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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} \textbf{PROBLEMS FOR CHAPTER 1 } \end{center}

\noindent      26 March 2005. 12 September 2006.  29 August 2012. 
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 2005. 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}
 \textbf{PROBLEMS FOR CHAPTER 1 }
 \end{center}

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


 

\noindent 
 {\bf 1.1. Nash and Iterated Dominance } (medium)

\begin{enumerate}

\item[(a)] Show that every iterated dominance equilibrium $s^*$ is Nash.

\noindent \underline{\textit{Answer.}} Suppose that $s^*$ is not Nash. This
means that there existS some $i$ and $s_i^{\prime}$ such that $i$ could
profitably deviate, i.e., $\pi_i(s^*) < \pi_i(s_i^{\prime}, s_{-i}^*)$. But that
means that there is no point during the iterated deletion at which  player $i$
could have eliminated strategy $s_i^{\prime}$ as being even weakly dominated for
him by $s_i^*$. Hence, iterated deletion could not possibly reach $s^*$ and we
have   a contradiction; it must be that every iterated dominance equilibrium is
Nash.

\item[(b)] Show by counterexample that not every Nash equilibrium can be
generated by iterated dominance.

\noindent \underline{\textit{Answer.}} In  Ranked Coordination (Table 7 of
Chapter 1) no strategy can be eliminated by dominance, and the boldfaced
strategies are Nash.

\item[(c)] Is every iterated dominance equilibrium made up of strategies that
are not weakly dominated?

\noindent
 \underline{\textit{Answer.}} \textrm{Yes.} As defined in Chapter 1, strategy $x$ is weakly dominated by strategy $y$  only if $y$ has a strictly higher payoff in some strategy profile and has a strictly lower payoff in no strategy profile. An iterated dominance equilibrium only exists if the iterative process results in a single strategy profile at the end. 

  In order for $x$ to be in the final surviving profile, it would have to weakly dominate the second-to-last surviving strategy for that player  (call it $x_2$).  Thus, it is strictly better than $x_2$ as a response to some profile of strategies of the other players:   $\pi_i (x, s_{-i}) >\pi_i (x_2, s_{-i})$ for some particular set of strategies for the other players $s_{-i}$ that has survived deletion so far.  But for $x_2$ to have survived deletion so far means that $x_2$ must be   at least a good a response to the profile $s_{-i}$  as the third-to-last surviving strategy: $\pi_i (x_2, s_{-i})  \geq \pi_i (x_3, s_{-i})$, and in turn none of the earlier deleted $x_i$ strategies could have done strictly  better as a response to $s_i$ or they would not have been weakly dominated.  Thus, $x$ must be a strictly better response in at least one strategy profile  than all the previously deleted strategies for that player   and it cannot have been weakly  dominated by any of them. 

 




If we define things a bit differently, we would get different answers to this question. Consider this: 

(i) Define ``quasi-weakly dominates'' as a strategy that is no worse than any other strategy. 


   \hspace*{16pt} A strategy that is in the
equilibrium strategy profile might be a bad reply to some strategies that
iterated deletion of quasi-dominated strategies removed from the original game. Consider the Iteration Path
Game in Table A1.1. The strategy profile  $(r_1, c_1)$ is a  
 iterated quasi-dominance equilibrium.  Delete $ r_2$, which  is weakly   dominated by $r_1$ (0,0,0 is beaten by 2,1,1). Then delete 
$c_3$, which is now  quasi-weakly dominated by $c_1$ (12,11 is equal to 12,11).  Then delete 
 $r_3$, which is weakly dominated by $r_1$ (0,1 is beaten by 2,1). Then delete $c_2$, which is strongly dominated by  $c_1$ (10 is beaten by 12). 




\begin{tabular}{lllccc}
 &  &  & \multicolumn{3}{c}{\textbf{Column}} \\
 &  &  &
$c_1$ & $c_2$ & $c_3$ \\
 \multicolumn{6}{l}{} \\ 
&  & $r_1$ & \textbf{2,12 } &1,10 &  1,12  \\ 
\multicolumn{6}{l}{} \\
 & \textbf{Row:} & $r_2$ & 0,7 &0,10 & 0,12 \\
 \multicolumn{6}{l}{} \\
 &  & $r_3$ & 0,11 & 1,10 & 0,11 \\
\multicolumn{6}{l}{} \\ 
\multicolumn{6}{l}{\textit{Payoffs to: (Row, Column)}}
\end{tabular} 
\end{enumerate}

\textbf{Table A1.1: The Iteration Path Game }
 
   
 
   

\bigskip

\noindent   
  {\bf 1.3. Pareto Dominance  (medium) (from  notes by Jong-Shin Wei)}
 
\begin{enumerate}

\item[(a)] If a strategy profile $s^*$ is a dominant strategy equilibrium,
does that mean it weakly pareto-dominates all other strategy profiles?

\noindent \underline{\textit{Answer.}} \textrm{No}--- think of The Prisoner's
Dilemma  in Table 1 of Chapter 1.  (\textit{Confess, Confess}) is a dominant
strategy equilibrium, but it does not weakly pareto-dominate (\textit{Deny,
Deny})

\item[(b)] If a strategy profile $s$ strongly pareto-dominates all other
strategy profiles, does that mean it is a dominant strategy equilibrium?

\noindent \underline{\textit{Answer.}} \textrm{No}--- think of  Ranked
Coordination  in Table  7 of Chapter 1. (\textit{Large, Large}) strongly pareto-
dominates all other strategy profiles, but is not a dominant strategy
equilibrium.

 The  Prisoner's Dilemma  is not a good example for this problem, because
\textit{(Deny, Deny)} does not pareto-dominate \textit{(Deny, Confess)}.

\item[(c)] If $s$ weakly pareto-dominates all other strategy profiles, then
must it be a Nash equilibrium?

\noindent \underline{\textit{Answer.}} \textrm{Yes.} (i) If and only if  $s$ is
weakly pareto-dominant, then   $\pi_i(s) \geq \pi_i(s^{\prime}), \forall
s^{\prime}, \forall i$. (ii) If and only if  $s$ is Nash, $\pi_i(s) \geq
\pi_i(s^{\prime}_i, s_{- i}), \forall s_i^{\prime}, \forall i$. Since
$\{s^{\prime}_i,s_{-i}\}$ is a subset of $ \{s^{\prime}\}$, if $s$ satisfies
condition (i)  to be weakly pareto-dominant, it must also  satisfy condition
(ii) and be  a Nash equilibrium.

\end{enumerate}

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


 
   

\bigskip \noindent {\bf 1.5. Drawing Outcome Matrices  } (easy)  \\
  It can be
surprisingly difficult to look at a game using  new notation.  In this
exercise, redraw the outcome matrix in a different form than   in the  main
text. In each case, read the description of the game and draw the outcome matrix
as instructed. You will learn more if you do this from the description, without
looking at the conventional outcome matrix.

 \begin{enumerate}

 \item[(a)]
  The {  Battle of the Sexes}  (Table  7). Put ({\it Prize Fight, Prize Fight})
in the northwest corner, but make the woman the row player.

\noindent 
 \underline{\textit{Answer.}} See Table  A1.2.

\textbf{Table  A1.2:  Rearranged Battle of the Sexes I }

\begin{tabular}{lllccc} &  &  & \multicolumn{3}{c}{\textbf{Man}} \\ &  &  &
\textit{Prize Fight} &  & $Ballet$ \\ &  & \textit{Prize Fight} & \textbf{1,2} &
$\leftarrow$ & - 5,- 5 \\ & \textbf{Woman:} &  & $\uparrow$ &  & $\downarrow$ \\
&  & $Ballet$ & -1,-1 & $\rightarrow$ & \textbf{2, 1} \\ \multicolumn{6}{l}
{\textit{Payoffs to: (Woman, Man).}} \end{tabular}

  \item[(b)]
 The  { Prisoner's Dilemma} (Table  2). Put ({\it Confess, Confess}) in the
northwest  corner.

\noindent 
\underline{\textit{Answer.}} See Table A1.3.

\textbf{Table  A1.3:  Rearranged Prisoner's Dilemma  }

\begin{tabular}{lllccc}
  &  &  & \multicolumn{3}{c}{\textbf{Column}} \\ &  &  &
$Confess$ &  & $Deny$ \\ &  & $Confess$ & \textbf{-8,-8 } & $\leftarrow$ & 0,-10
\\ & \textbf{Row:} &  & $\uparrow$ &  & $\uparrow$ \\ &  & $Deny$ & -10,0 &
$\leftarrow$ & -1,-1 \\
 \multicolumn{6}{l}{\textit{Payoffs to: (Row, Column).}}
\end{tabular}

 \item[(c)] 
The  {  Battle of the Sexes}  (Table  7). Make the man the row
player, but put ({\it Ballet, Prize Fight}) in the northwest corner.


\noindent 
\underline{\textit{Answer.}} See Table  A1.4.

\textbf{Table  A1.4:  Rearranged Battle of the Sexes II }

\begin{tabular}{lllccc}
 &  &  & \multicolumn{3}{c}{\textbf{Woman}} \\
 &  &  &
\textit{Prize Fight} &  & $Ballet$ \\
 &  & $Ballet$ & -5,-5 & $\rightarrow$ &
\textbf{1,2} \\
 & \textbf{Man:} &  & $\downarrow$ &  & $\uparrow$ \\
 &  &
\textit{Prize Fight} & \textbf{2,1} & $\leftarrow$ & - 1,- 1 \\ 
\multicolumn{6}
{l}{\textit{Payoffs to: (Man, Woman).}}
 \end{tabular}

 \end{enumerate}

 

 

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

 \noindent 
{\bf 1.7.  Finding More  Nash Equilibria}\\
 Find the Nash equilibria
of the   game illustrated in Table   12. Can any of them be reached by
iterated dominance?

\begin{center}
 {\bf Table  12: Flavor and Texture   }

 \begin{tabular}{lllccc} &       &             &\multicolumn{3}{c}{\bf Brydox}\\
&       &             &    $Flavor$   &   &  $Texture$      \\ &   &   $Flavor$
&     -2,0 & $ $  & 0,1 \\ & {\bf Apex:} &&$ $& & $ $ \\ &  &          $Texture$
&     -1,-1  & $ $  & 0,-2 \\ \multicolumn{6}{l}{\it Payoffs to: (Apex,
Brydox).} \end{tabular} \end{center}

\noindent \underline{\it Answer.} $(Flavor, Texture)$ and $(Texture, Flavor)$
are Nash equilibria.  $(Texture, Flavor)$ can be reached by iterated dominance.

\bigskip 

 \noindent
 {\bf 1.9.       Choosing Computers} (easy)\\ 
 The problem of deciding whether to
adopt IBM or HP computers by two offices in a company is most like which  game
that we have seen?
\noindent
 \underline{\it Answer.}  The Battle of the Sexes or Pure Coordination,
depending on whether the offices differ in their preferences.

\bigskip



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

  
  

\noindent
 {\bf 1.11. A Sequential Prisoner's Dilemma} (hard)\\
 Suppose Row moves first, then Column, in the Prisoner's Dilemma.  What are the
possible actions? What are the possible strategies?  Construct a normal form,
showing the relationship between strategy profiles and payoffs.

\noindent
Hint: The normal form is {\it not} a two-by-two matrix here.

 
\noindent
  \underline{\it Answer.}    The possible actions are $Confess$ and
$Deny$ for each player.

 For Column, the strategy set is: \begin{center} \noindent $\left\{
\begin{tabular}{l} ({\it C$|$C},  {\it C$|$D}),\\ ({\it C$|$C}, {\it D$|$D}),\\
({\it D$|$C}, {\it D$|$D}), \\ ({\it D$|$C}, {\it C$|$D}) \\ \end{tabular}
\right\}$ \end{center}

 For Row, the strategy set is simply $\{C, D\}$.

\noindent The normal form is:

\begin{center}

 \begin{tabular}{llcccc} &       &            \multicolumn{4}{c}{\bf Column}\\ &
&({\it C$|$C},  {\it C$|$D})   &  ({\it C$|$C}, {\it D$|$D})  & ({\it D$|$C},
{\it D$|$D}) & ({\it D$|$C}, {\it C$|$D})  \\ &    $Deny$   &   -10,0   &-1, -1
&-1, -1 &-10,0\\ {\bf Row} & &     &    & &   \\ &     $Confess$    &-8,-8  &
-8, -8&0, -10 & 0, -10 \\ \multicolumn{6}{l}{\it Payoffs to: (Row, Column)}
\end{tabular} \end{center}

   The question did not ask for the Nash equilibrium, but it is disappointing
not to know   it after all that work, so here it is:

\begin{tabular}{ccc} {\bf Equilibrium} &  {\bf  Strategies} & {\bf Outcome}\\
$E_1$ &\{{\it   Confess,  ( C$|$C)  (C$|$D)}\} & Both pick $Confess$.  \\
\end{tabular}

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