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\begin{Large}
\noindent
 June 26 1993\\
 Janaury 22, 1996
  
\section*{ 3 Mixed and Continuous Strategies} 
 
\newpage
 
    
    
 \begin{center}
{\bf Table 3.1  ``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 \\  
\multicolumn{6}{l}{\it Payoffs to: (Government, Pauper).}
\end{tabular}                
\end{center}


  
      
           In the mixed strategy equilibrium, the pauper selects {\it
Work} 20 percent of the time.  The way we obtained the number
might seem strange: to obtain the pauper's strategy, we
differentiated the government's payoff. Understanding why requires
several 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, then 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}

   \newpage
  \begin{large}

  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. 

  

  
         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.  


    One way to reinterpret ``The Welfare Game'' is  as  a many-pauper game with  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.
  
 
 Another interpretation of mixed strategies, which works even in the
single-pauper game, is 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): the less work-averse who pick {\it Work} if the
government picks $\theta_a =0.5$, and the more work-averse who pick {\it Loaf}.   It could even be that there is a continuum of types of   work-aversion, and that the split in actions is (0.2,0.8) only   when   $\theta_a =0.5$.     See Harsanyi (1973) for a careful exposition of this idea. 


\newpage
 
 
 \noindent
 { \bf Existence of Equilibrium} 
   
\noindent
 One of the strong points of Nash equilibria is that  they exist, in mixed strategies  if not in pure, in practically every game one is likely to encounter. 
 
  One  feature of a game that  favors existence    is continuity of the payoffs in the
strategies.  A payoff is continuous in a player's strategy if a small
change in his strategy causes a small  or zero change in the payoffs.
With continuity, the players' strategies can adjust more finely and
come closer to being best responses to each other. In `` The Welfare
Game'', the payoffs are discontinuous in pure strategies, so no
compromise between the players is possible and no pure strategy
equilibrium exists.  Once mixed strategies are incorporated, the
payoffs are continuous in the mixing probability, so an equilibrium
does exist.

  A second feature promoting existence is a closed and bounded
strategy set.  Suppose in a stock market game that Smith can borrow
money and buy as many shares of stock as he likes, so his strategy
set, the amount of stock he can buy, is $[0, \infty)$, which is
unbounded above (we assume he can buy fractional shares but cannot
sell short). If Smith knows that the price today is lower than it
will be tomorrow, he wants to buy an infinite number of shares, which
is not an equilibrium purchase.  If the amount he buys is restricted
to be strictly less than 1,000, then the strategy set is bounded (by
1,000)  but not closed (because 1,000 is not included), and no equilibrium purchase exists, because he
wants to buy 999.999 $\ldots$ shares. An open set like $\{x: 0 \leq x < 1000\}$ has no maximum, because there is no  closest real number to 1000.  If the strategy set is closed
and bounded, on the other hand,  its minimum and maximum  are well-defined. If Smith can buy up to but 
no more than 1,000 shares, then 1000 shares is his equilibrium purchase.
Sometimes, as in ``The  Cournot Game'' later in this chapter, the unboundedness of the strategy
sets does not matter because the optimum is an interior solution, but  in other games it is good modelling practice to use closed sets instead of open sets.   
 
 \end{large}





 \newpage
 
 \begin{center}
{\bf Table 3.2    ``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 \\  
\multicolumn{6}{l}{\it Payoffs to: (Smith, Jones).}
\end{tabular}                
\end{center}


    ``Chicken'' has two pure strategy Nash equilibria, $(Swerve, Continue)$
and $(Continue, Swerve)$, but they have the defect of asymmetry.    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 it does not
require any differences between the players. 

 
   
\newpage

\begin{normalsize}
 
\noindent
{\bf ``The  War of Attrition''}

 
  Smith and Jones  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 .
   
  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 he 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{e3.6}
  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{e3.7}
  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 (3.\ref{e3.7}) tells us that $\theta = 0.25$ in equilibrium, and that this is independent of the discount rate $r$.  



\newpage
 
   A{\bf preemption game}:  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. {\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.  



\end{normalsize}
 
  \begin{center}
{\bf Table 3.3 ``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}  & $\leftarrow$  & $0,0$ \\  
\multicolumn{6}{l}{\it Payoffs to: (Smith, Jones).}
\end{tabular}                
\end{center}

        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 the {\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) is an excellent discussion of games of timing in general.  



\newpage
  
   \begin{center}
{\bf Table 3.6 ``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}  & $\leftarrow$  & $7,7$ \\  
\multicolumn{6}{l}{\it Payoffs to: (Row, Column).}
\end{tabular}                
\end{center}

\begin{normalsize}
 
  ``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 equating-payoffs method of equilibrium calculation yields
\begin{equation}\label{e3.45a}
   \pi_{Smith} (Telephone) = 7 =    \pi_{Smith} (Ignore) =  \gamma^{N-1}  (0) +
 (1-\gamma^{N-1})  (10). 
 \end{equation}
  Equation (3.\ref{e3.45a})  tells us that    $
    \gamma^{N-1}  = 0.3$
  and 
$  \gamma^*   = 0.3^{\frac{1}{N-1}}.
$

     If $N=2$,  Smith chooses $Ignore$ with a probability of 0.30, and the probability that neither player phones the police is  $ \gamma^{*2}= 0.09$. As $N$ increases,   Smith's expected payoff remains constant  at 7, since his expected payoff always equals his payoff from the pure strategy of $Telephone$. The probability $\gamma^*$ of $Ignore$,  however,   increases in $N$. When there are  more players, each player  relies more on somebody else calling. The probability that nobody calls  is $\gamma^{*N} $. Equation 
(3.\ref{e3.45b}) shows that $\gamma^{*N-1}=0.3$, so  $\gamma^{*N} =0.3\gamma^*$, which is increasing in   $N$. When there are 38 players, the probability that nobody calls the police is about 0.29, because $\gamma^*$ is about 0.97. The  more people that watch a crime,  the less  likely  it is to be reported. 
  
   The expected payoff per player  is  7 whether there  is 1 player  or 38, whereas  if  one  and only  one  player called the police it  would rise from 7 with 1 player to about 9.9 with 38 (=[1(7) + 37(10]/38).   The problem is divided responsibility, lack of a focal point. One person must be         made responsible, 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!''). 

\end{normalsize}
%---------------------------------------------------------------
\newpage

 
 \begin{center}
{\bf Table 3.7 ``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    & $\leftarrow$  & $4,-1$ \\  
\multicolumn{6}{l}{\it Payoffs to: (IRS, Suspects) }
\end{tabular}                
\end{center}


``Auditing Game I'' is a discoordination game, with only a mixed strategy equilibrium. 
     

 A second way to model the situation is as a sequential game. Let us call this {\bf ``Auditing Game II''}.      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. 

 

  
\newpage
 \begin{large}

 \noindent
     AUDITING GAME III. 

 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 {\bf ``Auditing Game III''}, let 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{e3.50}
\pi_{suspect} (Obey)  \geq   \pi_{suspect} (Cheat),
  \end{equation}
i.e., 
\begin{equation}\label{e3.51}
-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 ``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.  

   
\end{large}
 
 

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




 
\newpage
  

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

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

\noindent 
 {\bf Payoffs}\\ 
 Production costs are zero. Demand is a function of the total
quantity sold, $Q= q_a + q_b$.
 \begin{equation} \label{e3.8}
P(Q) = 120-q_a - q_b .
  \end{equation}
 Payoffs are profits, which are given by a firm's price times its
quantity, i.e.,
 \begin{equation} \label{e3.9}
\begin{array}{l}
 \pi_{Apex}  = 120q_a  - q_a ^2 - q_a q_b ;\\
 \pi_{Brydox} = 120q_b  - q_a q_b  - q_b ^2.
\end{array}
\end{equation}

 




\end{Large}
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