,0$  &    & -2,-2 & & $-12,4$ \\
 &
&& & &  & \\ &  &    {\it Down }    &      $ 0,2$  &    & 823,$-1$  & & $ 2,0$
\\ &   && & &  & \\ \multicolumn{8}{l}{\it Payoffs to: (Row, Column).}
\end{tabular} \end{center}

\bigskip

 \noindent
 {\bf 3.8. Triopoly }  (easy) \\
 Three companies provide tires to the Australian market. The  total cost curve
for  a firm making $Q$  tires is    $TC =  5 + 20Q$, and the demand equation is
$P =  100-N$, where $N$ is the total number of tires on the market.

According to the Cournot model, in which the firms's simultaneously choose
quantities,  what will the total industry output   be?

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

\bigskip
 \noindent
 {\bf 3.9.    Cournot with Heterogeneous Costs } (hard)\\ 
  On a
 seminar  visit, Professor Schaffer of Michigan told me that in a Cournot model
with a
linear demand curve $P=\alpha -\beta Q$ and constant marginal cost $C_i$ for
firm $i$, the equilibrium industry  output $Q$ depends on $\Sigma_{i} C_i$, but
not on the individual levels of $C_i$.  I may have misremembered. Prove or
disprove this assertion.    Would your conclusion be altered if we made some
other assumption on demand? Discuss.

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

\bigskip

\noindent
{\bf 3.10. Alba and Rome: Asymmetric Information and Mixed Strategies } (medium)
\\
  A Roman, Horatius, unwounded, is fighting the three Curiatius brothers from
Alba, each of whom is wounded.  If Horatius continues fighting, he wins with
probability 0.1, and the payoffs are (10,-10) for (Horatius, Curiatii) if he
wins, and (-10,10) if he loses. With probability $\alpha = 0.5$, Horatius is
panic-stricken and runs away. If he runs and the Curiatii do not chase him, the
payoffs are (-20, 10).  If he runs and the Curiatius brothers chase and kill
him, the payoffs are (-21, 20). If, however, he is not panic-stricken, but he
runs anyway and the Curiatii give chase,   he is able to kill the fastest
brother first and  then  dispose of the other two, for payoffs of (10,-10).
Horatius is, in fact, not panic-stricken.

 \begin{enumerate} \item[(a)] With what probability $\theta$ would the Curiatii
give chase if Horatius were to run?

\item[(b)] With what probability $\gamma$ does Horatius run?

 \item[(c)] How would $\theta$ and $\gamma$ be affected if the Curiatii falsely
believed that the probability of Horatius being panic-stricken was 1?  What if
they believed it was 0.9?

 \end{enumerate}

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

\bigskip \noindent
 {\bf 3.11.  Finding Nash Equilibria } (easy) \\
 Find all  of the Nash equilibria for the game of Table 10.

  \begin{center} {\bf Table 10:   A  Takeover Game  }

 \begin{tabular}{lllccccc} &       &             &\multicolumn{5}{c}{\bf
Target}\\ &       &   &  {\it  Hard }   &   &  $Medium$   & &$Soft$   \\ &   &&
& &  & \\ &   &  $Hard $       &  -3, -3 &    &  $-1,0$&  & $4,0$\\ &  && & &  &
\\ & {\bf Raider:} &    {\it Medium}    &    $ 0, 0$  &    & {2,2} & & $3,1$ \\
&   && & &  & \\ &  &    {\it  Soft }    &     { 0,0}  &    &  $2,4$  & & $ 3,3$
\\ &   && & &  & \\ \multicolumn{8}{l}{\it Payoffs to: (Raider, Target).}
\end{tabular} \end{center}

\bigskip \noindent {\bf 3.12.    Risky Skating } (hard)\\
 Elena and  Mary  are the two leading figure skaters  in the world.  Each must
choose   during her training  what her routine is  going to look like.  They
cannot    change their minds later and try to  alter  any details of their
routines. Elena goes first in the Olympics, and Mary goes next. Each has five
minutes for her performance.  The judges will rate the routines on three
dimensions,  beauty,   how high they jump, and whether they stumble after they
jump.  A skater who stumbles is sure to lose, and if both Elena and Mary
stumble, one of the ten lesser skaters will win, though those ten skaters  have
no chance otherwise.

     Elena and  Mary are exactly equal in the  beauty of  their routines, and
both of them know this, but they are not  equal in their jumping ability.
Whoever jumps higher without stumbling will  definitely win.  Elena's
probability of  stumbling is  $P(h) $, where h is the height of the jump, and P
is increasing smoothly and continuously in $h$.  (In calculus terms, let  $P'$
and $P''$ both exist, and $P'$ be positive)  Mary's probability is  $P(h) -
0.1$--- that is, it is 10 percent less  for  equal heights.

Let us define as h=0 the maximum height that the lesser skaters can achieve, and
assume that P(0) = 0.

\begin{enumerate}

\item[(a)] Show that  it cannot be an equilibrium for both Mary and Elena to
choose the same value for $h$ (Call them $M$ and  $E $).

\item[(b)] Show for any pair of values $(M, E) $   that  it cannot be an
equilibrium for  Mary and Elena to choose   those values.

\item[(c)] Describe the optimal strategies to the best of your ability. (Do not
get hung up on trying to answer this question; my expectations are not high
here.)

\item[(d)] What is a business analogy?  Find some situation in business or
economics that could use this same model.

\end{enumerate}

\bigskip

\noindent
{\bf 3.13. The Kosovo War   } (easy) \\
Senator Robert Smith of New Hampshire said  of  the US policy in Serbia of
bombing but promising not to use ground forces, ``It's like saying we'll pass on
you but we won't run the football.''  ({ \it Human Events}, p. 1, April 16,
1999.)  Explain what he meant,  and why this is a strong criticism of U.S.
policy, using the concept of a mixed strategy equilibrium.   (Foreign students:
in American football,  a team can choose to throw the football (to pass it) or
to hold it and run with it to move towards the goal.)  Construct a  numerical
example to  compare the U.S. expected  payoff  in (a)  a mixed strategy
equilibrium in which it ends up not using ground forces, and (b) a pure strategy
equilibrium in which the U.S. has committed not to use ground forces.

\bigskip \noindent {\bf 3.14. IMF Aid } (easy) \\ 
Consider the game of Table 11.

\begin{center} {\bf Table 11: IMF Aid } 

 \begin{tabular}{lllccc} &          &
&\multicolumn{3}{c}{\bf Debtor}\\
 &      &       & Reform   &    &  Waste \\ 
&  &  Aid       &    3,2   &    &  -1,3 \\
 & {\bf IMF}    &          &    &     &   \\
 & & No Aid     & -1,1       &  &   0,0 \\
 &      &          &    &     &   \\
  \multicolumn{6}{l}{\it Payoffs to: (IMF, Debtor).}\\
 \end{tabular}\\
 \end{center}

\begin{enumerate}
 \item[(a)] What is the exact form of every  Nash equilibrium?

\item[(b)] For what story would this matrix be a good model?

\end{enumerate}

\bigskip \noindent
 {\bf 3.15. Coupon Competition }  (hard) \\ Two marketing executives
are arguing.  Smith  says that reducing our use of coupons will make us a less
aggressive competitor, and that will hurt our sales.  Jones says that reducing
our use of coupons will make us a less aggressive competitor,  but that will
end up helping  our sales.

Discuss, using the effect of reduced coupon use on your  firm's reaction curve,
under what circumstance each executive could be correct.









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

 \newpage

\begin{center} {\bf  The War of Attrition:  A Classroom Game for  Chapter 3}
\end{center}

   Each firm consists of 3 students. Each year a firm must decide whether to
stay in the industry or to exit. If it stays in, it incurs a fixed cost of 300
and a marginal cost of 2,  and it chooses an integer  price at which to sell.
The firms can lose unlimited amounts of money; they are backed by large
corporations who will keep supplying them with capital indefinitely.

Demand is inelastic at 60 up to a threshold price of \$10/unit,  above which the
quantity demanded  falls to zero.

 Each firm writes down its price (or the word ``EXIT'')  on a piece of paper and
gives it to the instructor. The instructor then writes the strategies of each
firm on the blackboard (EXIT or price). The firm charging the lowest price sells
to all 60 consumers. If there is a tie for the lowest price, the firms charging
that price split the consumers evenly.

The game then starts with a new year, but any firm that has exited is out
permanently and cannot re-enter. The game continues until only one firm is
active, in which case it is awarded a prize of \$2,000, the capitalized value of
being a monopolist.  This means the game can continue forever, in theory. The
instructor may wish to cut it off at some point, however.

 The game can then be restarted and continued for as long as class time permits.


\end{small}

\end{document}


