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


\noindent
 Eric Rasmusen, 	  Indiana University
School of Business, Rm. 456,
  1309 E 10th Street,
  Bloomington, Indiana, 47405-1701.
  Office: (812) 855-9219.  Fax: 812-855-3354.
Erasmuse@Indiana.edu.
   http://www.rasmusen.org/GI


\begin{center} {\bf PROBLEMS FOR CHAPTER 5 Reputation and Repeated Games}
\end{center}

\noindent     27 Sept. 2006. 10 June 2007. 
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 appeared in 2006. 
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  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 5 Reputation and Repeated Games}
\end{center}

\noindent 
 \textbf{5.1. Overlapping Generations (Samuelson [1958])}\\ 
 There is a
long sequence of players. One player is born in each period $t$, and he lives
for periods $t$ and $t+1$. Thus, two players are alive in any one period, a
youngster and an oldster. Each player is born with one unit of chocolate, which
cannot be stored. Utility is increasing in chocolate consumption, and a player
is very unhappy if he consumes less than 0.3 units of chocolate in a period: the
per-period utility functions are $U(C)=-1$ for $C < 0.3$ and $U(C)= C$ for $C
\geq 0.3$ , where $C$ is consumption. Players can give away their chocolate,
but, since chocolate is the only good, they cannot sell it. A player's action is
to consume $X$ units of chocolate as a youngster and give away $1-X$ to some
oldster. Every person's actions in the previous period are common knowledge, and
so can be used to condition strategies upon.

\begin{enumerate}

\item[(a)] If there is finite number of generations, what is the unique Nash
equilibrium?

\noindent \underline{\textit{Answer.}} \textrm{X=1}. The Chainstore Paradox
applies. Youngster $T$, the last one, has no incentive to give anything to
Oldster $ T-1$. Therefore, Youngster $T-1$ has no incentive either, and so for
for every $t$.

\item[(b)] If there are an infinite number of generations, what are two Pareto-
ranked perfect equilibria?

\noindent \underline{\textit{Answer.}} (i) \textit{($X=1$, regardless of what
others do)}, and (ii) \textit{($X=0.5$, unless some player has deviated, in
which case $X=1$)}. Equilibrium (ii) is pareto superior.

\item[(c)] If there is a probability $\theta$ at the end of each period (after
consumption takes place) that barbarians will invade and steal all the chocolate
(leaving the civilized people with payoffs of -1 for any $X $ ), what is the
highest value of $\theta$ that still allows for an equilibrium with $X=0.5$?

\noindent
 \underline{\textit{Answer.}} The payoff from the equilibrium strategy
is $ 0.5 + (1- \theta)0.5 + \theta (-1)=1-1.5\theta$. The payoff from deviating
to $X=1$ is $1 -1 = 0$. These are equal if $1-1.5 \theta = 0$; that is, if ${ \
\theta = \frac{2}{3}}$. Hence, $\theta$ can take values up to $\frac{2}{3}$ and
the $X=0.5$ equilibrium can still be maintained. \end{enumerate}

  \bigskip

 \noindent
 {\bf 5.3. Repeated Games (see Benoit \& Krishna
[1985]) }\\ Players Benoit and Krishna repeat the   game in Table  7  three
times, with discounting: \begin{center} {\bf Table  7:  A Benoit-Krishna Game }

 \begin{tabular}{lllccccc} &       &             &\multicolumn{5}{c}{\bf
Krishna}\\ &       &   &  {\it   Deny }   &   &  $Waffle $   & &$Confess$   \\ &
&& & &  & \\ &   &  $ Deny$       &  10,10 &    &  $-1,-12$&  & $-1, 15$\\ &  &&
& &  & \\ & {\bf Benoit:} &    {\it Waffle }    &      $ -12,-1$  &    & 8,8 & &
$-1,-1$ \\ &   && & &  & \\ &  &    {\it Confess }    &      $ 15,-1$  &    &
8,$-1$  & & $ 0,0$ \\ &   && & &  & \\ \multicolumn{8}{l}{\it Payoffs to:
(Benoit, Krishna).} \end{tabular} \end{center}

\begin{enumerate}

\item[(a)] Why is there no equilibrium in which the players play $Deny$ in all
three periods?

\noindent \underline{\textit{Answer.}} If  Benoit and Krishna  both chose $Deny$
in the third period, Krishna would get a payoff of 10 in that period. He  could
increase his payoff by deviating to $Confess$.

\item[(b)] Describe a perfect equilibrium in which both players pick $ Deny$ in
the first two periods.

\noindent \underline{\textit{Answer.}} In the last period, any equilibrium has
to have the players  either both choosing $Confess$ or both choosing $Waffle$
(which means to equivocate, to talk but neither to quite deny or quite confess).
Consider the following proposed equilibrium behavior for each player:

\noindent 1. Choose $Deny$ in the first period.

\noindent 2. Choose $Deny$ in the second period unless someone chose a different
action in the first period, in which case choose $Confess$.

\noindent 3. Choose $Waffle$ in the third period unless someone chose something
other than $Deny$   in the first or second period, in which case choose
$Confess$.

This is an equilibrium. In the third period, a deviator to  either $Deny$ or
$Confess$ would have a payoff of -1 instead of 8 in that period. If, however,
someone has already deviated in an earlier period, each player expects the other
to choose $Confess$, in which case $Confess$ is his best response.

In the second period,    if a player deviates to $Deny$ he will have a payoff of
15 instead of 10 in that period. In the third period, however, his payoff will
then  be 0 instead of 8, because the actions will be ($Confess, Confess$)
instead of ($Waffle, Waffle$). If the discount rate is low  enough (for example
$r=0$), then deviation in the second period is not profitable. If some other
player has deviated in the first period, however, the players expect each other
to choose $Confess$ in the second period and that is self- confirming.

 In the first period,   if a player deviates to $Deny$ he will have a payoff of
15 instead of 10 in that period. In the second period, however, his payoff will
then be    0 instead of 10, because the actions will be ($Confess, Confess$)
instead of ($Deny, Deny$).  And in the third period his payoff will then be 0
instead of 8,  because the actions will be ($Confess, Confess$) instead of
($Waffle, Waffle$).   If the discount rate is low  enough (for example $r=0$),
then deviation in the first period is not profitable.

\item[(c)] Adapt your equilibrium to the twice-repeated game.

\noindent \underline{\textit{Answer.}}  Simply leave out the middle period of
the three-period model:

\noindent 1. Choose $Deny$ in the first period.

\noindent 2. Choose $Waffle$ in the second period unless someone chose something
other than $Deny$   in the first   period, in which case choose $Confess$.

\item[(d)] Adapt your equilibrium to the $T$-repeated game.

\noindent \underline{\textit{Answer.}} Now we just add extra middle periods:

\noindent 1. Choose $Deny$ in the first period.

\noindent 2. Choose $Deny$ in the second period unless someone chose a different
action in the first period, in which case choose $Confess$.

\noindent $t$. Choose $Deny$ in the $t$'th period for $t=3,...,T-1$ unless
someone chose a different action previously, in which case choose $Confess$.

\noindent $T$. Choose $Waffle$ in the third period unless someone chose
something other than $Deny$   previously, in which case choose $Confess$.

\item[(e)]  What is the greatest discount rate for which your equilibrium still
works in the 3-period game?

\noindent \underline{\textit{Answer.}} It is harder to prevent deviation in the
second period than in the first period, because deviation in the first period
leads to lower payoffs in two future periods instead of one. So if a discount
rate is low enough to prevent deviation in the second period, it is low enough
to prevent deviation in the first period.

The equilibrium payoff in the subgame starting with the second period is, if the
discount rate is $\rho$, $$ 10 + \frac{1}{1+\rho} \left( 8 \right) $$ The payoff
to deviating to $Confess$ in the second period and then choosing $Confess$ in
the third period is $$ 15 + \frac{1}{1+\rho} \left( 0 \right). $$ Equating these
two payoffs yields $10 + \frac{8}{1+\rho} = 15$, so $8 = 5(1+\rho)$, $3 =
5\rho$,  and $\rho = .6$.  This is the greatest discount rate for which the
strategy combination in part (a) remains an equilibrium.

\end{enumerate}

\bigskip

 \noindent
 \textbf{5.5. The Repeated {\ Prisoner's Dilemma}  } \\ Set $
P=0$ in the general {\ Prisoner's Dilemma} in Table  1.10 of Chapter 1, and assume
that $2R > S+T$.

\begin{enumerate}

 \item[(a)]
 Show that the Grim Strategy, when played by both players, is a
perfect equilibrium for the infinitely repeated game. What is the maximum
discount rate for which the Grim Strategy remains an equilibrium?

\noindent
\underline{\textit{Answer.}} The grim strategy is a perfect
equilibrium because the payoff from continued cooperation is $R + \frac{R}{r}$,
which for low discount rates is greater than the payoff from $(Confess, Deny)$
once and $(Confess, Confess)$ forever after, which is $T + \frac{0}{r} $. To
find the maximum discount rate, equate these two payoffs: $R + \frac{R}{r} = T$.
This means that ${r = \frac{R}{T-R}  }$ is the maximum.

\item[(b)]
  Show that Tit-for-Tat is not a perfect equilibrium in the infinitely
repeated {\ Prisoner's Dilemma} with no discounting.\footnote{xxx Add: The idea
is informally explained on page 112).}

\noindent \underline{\textit{Answer.}} Suppose Row has played $Confess$. Will
Column retaliate? If both follow tit-for-tat after the deviation, retaliation
results in a cycle of $(Confess, Deny)$, $(Deny, Confess) $, forever. Row's
payoff is $T+S+T+S+...$. If Column forgives, and they go back to cooperating, on
the other hand, his payoff is $R+R+R+R+ ...$. Comparing the first four periods,
forgiveness has the higher payoff because $4R > 2S + 2T$. The payoffs of the
first four periods simply repeat an infinite number of times to give the total
payoff, so forgiveness dominates retaliation, and tit-for-tat is not perfect.

  Kalai, Samet \& Stanford (1988)  pointed this out.

\end{enumerate}

\bigskip 

\noindent 
\textbf{5.7.  Grab the Dollar }\\
  Table  5.10 shows the payoffs
for the simultaneous-move game of Grab the Dollar. A silver dollar is put on the
table between Smith and Jones. If one grabs it, he keeps the dollar, for a
payoff of 4 utils. If both grab, then neither gets the dollar, and both feel
bitter. If neither grabs, each gets to keep something.

\begin{center} \textbf{Table 5.10: Grab the Dollar}

\begin{tabular}{lllccc} &  &  & \multicolumn{3}{c}{\textbf{Jones}} \\ &  &  &
\textit{Grab} ($\theta$) &  & \textit{Wait} ($1- \theta$) \\ &  & $Grab$
($\theta$) & $-1,-1$ &  & $4,0$ \\ & \textbf{Smith:} &  &  &  &  \\ &  &
\textit{Wait } ($1-\theta$) & $0,4$ &  & 1,1 \\ \multicolumn{6}{l}
{\textit{Payoffs to: (Smith, Jones).}} \end{tabular} \end{center}

\begin{enumerate} 

\item[(a)] 
  What are the evolutionarily stable strategies?

\noindent
  \underline{\textit{Answer.}} The ESS is mixed and unique. Let
$Prob(Grab) = \theta$. Then $\pi(Grab) = -1(\theta) + 4(1-\theta) = \pi(Wait) =
0(\theta) + 1(1-\theta)$, which solves to $\theta = 3/4$. Three fourths of the
population plays \textit{Grab}.

\item[(b)] Suppose each player in the population is a point on a continuum, and
that the initial amount of players is 1, evenly divided between \textit{Grab}
and \textit{Wait}. Let $N_t(s)$ be the amount of players playing a particular
strategy in period $t$ and let $\pi_t(s)$ be the payoff. Let the population
dynamics be $N_{t+1} (i) = \left(2 N_t(i) \right)\left(\frac{ \pi_t(i)}{\sum_j
\pi_t(j) } \right) $. Find the missing entries in Table  5.11.

\textbf{Table  5.11: Grab the Dollar, Dynamics }

\begin{tabular}{|l|l|l|l|l|l|l|} \hline t & $N_t(G)$ & $N_t(W)$ & $N_t(total)$ &
$\theta$ & $\pi_t(G)$ & $\pi_t(w)$ \\ \hline 0 & 0.5 & 0.5 & 1 & 0.5 & 1.5 & 0.5
\\ \hline 1 &  &  &  &  &  &  \\ \hline 2 &  &  &  &  &  &  \\ \hline
\end{tabular}\\

\noindent \underline{\textit{Answer.}} See Table A5.1.

\textbf{Table A5.1:   Grab the Dollar,  Dynamics I}

\begin{tabular}{|l|l|l|l|l|l|l|} \hline t & $N_t(G)$ & $N_t(W)$ & $N_t(total)$ &
$\theta$ & $\pi_t(G)$ & $\pi_t(w)$ \\ \hline 0 & 0.5 & 0.5 & 1 & 0.5 & 1.5 & 0.5
\\ \hline 1 & 0.75 & 0.25 & 1 & 0.75 & 0.25 & 0.25 \\ \hline 2 & 0.75 & 0.25 & 1
& 0.75 & 0.25 & 0.25 \\ \hline \end{tabular}

\item[(c)] Repeat part (b), but with the dynamics $N_{t+t}(s) = [1 + \frac{
\pi_t(s)}{\sum_j \pi_t(j) }][2N_t(s)]$.

\noindent \underline{\textit{Answer.}} See Table A5.2.

\textbf{Table A5.2:   Grab the Dollar,  Dynamics II}

\begin{tabular}{|l|l|l|l|l|l|l|} \hline t & $N_t(G)$ & $N_t(W)$ & $N_t(total)$ &
$\theta$ & $\pi_t(G)$ & $\pi_t(w)$ \\ \hline 0 & .5 & 0.5 & 1 & .5 & 1.5 & 0.5
\\ \hline 1 & 1.75 & 1.25 & 3 & 0.58 & 1.1 & 0.42 \\ \hline 2 & 6.03 & 3.19 &
9.22 & 0.65 & 0.75 & 0.35 \\ \hline \end{tabular}

\item[(d)] Which three games that have appeared so far in the book resemble
\textit{Grab the Dollar}?

\noindent \underline{\textit{Answer.}}  Chicken, the Battle of the Sexes, and
the
Hawk-Dove Game.

\end{enumerate}

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