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\begin{LARGE}


  \parindent 24pt \parskip 10pt



\noindent
January 30, 2014 \\
   Eric Rasmusen, Erasmuse@indiana.edu. Http://www.rasmusen.org  
 
  \begin{center} {\bf  3 Mixed and Continuous Strategies} \end{center}
 


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

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

\newpage
 %-----------------------------------
 
 \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.   }



  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}


Differentiate the payoff function with respect to the choice variable to obtain the first-order
condition. \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}

We obtained the pauper's strategy by differentiating  the government's payoff!

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

\newpage
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\noindent THE LOGIC

  \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.

 %-----------------------------------
 %\vspace{1in}
  To obtain the probability of the government choosing $Aid$: \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}

 %-----------------------------------
\newpage
\noindent
  {\bf  The Payoff-Equating Method}

    In equilibrium, each player is willing to mix only because he is indifferent between the pure
strategies he is mixing over.  This gives us a better way to find mixed strategies.

First, guess which strategies are being mixed between.

Then, see what mixing probability for the other player makes a given player indifferent.

 \begin{center} {\bf    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}

Here, $$ \pi_g(Aid) =  \gamma_w (3) + (1-\gamma_w) (-1) = \pi_g(No\; aid)  =  \gamma_w  (-1) +
(1-\gamma_w) (0) $$ So $ \gamma_w (3+1+1) =  1$, so $\gamma_w = .2$. $$ \pi_p(Work ) = \theta_a (2)
+ (1-\theta_a)(1) = \pi_p(Loaf) = \theta_a (3) + (1-\theta_a)(0) $$ so $\theta_a (2-1-3 ) = -1$ and
$\theta_a=.5$.

 %-----------------------------------
\newpage
 \noindent
 {\bf Interpreting  Mixed Strategies }

A player who selects a mixed strategy is always indifferent between two pure strategies and an
entire continuum of mixed strategies.

 What matters is that a player's strategy appear random to other players, not that it really be
random.

It could be based on time of day, temperature, etc.

It could be there is a population of identical players, each of whom picks a pure strategy. But each
would still be indifferent about his strategy.

Harsanyi based an interpretation on this: model it as an incomlpete info game,a nd let the
incomplete
info shrink to zero.

 Here do the Gintis example of mixing. 

Then do the soccer example, which is true randomization. 

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

  %-----------------------------------
\newpage

\begin{normalsize}

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

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Column}\\ &       &
&    $North$  & & $South$  \\ &   &  $North$   &          0,0 &   & 4,-9 \\ &  &     &    & &   \\ &
{\bf Row} &   $South$  &   4,-6 &    & 0,0 \\
 & & & &\\ &  &       $Defense$     &  1,-1 &    & 1,-1
\\ & & & &\\ \multicolumn{6}{l}{\it Payoffs to: (Row, Column)} \end{tabular} \end{center}

 
 For Row,  {\it  Defense } is strictly dominated  by (0.5 {\it  North}, 0.5 {\it  South}), though that is not the Nash equilibrium. 

Row's expected payoff from  (.5,.5) if Column plays {\it  North} is .5(0) + .5(4) = 2. Row's
expected payoff from it if Column plays {\it South} is .5(4) + .5(0)= 2. 

Row's 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}.
 
 Column's strategy must make Row willing to randomize, for a Nash equilibrium. Thus, if $c$ is Column's probability of North, we need  $\pi_r(North) = c(0) + (1-c)4 = \pi_r(South) = c(4) + (1-c)(0)$, so $4-4c= 4c$ so $c= 1/2. $

Row's strategy must make Column willing to randomize, for a Nash equilibrium. Thus, if $r$ is Row's probability of North, we need  $\pi_c(North) = r(0) + (1-c)(-6) = \pi_c(South) = c(-9) + (1-c)(0)$, so $-6-6c= -9c$ so $c= 3/5. $

 Note that $c= 3/5$ is even better  than  $c=.5$ for Row in equilibrium. He gets a payoff of $ \pi_r(South) = c(4) + (1-c)(0) = 2.4$

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


 \begin{center} {\bf       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}

\bigskip



  \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 $\theta$.

The two teenagers will survive with probability $ 1- (\theta \cdot \theta) = 0.9375.$

How can we prove there is no asymmetric mixed-strategy equilibrium, with unequal mixing probabilities? 

  %-----------------------------------
\newpage

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Jones}\\ &       &
&    $Continue$ ($\theta$)  &   &  $Swerve$ ($1- \theta$) \\ &   &  $Continue$ ($\theta$)      &
$x,x$ & $\rightarrow$  & {\bf 2, 0} \\ & {\bf Smith:} &&$\downarrow$& & $\uparrow$ \\ &  &
$Swerve$ ($1-\theta$)       &    {\bf 0, 2}  & $\leftarrow$  & 1, 1 \\ \end{tabular}


  \begin{equation} \label{e3.6} \theta =  \frac{1}{1-x}. \end{equation}

If $x=-3$, this yields $\theta = 0.25$, as was just calculated.

If $x= -9$,  it yields $\theta = 0.10 $.


If $x= 0.5$,  the equilibrium probability of continuing appears to be $\theta =  2 $. What is going
on?

In the mixed-strategy equilibrium, the  expected payoff is  $\pi(swerve) = \theta (0) + (1-\theta) (1)$. 

 Note that this is decreasing in $\theta$. 

 %-----------------------------------
\newpage
  \noindent {\bf The  War of Attrition}


  The possible  actions are  $Exit$  and  $Continue$. 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 $r > 0$.


(1)  Continue in each period, Exit in each period

(2) Each exits with probability $\theta$ if it hasn't yet.



 Let Smith's payoffs be $V_{stay}$ if he stays and $V_{exit}$ if he exits.

 $V_{exit} =0$.

  \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} Thus, $\theta = 0.25$.

This does not have to solved with the dynamic programming/Bellman equation method. 

  %-----------------------------------
\newpage
 {\bf Timing games}

 {\bf Pre-emption games}: 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}.  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,  each is 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.

  \begin{center} {\bf   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}

\newpage


 \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} 



Let$ s$ be Smith's probability of grabbing  and $j$  be Jones's.  If Smith grabs, that ends the game:
$$
 \pi_s(grab) = j (-1) + (1-j)(1)  
$$
 If he chooses not to grab, then the game continues, and if Jones does not grab either, he remains in the same position as at the start: 
$$
 \pi_s(not\; grab) = j (0) + (1-j)(\frac{1}{1+r}   \pi_s(not\; grab)  )
$$
  The only value that solves this second equation is $ \pi_s(not\; grab)=0$. Equating that to  $\pi_s(grab)$ gives us \\
       $0 =  j (-1) + (1-j)(1)$, so $ j=.5$  

\newpage


 \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} 


Suppose we had an equilibrium where if the second period is reached, Smith grabs with probability one. What will happen to the mixed strategy in the first period? 

  Smith would have to equate  his first period payoffs thus:
$$
 \pi_s(grab) = j (-1) + (1-j)(1) = \pi(don't) = j(0) + (1-j) (\frac{1}{1+r} (1))
$$
If $r=0$,  these are equal only if $j=0$ or $j=1$. So there can't be an equilibrium with mixed strategies in the first period and pure strategies in the second. 

 If $r \neq 0$ then some algebra shows that $j = \frac{r}{1+2r}$. 

 As for Jones: 
$$
 \pi_j(grab) = s (-1) + (1-s)(1) = \pi(don't) = 0,
$$
 so $s = 1/2$. 

Smith probably wins in the first period because of the forecast that he would otherwise win in the second period. 

\newpage


 \begin{center} {\bf  Asymmetric Grab the Dollar }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Jones}\\ &       &
&  {\it  Grab}    &   &  {\it Don't Grab } \\
 &   & {\it Grab} &      $-2, -3$ & $\rightarrow$  &
{\bf 1,0} \\
 & {\bf Smith:} &&$\downarrow$& & $\uparrow$ \\
 &  &    {\it Don't Grab }       &
{\bf 0,1}  & $\leftarrow$  & $0, 0$ \\
 \end{tabular} \end{center} \vspace{-24pt}

Let$ s$ be Smith's probability of grabbing,  and $j$  be Jones's. Then Smith equates his payoffs thus (remember: the continuation payoff is zero): 
$$
 \pi(grab) = j (-2) + (1-j)(1) = \pi(don't) = j(0) + (1-j) (0)
$$
 so $ -2j + 1-j = 0$ and $j =1/3$.

Jones equates his payoffs thus:
$$
 \pi(grab) = s (-3) + (1-s)(1) = \pi(don't) = s(0) + (1-s) (0)
$$
 Then  $-3s + 1-s = 0$, and s =1/4.   

 Smith has a smaller probability of grabbing even tho
his
penalty from Grab, Grab is less.  That is because for Smith, with his smaller penalty, not to always
want to grab requires that Jones have a bigger probability of grabbing.



 

  %-----------------------------------
\newpage

\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. 

 \begin{tabular}{ll} $ \pi_i =$& $\left\{ \begin{tabular}{lll} $V-x_i$ & if $T(x_i) < Min\{ T(x_j,
T(x_k) \} $  & (wins)\\ & & \\ $\frac{V}{2} - x_i$ & if $T(x_i) = Min \{T(x_j),T(x_k)\}    $ & (
shares with 1)  \\ & $\;\;\;\;\;\;\;\;\;\;\; <Max \{T(x_j),T(x_k)\}$ & \\ & & \\ $\frac{V}{3} - x_i$
& if $T(x_i) = T(x_j=T(x_k)  $ & (shares   with 2) \\ & & 2 other firms)\\ & & \\ $- x_i$ & if
$T(x_i) > Min\{ T(x_j, T(x_k) \}$  & (loses) \\ \end{tabular} \right.$ \end{tabular}\\


 %-----------------------------------
\newpage
No  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
 the figure   for fixed values of $x_b$ and $x_c$. (The research levels shown    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}

 %-----------------------------------
\newpage

 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.

Since we know that the pure strategies $x_a=0$ and $x_a= V$ yield zero payoffs, if Apex mixes over
$[0,V]$ then the expected payoff for every strategy mixed between must also equal zero.

 \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}

 This is an  ``all-pay auction.'' 

 %-----------------------------------
\newpage
  \noindent {\bf Correlated Strategies}



  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, these are
{\bf correlated strategies}.

In   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. The probability of each choosing $Continue$ is 0.5, and the expected payoff for each is
1.0.

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

     {\bf Cheap talk} (Crawford \& Sobel [1982]). Cheap talk refers to costless communication when
players can lie without penalty.

In { Ranked Coordination}, cheap talk instantly allows the players to make the desirable outcome a
focal point, though it does not get rid of the other equilibria.



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

\newpage


   \begin{center} {\bf  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.   }

 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.

  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 $1- \gamma^{N-1}$.


  \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} so \begin{equation}
\label{e3.18} \gamma^*   = 0.3^{\frac{1}{N-1}}. \end{equation}

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

\newpage

\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  value $\gamma^*$, the   probability  of $Ignore$ for each player, 
rises with $N$. 

If $N=38$, the value of $\gamma^*$ is about 0.97. (Kitty Genovese case) (think of juries) 

 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.


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

\newpage

  \noindent {\bf   Randomizing Is Not Always Mixing:       }




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 $F >1$.


 \begin{center} {\bf   { Auditing Game I, II}  }

 \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}  

$$
\begin{array}{l}
 \pi(Audit) = \theta (4-C) + (1-\theta)(4-C) = \\
\pi(Trust) =  \theta (0) + (1-\theta)(4 ),  4-C=4-4\theta,  \theta^*=C/4
\end{array}
$$
$$
\begin{array}{l}
 \pi(Cheat) = \gamma( -F) + (1-\gamma)(0) = \\
\pi(Obey) =  \gamma(-1) + (1-\gamma)(-1 ),   -\gamma F = -1, \gamma^*=1/F. 
\end{array}
$$

The payoffs are $\pi_{gov} = 4-C$ and $\pi_{suspect}= -1$. 

 
Auditing Game II makes this sequential:  The government moves first.  The payoffs are identical, but there is always auditing and never cheating.   

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

\newpage

  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 moves  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.



%---------------------------------------------------------------
 
\newpage

\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}

 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.


  %-----------------------------------
\newpage
   To find the ``Cournot-Nash'' equilibrium, we need to refer to the {\bf best-response functions}
or {\bf reaction functions}.  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,   $\pi_{Apex}  =   (120-c)q_a  - q_a^2 - q_a q_b$, 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}

\newpage

\begin{equation} \label{e3.28} q_a  =   54 - \left( \frac{1 }{2}\right) q_b. \end{equation}

\includegraphics[width=6in]{fig03-02.jpg} 



 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},   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).




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

\newpage

\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}

The actions and payoffs are identical to the Cournot Game. All that has changed is that it is sequential now.  And--- that we  act as if the sequence mattered. We will look for an asymmetric equilibrium in quantities now. 

\newpage

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

If  Apex forecasts Brydox's output to be $q_b  = 60 - \frac{q_a +c}{2}$, 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} so $q_a = 60-   c/2 =
54$. Once Apex chooses this output, Brydox chooses his output to be $q_b = 27$.

 %-----------------------------------
\newpage
\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}




 %-----------------------------------
\newpage

 {\bf The Differentiated Bertrand Game}

  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 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}

  Apex  and Brydox simultaneously choose prices  $p_a$    and $p_b$ from the  set $[0, \infty)$.


  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}
 %-----------------------------------
\newpage

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

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

Brydox's reaction curve slopes upwards too. 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. 




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

\newpage

\includegraphics[width=8in ]{fig03-05.jpg}
 
     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.

   Supermodularity is  a related concept.  With only one choice variable, as here, it boils down to, for players i and j with strategies $s_i, s_j$: 
$$
\frac{\partial^2  \pi_i}{\partial s_i s_j} \geq 0. 
$$


   %-----------------------------------
\newpage
  \noindent { \bf Four common reasons why an equilibrium might not exist}



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

  Smith can borrow money and buy as much tin as he wants  for \$6/pound.  He knows that the price
will be \$7/pound tomorrow. What quantity $x$  will he buy, if his borrowing is unlimited?

Choosing $x$ in the strategy set $[0, \infty)$  when his payoff function is $\pi = (1) x$, there is
no best strategy.


\newpage

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

 Now say that government regulations constrain him to buy less than 1,000 pounds.
His strategy is $x \in [0, 1,000)$, which is bounded by 1000.



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

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

  If 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  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.

This 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.

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

 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.

\includegraphics[width=7in]{fig03-06.jpg}

In Panel (a) a Nash equilibrium exists, at the point, $E$,  where the two reaction functions
intersect.  In Panel (b)  no Nash equilibrium exists.   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
$s_2(0.50001)=0.4$. 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).


 %-----------------------------------
\newpage
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.


\newpage

 

  If   firms are Cournot competitors with different marginal costs, they will have different market shares. The Herfindahl Index equals the weighted average  of their price-cost margins  multiplied  by the industry elasticity of demand (and multiplied by -10,000).  

 
 
\end{LARGE}
  }

\end{document}














