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 ODD

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\noindent \textbf{Answers to Odd-Numbered Problems, 4th Edition of Games and
Information, Rasmusen }

\begin{center} {\bf PROBLEMS FOR CHAPTER 3: Mixed and Continuous Strategies}
\end{center}

\noindent  22 November 2005. 14 September 2006. Erasmuse@indiana.edu. Http://www.rasmusen.org.

\noindent This appendix contains answers to the odd-numbered problems in the
gourth edition of \textit{Games and Information} by Eric Rasmusen, which I am
working on now and perhaps will come out in 2006. The answers to the even-
numbered problems are available to instructors or self-studiers on request to me
at Erasmuse@indiana.edu.

Other books which contain exercises with answers include Bierman \& Fernandez
(1993), Binmore (1992), Fudenberg \& Tirole (1991a), J. Hirshleifer \& Riley
(1992), Moulin (1986), and Gintis (2000). I must ask pardon of any authors from
whom I have borrowed without attribution in the problems below; these are the
descendants of problems that I wrote for teaching without careful attention to
my sources.

\newpage

\begin{center} {\bf PROBLEMS FOR CHAPTER 3: Mixed and Continuous Strategies}
\end{center}

\bigskip \noindent \textbf{3.1. Presidential Primaries }\\ 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?

\noindent \underline{\textit{Answer.}} The value of exiting is zero. The value
of staying in is $V = \theta(10) + (1-\theta) (-1 + \frac{V} {1+r})$. Thus, $V -
(1-\theta)\frac{V}{1+r} = 10 \theta -1 + \theta$, and $V= \frac{(11\theta-1)
(1+r)}{(r+\theta)}$. As a result, ${\theta = 1/11}$ in equilibrium.

\hspace*{16pt} The discount rate does not affect the equilibrium outcome, so a
change in $r$ produces no observable effect.

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

\noindent \underline{\textit{Answer.}} (Smith drops out, Jones stays in no
matter what) and (Jones drops out, Smith stays in no matter what).

\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?

\noindent \underline{\textit{Answer.}} The payoff matrix is shown in Table A3.1.

\textbf{Table A3.1: Fighting Democrats}

\begin{tabular}{llll} &  & \multicolumn{2}{c}{Jones} \\ &  & $Exit$ ($\gamma$) &
$Stay$ ($1-\gamma$) \\ & $Exit$ ($\gamma$) & 0,0 & 0, 10 \\ Smith &  &  &  \\ &
$Stay$ ($1-\gamma$) & 10,0 & -1,-1 \end{tabular}

The value of exiting is $V(exit)=0$. The value of staying in is $V(Stay) = 10
\gamma + (-1) (1-\gamma) = 11\gamma -1$. Hence, each player stays in with
probability ${\gamma = 1/11}$ --- the same as in the war of attrition of part
(a).

\end{enumerate}

\bigskip \noindent \textbf{3.3. Uniqueness in Matching Pennies  }\\ 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.

\textbf{Table A3.2: ``Matching Pennies''}

\begin{tabular}{lllccc} &  &  & \multicolumn{3}{c}{\textbf{Jones}} \\ &  &  &
\textit{Heads} ($\theta$) &  & $Tails$ ($1- \theta$) \\ &  & $Heads$ ($\gamma$)
& $1,-1$ &  & $-1,1$ \\ & \textbf{Smith:} &  &  &  &  \\ &  & $Tails$
($1-\gamma$) & $-1,1$ &  & $1,-1$ \\ \multicolumn{6}{l}{\textit{Payoffs to:
(Smith, Jones).}} \end{tabular}

 \item[(b)] Show that there is no Nash equilibrium in pure strategies.

\noindent \underline{\textit{Answer.}} ($Heads, Heads$) is not Nash, because
Jones would deviate to $Tails$. \textit{Heads, Tails} is not Nash, because Smith
would deviate to $Tails$. \textit{(Tails, Tails)} is not Nash, because Jones
would deviate to $Heads$. \textit{(Tails, Heads)} is not Nash, because Smith
would deviate to $Heads$.

 \item[(c)] Find the mixed-strategy equilibrium, denoting Smith's probability of
$Heads$ by $\gamma$ and Jones' by $\theta$.

\noindent \underline{\textit{Answer.}} Equate the pure strategy payoffs. Then
for Smith, $\pi (Heads) = \pi(Tails)$, and \begin{equation}  \label{e81}
\theta(1) + (1-\theta) (-1) = \theta(-1) + (1-\theta) (1), \end{equation} which
tells us that $2\theta -1 = -2\theta+1$, and ${\theta= 0.5}$. For Jones, $\pi
(Heads) = \pi(Tails)$, so \begin{equation}  \label{e82} \gamma(-1) + (1-\gamma)
(1) = \gamma(1) + (1-\gamma) (-1), \end{equation} which tells us that $1-2\gamma
= 2\gamma -1$ and ${\gamma = 0.5}$.

 \item[(d)] Prove that there is only one mixed-strategy equilibrium.

\noindent \underline{\textit{Answer.}} Suppose $\theta >0.5$. Then Smith will
choose $ Heads$ as a pure strategy. Suppose $\theta < 0.5$. Then Smith will
choose $ Tails$ as a pure strategy. Similarly, if $\gamma > 0.5$, Jones will
choose $ Tails$ as a pure strategy, and if $\gamma < 0.5$, Jones will choose
$Heads$ as a pure strategy. This leaves (0.5, 0.5) as the only possible mixed-
strategy equilibrium.

Compare this with the multiple equilibria in problem 3.5. In that problem, there
are three players, not two. Should that make a difference?

\end{enumerate}

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


\bigskip 


\noindent
 \textbf{3.5. A Voting Paradox } \\
 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 $\gamma$?

\noindent
 \underline{\textit{Answer.}} The probability that Adam loses can be
decomposed into three probabilities--- that all three vote, that Adam does not
vote but one other does, and that Adam does not vote but both others do. These
sum to $\theta \gamma^2 + (1- \theta)2\gamma(1-\gamma) + (1- \theta) \gamma^2$,
which is, rearranged, $ \gamma^2 + 2\gamma(1-\gamma)(1- \theta) $  or${\ \gamma (2\gamma \theta - 2\theta + 2 - \gamma)}$.

 \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?

\noindent
 \underline{\textit{Answer.}} The equilibrium is in mixed strategies,
so each player must have equal payoffs from his pure strategies. Let us start
with Adam's payoffs. If he votes, he loses 20 immediately, and 120 more if both
Karl and Vladimir have voted.
 \begin{equation}  \label{e70}
 \pi_a(Vote) = -20 +
\gamma^2(-120).
  \end{equation}
 If Adam does not vote, then he loses 120 if
either Karl or Vladimir vote, or if both vote: 
\begin{equation}  \label{e71}
\pi_a(Not\; Vote) = (2\gamma(1-\gamma) + \gamma^2) (-120)
 \end{equation}
Equating $\pi_a(Vote)$ and $\pi_a(Not\; Vote)$ gives
 \begin{equation}
\label{e72} 0 = 20 - 240 \gamma + 240 \gamma^2. 
\end{equation}
 The quadratic
formula solves for $\gamma$:
 \begin{equation}  \label{e73} \gamma= \frac{12 \pm
\sqrt{144- 4\cdot 1 \cdot 12}}{24}.
 \end{equation} 
This equations has two
solutions,${\ \gamma_1=0.09 }$ (rounded) and ${\ \gamma_2=0.91 }$(rounded).

\hspace*{16pt} Why are there two solutions? If  Karl and Vladimir are sure not
to vote, Adam will not vote, because if he does not vote he will win, 0-0. If
Karl and Vladimir are sure to vote, Adam will not vote, because if he does not
vote he will lose, 2-0, but if he does vote, he will lose anyway, 2-1. Adam only
wants to vote if  Karl and Vladimir vote with moderate probabilities. Thus, for
him to be indifferent between voting and not voting, it suffices either for
$\gamma$ to be low or to be high-- it just cannot be moderate.

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

\noindent
 \underline{\textit{Answer.}} Now use the payoffs for Karl, which
depend on whether Adam and Vladimir vote.
 \begin{equation}  \label{e74}
\pi_c(Vote)= -20 + 60 [\gamma + (1-\gamma)(1-\theta)] \end{equation}
\begin{equation}  \label{e75} \pi_c(Not\; Vote)= 60 \gamma (1-\theta).
\end{equation} 
Equating these and using $\gamma^*=0.09$ gives ${\ \theta=}$ {\
\textrm{0.70 }} (rounded). Equating these and using $\gamma^*=0.91$ gives ${\
\theta=}$ \textrm{0.30 } (rounded).

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

\noindent
 \underline{\textit{Answer.}} The probability that Adam will lose his
property is, using the equation in part (a) and the values already discovered,
either \textrm{0.06 } (rounded) ($=(0.7)( 0.09) ^2 + (0.3) ( 2(0.09) (0.91) +
(0.09)^2))$ or \textrm{0.94} (rounded ($= (0.3)( 0.91) ^2 + (0.7) ( 2(0.91)
(0.09) + (0.91)^2)) $.
 \end{enumerate}

\bigskip
 
\noindent
 {\bf 3.7.  Nash Equilibrium} \\
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}

\noindent
 \underline{\textit{Answer.}}  The equilibrium is in mixed strategies.
Denote Row's probability of $Up$ by $\gamma$ and Column's probability of $Left$
by $\theta$.   Strategies $Sideways$ and $Middle$ are strongly dominated
strategies, so we can forget about them.   Row has no reason ever to choose
$Sideways$, and Column has no reason ever to choose $Middle$.

In equilibrium, Row must be indifferent between $Up$ and $Down$, so $$ \pi_R(Up)
=  \theta (1) + (1-\theta)(0) =  \pi_R(Down) =  \theta (0) + (1-\theta)(2) $$
This yields  $\theta^*=  2/3 $. Column must be indifferent between $Left$ and
$Right$, so $$
 \pi_R(Left) =  \gamma (0) + (1-\gamma)(2) =  \pi_R(Right) = \gamma (1) +
(1-\gamma)(0) $$
 This yields  $\gamma^*=  2/3$.

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

\bigskip
 \noindent
 {\bf 3.9 (hard).  Cournot with Heterogeneous Costs }\\ 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.


\noindent
 \underline{\textit{Answer.}}   A good approach when stymied is to start with a
simple case. Here, the two-firm problem is the obvious simple  case.  Prove the
proposition for the simple case, and then use that as a pattern to extend it.
(Also, you can disprove a general proposition using a simple counterexample,
though you cannot prove  one  using a simple example.)

 Note that you cannot assume symmetry of strategies in this game. It is
plausible, though not always correct (remember Chicken), when players  are
identical,  but they are not here--- firms have different costs. So we would
expect their equilibrium outputs to differ.


The proposition is true.
 $$
\pi_j =  (\alpha -\beta  \Sigma_iQ_i - C_j) Q_j, $$ so $$
 \frac{d  \pi_j }{dQ_j} =  \alpha -   \beta  \Sigma_{i \neq j} Q_i  - 2\beta
Q_j  - C_j =0,  $$
and
$$
 Q_j = \frac{ \alpha  - C_j -   \beta  \Sigma_{i \neq j}Q_i }{2\beta}.
 $$
Industry output is
 $$
 \Sigma_j Q_j =  \Sigma_j \frac{\alpha  - C_j  -   \beta  \Sigma_{i \neq j}Q_i }
{2\beta} = \Sigma_j \frac{\alpha  - C_j  } {2 \beta}   -   \Sigma_j  \frac{
\Sigma_{i \neq j}Q_i }{2 }.
 $$
 The first term of this last expression depends on the sum of the firms' cost
parameters, but not on their individual levels.  The second term  adds up the
outputs of all but one firm $N$ times, and so equals $(N-1)$ times the sum of
the output, $(N-1) \Sigma_j Q_j$. Thus,
 $$
\begin{array}{ll}
 \Sigma_j Q_j &  = \Sigma_j \frac{\alpha  - C_j  } {2 \beta}   -   \left(N-1
\right) \left( \frac{ \Sigma_{i \neq j}Q_i }{2 }\right)\\
 & \\
 & = \Sigma_j \frac{\alpha  - C_j  } {  \beta (N+1)}.\\
\end{array}
  $$
  This does not depend on the  cost parameters except through their sum. Q.E.D.

 A caveat:  This proof implicitly assumed that  every firm had low enough costs
that it would produce positive output. If it produces zero output, it is at a
corner solution, and the first order condition does not hold, so the proof
fails.  Thus, the validity of the proposition depends on the following being
true for every $j$:
$$
Q_j = \frac{C_j-\alpha  -   \beta  \Sigma_{i \neq j}Q_i }{2\beta}>0. $$ This
condition is not stated in terms of the  primitive parameters (it depends on $
\Sigma_{i \neq j}Q_i $), so to be quite proper I ought to solve it out further,
but I will not do that here.

The result  does depend on linear demand.  This can be shown by counterexample.
Suppose  $P=\alpha -\beta Q^2$.  Then, attempting the construction above, $$
\pi_j =  (\alpha -\beta  (\Sigma_iQ_i)^2 - C_j) Q_j, $$ so $$ \frac{d  \pi_j }
{dQ_j} =  \alpha - 3\beta    Q_j^2 - 2\beta \Sigma_{i \neq j} Q_i  Q_j - C_j =0.
$$ Solving this for $Q_j$ will involve taking a square root of $C_j$.  But if
$Q_j$ is  a function  of the square root of $C_j$, then  increasing $C_j$ by a
given amount and decreasing $C_l$ by the same amount will {\it not} keep the sum
of $Q_j$ and $Q_l$ the same, unlike before, where  $Q_j$ was a linear function
of $C_j$.  So the proposition fails for quadratic demand, and, more generally,
whenever demand is nonlinear.

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

\bigskip

 \noindent
 {\bf 3.11.  Finding Nash Equilibria }  \\ 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}

\noindent \underline{\textit{Answer.}}  The three equilibria are in pure
strategies: (Hard, Soft), (Medium, Medium), and  (Soft, Medium).

     There  are two   mixed strategy equilibria.

(1)  ( Raider: Hard. Target:  Mix between  Medium and Soft.) Notice that for
Target, Hard is a dominated strategy. That means it will not be part of any
mixed strategy equilibrium.  Next, notice that  Soft is weakly dominated. Thus,
if Raider ever plays anything but Hard, Target will want strongly  to   play
Medium.   What if Raider plays Hard? Then Target would be willing to mix between
Medium and Soft.  If Target plays  Medium  with a probability of $\gamma$,
Raider's payoff from Hard is  $(-1) \gamma + 4 (1-\gamma) $, whereas his payoff
from Medium or Soft is   $(2) \gamma + 3 (1- \gamma)$.  Equating these yields
$\gamma^* = $. 25. If  $\gamma$ is no bigger than .25, we have a Nash
equilibrium.

(2)  ( Raider: Mix between Medium and Soft. Target: Medium. )  How about if
Target plays Medium and Raider mixes?  Raider would only want to mix between
Medium and Soft.  But that would generate a Nash equilibrium, for any mixing
probability, since  Raider gets 2 no matter what, and  Target prefers Medium no
matter what the mixing probability may be.

\bigskip


 \noindent 
{\bf 3.13. The Kosovo War  }\\
 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.

\noindent
 \underline{\textit{Answer.}} Senator Smith meant that  by declaring our action,
we have allowed the Yugoslavs to choose a better response (for them) than if we
left them uncertain.  Thus, the declaration reduces  the expected  U.S. payoff.
Rather than mixing-- which means to be unpredictable-- we chose a pure strategy.

Ane example can show this. Suppose that the US has the two alternatives of Air
and Ground, and the Yugoslavs have the two alternatives of  Air Defense and
Ground Defense. Air and Air Defense represent policies of just positioning
forces for an air war; Ground and Ground Defense represent policies that also
prepare for ground war.

 Let the payoffs be as in Table A3.3.

  \begin{center}
{\bf Table A3.3: The Kosovo War}\\
 \begin{tabular}{lllccc} &          &           &\multicolumn{3}{c}{\bf
Yugoslavia}\\ &          &                & Air Defense ($\gamma$)  &  & Ground
Defense \\ &          &  Air  ($\theta$)     &   0,0 &   &   1,-1 \\ & {\bf US}
&   &      & &        \\ & & Ground  & 2,-5&  & -2,-2 \\
\multicolumn{6}{l}{\it Payoffs to: (U.S., Yugoslavia).}\\
 \end{tabular}\\
\end{center}

 (a)  In the mixed strategy equilibrium,  Yugoslavia chooses  its probability of
Air Defense to equate the US payoffs from Air and Ground.  Thus,
\begin{equation} \label{ee3} \pi_{US}(Air) =    \gamma (0) + (1-\gamma)(1)   =
2\gamma + (1- \gamma) (-2)  =   \pi_{US}(Ground) . \end{equation} This reduces
to  $1-\gamma = 2\gamma -2 +  2 \gamma$, so  $3 = 5 \gamma$, and $\gamma = 3/5$.
The U.S. expected payoff  from choosing Air is then $\pi_{US}(Air) =    \gamma
(0) + (1-\gamma)(1)= 1-3/5  = .4. $

 (b)   If the U.S. instead moves first and chooses Air, Yugoslavia will respond
with Air Defense, and the U.S. expected payoff is 0.

 Thus,  by volunteering to move first, the U.S. reduces its payoff.

\bigskip

 \noindent
{\bf 3.15.  Coupon Competition  } \\
  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.

\noindent
\underline{\textit{Answer.}} There are a couple of ways to look at this problem.

 (1)   One way is   that    the  important  strategy is coupon use directly.
Smith thinks that coupons are strategic  substitutes, so when we reduce our use
of coupons, our rival will increase their use, and we will be hurt.  Jones
thinks that coupons are  strategic complements, so when we reduce our use of
coupons, our rival will reduce their use too, to the benefit of both of us.

(2) A second way is in terms of how coupon use affects how the two companies
play a game in the  consumer market.

 Smith thinks that our firm is in a market with downward sloping reaction curves
in the important strategy-- strategic substitutes, as with Cournot competition.
If we  use fewer coupons, that will shift in our reaction curve, and  we will
end up with lower sales. We need to be ``lean and hungry'', because if we use
coupons to make us  softer   in the product market, our rival will react by
being  tougher.

The important strategy might be, for example,  output, and if we use more
coupons,   that will  make us less willing to produce high output in reaction to
what our rival does, because each sale will be profitable.  In the end, we will
contract our output and our rival will increase his.

Jones thinks  that our firm is in a market with upward  sloping reaction curves
in the important strategy-- strategic complements, as with Bertrand competition.
If the important variable is price, and   we  use fewer coupons, that will
shift  out  our reaction curve, and  we will increase our price. So will our
rival, and we will both end up with higher profits.

    We  thus  adopt a ``fat cat'' strategy-- we use more coupons to make us
softer in the product market, and our rival becomes softer in response.

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

\end{document}


