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\noindent \textbf{Answers to Odd-Numbered Problems, 3rd
Edition of
Games and
Information, Rasmusen }

\noindent   March 21, 2002.

\noindent 
 This appendix contains answers to the odd-numbered problems in the
third edition of \textit{Games and Information} by Eric Rasmusen,
published in 2001. 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.

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

\noindent \textbf{1.1: Nash and Iterated Dominance.}

\begin{enumerate}
\item[(1.1a)]  Show that every iterated dominance
equilibrium $s^*$ is
Nash.

\underline{\textit{Answer.}} Suppose that $s^*$ is not Nash.
This
means that
there exist 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 that
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 have a contradiction; it must be that every iterated
dominance
equilibrium is Nash.



\item[(1.1b)]  Show by counterexample that not every Nash
equilibrium
can be
generated by iterated dominance.

\underline{\textit{Answer.}} In ``Ranked Coordination''
(Table 1.7) no
strategy can be eliminated by dominance, and the boldfaced
strategies
are
Nash.

\item[(1.1c)]  Is every iterated dominance equilibrium made
up of
strategies
that are not weakly dominated?

\underline{\textit{Answer.}} \textrm{No.} A strategy that is
in the
equilibrium strategy profile might be a bad reply to some
strategies
that
iterated deletion removed from the original game. Consider
the
Iteration
Path Game below. The strategy profiles $(r_1, c_1)$ and
$(r_1, c_3)$
are
both iterated dominance equilibria, because each of those
strategy
profiles
can be found by iterated deletion. The deletion can proceed
in the
order $
(r_3, c_3, c_2, r_2)$ or in the order $(r_2, c_2, c_1, r_3)
$. But
$c_3$,
which is a part of the $(r_1, c_3)$ equilibrium, is weakly
dominated
by $c_1$.

\textbf{The Iteration Path Game }

\begin{tabular}{lllccc}
&  &  & \multicolumn{3}{c}{\textbf{Column}} \\
&  &  & $c_1$ & $c_2$ & $c_3$ \\
\multicolumn{6}{l}{} \\
&  & $r_1$ & \textbf{2,12 } & 1,10 & \textbf{1,12} \\
\multicolumn{6}{l}{} \\
& \textbf{Row:} & $r_2$ & 0,12 & 0,10 & 0,11 \\
\multicolumn{6}{l}{} \\
&  & $r_3$ & 0,12 & 1,10 & 0,13 \\
\multicolumn{6}{l}{} \\
\multicolumn{6}{l}{} \\
\multicolumn{6}{l}{\textit{Payoffs to: (Row, Column)}}
\end{tabular}
\end{enumerate}

\bigskip

\bigskip

\noindent \textbf{1.3: Pareto Dominance (based on notes by
Jong-shin
Wei)}

\begin{enumerate}
\item[ (1.3a)]  If a strategy combination $s^*$ is a
dominant strategy
equilibrium, does that mean it weakly pareto-dominates all
other
strategy
combinations?

\underline{\textit{Answer.}} \textrm{No}--- think of the
``Prisoner's
Dilemma'', in Table 1.1. (\textit{Confess, Confess}) is a
dominant
strategy
equilibrium, but it does not weakly pareto-dominate
(\textit{Deny,
Deny})

\item[ (1.3b)]  If a strategy combination $s$ strongly
pareto-
dominates all
other strategy combinations, does that mean it is a dominant
strategy
equilibrium?

\underline{\textit{Answer.}} \textrm{No}--- think of ``Ranked
Coordination'' in Table 1.7. (\textit{Large, Large}) strongly pareto-
dominates all other strategy profiles, but is not a dominant strategy
equilibrium.\footnote{ The  Prisoner's Dilemma  is not a good example
for this problem, because \textit{(Deny, Deny)} does not pareto-
dominate \textit{(Deny, Confess)}.}

\item[ (1.3c) ]  If $s$ weakly pareto-dominates all other
strategy
combinations, then must it be a Nash equilibrium?

\underline{\textit{Answer.}} \textrm{Yes.} If $s$ is weakly
pareto-
dominant, then $\pi_i(s) \geq \pi_i(s^{\prime}), \forall
s^{\prime},
\forall
i$. 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 the condition to be weakly
pareto-dominant, it must also be a Nash equilibrium.
\end{enumerate}

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

\bigskip \noindent \textbf{1.5: Drawing Outcome Matrices. } 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[(1.5a)]  The {\ Battle of the Sexes} (Table 1.8). Put
(\textit{Prize
Fight, Prize Fight}) in the northwest corner, but make the
woman the
row
player.

\underline{\textit{Answer.}} See Table C.1.

\textbf{Table C.1 ``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[ (1.5b)]  The {\ Prisoner's Dilemma} (Table 1.2). Put
(\textit{%
Confess, Confess}) in the northwest corner.

\underline{\textit{Answer.}} See Table C.2.

\textbf{Table C.2 ``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[ (1.5c) ]  The {\ Battle of the Sexes} (Table 1.8).
Make the man
the
row player, but put (\textit{Ballet, Prize Fight}) in the
northwest
corner.

\underline{\textit{Answer.}} See Table C.3.

\textbf{Table C.3 ``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}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 2: INFORMATION}
\end{center}

\bigskip \noindent \textbf{2.1: The Monty Hall Problem.} You
are a
contestant on the TV show, ``Let's Make a Deal.'' You face
three
curtains,
labelled A, B and C. Behind two of them are toasters, and
behind the
third
is a Mazda Miata car. You choose A, and the TV showmaster
says,
pulling
curtain B aside to reveal a toaster, ``You're lucky you
didn't choose
B, but
before I show you what is behind the other two curtains,
would you
like to
change from curtain A to curtain C?'' Should you switch?
What is the
exact
probability that curtain C hides the Miata?

\underline{\textit{Answer.}} You should switch to curtain C,
because

\begin{center}
\begin{tabular}{ll}
Prob (Miata behind C $|$ Host chose B) & =
$\frac{\mathrm{Prob( Host\;
chose
\;B\; | \;Miata \; behind \;C) Prob(Miata \; behind\; C)}}
{\mathrm{Prob(
Host\; chose \;B)}}$ \\
&  \\
& $= \frac{(1) (\frac{1}{3})}{(1)(\frac{1}{3}) + (\frac{1}
{2})
(\frac{1}{3})
}$. \\
&  \\
& = \textrm{$\frac{2}{3}$.}
\end{tabular}
\end{center}
    The key is to remember that this is a game. The host's
action has
revealed
more than that the Miata is not behind B; it has also
revealed that
the host
did not want to choose curtain C. If the Miata were behind B
or C, he
would
pull aside the curtain it was not behind. Otherwise, he
would pull
aside a
curtain randomly. His choice tells you nothing new about the
probability
that the Miata is behind curtain A, which remains $\frac{1}
{3}$, so
the
probability of it being behind C must rise to $\frac{2}{3}$
(to make
the
total probability equal one).

 What would be the best choice if curtain B simply was blown
aside by
the wind, revealing a toaster, and  the host, Monty Hall,
asked if you
wanted to
switch to curtain C?  In that case you should be
indifferent. Just as
easily, curtain C might have blown aside, possibly
revealing a Miata,
but though the wind's random choice is informative-- your
posterior on
the probability that the Miata is behind curtain C rises
from 1/3 to
1/2--- it does not convey as much information as Monty
Hall's
deliberate choice.


\bigskip
 \noindent
\textbf{2.3: Cancer Tests.} Imagine that you are
being
tested for cancer, using a test that is 98\% accurate. If
you indeed
have
cancer, the test shows positive (indicating cancer) 98\% of
the time.
If you
do not have cancer, it shows negative 98\% of the time. You
have heard
that
1 in 20 people in the population actually have cancer. Now
your doctor
tells
you that you tested positive, but you shouldn't worry
because his last
19
patients all died. How worried should you be? What is the
probability
you
have cancer?

\underline{\textit{Answer.}} Doctors, of course, are not
mathematicians.
Using Bayes' Rule:
\begin{equation}  \label{e58}
\begin{array}{ll}
Prob(Cancer|Positive) & = \frac{Prob(Positive|Cancer)
Prob(Cancer)}{%
Prob(Positive)} \\
&  \\
& = \frac{0.98(0.05)}{0.98(0.05) + 0.02(0.95)} \\
&  \\
& {\approx 0.72}.
\end{array}
\end{equation}
With a 72 percent chance of cancer, you should be very
worried. But at
least it is not 98 percent. \newline
    \hspace*{16pt} Here is another way to see the answer.
Suppose
10,000 tests are done. Of these, an average of 500 people
have cancer.
Of these, 98\% test positive on average--- 490 people. Of
the 9,500
cancer-free people, 2\% test positive on average---190
people. Thus
there are 680 positive tests, of which 490 are true
positives. The
probability of having cancer if you test positive is
490/680, about
72\%. \newline
    \hspace*{16pt} This sort of analysis is one reason why
HIV testing
for the entire population, instead of  for  high-risk
subpopulations,
would not be very informative--- there  would be more false
positives
than true positives.

\bigskip \noindent \textbf{2.5: Joint Ventures.} Software
Inc. and
Hardware
Inc. have formed a joint venture. Each can exert either high
or low
effort,
which is equivalent to costs of 20 and 0. Hardware moves
first, but
Software
cannot observe his effort. Revenues are split equally at the
end, and
the
two firms are risk neutral. If both firms exert low effort,
total
revenues
are 100. If the parts are defective, the total revenue is
100;
otherwise, if
both exert high effort, revenue is 200, but if only one
player does,
revenue
is 100 with probability 0.9 and 200 with probability 0.1.
Before they
start,
both players believe that the probability of defective parts
is 0.7.
Hardware discovers the truth about the parts by observation
before he
chooses effort, but Software does not.

\begin{enumerate}
\item[ (2.5a)]  Draw the extensive form and put dotted lines
around
the
information sets of Software at any nodes at which he moves.

\underline{\textit{Answer.}} See Figure A.1. To understand
where the
payoff numbers come from, see the answer to part (b).

\textbf{Figure A.1 The Extensive Form for the Joint Ventures
Game}

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\item[(2.5b)]  What is the Nash equilibrium?

\underline{\textit{Answer.}} (Hardware: $Low$ if defective
parts,
$Low$ if not
defective parts; Software:  $Low$).

$$
 \pi_{Hardware} (Low|Defective) = \frac{100}{2}  = 50.
 $$
 Deviating would yield Hardware a lower payoff:
 $$
 \pi_{Hardware} (High|Defective) =  \frac{100}{2} -20 = 30.
 $$

$$
 \pi_{Hardware} (Low|Not\;Defective) = \frac{100}{2}  = 50.
 $$
 Deviating would yield Hardware a lower payoff:
 $$
 \pi_{Hardware} (High| Not\; Defective) = .9\left(
\frac{100}{2}
\right)+ .1 \left( \frac{200}{2}  \right)  -20 = 45 + 10 -20
= 35.
 $$


 $$
 \pi_{Software} (Low) = \frac{100}{2}  = 50.
 $$
 Deviating would yield Software a lower payoff:
 $$
 \pi_{Software} (High) = .7  \left( \frac{100}{2}  \right)
+ .3
\left[  .9\left( \frac{100}{2}  \right)+ .1 \left(
\frac{200}{2}
\right)    \right]  -20 =  35 + .3 (45+10) - 20.
 $$
 This equals $15 + .3 (35) =31.5$, less than the equilibrium
payoff of
50.


\noindent
 {\it Elaboration}.  A strategy profile that is {\it not} an
equilibrium (because Software would deviate) is:

\noindent
(Hardware: $Low$ if defective parts, $High$ if not
defective parts; Software:  $High$).

$$
 \pi_{Hardware} (Low|Defective) = \frac{100}{2}  = 50.
 $$
 Deviating would indeed yield Hardware a lower payoff:
 $$
 \pi_{Hardware} (High|Defective) =  \frac{100}{2} -20 = 30.
 $$


 $$
 \pi_{Hardware} (High| Not\; Defective) = \frac{200}{2} -20
= 100-20=
80.
 $$
 Deviating would  indeed yield Hardware a lower payoff:
 $$
 \pi_{Hardware} (Low|Not\;Defective) = .9\left( \frac{100}
{2}  \right)
+ .1 \left( \frac{200}{2}  \right) =55.
 $$


 $$
 \pi_{Software} (High) = .7  \left( \frac{100}{2}  \right)
+ .3
\left(  \frac{200}{2}  \right)  -20 =  35 +  30-20 = 45.
 $$
 Deviating would yield Software a higher payoff, so the
strategy
profile we are testing is not a Nash equilibrium:
$$
 \pi_{Software} (Low) =  .7  \left( \frac{100}{2}  \right)
+ .3
\left[  .9\left( \frac{100}{2}  \right)+ .1 \left(
\frac{200}{2}
\right)    \right]    =  35 + .3 (45+10)  = 35 + .16.5 =
51.5.
 $$

\noindent
 {\it More  Elaboration}.   Suppose the probability of
revenue of 100
if one player choose High and the other chooses Low were $z$
instead
of .9.  If $z$ is too low, the equilibrium  described above
breaks
down because Hardware finds it profitable to deviate to
$High|Not\;
Defective$.
$$
 \pi_{Hardware} (Low|Not\;Defective) = \frac{100}{2}  = 50.
 $$
 Deviating would yield Hardware a lower payoff:
 $$
 \pi_{Hardware} (High| Not\; Defective) = z\left( \frac{100}
{2}
\right)+  (1-z) \left( \frac{200}{2}  \right)  -20 =  50z +
100-100z
-20.
 $$
 This comes to be  $\pi_{Hardware} (High| Not\; Defective) =
80-50z$,
so if  $z<.6$  then the payoff from $(High| Not\; Defective)
$ is
greater
than 50, and so Hardware would be willing to unilaterally
supply High
effort even though Software is providing Low effort.

 You might wonder whether Software would deviate from the
equilibrium
for some value of $z$ even greater than .6. To see that he
would not,
note that
 $$
\pi_{Software} (High) = .7  \left( \frac{100}{2}  \right)
+ .3
\left[  z\left( \frac{100}{2}  \right)+ (1-z) \left(
\frac{200}{2}
\right)    \right]  -20.
 $$
  This takes its greatest value at  $z=0$,  but even then
the payoff
from $High$ is just $.7 (50) + .3 (100) - 20 = 45$, less
than the
payoff of 50 from $Low$. The chances of non-defective parts
are just
too low for Software to want to take the  risk of playing
$High$ when
Hardware is sure to play $Low$.


\item[ (2.5c)]  What is Software's belief, in equilibrium,
as to the
probability that Hardware chooses low effort?

\underline{\textit{Answer.}} \textrm{One.} In equilibrium,
Hardware
always
chooses $Low$.

\item[(2.5d)]  If Software sees that revenue is 100, what
probability
does
he assign to defective parts if he himself exerted high
effort and he
believes that Hardware chose low effort?

\underline{\textit{Answer.}} \textrm{0.72} ($= (1)
\frac{0.7}{(1)(0.7)
+(0.9)
(0.3)}$).
\end{enumerate}

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


\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[ (3.1a) ]  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?

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

\item[ (3.1b) ]  What are the two pure-strategy equilibria?

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

\item[ (3.1c)]  If the game only lasts one period, and the
Republican
wins
the general election (for Democrat payoffs of zero) if both
Democrats
refuse
to exit, what is the probability $\gamma$ with which each
candidate
exits in
a symmetric equilibrium?

\underline{\textit{Answer.}} The payoff matrix is shown in
Table A.5.

\textbf{Table A.5 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  [(3.3a)] Draw the outcome matrix for Matching
Pennies.

\textbf{Table A.6 ``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[(3.3b)]  Show that there is no Nash equilibrium in
pure
strategies.

\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[(3.3c)]  Find the mixed-strategy equilibrium, denoting
Smith's
probability of $Heads$ by $\gamma$ and Jones' by $\theta$.

\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[ (3.3d) ]  Prove that there is only one mixed-strategy
equilibrium.

\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,
Charles, 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 Charles and Vladimir each vote with the same probability
$\gamma$,
but
they decide to vote independently of each other.

\begin{enumerate}
\item[(3.5a)]  What is the probability that the proposition
will pass,
as a
function of $\theta$ and $\gamma$?

\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, ${\ \gamma (2\gamma \theta
- 2\theta
+ 2 -
\gamma)}$.

\item[ (3.5b)]  What are the two possible equilibrium
probabilities $%
\gamma_1 $ and $\gamma_2$ with which Charles might vote?
Why,
intuitively,
are there two symmetric equilibria?

\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
Charles 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 Charles
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).\newline
\hspace*{16pt} Why are there two solutions? If Charles and
Vladimir
are sure
not to vote, Adam will not vote, because if he does not vote
he will
win, 0-
0. If Charles 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 Charles 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[ (3.5c)]  What is the probability $\theta$ that Adam
will vote
in each
of the two symmetric equilibria?

\underline{\textit{Answer.}} Now use the payoffs for
Charles, 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[(3.5d) ]  What is the probability that the proposition
will
pass?

\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{0.37} (rounded ($=
(0.3)( 0.91)
^2 +
(0.7) ( 2(0.91) (0.09) + (0.91)^2)) $.
\end{enumerate}

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

\begin{center}
\textbf{PROBLEMS FOR CHAPTER 4}
\end{center}

\noindent \textbf{4.1: Repeated Entry Deterrence.} Consider
two
repetitions
without discounting of the game {\ Entry Deterrence I } from
Section
4.2.
Assume that there is one entrant, who sequentially decides
whether to
enter
two markets that have the same incumbent.

\begin{enumerate}
\item[(4.1a)]  Draw the extensive form of this game.

\underline{\textit{Answer.}} See Figure A.2. If the entrant
does not
enter,
the incumbent's response to entry in that period is
unimportant.

\textbf{Figure A.2 ``Repeated Entry Deterrence'' }

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\item[(4.1b)]  What are the 16 elements of the strategy sets
of the
entrant?

\underline{\textit{Answer.}} The entrant makes a binary
decision at
four
nodes, so his strategy must have four components, strictly
speaking,
and the
number of possible arrangements is (2)(2)(2)(2) = 16. Table
A.7 shows
the
strategy space, with $E$ for $Enter$ and $S$ for
\textit{Stay out}.

\textbf{Table A.7 The Entrant's Strategy Set }
\begin{tabular}{|l|llll|}
\hline
Strategy & $E_{1}$ & $E_{2}$ & $E_{3}$ & $E_{4}$ \\ \hline
1 & E & E & E & E \\
2 & E & E & E & E \\
3 & E & E & E & S \\
4 & E & E & S & S \\
5 & E & S & S & S \\
6 & E & S & E & E \\
7 & E & S & S & E \\
8 & E & S & E & S \\
9 & S & E & E & E \\
10 & S & S & E & E \\
11 & S & S & S & E \\
12 & S & S & S & S \\
13 & S & E & S & S \\
14 & S & E & S & E \\
15 & S & E & E & S \\
16 & S & S & E & S \\ \hline
\end{tabular}

Usually modellers are not so careful. Table A.7 includes
action rules
for
the Entrant to follow at nodes that cannot be reached unless
the
Entrant
trembles, somehow deviating from its own strategy. If the
Entrant
chooses
Strategy 16, for example, nodes $E_{3}$ and $E_{4}$ cannot
possibly be
reached, even if the Incumbent deviates, so one might think
that the
parts
of the strategy dealing with those nodes are unimportant.
Table A.8
removes
the unimportant parts of the strategy, and Table A.16
condenses the
strategy
set down to its six importantly distinct strategies.
\pagebreak

\textbf{Table A.8 The Entrant's Strategy Set, Abridged
Version I }

\begin{tabular}{|l|llll|}
\hline
Strategy & $E_1$ & $E_2$ & $E_3$ & $E_4$ \\ \hline
1 & E & - & E & E \\
2 & E & - & E & E \\
3 & E & - & E & S \\
4 & E & - & S & S \\
5 & E & - & S & S \\
6 & E & - & E & E \\
7 & E & - & S & E \\
8 & E & - & E & S \\
9 & S & E & - & - \\
10 & S & S & - & - \\
11 & S & S & - & - \\
12 & S & S & - & - \\
13 & S & E & - & - \\
14 & S & E & - & - \\
15 & S & E & - & - \\
16 & S & S & - & - \\ \hline
\end{tabular}

\textbf{Table A.9 The Entrant's Strategy Set, Abridged
Version II }

\begin{tabular}{|l|llll|}
\hline
Strategy & $E_1$ & $E_2$ & $E_3$ & $E_4$ \\ \hline
1 & E & - & E & E \\
3 & E & - & E & S \\
4 & E & - & S & S \\
7 & E & - & S & E \\
9 & S & E & - & - \\
10 & S & S & - & - \\ \hline
\end{tabular}

\item[(4.1c)]  What is the subgame perfect equilibrium?

\underline{\textit{Answer.}} The entrant always enters and
the
incumbent
always colludes.

\item[(4.1d) ]  What is one of the nonperfect Nash
equilibria?

\underline{\textit{Answer.}} The entrant stays out in the
first
period, and
enters in the second period. The incumbent fights any entry
that might
occur
in the first period, and colludes in the second period.
\end{enumerate}

\bigskip \noindent \textbf{4.3: Heresthetics in Pliny and
the
Freedmens'
Trial.} (Pliny, 1963, pp. 221-4, Riker, 1986, pp.78-88).
Afranius
Dexter
died mysteriously, perhaps dead by his own hand, perhaps
killed by his
freedmen (servants a step above slaves), or perhaps killed
by his
freedmen
by his own orders. The freedmen went on trial before the
Roman Senate.
Assume that 45 percent of the senators favor acquittal, 35
percent
favor
banishment, and 20 percent favor execution, and that the
preference
rankings
in the three groups are $A \succ B \succ E$, $B \succ A
\succ E$, and
$E
\succ B \succ A$. Also assume that each group has a leader
and votes
as a
bloc.

\begin{enumerate}
\item[ (4.3a)]  Modern legal procedure requires the court to
decide
guilt
first and then assign a penalty if the accused is found
guilty. Draw a
tree
to represent the sequence of events (this will not be a game
tree,
since it
will represent the actions of groups of players, not of
individuals) .
What
is the outcome in a perfect equilibrium?

\underline{\textit{Answer.}} Guilt would win in the first
round by a
vote of
55 to 45, and banishment would win in the second by 80 to
20. See
Figure A.3.

\textbf{Figure A.3 Modern Legal Procedure}

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\item[ (4.3b) ]  Suppose that the acquittal bloc can pre-
commit to how
they
will vote in the second round if guilt wins in the first
round. What
will
they do, and what will happen? What would the execution bloc
do if
they
could control the second-period vote of the acquittal bloc?

\underline{\textit{Answer.}}The acquittal bloc would commit
to
execution,
inducing the Banishment bloc to vote for Acquittal in the
first round,
and
acquittal would win. The execution bloc would order the
acquittal bloc
to
choose banishment in the second round to avoid making the
banishment
bloc
switch to acquittal.\footnote{
Note that preferences do not always work out this way. In
Athens, six
centuries before the Pliny episode, Socrates was found
guilty in a
first
round of voting and then sentenced to death (instead of a
lesser
punishment
like banishment) by a bigger margin in the second round.
This would
imply
the ranking of the acquittal bloc there was AEB, except for
the
complicating
factor that Socrates was a bit insulting in his sentencing
speech.}

\item[(4.3c)]  The normal Roman procedure began with a vote
on
execution
versus no execution, and then voted on the alternatives in a
second
round if
execution failed to gain a majority. Draw a tree to
represent this.
What
would happen in this case?

\underline{\textit{Answer.}} Execution would fail by a vote
of 20 to
80, and
banishment would then win by 55 to 45. See Figure A.4.

\textbf{Figure A.4 Roman Legal Procedure}

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\item[ (4.3d)]  Pliny proposed that the Senators divide into
three
groups,
depending on whether they supported acquittal, banishment,
or
execution, and
that the outcome with the most votes should win. This
proposal caused
a roar
of protest. Why did he propose it?

\underline{\textit{Answer.}} It must be that Pliny favored
acquittal
and
hoped that every senator would vote for his preference,.
Acquittal
would
then win 45 to 35 to 25.

\item[(4.3e)]  Pliny did not get the result he wanted with
his voting
procedure. Why not?

\underline{\textit{Answer.}} Pliny said that his arguments
were so
convincing that the senator who made the motion for the
death penalty
changed his mind, along with his supporters, and voted for
banishment,
which
won (by 55 to 45 in our hypothesized numbers). He forgot
that people
do not
always vote for their first preference. The execution bloc
saw that
acquittal would win unless they switched to banishment.

\item[ (4.3f)]  Suppose that personal considerations made it
most
important
to a senator that he show his stand by his vote, even if he
had to
sacrifice
his preference for a particular outcome. If there were a
vote over
whether
to use the traditional Roman procedure or Pliny's procedure,
who would
vote
with Pliny, and what would happen to the freedmen?

\underline{\textit{Answer.}} Traditional procedure would win
by
capturing
the votes of the execution bloc and the banishment bloc, and
the
freedmen
would be banished. In this case, the voting procedure would
matter to
the
result, because each senator would vote for his preference.
\end{enumerate}

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

\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[ (5.1a)]  If there is finite number of generations,
what is the
unique
Nash equilibrium?

\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[(5.1b)]  If there are an infinite number of
generations, what
are two
Pareto-ranked perfect equilibria?

\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[(5.1c) ]  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 $X=0.5$?

\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
\textbf{5.3: Repeated Games.}\footnote{
See Benoit \& Krishna (1985).} Players Benoit and Krishna
repeat the
game in
Table 5.7 three times, with discounting:

\begin{center}
\textbf{Table 5.7 A Benoit-Krishna Game }

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

\begin{enumerate}
\item[(5.3a)]  Why is there no equilibrium in which the
players play
$Deny$
in all three periods?

\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[(5.3b)]  Describe a perfect equilibrium in which both
players
pick $
Deny$ in the first two periods.

\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[ (5.3c) ]  Adapt your equilibrium to the twice-
repeated game.

\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[(5.3d)]  Adapt your equilibrium to the $T$-repeated
game.

\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[(5.3e)]  What is the greatest discount rate for which
your
equilibrium
still works in the 3-period game?


\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 profile 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 1.11, and
assume
that $2R
> S+T$.

\begin{enumerate}
\item[(5.5a)]  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?

\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 ${r = \frac{T-R} {R}}$ is the maximum.

\item[ (5.5b)]  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).}

\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.


See Kalai, Samet \& Stanford (1988), which 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 5.9 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[ (5.7a)]  What are the evolutionarily stable
strategies?

\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[(5.7b)]  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. \newline

\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}

\underline{\textit{Answer.}} See Table C.7.

\textbf{Table C.7 ``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[(5.7c)]  Repeat part (b), but with the dynamics
$N_{t+t}(s) = [1
+
\frac{ \pi_t(s)}{\sum_j \pi_t(j) }][2N_t(s)]$.

\underline{\textit{Answer.}} See Table C.8.

\textbf{Table C.8 ``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[(5.7d)]  Which three games that have appeared so far
in the book
resemble \textit{Grab the Dollar}?

\underline{\textit{Answer.}} ``Chicken'', ``The Battle of
the Sexes'',
and
``The Hawk-Dove Game''.
\end{enumerate}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 6 Dynamic Games with Asymmetric
Information}
\end{center}

\noindent \textbf{6.1: Cournot Duopoly Under Incomplete
Information
About
Costs.} This problem introduces incomplete information into
the
Cournot
model of Chapter 3 and allows for a continuum of player
types.

\begin{enumerate}
\item[(6.1a)]  Modify the Cournot Game of Chapter 3 by
specifying that
Apex' average cost of production is $c$ per unit, while
Brydox'
remains zero. What are the outputs of each firm if the costs
are
common knowledge? What are the numerical values if $c=10$?

\underline{\textit{Answer.}} The payoff functions are
\begin{equation}  \label{e0e2}
\begin{array}{l}
\pi_{Apex} = (120- q_a - q_b- c) q_a \\
\pi_{Brydox} = (120- q_a - q_b- c) q_b
\end{array}
\end{equation}
The first order conditions are then
\begin{equation}  \label{e0a1}
\begin{array}{l}
\frac{ \partial \pi_{Apex} }{\partial q_a} = 120 - 2q_a -
q_b -c=0 \\
\frac{ \partial\pi_{Brydox} }{\partial q_b} = 120- q_a -
2q_b =0
\end{array}
\end{equation}
Solving the first order conditions together gives
\begin{equation}  \label{e0b3}
\begin{array}{l}
q_a = 40-\frac{2c}{3} \\
q_b = 40 + \frac{c}{3}
\end{array}
\end{equation}

If $c=10$, Apex produces 33 1/3 and Brydox produces 43 1/3.
Apex's
higher
costs make it cut back its output, which encourages Brydox
to produce
more.

\item[(6.1b)]  Let Apex's cost $c$ be $c_{max}$ with
probability
$\theta$ and
0 with probability $(1-\theta)$, so Apex is one of two
types. Brydox
does not
know Apex' type. What are the outputs of each firm?

\underline{\textit{Answer.}} Apex's payoff function is the
same as in
part
(a), because
\begin{equation}  \label{e86d}
\pi_{Apex} = (120- q_a - q_b- c) q_a ,
\end{equation}
which yields the reaction function
\begin{equation}  \label{e86e}
q_a= 60 - \frac{q_b + c}{2}.
\end{equation}

Brydox's expected payoff is
\begin{equation}  \label{e85a}
\pi_{Brydox} = (1-\theta) (120- q_a(c=0) - q_b ) q_b +
\theta (120-
q_a(c=c_{max}) - q_b ) q_b .
\end{equation}
The first order condition is
\begin{equation}  \label{e85b}
\frac{ \partial \pi_{Brydox}}{\partial q_b}= (1-\theta)
(120- q_a(c=0)
-
2q_b ) + \theta (120- q_a(c=c_{max}) - 2q_b ) =0 .
\end{equation}
Now substitute the reaction function of Apex, equation
(\ref{e86e}),
into (%
\ref{e85b}) and condense a few terms to obtain
\begin{equation}  \label{e85c}
120- 2q_b - [1-\theta] [60 - \frac{q_b +0}{2}] - \theta [60
-
\frac{q_b
+c_{max}}{2}] =0 .
\end{equation}
Solving for $q_b$ yields
\begin{equation}  \label{e85d}
q_b = 40 + \frac{ \theta c_{max}}{3}
\end{equation}
One can then use equations (\ref{e86e}) and (\ref{e85d}) to
find
\begin{equation}  \label{e85e}
q_a= 40 - \frac{ \theta c_{max}}{6} - \frac{ c}{2}.
\end{equation}
Note that the outputs do not depend on $\theta$ or $c_{max}$
separately,
only on the expected value of Apex's cost, $\theta c_{max}
$.

\item[(6.1c)]  Let Apex' cost $c$ be drawn from the interval
$[0,
c_{max}]$
using the uniform distribution, so there is a continuum of
types.
Brydox
does not know Apex' type. What are the outputs of each firm?

\underline{\textit{Answer.}} Apex's payoff function is the
same as in
parts
(a) and (b),
\begin{equation}  \label{e86g}
\pi_{Apex} = (120- q_a - q_b- c) q_a ,
\end{equation}
which yields the reaction function
\begin{equation}  \label{e86h}
q_a= 60 - \frac{q_b + c}{2}.
\end{equation}

Brydox's expected payoff is (letting the density of possible
values of
$c$
be $f(c)$)
\begin{equation}  \label{e86i}
\pi_{Brydox} = \int_0^{c_{max}} (120- q_a(c) - q_b ) q_b
f(c)dc .
\end{equation}
The probability density is uniform, so $f(c)= \frac{1}
{c_{max}}$.
Substituting this into (\ref{e86i}), the first order
condition is
\begin{equation}  \label{e86j}
\frac{ \partial \pi_{Brydox}}{\partial q_b}=
\int_0^{c_{max}} (120-
q_a(c) -
2q_b ) \left( \frac{1}{c_{max}} \right)dc=0 .
\end{equation}
Now substitute in the reaction function of Apex, equation
(\ref{e86h}),
which gives
\begin{equation}  \label{e86k}
\int_0^{c_{max}} (120- [60 - \frac{q_b + c}{2}] - 2q_b )
\left(\frac{1}{%
c_{max}} \right)dc=0 .
\end{equation}
Simplifying by integrating out the terms in (\ref{e86k})
which depend
on $c$
only through the probability density yields
\begin{equation}  \label{e86l}
60 - \frac{3q_b}{2} + \int_0^{c_{max}} \left( \frac{ c}
{2c_{max}}
\right)
dc =0.
\end{equation}
Integrating and rearranging yields
\begin{equation}  \label{e86m}
q_b = 40 + \frac{c_{max}}{6}
\end{equation}
One can then use equations (\ref{e86h}) and (\ref{e86m}) to
find
\begin{equation}  \label{e86n}
q_a= 40 - \frac{ c_{max}}{12} - \frac{ c}{2}.
\end{equation}

\item[(6.1d)]  Outputs were 40 for each firm in the zero-
cost game in
Chapter 3. Check your answers in parts (b) and (c) by seeing
what
happens if
$c_{max}=0$.

\underline{\textit{Answer.}} If $c_{max}=0$, then in part
(b), $q_a =
40 -
\frac{ 0}{6} - \frac{ 0}{2}=40$ and $q_b = 40 + \frac{0}
{3}=40$,
which is
as it should be.

If $c_{max}=0$, then in part (c), $q_a =40 - \frac{ 0}{12} -
\frac{ 0}
{2}=40$
and $q_b = 40 + \frac{0} {6}=40$, which is as it should be.

\item[(6.1e)]  Let $c_{max}=20$ and $\theta =0.5$, so the
expectation
of
Apex' average cost is 10 in parts (a), (b), and (c) . What
are the
average
outputs for Apex in each case?

\underline{\textit{Answer.}} In part (a), under full
information, the
outputs were $q_a =$ 33 1/3 and $q_b =$ 43 1/3 . In part
(b), with two
types, $q_b=$ 43 1/3 from equation (\ref{e85d}), and the
average value
of $%
q_a$ is
\begin{equation}  \label{e86o}
E q_a= (1-\theta)( 40 - \frac{ 0.5 (20) }{6} - \frac{ 0}{2})
+ \theta
( 40 -
\frac{ 0.5 (20) }{6} - \frac{20}{2})= 33 \;\; 1/3.
\end{equation}
In part (c), with a continuum of types, $q_b =43\;\; 1/3$
and $q_a$ is
found
from
\begin{equation}  \label{e86p}
\begin{array}{ll}
Eq_a= & \int_0^{c_{max}} (40 - \frac{ c_{max}}{8} - \frac{
c} {2})
\left(
\frac{1}{c_{max}} \right) dc \\
& =40 - \frac{20}{8} - \frac{c_{max}^2}{4 c_{max}} = 33 \;\;
1/3.
\end{array}
\end{equation}

\item[(6.1f)]  Modify the model of part (b) so that
$c_{max}= 20$ and
$%
\theta=0.5$, but somehow $c= 30$. What outputs do your
formulas from
part
(b) generate? Is there anything this could sensibly model?

\underline{\textit{Answer.}} The purpose of Nature's move is
to
represent
Brydox's beliefs about Apex, not necessarily to represent
reality.
Here,
Brydox believes that Apex's costs are either 0 or 20 but he
is wrong
and
they are actually 30. In this game that does not cause
problems for
the
analysis. Using equations (\ref{e85d}) and (\ref{e85e}), the
outputs
are $%
q_b $ = 43 1/3 ($=40 + \frac{ 0.5(20)}{3}$) and $q_a$ = 26
2/3 ($= 40
-
\frac{0.5( 20)}{6} - \frac{ 30}{2} $ ).

If the game were dynamic, however, a problem would arise.
When Brydox
observes the first-period output of $q_a=$ 24 1/6, what is
he to
believe
about Apex's costs? Should he deduce that $c=30$, or
increase his
belief
that $c=20$, or believe something else entirely? This
departs from
standard
modelling.
\end{enumerate}

\bigskip \noindent \textbf{6.3: Symmetric Information and
Prior
Beliefs. }
In the \textit{Expensive-Talk Game} of Table 6.1, the
\textit{Battle
of the
Sexes} is preceded by by a communication move in which the
man chooses
$%
Silence$ or $Talk$. $Talk$ costs 1 payoff unit, and consists
of a
declaration by the man that he is going to the prize fight.
This
declaration
is just talk; it is not binding on him.

\begin{center}
\textbf{Table 6.1 Subgame Payoffs in The Expensive-Talk Game
}

\begin{tabular}{lllccc}
&  &  & \multicolumn{3}{c}{\textbf{Woman}} \\
&  &  & \textit{Fight} &  & $Ballet$ \\
&  & $Fight$ & 3,1 &  & $0,0$ \\
& \textbf{Man:} &  &  &  &  \\
&  & \textit{Ballet } & $0,0$ &  & 1,3 \\
\multicolumn{6}{l}{\textit{Payoffs to: (Man, Woman).}}
\end{tabular}
\end{center}

\begin{enumerate}
\item[ (6.3a)]  Draw the extensive form for this game,
putting the
man's
move first in the simultaneous-move subgame.

\underline{\textit{Answer.}} See Figure A.5.

\textbf{Figure A.5 The Extensive Form for the ``Expensive
Talk Game''
}

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\item[ (6.3b)]  What are the strategy sets for the game?
(start with
the
woman's)

\underline{\textit{Answer.}} The woman has two information
sets at
which to
choose moves, and the man has three. Table A.10 shows the
woman's four
strategies.

\textbf{Table A.10 The Woman's Strategies in ``The Expensive
Talk
Game''}

\begin{tabular}{|l|ll|}
\hline
Strategy & $W_{1},W_{2}$ & $W_{3},W_{4}$ \\ \hline
1 & F & F \\
2 & F & B \\
3 & B & F \\
4 & B & B \\ \hline
\end{tabular}

Table A.11 shows the man's eight strategies, of which only
the
boldfaced
four are important, since the others differ only in portions
of the
game
tree that the man knows he will never reach unless he
trembles at
$M_{1}$.

\textbf{Table A.11 The Man's Strategies in ``The Expensive
Talk
Game''}

\begin{tabular}{|l|lll|}
\hline
Strategy & $M_{1}$ & $M_{2}$ & $M_{3}$ \\ \hline
\textbf{1} & \textbf{T} & \textbf{F} & F \\
2 & T & F & B \\
\textbf{3} & \textbf{T} & \textbf{B} & B \\
4 & T & B & F \\
\textbf{5} & \textbf{S} & F & \textbf{F} \\
6 & S & B & F \\
\textbf{7} & \textbf{S} & B & \textbf{B} \\
8 & S & F & B \\ \hline
\end{tabular}

\item[( 6.3c)]  What are the three perfect pure-strategy
equilibrium
outcomes in terms of observed actions? (Remember: strategies
are not
the
same thing as outcomes.)

\underline{\textit{Answer.}} SFF, SBB, TFF.\footnote{%
The equilibrium that supports SBB is [($S, B$), $(B|S, B|T)]
$.}

\item[(6.3d) ]  Describe the equilibrium strategies for a
perfect
equilibrium in which the man chooses to talk.

\underline{\textit{Answer.}} Woman: $(F|T, B|S)$ and Man:
$(T, F|T,
B|S)$.

\item[(6.3e) ]  The idea of ``forward induction'' says that
an
equilibrium
should remain an equilibrium even if strategies dominated in
that
equilibrium are removed from the game and the procedure is
iterated.
Show
that this procedure rules out SBB as an equilibrium
outcome.\footnote{%
See Van Damme (1989). In fact, this procedure rules out TFF
\textit{(Talk,
Fight, Fight)} also.}

\underline{\textit{Answer.}} First delete the man's strategy
of $(T,
B)$,
which is dominated by $(S, B)$ whatever the woman's strategy
may be.
Without
this strategy in the game, if the woman sees the man deviate
and
choose $Talk
$, she knows that the man must choose $Fight$. Her
strategies of
$(B|T, F|S)$
and $(B|T, B|S)$ are now dominated, so let us drop those.
But then the
man's
strategy of $(S, B)$ is dominated by $(T, F|T, B|S)$. The
man will
therefore
choose to $Talk$, and the SBB equilibrium is broken.\newline
\hspace*{16pt} This is a strange result. More intuitively:
if the
equilibrium is SBB, but the man chooses $Talk$, the argument
is that
the
woman should think that the man would not do anything
purposeless, so
it
must be that he intends to choose $Fight$. She therefore
will choose
$Fight$
herself, and the man is quite happy to choose $Talk$ in
anticipation
of her
response. Taking forward induction one step further: TFF is
not an
equilibrium, because now that SBB has been ruled out, if the
man
chooses $
Silence$, the woman should conclude it is because he thinks
he can
thereby
get the $SFF$ payoff. She decides that he will choose
$Fight$, and so
she
will choose it herself. This makes it profitable for the man
to
deviate to $
SFF$ from $TFF$.
\end{enumerate}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 7: Moral Hazard: Hidden Actions}
\end{center}

\noindent \textbf{7.1: First-Best Solutions in a Principal-
Agent
Model.}
Suppose an agent has the utility function of $U= \sqrt {w} -
e $, where
$e$
can assume the levels 0 or 1. Let the reservation utility
level be $%
\overline{U} = 3$. The principal is risk neutral. Denote the
agent's
wage,
conditioned on output, as $\underline{w}$ if output is 0 and
$\overline{w}$
if output is 100. Table 7.5 shows the outputs.

\begin{center}
\textbf{Table 7.5 A Moral Hazard Game}

\begin{tabular}{|l|cc|c|}
\hline
& \multicolumn{2}{c|}{\textbf{Probability of Output of}} &
\\
\textbf{Effort } & 0 & 100 & Total \\ \hline
&  &  &  \\
$Low$ ($e=0$) & 0.3 & 0.7 & 1 \\
&  &  &  \\
$High $ ($e=1$) & 0.1 & 0.9 & 1 \\
&  &  &  \\ \hline
\end{tabular}
\end{center}

\begin{enumerate}
\item[(7.1a)]  What would the agent's effort choice and
utility be if
he
owned the firm?

\hspace*{32pt} \underline{\textit{Answer}}. The agent gets
everything
in
this case. His utility is either
\begin{equation}  \label{e86a}
U(High) = 0.1 (0) + 0.9 \sqrt{100} - 1 = 8
\end{equation}
or
\begin{equation}  \label{e86b}
U(Low) = 0.3 (0) + 0.7 \sqrt{100} -0 = 7.
\end{equation}
So the agent chooses \textrm{high effort} and a utility of
\textrm{8}.

\item[(7.1b) ]  If agents are scarce and principals compete
for them,
what
will the agent's contract be under full information? His
utility?

\underline{\textit{Answer}.} The efficient effort level is
$High$,
which
produces an expected output of 90. The principal's profit is
zero,
because
of competition. Since the agent is risk averse, he should be
fully
insured
in equilibrium: $\overline{w}=\underline{w}=90$ But he
should get this
only
if his effort is high. Thus, the contract is \textrm{w=90 if
effort is
high,
w=0 if effort is low.} \textrm{The agent's utility is 8.5 }
($=
\sqrt{90} -1,
$ rounded).

\item[(7.1c)]  If principals are scarce and agents compete
to work for
them,
what would the contract be under full information? What will
the
agent's
utility and the principal's profit be in this situation?

\underline{\textit{Answer}}. The efficient effort level is
high. Since
the
agent is risk averse, he should be fully insured in
equilibrium:
$\overline{w%
}=\underline{w}=w$. The contract must satisfy a
participation
constraint for
the agent, so $\sqrt{w} - 1 = 3$. This yields $w=16$, and a
\textrm{utility
of 3} for the agent. The actual contract specified a
\textrm{wage of
16 for
high effort and 0 for low effort}. This is incentive
compatible,
because the
agent would get only 0 in utility if he took low effort.
\textrm{The
principal's profit is 74 } (= 90-16).

\item[(7.1d)]  Suppose that $U = w-e$. If principals are the
scarce
factor
and agents compete to work for principals, what would the
contract be
when
the principal cannot observe effort? (Negative wages are
allowed.)
What will
be the agent's utility and the principal's profit be in this
situation?

\underline{\textit{Answer}}. The contract must satisfy a
participation
constraint for the agent, so \textrm{U = 3}. Since effort is
1, the
expected
wage must equal 4. One way to produce this result is to
allow the
agent to
keep all the output, plus 4 extra for his labor, but to make
him pay
the
expected output of 90 for this privilege (``selling the
store''). Let
${%
\overline{w} =14}$ and ${\underline{w} = -86 }$ (other
contracts also
work).
Then expected utility is 3 $(= 0.1(-86) + 0.9 (14)-1 = -8.6
+ 12.6 -1)
.$
\textrm{Expected profit is 86} ($=0.1(0- -86) + 0.9(100-14)
= 8.6+
77.4$).
\end{enumerate}

%-----------------------------------------------------------
----
\bigskip \noindent \textbf{7.3: Why Entrepreneurs Sell Out.}
Suppose
an
agent has a utility function of $U= \sqrt {w} -e$, where $e$
can
assume the
levels 0 or 2.4, and his reservation utility is
$\overline{U} = 7$.
The
principal is risk neutral. Denote the agent's wage,
conditioned on
output,
as $w(0)$, $w(49)$, $w(100)$, or $w(225)$. Table 7.7 shows
the output.

\begin{center}
\textbf{Table 7.7 Entrepreneurs Selling Out}

\begin{tabular}{|l|cccc|c|}
\hline
& \multicolumn{4}{c|}{\textbf{Probability of Output of}} &
\\
\textbf{Method } & 0 & 49 & 100 & 225 & Total \\ \hline
&  &  &  &  &  \\
$Safe$ ($e=0$) & 0.1 & 0.1 & 0.8 & 0 & 1 \\
&  &  &  &  &  \\
$Risky$ ($e=2.4$) & 0 & 0.5 & 0 & 0.5 & 1 \\
&  &  &  &  &  \\ \hline
\end{tabular}
\end{center}

\begin{enumerate}
\item[(7.3a)]  What would the agent's effort choice and
utility be if
he
owned the firm?

\underline{\textit{Answer}.} $U(safe) = 0+ 0.1 \sqrt{49} +
0.8
\sqrt{100}+0
- 0 = 0.7 + 8= 8.7$. $U(risky) = 0 + 0.5 \sqrt{49} + 0.5
\sqrt{225} -
2.4
=3.5 +7.5-2.4=8.6. $ Therefore he will choose the
\textrm{safe method,
e=0,
and utility is 8.7.}

\item[(7.3b)]  If agents are scarce and principals compete
for them,
what
will the agent's contract be under full information? His
utility?

\underline{\textit{Answer}.} Agents are scarce, so $\pi =0
$. Since
agents
are risk averse, it is efficient to shield them from risk.
If the
risky
method is chosen, then $w=0.5(49) + 0.5 (225)= 24.5 + 112.5=
137.$
\textrm{%
Utility is 9.3} ($\sqrt{137} -2.4 = 11.7-2.4$). If the safe
method is
chosen, then $w=0.1(49) + 0.8 (100) = 84.9.$ Utility is $U=
\sqrt{84.9}= 9.21$%
. Therefore, the optimal contract specifies a wage of
\textrm{137 if
the
risky method is used and 0 (or any wage less than 49) if the
safe}
method is
used. This is better for the agent than if he ran the firm
by himself
and
used the safe method.

\item[(7.3c)]  If principals are scarce and agents compete
to work for
principals, what will the contract be under full
information? What
will the
agent's utility and the principal's profit be in this
situation?

\underline{\textit{Answer}.} Principals are scarce, so $U =
\overline{U} = 7$%
, but the efficient effort level does not depend on who is
scarce, so
it is
still high. The agent is risk averse, so he is paid a flat
wage. The
wage
satisfies the participation constraint $\sqrt{w} - 2.4 = 7$,
if the
method
is risky. \textrm{The contract specifies a wage of 88.4
(rounded) for
the
risky method and 0 for the safe}. \textrm{Profit is 48.6 }
(= $0.5(49)
+
0.5(225) - 88.4$).

\item[(7.3d)]  If agents are the scarce factor, and
principals compete
for
them, what will the contract be when the principal cannot
observe
effort?
What will the agent's utility and the principal's profit be
in this
situation?

\underline{\textit{Answer}.} A boiling in oil contract can
be used.
\textrm{%
Set either w(0) = -1000 or w(100) = -1000}, which induces
the agent to
pick
the risky method. In order to protect the agent from risk,
the wage
should
be flat except for those outputs, so $w(49)=w(225) = 137$.
${\pi=0}$,
since
agents are scarce. \textrm{U= 9.3}, from part (b).
\end{enumerate}

%-----------------------------------------------------------
----
\noindent \textbf{7.5: Worker Effort.} A worker can be
$Careful$ or
$Careless
$, efforts which generate mistakes with probabilities 0.25
and 0.75.
His
utility function is $U= 100 - 10/w - x$, where $w$ is his
wage and $x$
takes
the value 2 if he is careful, and 0 otherwise. Whether a
mistake is
made is
contractible, but effort is not. Risk-neutral employers
compete for
the
worker, and his output is worth 0 if a mistake is made and
20
otherwise. No
computation is needed for any part of this problem.

\begin{enumerate}
\item[(7.5a)]  Will the worker be paid anything if he makes
a mistake?

\underline{\textit{Answer}}. Yes. He is risk averse, unlike
the
principal,
so his wage should be even across states.

\item[(7.5b)]  Will the worker be paid more if he does not
make a
mistake?

\underline{\textit{Answer}.} \textrm{Yes.} Careful effort is
efficient, and
lack of mistakes is a good statistic for careful effort,
which makes
it
useful for incentive compatibility.

\item[ (7.5c)]  How would the contract be affected if
employers were
also
risk averse?

\underline{\textit{Answer}}. The wage would vary more across
states,
because
the workers should be less insured--- and perhaps should
even be
insuring
the employer.

\item[(7.5d)]  What would the contract look like if a third
category,
``slight mistake,'' with an output of 19, occurs with
probability 0.1
after $%
Careless$ effort and with probability zero after $Careful$
effort?

\underline{\textit{Answer}}. The contract would pay equal
amounts
whether or
not a mistake was made, but zero if a slight mistake was
made, a
``boiling
in oil'' contract.
\end{enumerate}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 8: Further Topics in Moral Hazard}
\end{center}

\noindent \textbf{8.1: Monitoring with Error}. An agent has
a utility
function $U= \sqrt ({w}) - \alpha e$, where $\alpha = 1$ and
$e$ is
either 0
or 5. His reservation utility level is $\overline{U} = 9$,
and his
output is
100 with low effort and 250 with high effort. Principals are
risk
neutral
and scarce, and agents compete to work for them. The
principal cannot
condition the wage on effort or output, but he can, if he
wishes,
spend five
minutes of his time, worth 10 dollars, to drop in and watch
the agent.
If he
does that, he observes the agent $Daydreaming$ or $Working$,
with
probabilities that differ depending on the agent's effort.
He can
condition
the wage on those two things, so the contract will be
$\{\underline{w},%
\overline{w}\} $. The probabilities are given by Table 8.1.

\begin{center}
\textbf{Table 8.1 Monitoring with Error}

\begin{tabular}{l|cc}
& \multicolumn{2}{c|}{\textbf{Probability of }} \\
\textbf{Effort} & \textbf{$Daydreaming$} &
\textbf{$Working$} \\
\hline
$Low $($e=0$) & 0.6 & 0.4 \\ \hline
$High $($e=5$) & 0.1 & 0.9
\end{tabular}
\end{center}

\begin{enumerate}
\item[(8.1a)]  What are profits in the absence of
monitoring, if the
agent
is paid enough to make him willing to work for the
principal?

\underline{\textit{Answer}.} Without monitoring, effort is
low. The
participation constraint is $9 \geq \sqrt{w} - 0$, so $w=
81$. Output
is 100,
so \textrm{profit is 19}.

\item[(8.1b)]  Show that high effort is efficient under full
information.

\underline{\textit{Answer}}. High effort yields output of
250.
$\overline{U}
\geq \sqrt{w} - \alpha e$ or $9= \sqrt{w} - 5$ is the
participation
constraint, so $14 = \sqrt{w}$ and $w=196$. Profit is then
54. This is
superior to the profit of 19 from low effort (and the agent
is no
worse
off), so high effort is more efficient.

\item[(8.1c)]  If $\alpha = 1.2$, is high effort still
efficient under
full
information?

\underline{\textit{Answer}}. If $\alpha = 1.2$, then the
wage must
rise to
225, for profits of 25, so \textrm{high effort is still
efficient}.
The wage
must rise to 225 because the participation constraint
becomes $9 \geq
\sqrt{w%
} - 1.2(5)$.

\item[(8.1d)]  Under asymmetric information, with $\alpha =
1$, what
are the
participation and incentive compatibility constraints?

\underline{\textit{Answer}}. \textrm{The incentive
compatibility
constraint
is}
\[
{0.6 \sqrt{\underline{w}} + 0.4 \sqrt {\overline{w}} \leq
0.1 \sqrt{%
\underline{w}} + 0.9 \sqrt{\overline{w}} - 5}.
\]

\textrm{The participation constraint is} ${9 \leq
0.1\sqrt{\underline{w}} +
0.9 \sqrt{\overline{w}}-5}$.

\item[(8.1e)]  Under asymmetric information, with $\alpha =
1$, what is
the
optimal contract?

\underline{\textit{Answer}.} From the participation
constraint, $14 =
0.1
\sqrt{\underline{w}} + 0.9 \sqrt{\overline{w}} $, and
$\sqrt{\overline{w}} =
\frac{14}{0.9} - (\frac{1}{9}) \sqrt{\underline{w}}$. The
incentive
compatibility constraint tells us that $0.5
\sqrt{\overline{w}} = 5 +
0.5
\sqrt{\underline{w}}$, so $\sqrt{\overline{w}} = 10 +
\sqrt{\underline{w}}$.
Thus,
\begin{equation}  \label{e89}
10 + \sqrt{\underline{w}} = 15.6 - 0.11 \sqrt{\underline{w}}
\end{equation}
and $\sqrt{\underline{w}} = 5.6/1.11 = 5.05.$ Thus, ${\
\underline{w}
= 25.5}
$. It follows that $\sqrt{\overline{w}} = 10 + 5.05$, so ${\
\overline{w} =
226.5}$ .
\end{enumerate}

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

\noindent \textbf{8.3: Bankruptcy Constraints.} A risk-
neutral
principal
hires an agent with utility function $U= w-e$ and
reservation utility
$%
\overline{U} = 7$. Effort is either 0 or 20. There is a
bankruptcy
constraint: $w \geq 0$. Output is given by Table 8.4.

\begin{center}
\textbf{Table 8.4 Bankruptcy}

\begin{tabular}{l|cc|c}
& \multicolumn{2}{c}{\textbf{Probability of Output of}} &
\\
\textbf{Effort} & 0 & 400 & Total \\ \hline
$Low$ ($e=0$) & 0.5 & 0.5 & 1 \\
$High$ ($e=10$) & 0.2 & 0.8 & 1
\end{tabular}
\end{center}

\begin{enumerate}
\item[(8.3a)]  What would the agent's effort choice and
utility be if
he
owned the firm?

\item[(8.3b)]  If agents are scarce and principals compete
for them,
what
will the agent's contract be under full information? His
utility?

\item[ (8.3c)]  If principals are scarce and agents compete
to work
for
them, what will the contract be under full information? What
will the
agent's utility be?

\item[(8.3d)]  If principals are scarce and agents compete
to work for
them,
what will the contract be when the principal cannot observe
effort?
What
will the payoffs be for each player?

\item[(8.3e)]  Suppose there is no bankruptcy constraint. If
principals are
the scarce factor and agents compete to work for them, what
will the
contract be when the principal cannot observe effort? What
will the
payoffs
be for principal and agent?
\end{enumerate}

\bigskip \bigskip

\noindent \textbf{8.5: Efficiency Wages and Risk Aversion}
.\footnote{%
See Rasmusen (1992c).} In each of two periods of work, a
worker
decides
whether to steal amount $v$, and is detected with
probability $\alpha$
and
suffers legal penalty $p$ if he, in fact, did steal. A
worker who is
caught
stealing can also be fired, after which he earns the
reservation wage
$w_0$.
If the worker does not steal, his utility in the period is
$U(w)$; if
he
steals, it is $U(w+v) - \alpha p$, where $U(w_0 +v) - \alpha
p >
U(w_0)$.
The worker's marginal utility of income is diminishing:
$U^{\prime}>
0$, $%
U^{\prime\prime}<0$, and $lim_{x \rightarrow \infty}
U^{\prime}(x) =
0.$
There is no discounting. The firm definitely wants to deter
stealing
in each
period, if at all possible.

\begin{enumerate}
\item[ (8.5a)]  Show that the firm can indeed deter theft,
even in the
second period, and, in fact, do so with a second-period wage
$w_2^*$
that is
higher than the reservation wage $w_0$.

\underline{\textit{Answer}.} It is easiest to deter theft in
the first
period, since a high second-period wage increases the
penalty of being
fired. If $w_2$ is increased enough, however, the marginal
utility of
income
becomes so low that $U(w_2+v)$ and $U(w_2)$ become almost
identical,
and the
difference is less than $\alpha P$, so theft is deterred
even in the
second
period.

\item[(8.5b)]  Show that the equilibrium second-period wage
$w_2^*$ is
higher than the first-period wage $w_1^*$.

\underline{\textit{Answer}.} We already determined that $w_2
> w_0$.
Hence,
the worker looks hopefully towards being employed in period
2, and in
Period
1 he is reluctant to risk his job by stealing. This means
that he can
be
paid less in Period 1, even though he may still have to be
paid more
than
the reservation wage.
\end{enumerate}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 9 Adverse Selection}
\end{center}

\bigskip \noindent \textbf{9.1: Insurance with Equations and
Diagrams.}

The text analyzes Insurance Game III using diagrams. Here,
let us use
equations too. Let $U(t)= log(t)$.

\begin{enumerate}
\item[(9.1a)]  Give the numeric values $(x,y)$ for the full-
information
separating contracts $C_3$ and $C_4$ from Figure 9.6. What
are the
coordinates for $C_3$ and $C_4$?

\hspace*{24pt}\underline{\textit{Answer}.} $C_3: 0.25x +
0.75(y-x)=0$,
and $%
12 -x=y-x$. Put together, these give $y=4x/3$ and $y=12$, so
${\ x^*=9
}$
and ${\ y^*=12 }$.

${\ C_3 = (3,3) }$ because 12-9 = 3.

$C_4$ is such that $0.5x + 0.5(y-x)=0$, and $12 -x=y-x$. Put
together,
these
give $y = 2x$ and $y=12$, so ${\ x^*=6 }$ and ${\ y^*=12 }$.

${\ C_4 = (6, 6) }$ because 12-6 = 6.

\item[(9.1b)]  Why is it not necessary to use the $U(t)=log
(t)$
function to
find the values?

\hspace*{24pt}\underline{\textit{Answer}.} We know there is
full
insurance
at the first-best with any risk-averse utility function, so
the
precise
function does not matter.

\item[(9.1c)]  At the separating contract under incomplete
information, $C_5$%
, $x =2.01$. What is $y$? Justify the value 2.01 for $x$.
What are the
coordinates of $C_5$?

\hspace*{36pt} \underline{\textit{Answer}.} At $C_5$, the
incentive
compatibility constraints require that $0.5x + 0.5(x-y) =
0$, so $y =
2x$;
and $\pi_u(C_5) = \pi_u(C_3)$, so $0.25 log(12-x) + 0.75
log(y-x) =
0.25
log(3) + 0.75 log(3)$. Solving these equations yields ${x^*=
2.01}$ and
${y =
4.02}$.

${\ C_5 = (9.99, 2.01) }$ because 9.99=12-2.01 and $2.01=
4.02 - 2.01$.

\item[(9.1d)]  What is a contract $C_6$ that might be
profitable and
that
would lure both types away from $C_3$ and $C_5$?

\underline{\textit{Answer}.} One possibility is ${\ C_6 =
(8, 3) }$,
or ${\
x=4, y=7) }$. The utility of this to the Highs is 1.59 (=
$0.5 log(8) +
0.5
log(3)$), compared to 1.57 (=$0.5 log(10.99) + 0.5
log(2.01))$, so the
High's prefer it to $C_5$, and that means the Lows will
certainly
prefer it.
If there are not many Lows, the contract can make a profit,
because if
it is
only Highs, the profit is 0.5 (=$0.5(4) + 0.5 (4-7))$.
\end{enumerate}

\bigskip

%-----------------------------------------------------------
----
\noindent \textbf{9.3: Finding the Mixed-Strategy
Equilibrium in a
Testing
Game.} \footnote{%
This is the slightly modified July 23, 2001 version.} Half
of high
school
graduates are talented, producing output $a=x$, and half are
untalented,
producing output $a=0$. Both types have a reservation wage
of 1 and
are risk
neutral. At a cost of 2 to himself and 1 to the job
applicant, an
employer
can test a graduate and discover his true ability. Employers
compete
with
each other in offering wages but they cooperate in revealing
test
results,
so an employer knows if an applicant has already been tested
and
failed.
There is just one period of work. The employer cannot commit
to
testing
every applicant or any fixed percentage of them.

\begin{enumerate}
\item[(9.3a)]  Why is there no equilibrium in which either
untalented
workers do not apply or the employer tests every applicant?

\underline{\textit{Answer}.} If no untalented workers apply,
the
employer
would deviate and save 2 by skipping the test and just
hiring
everybody who
applies. Then the untalented workers would start to apply.
If the
employer
tests every applicant, however, and pays only $w_H$, then no
untalented
worker will apply. Again, the employer would deviate and
skip the
test.

\item[(9.3b)]  In equilibrium, the employer tests workers
with
probability $%
\gamma$ and pays those who pass the test $w$, the talented
workers all
present themselves for testing, and the untalented workers
present
themselves with probability $\alpha$, where possibly
$\gamma=1$ or
$\alpha=1$%
. Find an expression for the equilibrium value of $\alpha$
in terms of
$w$.
Explain why $\alpha$ is not directly a function of $x$ in
this
expression,
even though the employer's main concern is that some workers
have a
productivity advantage of $x$.

\underline{\textit{Answer}.} Using the payoff-equating
method of
calculating
a mixed strategy, and remembering that one must equate
player 1's
payoffs to
find player 2's mixing probability, we must focus on the
employer's
profits.
In the mixed-strategy equilibrium, the employer's profits
are the same
whether it tests a particular worker or not. Fraction $0.5 +
0.5
\alpha$ of
the workers will take the test, and the employer's cost for
each one
that
applies is 2, whether he is hired or not, so
\begin{equation}  \label{e95}
\begin{array}{l}
\pi(test) = \left( \frac{0.5}{0.5 + 0.5 \alpha} \right) (x -
w) - 2 \\
\\
= \pi(no\;test) = \left( \frac{0.5}{0.5 + 0.5 \alpha}\right)
(x -w)
+\left(%
\frac{0.5 \alpha } {0.5 + 0.5 \alpha}\right) (0-w),
\end{array}
\end{equation}
which yields
\begin{equation}  \label{e95a}
\alpha = \frac{2}{w-2}.
\end{equation}
The naive answer to why expression (\ref{e95a}) does not
depend on $x$
is
that $\alpha$ is the worker's strategy, so there is no
reason why it
should
depend on a parameter that enters only into the employer's
payoffs.
That is
wrong, because usually in mixed strategy equilibria that is
precisely
the
case, because the worker is choosing his probability in a
way that
makes the
employer indifferent between his payoffs. Rather, what is
going on
here is
that a talented worker's productivity is irrelevant to the
decision of
whether to test or not. The employer already knows he will
hire all
the
talented workers, and the question for him in deciding
whether to test
is
how costly it is to test and how costly it is to hire
untalented
workers.

\item[(9.3c)]  If $x=9$, what are the equilibrium values of
$\alpha$,
$%
\gamma $, and $w$?

\underline{\textit{Answer}.} We already have an expression
for
$\alpha$ from
part (b). The next step is to find the the wage. Profits are
zero in
equilibrium, which requires that
\begin{equation}  \label{e96}
\pi(no \;test) = \left( \frac{0.5}{0.5 + 0.5 \alpha} \right)
(x -w)
+\left(%
\frac{0.5 \alpha }{0.5 + 0.5 \alpha}\right)(0-w) = 0.
\end{equation}
This implies that
\begin{equation}  \label{e96a}
\alpha = \frac{x -w}{w}.
\end{equation}
Solving (\ref{e95a}) and (\ref{e96}) together yields
$\frac{2}{w- 2}=
\frac{x
-w}{w}$, so
\begin{equation}  \label{e96b}
2w = (w-2)(x-w)
\end{equation}
Substituting $x=9 $ and solving equation (\ref{e96b}) for
$w$ yields

${w^*=6}$ and ${\alpha^*= \frac{9-6}{6}= .5 }$.

There is also a root $w=3$ to equation (\ref{e96b}), but it
would
violate an
implicit assumption: that $\alpha \leq 1$, since it would
make $\alpha
=
\frac{9-3}{3} =2$.

We still need to find $\gamma^*$. In the mixed-strategy
equilibrium,
the
untalented worker's profits are the same whether he applies
or not, so
\begin{equation}  \label{e97}
\pi(apply ) = \gamma (-1 + 1) + (1-\gamma) (-1 + w) = \pi
(not \;
apply)=1.
\end{equation}
Substituting $w=6$ and solving for $\gamma$ yields $(1-
\gamma)(-1+6) =
1$, so
$(1- \gamma) =.2$ and

${\gamma^*= .8.}$

\item[(9.3d)]  If $x=8$, what are the equilibrium values of
$\alpha$,
$%
\gamma $, and $w$?

\underline{Answer}. Substituting $x=8 $ and solving equation
(\ref{e96b})
for $w$ yields

${w^*=4}$ and ${\alpha^*= \frac{8-4}{4}= 1 }$.

Thus, now all the untalented workers apply in equilibrium.

Now let us find $\gamma^*$. We need to make all the
untalented workers
want
to apply, so we need
\begin{equation}  \label{e97a}
\pi(apply ) = \gamma (-1 + 1) + (1-\gamma) (-1 + w) \geq \pi
(not \;
apply)=1.
\end{equation}
Making equation (\ref{e97a}) an equality, substituting $w=6$
and
solving for
$\gamma$ yields $(1-\gamma)(-1+4) =1$, so $(1- \gamma) =1/3$
and

${\gamma^* \leq 2/3 }$.

There is not a single equilibrium when $x=8$, because the
employer is
indifferent over all values of $\gamma$ (that is how we
calculated
$\alpha$
and $w$), and values over the entire range $\gamma \in [0,
2/3]$ will
induce
all the untalented workers to apply.
\end{enumerate}

\bigskip \noindent \textbf{9.5: Insurance and State-Space
Diagrams.}
Two
types of risk-averse people, clean-living and dissolute,
would like to
buy
health insurance. Clean-living people become sick with
probability
0.3, and
dissolute people with probability 0.9. In state-space
diagrams with
the
person's wealth if he is healthy on the vertical axis and if
he is
sick on
the horizontal, every person's initial endowment is (5,10),
because
his
initial wealth is 10 and the cost of medical treatment is 5.

\begin{enumerate}
\item[(9.5a)]  What is the expected wealth of each type of
person?

\textit{Answer}. ${E (W_c)= 8.5} (=0.7 (10)+ 0.3(5)).$ ${\ E
(W_d)=
5.5} (=
0.1 (10)+ 0.9(5))$.

\item[(9.5b)]  Draw a state-space diagram with the
indifference curves
for a
risk-neutral insurance company that insures each type of
person
separately.
Draw in the post-insurance allocations $C_1$ for the
dissolute and
$C_2$ for
the clean-living under the assumption that a person's type
is
contractible.

\underline{\textit{Answer}.} See Figure A.6.

\textbf{Figure A.6 A State-Space Diagram Showing
Indifference Curves
for the
Insurance Company }

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\item[(9.5c)]  Draw a new state-space diagram with the
initial
endowment and
the indifference curves for the two types of people that go
through
that
point.

\underline{\textit{Answer}.} See Figure A.7.

\textbf{Figure A.7 A State-Space Diagram Showing
Indifference Curves
for the
Customers }

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\item[(9.5d)]  Explain why, under asymmetric information, no
pooling
contract $C_{3}$ can be part of a Nash equilibrium.

\underline{\textit{Answer}.} Call the pooling contract
$C_{3}$.
Because
indifference curves for the the clean-living are flatter
than for the
dissipated, a contract $C_{4}$ can be found which yields
positive
profits
because it attracts the clean-living but not the dissipated.
See
Figure A.8.

\textbf{Figure A.8 Why A Pooling Contract Cannot be Part of
an
Equilibrium }

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\item[(9.5e)]  If the insurance company is a monopoly, can a
pooling
contract be part of a Nash equilibrium?

\underline{\textit{Answer}.} \textrm{Yes.} If the insurance
company is
a
monopoly, then a pooling contract can be part of a Nash
equilibrium,
because
there is no other player who might deviate by offering a
cream-
skimming
contract.
\end{enumerate}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 10: Mechanism Design in Adverse
Selection
and in
Moral
Hazard with Hidden Information}
\end{center}

\noindent \textbf{10.1: Unravelling} (formerly Problem 8.3).
An
elderly
prospector owns a gold mine worth an amount $\theta$ drawn
from the
uniform
distribution $U[0,100]$ which nobody knows, including
himself. He will
certainly sell the mine, since he is too old to work it and
it has no
value
to him if he does not sell it. The several prospective
buyers are all
risk
neutral. The prospector can, if he desires, dig deeper into
the hill
and
collect a sample of gold ore that will reveal the value of
$\theta$.
If he
shows the ore to the buyers, however, he must show genuine
ore, since
an
unwritten Law of the West says that fraud is punished by
hanging
offenders
from joshua trees as food for buzzards.

\begin{enumerate}
\item[(10.1a)]  For how much can he sell the mine if he is
clearly too
feeble to have dug into the hill and examined the ore? What
is the
price in
this situation if, in fact, the true value is $\theta=70$?

\underline{\textit{Answer}.} The price is 50 -- the expected
value of
the
uniform distribution from 0 to 100. Even if the mine is
actually worth
$%
\theta=70 $, the price remains at 50.

\item[(10.1b)]  For how much can he sell the mine if he can
dig the
test
tunnel at zero cost? Will he show the ore? What is the price
in this
situation if, in fact, the true value is $\theta=70$?

\underline{\textit{Answer}}. The expected price is 50.
Unravelling
occurs,
so \textrm{he will show the ore}, and the buyer can discover
the true
value,
which is 50 on average. If the true value is $\theta=70$,
the buyer
receives
70.

\item[(10.1c)]  For how much can he sell the mine if, after
digging
the
tunnel at zero cost and discovering $\theta$, it costs him
an
additional 10
to verify the results for the buyers? What is his expected
payoff?

\underline{\textit{Answer}.} He shows the ore iff $\theta
\in [20,100]
$.
This is because if the minimum quality ore he shows is $b$,
then the
price
at which he can sell the mine without showing the ore is
$\frac{b}{2}
$. If $%
b=20$ and the true value is 20, then he can sell the mine
for 10, and
showing the ore to raise the price to 20 would not increase
his net
profit,
given the display cost of 10.

With probability 0.2, his price is 10, and with probability
0.8, it is
an
average price of 60 but he pays 10 to display the ore. Thus,
the
prospector's expected payoff is 42 (= $0.2(10) +0.8 (60-
10)= 2+40 =
42.)$

\item[(10.1d)]  What is the prospector's expected payoff if
with
probability
0.5 digging the tunnel is costless, but with probability 0.5
it costs
120?
(Assume, as usual, that all these parameters are common
knowledge,
although
only the prospector learns whether the cost is actually 0 or
120.)

\underline{\textit{Answer}.} In equilibrium there exists
some number
$c$
such that if the prospector has dug the tunnel and found the
value of
the
mine to be $\theta \geq c$ he will show the ore. If he does
not show
any
ore, the buyers' expected value for the mine is $0.5 \left(
\frac{
100- 0}{2}
\right) + 0.5 \left( \frac{c- 0}{2} \right) = \frac{c}{4} +
25$.
Having dug
the tunnel, he will therefore show the ore if $\theta \geq
\frac{c}
{4} + 25$%
, because then he can get a price of $\theta$ instead. Since
$c$ is
defined
as the minimal level he will disclose, it follows that $c =
\frac{c}
{4} + 25$%
, which implies that $c= 33\; \frac{1}{3}$ (and the price is
$(\frac{1}{4}%
)(33\; \frac{1} {3} )+ 25 = 33\; \frac{1}{3}$ if he does not
show the
ore).

With probability 0.5, the prospector will not dig the
tunnel, and will
receive a price of $33\; \frac{1}{3}$. With probability 0.5
he will
dig the
tunnel, and will refuse to disclose with probability
$\frac{1}{3}$,
for a
price of $33\; \frac{1}{3}$, and disclose with probability
$\frac{2}
{3}$,
for an average price of $66\; \frac{2}{3} $, for an expected
payoff of
about
44.4.
\end{enumerate}

%-----------------------------------------------------------
----
\bigskip \noindent \textbf{10.3: Agency Law.} Mr. Smith is
thinking of
buying a custom-designed machine from either Mr. Jones or
Mr. Brown.
This
machine costs 5000 dollars to build, and it is useless to
anyone but
Smith.
It is common knowledge that with 90 percent probability the
machine
will be
worth 10,000 dollars to Smith at the time of delivery, one
year from
today,
and with 10 percent probability it will only be worth 2,000
dollars.
Smith
owns assets of 1,000 dollars. At the time of contracting,
Jones and
Brown
believe there is there is a 20 percent chance that Smith is
actually
acting
as an ``undisclosed agent'' for Anderson, who has assets of
50,000
dollars.

Find the price be under the following two legal regimes: (a)
An
undisclosed
principal is not responsible for the debts of his agent; and
(b) even
an
undisclosed principal is responsible for the debts of his
agent. Also,
explain (as part [c]) which rule a moral hazard model like
this would
tend
to support.

\underline{\textit{Answer}.} (a) The zero profit condition,
arising
from
competition between Jones and Brown, is
\begin{equation}  \label{e1}
-5000 + .9P + .1 (1000) =0,
\end{equation}
because Smith will only pay for the machine with probability
0.9, and
otherwise will default and only pay up to his wealth, which
is 1. This
yields $P \approx 5,444$.

(b) If Anderson is responsible for Smith's debts, then Smith
will pay
the
5,000 dollars. Hence, zero profits require
\begin{equation}  \label{e2}
-5000 + .9P + .1 (.2) P + .1(.8)(1000) =0,
\end{equation}
which yields $P \approx 5,348$.

(c) Moral hazard tends to support rule (b). This is because
it reduces
bankruptcy and the agent will be more reluctant to order the
machine
when
there is a high chance it is unprofitable. In the model as
constructed, this
does not arise, because there is only one type of agent, but
more
generally
it would, because there would be a continuum of types of
agents, and
some
who would buy the machine under rule (b) would find it too
expensive
under
rule (a).

Even in the model as it stands, rule (a) leads to the
inefficient
outcome
that a machine worth 2,000 to Smith is not give to Smith.
Rather, he
pays
his wealth and lets the seller keep the machine, which is
inefficient
since
the machine really is worth 2000 to Smith.

Nobody in my class answered this question correctly, which
surprised
me. It
basically is a question about zero-profit prices. Guessing
would have
been a
good idea here: it is very intuitive that the price would
always be
above
\$5,000, and that it would be higher if the principal never
had to
cover the
agent's debts. You should be able to tell that $P>10,000$ is
impossible,
because Smith would never pay it. Also, the sellers compete,
so it is
their
profits that provide a participation constraint, not the
benefit to
the
buyer.

\bigskip

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 11: Signalling}
\end{center}

\noindent \textbf{11.1: Is Lower Ability Better?} Change
Education I
so that
the two possible worker abilities are $a \in \{1,4\}$.

\begin{enumerate}
\item[(11.1a)]  What are the equilibria of this game? What
are the
payoffs
of the workers (and the payoffs averaged across workers) in
each
equilibrium?

\textit{Answer}. The pooling equilibrium is
\begin{equation}  \label{e98}
{\ s_L = s_H = 0, w_0=w_1 =2.5, Pr(L|s=1) =0.5},
\end{equation}
which uses passive conjectures. The payoffs are ${U_L = U_H
= 2.5}$,
for an
average payoff of \textrm{2.5}.

The separating equilibrium is
\begin{equation}  \label{e99}
{\ s_L = 0, s_H = 1, w_0=1, w_1 =4. }
\end{equation}
The payoffs are ${U_L =1}$ and ${U_H = 2}$, for an
\textrm{average
payoff of
1.5 }. This equilibrium can be justified by the self
selection
constraints
\begin{equation}  \label{e100}
U_L(s=0) = 1 > U_L(s=1)= 4 - 8/1 =-4
\end{equation}
and
\begin{equation}  \label{e101}
U_H(s=0) = 1 < U_H(s=1)= 4 - 8/4= 2.
\end{equation}

\item[(11.1b)]  Apply the Intuitive Criterion (see N6.2).
Are the
equilibria
the same?

\textit{Answer}. \textrm{Yes.} The intuitive criterion does
not rule
out the
pooling equilibrium in the game with $a_h=4$. There is no
incentive
for
\textit{either} type to deviate from $s=0$ even if the
deviation makes
the
employers think that the deviator is high-ability. The
payoff to a
persuasive high-ability deviator is only 2, compared the 2.5
that he
can get
in the pooling equilibrium.

\item[(11.1c)]  What happens to the equilibrium worker
payoffs if the
high-ability is 5 instead of 4?

\textit{Answer}. The pooling equilibrium is
\begin{equation}  \label{e102}
{\ s_L = s_H = 0, w_0=w_1 =, Pr(L|s=1) =0.5},
\end{equation}
which uses passive conjectures. The payoffs are ${U_L = U_H
= 3}$,
with an
\textrm{average payoff of 3}.

The separating equilibrium is
\begin{equation}  \label{e103}
{\ s_L =0, s_H = 1, w_0= 1, w_1 =5.}
\end{equation}
The payoffs are ${U_L=1}$ and ${U_H=3.4}$ with an
\textrm{average
payoff of
2.2}. The self selection constraints are
\begin{equation}  \label{e104}
U_H(s=0) = 1 < U_H(s=1)= 5 - \frac{8}{5} =3.4
\end{equation}
and
\begin{equation}  \label{e105}
U_L(s=0) = 1 > U_L(s=1)= 5 - \frac{8}{1} =-3.
\end{equation}

\item[(11.1d)]  Apply the Intuitive Criterion to the new
game. Are the
equilibria the same?

\textit{Answer}. \textbf{No.} The strategy of choosing $s=1$
is
dominated
for the Lows, since its maximum payoff is $-3$, even if the
employer
is
persuaded that he is High. So only the separating
equilibrium
survives.

\item[(11.1e)]  Could it be that a rise in the maximum
ability reduces
the
average worker's payoff? Can it hurt all the workers?

\underline{\textit{Answer}.} \textbf{Yes.} Rising ability
would reduce
the
average worker payoff if the shift was from a pooling
equilibrium when
$%
a_h=4 $ to a separating equilibrium when $a_h=5$. Since the
Intuitive
Criterion rules out the pooling equilibrium when $a_h=5$, it
is
plausible
that the equilibrium is separating when $a_h=5$. Since the
pooling
equilibrium is pareto-dominant when $a_h=4$, it is plausible
that it
is the
equilibrium played out. So the average payoff may well fall
from 2.5
to 2.2
when the high ability rises from 4 to 5. \textbf{This cannot
make
every
player worse off}, however; the high-ability workers see
their payoffs
rise
from 2.5 to 3.4.
\end{enumerate}

\bigskip \noindent \textbf{11.3: Price and Quality.}
Consumers have
prior
beliefs that Apex produces low-quality goods with
probability 0.4 and
high-
quality with probability 0.6. A unit of output costs 1 to
produce in
either
case, and it is worth 10 to the consumer if it is high-
quality and 0
if
low-quality. The consumer, who is risk neutral, decides
whether to buy
in
each of two periods, but he does not know the quality until
he buys.
There
is no discounting.

\begin{enumerate}
\item[(11.3a)]  What is Apex' price and profit if it must
choose one
price, $%
p^*$, for both periods?

\underline{\textit{Answer}.} A consumer's expected consumer
surplus is
\begin{equation}  \label{e109}
CS = 0.4 (0-p^*) + 0.6 (10-p^*) + 0.6(10-p^*) = -1.6p^* +
12.
\end{equation}
Apex maximizes its profits by setting $CS=0$, in which case
${\ p^* =
7.5 }$
and profit is ${\ \pi_H = 13 }$ ( = 2(7.5 - 1)) or ${\
\pi_L= 6.5}$ (
(=7.5-1).

\item[(11.3b)]  What is Apex' price and profit if it can
choose two
prices, $%
p_1$ and $p_2$, for the two periods, but it cannot commit
ahead to
$p_2$?

\textit{Answer}. If Apex is high quality, it will choose ${\
p_2 = 10
}$,
since the consumer, having learned the quality first period,
is
willing to
pay that much. Thus consumer surplus is
\begin{equation}  \label{e110}
CS = 0.4 (0-p_1) + 0.6 (10-p_1) + 0.6(10-10) = -p_1 + 6,
\end{equation}
and, setting this equal to zero, ${\ p_1=6}$, for a profit
of ${\
\pi_H =14}$
(= (6-1) + (10-1) ) or ${\ \pi_L =5}$ (= 6-1 ).

\item[(11.3c)]  What is the answer to part (b) if the
discount rate is
$r =
0.1$?

\textit{Answer}. Apex cannot do better than the prices
suggested in
part (b).

\item[(11.3d)]  Returning to $r=0$, what if Apex can commit
to $p_2$?

\underline{\textit{Answer}.} Commitment makes no difference
in this
problem,
since Apex wants to charge a higher price in the second
period anyway
if it
has high quality--- a high price in the first period would
benefit the
low-quality Apex too, at the expense of the high-quality
Apex.

\item[(11.3e)]  How do the answers to (a) and (b) change if
the
probability
of low quality is 0.95 instead of 0.4? (There is a twist to
this
question.)

\underline{\textit{Answer}.} With a constant price, a
consumer's
expected
consumer surplus is
\begin{equation}  \label{e111}
CS = 0.95 (0-p^*) + 0.05 (10-p^*) + 0.05(10-p^*) = -1.05p^*
+ 0.5
\end{equation}
Apex would set $CS=0$, in which case $p^* =\frac{5}{21}$,
but since
this is
less than cost, Apex in fact would not sell anything at all,
and would
earn
zero profit.

With changing prices, high-quality Apex will choose ${\ p_2
= 10}$,
since
the consumer, having learned the quality first period, is
willing to
pay
that much. Thus consumer surplus is
\begin{equation}  \label{e112}
CS = 0.95 (0-p_1) + 0.05 (10-p_1) + 0.05(10-10) = -p_1 +
0.5.
\end{equation}
and, setting this equal to zero, you might think that $p_1=
0.5$, for a
profit of $\pi_H = 8.5 (= (0.5-1) + (10-1))$. But notice
that if the
low-quality Apex tries to follow this strategy, his payoff
is $\pi_L =
0.5-1
<0$. Hence, only the high-quality Apex will try it. But then
the
consumers
know the product is high-quality, and they are willing to
pay 10 even
in the
first period. What the high-quality Apex can do is charge up
to ${\
p_1 = 1}$
in the first period, for profits of 9 (=$(1-1) + (10-1))$.
\end{enumerate}

\bigskip \noindent \textbf{11.5: Advertising.} Brydox
introduces a new
shampoo which is actually very good, but is believed by
consumers to
be good
with only a probability of 0.5. A consumer would pay 10 for
high
quality and
0 for low quality, and the shampoo costs 6 per unit to
produce. The
firm may
spend as much as it likes on stupid TV commercials showing
happy
people
washing their hair, but the potential market consists of 100
cold-
blooded
economists who are not taken in by psychological tricks. The
market
can be
divided into two periods.

\begin{enumerate}
\item[(11.5a)]  If advertising is banned, will Brydox go out
of
business?

\textit{Answer}. \textrm{No.} It can sell at a price of 5 in
the first
period and 10 in the second period. This would yield profits
of 300
(=(100)(5-6) +(100) (10-6)).

\item[(11.5b)]  If there are two periods of consumer
purchase, and
consumers
discover the quality of the shampoo if they purchase in the
first
period,
show that Brydox might spend substantial amounts on stupid
commercials.

\underline{\textit{Answer}.} If the seller produces high
quality, it
can
expect repeat purchases. This makes expenditure on
advertising useful
if it
increases the number of initial purchases, even if the firm
earns
losses in
the first period. If the seller produces low quality, there
will be no
repeat purchases. Hence, \textrm{advertising expenditure can
act as a
signal}
of quality: consumers can view it as a signal that the
seller intends
to
stay in business two periods.

\item[(11.5c)]  What is the minimum and maximum that Brydox
might
spend on
advertising, if it spends a positive amount?

\underline{\textit{Answer}.} If there is a separating
signalling
equilibrium, it will be as follows. Brydox would spend
nothing on
advertising if its shampoo is low quality, and consumers
will not buy
from
any company that advertises less than some amount X, because
such a
company
is believed to produce low quality. Brydox would spend X on
advertising if
its quality is high, and charge a price of 10 in both
periods.

\textrm{Amount X is between 400 and 500. } If a low-quality
firm
spends X on
advertising, consumers do buy from it for one period, and it
earns
profits
of (100)(10-6)-X = 400-X. Thus, the high-quality firm must
spend at
least
400 to distinguish itself. If a high-quality firm spends X
on
advertising,
consumers buy from it for both periods, and it earns profits
of (2)
(100)(10-6)-X = 800-X. Since it can make profits of 300 even
without
advertising, a high-quality firm will spend up to 500 on
advertising.
\end{enumerate}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 12: Bargaining}
\end{center}

\noindent \textbf{12.1: A Fixed Cost of Bargaining and
Grudges.} Smith
and
Jones are trying to split 100 dollars. In bargaining round
1, Smith
makes an
offer at cost $0$, proposing to keep $S_1$ for himself and
Jones
either
accepts (ending the game) or rejects. In round 2, Jones
makes an offer
at
cost $10$ of $S_2$ for Smith and Smith either accepts or
rejects. In
round
3, Smith makes an offer of $S_3$ at cost $c$, and Jones
either accepts
or
rejects. If no offer is ever accepted, the 100 dollars goes
to a third
player, Dobbs.

\begin{enumerate}
\item[(12.1a)]  If $c=0$, what is the equilibrium outcome?

\underline{\textit{Answer}.} ${S_1=100}$ \textbf{and Jones
accepts
it.} If
Jones refused, he would have to pay 10 to make a proposal
that Smith
would
reject, and then Smith would propose $S_3 =100$ again.
$S_1<100$ would
not
be an equilibrium, because Smith could deviate to $S_1=100$
and Jones
would
still be willing to accept.

\item[(12.1b)]  If $c=80$, what is the equilibrium outcome?

\underline{\textit{Answer}.} If the game goes to Round 3,
Smith will
propose
$S_3=100$ and Jones will accept, but this will cost Smith
80. Hence,
if
Jones proposes $S_2 =20$, Smith will accept it, leaving 80
for
Jones---who
would, however pay 10 to make his offer. Hence, in Round 1
\textrm{Smith
must offer} ${S_1 =30}$ \textrm{to induce Jones to accept},
and that
will be
the equilibrium outcome.

\item[(12.1c)]  If $c=10$, what is the equilibrium outcome?

\underline{\textit{Answer}.} If the game goes to Round 3,
Smith will
propose
$S_3=100$ and Jones will accept, but this will cost Smith
10. Hence,
if
Jones proposes $S_2 =90$, Smith will accept it, leaving 10
for
Jones---who
would, however pay 10 to make his offer. Hence, in Round 1
\textrm{Smith
need only offer} ${S_1 =100}$ \textrm{to induce Jones to
accept, and
that
will be the equilibrium outcome.}

\item[(12.1d)]  What happens if $c=0$, but Jones is very
emotional and
would
spit in Smith's face and throw the 100 dollars to Dobbs if
Smith
proposes $%
S=100$? Assume that Smith knows Jones's personality
perfectly.

\underline{\textit{Answer}.} However emotional Jones may be,
there is
some
minimum offer $M$ that he would accept, which probably is
less than 50
(but
you never know---some people think they are entitled to
everything,
and one
could imagine a utility function such that Jones would
refuse $S=5$
and
prefer to bear the cost 10 in the second round in order to
get the
whole 100
dollars). \textrm{The equilibrium will be for Smith to
propose exactly
S-M
in Round 1, and for Jones to accept}.
\end{enumerate}

\bigskip \noindent \textbf{12.3: The Nash Bargaining
Solution.} Smith
and
Jones, shipwrecked on a desert island, are trying to split
100 pounds
of
cornmeal and 100 pints of molasses, their only supplies.
Smith's
utility
function is $U_s = C + 0.5M$ and Jones' is $U_j = 3.5C + 3.5
M$. If
they
cannot agree, they fight to the death, with $U=0$ for the
loser. Jones
wins
with probability 0.8.

\begin{enumerate}
\item[(12.3a)]  What is the threat point?

\underline{\textit{Answer}.} The threat point gives the
expected
utility for
Smith and Jones if they fight. This is \textrm{560} for
Jones (=
0.8(350 +
350) + 0), and \textrm{30} for Smith (=0.2(100+50) + 0).

\item[(12.3b)]  With a 50-50 split of the supplies, what are
the
utilities
if the two players do not recontract? Is this efficient?

\underline{\textit{Answer}.} The split would give the
utilities ${\
U_s =75}$
(= 50 + 25) and ${U_j = 350}$. If Smith then traded 10 pints
of
molasses to
Jones for 8 pounds of cornmeal, the utilities would become
$U_s = 78$
(=
58+20) and $U_j = 357$ (=3.5(60) + 3.5(42)), so both would
have
gained.
\textrm{The 50-50 split is not efficient}.

\item[(12.3c)]  Draw the threat point and the Pareto
frontier in
utility
space (put $U_s$ on the horizontal axis).

\underline{\textit{Answer}.} See Figure A.9.

\textbf{Figure A.9 The Threat Point and Pareto Frontier }

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To draw the diagram, first consider the extreme points. If
Smith gets
everything, his utility is 150 and Jones's is 0. If Jones
gets
everything,
his utility is 700 and Smith's is 0. If we start at (150,0)
and wish
to
efficiently help Jones at the expense of Smith, this is done
by giving
Jones
some molasses, since Jones puts a higher relative value on
molasses.
This
can be done until Jones has all the molasses, at utility
point (100,
350).
Beyond there, one must take cornmeal away from Smith if one
is to help
Jones
further, so the Pareto frontier acquires a flatter slope.

\item[(12.3d)]  According to the Nash bargaining solution,
what are
the
utilities? How are the goods split?

\underline{\textit{Answer}.} To find the Nash bargaining
solution,
maximize $%
(U_s - 30)(U_j - 560)$. Note from the diagram that it seems
the
solution
will be on the upper part of the Pareto frontier, above
(100,350),
where
Jones is consuming all the molasses, and where if Smith
loses one
utility
unit, Jones gets 3.5. If we let $X$ denote the amount of
cornmeal that
Jones
gets, we can rewrite the problem as
\begin{equation}  \label{e121}
\stackrel{ Maximize}{X} (100 - X - 30)(350 +3.5X - 560)
\end{equation}
This maximand equals $(70-X)( 3.5X-210)= -14,700+455X -
3.5X^2$. The
first
order condition is $455-7X =0$, so $X^*=65$. Thus,
\textrm{Smith gets
35
pounds of cornmeal, Jones gets 65 pounds of cornmeal and 100
of
molasses},
and ${U_s = 35}$ and ${U_j = 577.5} $.

\item[ (12.3e)]  Suppose Smith discovers a cookbook full of
recipes
for a
variety of molasses candies and corn muffins, and his
utility function
becomes $U_s= 10C + 5M$. Show that the split of goods in
part (d)
remains
the same despite his improved utility function.

\underline{\textit{Answer}.} The utility point at which
Jones has all
the
molasses and Smith has the molasses is now (1000, 350),
since Smith's
utility is (10) (100). Smith's new threat point utility is
$300 (=
0.2(
(10)(100) + (5)(100)) $. Thus, the Nash problem of equation
(\ref{e121})
becomes
\begin{equation}  \label{e121a}
\stackrel{ Maximize}{X} (1000 - 10X - 300)(350 +3.5X - 560).
\end{equation}
But this maximand is the same as $(10)(100 - X - 30 )(350
+3.5X - 560)
$, so
it must have the same solution as was found in part (d).
\end{enumerate}

\bigskip

\noindent \textbf{12.5: A Fixed Cost of Bargaining and
Incomplete
Information. } Smith and Jones are trying to split 100
dollars. In
bargaining round 1, Smith makes an offer at cost $c$,
proposing to
keep $S_1$
for himself. Jones either accepts (ending the game) or
rejects. In
round 2,
Jones makes an offer of $S_2$ for Smith, at cost $10$, and
Smith
either
accepts or rejects. In round 3, Smith makes an offer of
$S_3$ at cost
$c$,
and Jones either accepts or rejects. If no offer is ever
accepted, the
100
dollars goes to a third player, Parker.

\begin{enumerate}
\item[(12.5a)]  If $c=0$, what is the equilibrium outcome?

\underline{\textit{Answer}.} \underline{\textit{Answer}.}
${S_1= 100}
$%
\textrm{and Jones accepts it.} If Jones refused, he would
have to pay
10 to
make a proposal that Smith would reject, and then Smith
would propose
$S_3
=100$ again. $S_1<100$ would not be an equilibrium, because
Smith
could
deviate to $S_1=100$ and Jones would still be willing to
accept .

\item[(12.5b)]  If $c=80$, what is the equilibrium outcome?

\underline{\textit{Answer}.} If the game goes to Round 3,
Smith will
propose
$S_3=100$ and Jones will accept, but this will cost Smith
80. Hence,
if
Jones proposes $S_2 =20$, Smith will accept it, leaving 80
for
Jones---who
would, however, pay 10 to make his offer. Hence, \textrm{in
Round 1
Smith
must offer ${S_1 =30}$ to induce Jones to accept, which will
be the
equilibrium outcome.}

\item[(12.5c)]  If Jones' priors are that $c=0$ and $c=80$
are equally
likely, but only Smith knows the true value, what is the
equilibrium
outcome? (Hint: the equilibrium uses mixed strategies.)

\underline{\textit{Answer}.} Smith's equilibrium strategy is
to offer
$%
S_1=100$ with probability 1 if $c=0$ and probability
$\frac{1}{7}$ if
$c=80$%
; to offer $S_1 = 30$ with probability 6/7 if $c=80$. He
accepts $S_2
\geq
20 $ if $c=80$ and $S_2=100$ if $c= 0$, and proposes $S_3=
100$
regardless of
$c$. Jones accepts $S_1 = 100$ with probability $\frac{1}{8}
$, rejects
$S_1
\in (30, 100) $, and accepts $S_1 \leq 30$. He proposes $S_2
= 20$ and
accepts $S_3 = 100$. Out of equilibrium, a supporting belief
for Jones
to
believe that if $S_1$ equals neither 30 nor 100, then
$Prob(c= 80)=1$.

If $c=0$, the equilibrium outcome is for Smith to propose
$S_1= 100$,
for
Jones to accept with probability $\frac{1}{8}$ and to
propose $S_2 =
20$
otherwise and be rejected, and for Smith to then propose
$S_3 =100$
and be
accepted. If $c=80$, the equilibrium outcome is with
probability 6/7
for
Smith to propose $S_1=30$ and be accepted, with probability
($\frac{1}
{7} ) (%
\frac{1}{8})$ to propose $S_1=100$ and be accepted, and with
probability ($%
\frac{1}{7}$) ($\frac{7}{8}$) to propose $S_1=100$, be
rejected, and
then to
be proposed $S_2=20$ and to accept.

The rationale behind the equilibrium strategies is as
follows. In
Round 3,
either type of Smith does best by proposing a share of 100,
and Jones
might
as well accept. In Round 2, anything but $S_2=100$ would be
rejected
by
Smith if $c=0$, so Jones should give up on that and offer
$S_2=20$,
which
would be accepted if $c=80$ because if that type of Smith
were to
wait, he
would have to pay 80 to propose $S_3=100$. In Round 1, if
$c=0$, Smith
should propose $S_1= 100$, since he can wait until Round 3
and get
that
anyway at zero extra cost. There is no pure strategy
equilibrium,
because if
$c=80$, Smith would pretend that $c=0$ and propose $S_1=100$
if Jones
would
accept that. But if Jones accepts only with probability
$\theta$, then
Smith
runs the risk of only getting 20 in the second period, less
than $S_1=
30$,
which would be accepted by Jones with probability 1.
Similarly, if
Smith
proposes $S_1=100$ with probability $\gamma$ when $c=80$,
Jones can
either
accept it, or wait, in which case Jones might either pay a
cost of 10
and
end up with $S_3=100$ anyway, or get Smith to accept $S_2=
20$.

The probability $\gamma $ must equate Jones's two pure-
strategy
payoffs.
Using Bayes's Rule for the probabilities in (\ref{e124}),
the payoffs
are
\begin{equation}  \label{e123}
\pi_j(accept\;S_1=100) = 0
\end{equation}
and
\begin{equation}  \label{e124}
\pi_j(reject \;S_1=100) = -10 + \left( \frac{0.5 \gamma}{0.5
\gamma +
0.5}
\right)\left( 80 \right) + \left( \frac{0.5 }{0.5 \gamma +
0.5}
\right)
\left( 0 \right),
\end{equation}
which yields $\gamma = \frac{1}{7}$.

The probability $\theta $ must equate Smith's two pure-
strategy
payoffs:
\begin{equation}  \label{e125}
\pi_s( S_1=30) = 30
\end{equation}
and
\begin{equation}  \label{e126}
\pi_s( S_1=100) = \theta 100 + (1-\theta)20,
\end{equation}
which yields $\theta = \frac{1}{8}$.
\end{enumerate}

\bigskip

\noindent \textbf{12.7: Myerson-Satterthwaite. } The owner
of a tract
of
land values his land at $v_s$ and a potential buyer values
it at
$v_b$. The
buyer and seller do not know each other's valuations, but
guess that
they
are uniformly distributed between 0 and 1. The seller and
buyer
suggest $p_s$
and $p_b$ simultaneously, and they have agreed that the land
will be
sold to
the buyer at price $p =\frac{(p_b + p_s)}{2}$ if $p_s \leq
p_b$.

The actual valuations are $v_s=.2$ and $v_b= .8$. What is
one
equilibrium
outcome given these valuations and this bargaining
procedure? Explain
why
this can happen.

\underline{\textit{Answer}.} This game is Bilateral Trading
III. It
has
multiple equilibria, even for this one pricing mechanism.

The One Price Equilibrium described in Chapter 12 is one
possibility.
The
Buyer offers $p_b=x$ and the Seller offers $p_s= x$, with $x
\in [.2,
.8]$,
so that $p=x$. If either player tries to improve the price
from his
point of
view, he will lose all gains from trade. And he of course
will not
want to
give the other player a better price when that does not
increase the
probability of trade.

A degenerate equilibrium is for the Buyer to offer $p_b=0$
and the
Seller to
offer $p_s= 1$, in which case trade will not occur. Neither
player can
gain
by unilaterally altering his strategy, which is why this is
a Nash
equilibrium. You will be able to think of other degenerate
no-trade
equilibria too.

The Linear Equilibrium described in Chapter 12 uses the
following
strategies:
\[
p_b = \frac{2}{3} v_b + \frac{1}{12}
\]
and
\[
p_s = \frac{2}{3} v_s + \frac{1}{4}.
\]

Substituting in our $v_b$ and $v_s$ yields a buyer price of
$p_b=
(2/3) (.8) + 1/12 = 192/360 + 30/360 = 222/360$ and a seller
price of
$p_s= (2/3) (.2) + 1/4 = 16/120 + 30/120 = 23/60= 138/360$.
Trade will
occur, and at a price halfway between these values, which is
$p= (1/2)
(222+138)/360 = 1/2$.

This will be an equilibrium because although we have
specified
$v_s$and $v_b$ , the players do not both know those values
till after
the mechanism is played out.

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 13: Auctions}
\end{center}

\bigskip \noindent \textbf{13.1: Rent-Seeking}. Two risk-
neutral
neighbors
in 16th century England, Smith and Jones, have gone to court
and are
considering bribing a judge. Each of them makes a gift, and
the one
whose
gift is the largest is awarded property worth \pounds 2,000.
If both
bribe
the same amount, the chances are 50 percent for each of them
to win
the
lawsuit. Gifts must be either \pounds 0, \pounds 900, or
\pounds 2,
000.

\begin{enumerate}
\item[13.1a)]  ( What is the unique pure-strategy
equilibrium for this
game?

\underline{\textit{Answer}.} \text{Each bids \pounds 900,}
for
expected
profits of 100 each (=-900 + 0.5(2000)). Table A.12 shows
the payoffs
(but
also includes the payoffs for when the strategy of a bid of
1,500 is
allowed). A player who deviates to 0 has a payoff of 0; a
player who
deviates to 2,000 has a payoff of 0. (0,0) is not an
equilibrium,
because the
expected payoff is 1,000, but a player who deviated to 900
would have
a
payoff of 1,100.

\textbf{Table A.12 ``Bribes I$^{\prime\prime}$}

\begin{tabular}{lll|llll}
&  &  & \multicolumn{4}{c}{\textbf{Jones}} \\
&  &  & \pounds 0 & \pounds 900 & \pounds 1500 & \pounds
2000 \\
&  &  & \multicolumn{4}{l}{} \\ \hline
&  & \pounds 0 & 1000,1000 & 0,1100 & 0,500 & 0,0 \\
& \textbf{Smith:} & \pounds 900 & 1100,0 & 100,100 & -900,
500 & -
900,0 \\
&  & \pounds 1500 & 500,0 & $500,-900$ & $-500,-500$ & $-
1500, 0$ \\
&  & \pounds 2000 & 0,0 & $0,-900$ & $0,-1500$ & $-1000, -
1000$ \\
\multicolumn{7}{l}{} \\
\multicolumn{7}{l}{\textit{Payoffs to: (Smith, Jones).}}
\end{tabular}

\item[(13.1b)]  Suppose that it is also possible to give a
\pounds
1500
gift. Why does there no longer exist a pure-strategy
equilibrium?

\underline{\textit{Answer}.} If Smith bids 0 or 900, Jones
would bid
1500.
If Smith bids 1500, Jones would bid 2000. If both bid 2000,
then one
can
profit by deviating to 0. But if Smith bids 2000 and Jones
bids 0,
Smith
will deviate to 900. This exhausts all the possibilities.

\item[(13.1c)]  What is the symmetric mixed-strategy
equilibrium for
the
expanded game? What is the judge's expected payoff?

\underline{\textit{Answer}.} Let $(\theta_0$, $\theta_{900}
$,
$\theta_{1500}$
,$\theta_{2000})$ be the probabilities. It is pointless ever
to bid
2000,
because it can only yield zero or negative profits, so
$\theta_{2000}=
0$. In
a symmetric mixed-strategy equilibrium, the return to the
pure
strategies
is equal, and the probabilities add up to one, so
\begin{equation}  \label{e127}
\begin{array}{l}
\pi_{Smith} (0) = \pi_{Smith} (900)= \pi_{Smith} (1500) \\
\\
0.5\theta_0(2000) = -900+ \theta_0(2000)+0.5 \theta_{900}
(2000) = -
1500 +
\theta_0(2000)+ \theta_{900} (2000) + 0.5\theta_{1500}
(2000),
\end{array}
\end{equation}
and
\begin{equation}  \label{e128}
\theta_0 + \theta_{900} + \theta_{1500} = 1.
\end{equation}
Solving out these three equations for three unknowns,
\textrm{the
equilibrium is} ${(0.4, 0.5, 0.1,0.0)}$.

\textrm{The judge's expected payoff is 1200 } (=2(0.5(900) +
0.1(1500)))

\underline{Note:} The results are sensitive to the bids
allowed. Can
you
speculate as to what might happen if the strategy space were
the whole
continuum from 0 to 2000?

\item[(13.1d)]  In the expanded game, if the losing litigant
gets back
his
gift, what are the two equilibria? Would the judge prefer
this rule?

\underline{\textit{Answer}.} Table A.13 shows the new
outcome matrix.
\textrm{There are three equilibria: ${\ x_1 = (900,900), x_2
= 1500,
1500)}$,
and } ${x_3 = (2000,2000)}$.

\textbf{Table A.13 ``Bribes II''}

\begin{tabular}{lll|llll}
&  &  & \multicolumn{4}{c}{\textbf{Jones}} \\
&  &  & \pounds 0 & \pounds 900 & \pounds 1500 & \pounds
2000 \\
&  &  & \multicolumn{4}{l}{} \\ \hline
&  & \pounds 0 & \fbox{1000}, \fbox{1000} & 0,1100 & 0,500 &
0,0 \\
& \textbf{Smith:} & \pounds 900 & 1100 ,0 & \textrm{\fbox{
550},
\fbox{550}}
& 0, 500 & 0,0 \\
&  & \pounds 1500 & 500,0 & $500,0$ & \textbf{\fbox{ 250},
\fbox{250}}
& $0,
0$ \\
&  & \pounds 2000 & 0,0 & $0,0$ & $0,0$ & \textbf{\fbox{ 0},
\fbox{0}
} \\
\multicolumn{7}{l}{} \\
\multicolumn{7}{l}{\textit{Payoffs to: (Smith, Jones).}}
\end{tabular}

The judge's payoff was 1200 under the unique mixed-strategy
equilibrium in
the original game. Now, his payoff is either 900, 1500, or
2000. Thus,
\textrm{whether he prefers the new rules depends on which
equilibrium
is
played out in it.}
\end{enumerate}

\bigskip \noindent \textbf{13.3: Government and Monopoly. }
Incumbent
Apex
and potential entrant Brydox are bidding for government
favors in the
widget
market. Apex wants to defeat a bill that would require it to
share its
widget patent rights with Brydox. Brydox wants the bill to
pass.
Whoever
offers the chairman of the House Telecommunications
Committee more
campaign
contributions wins, and the loser pays nothing. The market
demand
curve is $%
P = 25-Q$, and marginal cost is constant at 1.

\begin{enumerate}
\item[(13.3a)]  Who will bid higher if duopolists follow
Bertrand
behavior?
How much will the winner bid?

\underline{\textit{Answer}.} Apex bids higher, because it
gets
monopoly
profits from winning, and Bertrand profits equal zero. {\
\textrm{Apex
can
bid some small ${\ \epsilon} $ and win. }}

\item[(13.3b)]  Who will bid higher if duopolists follow
Cournot
behavior?
How much will the winner bid?

\underline{\textit{Answer}.} Monopoly profits are found from
the
problem
\begin{equation}  \label{e0c1}
Maximize_{Q_a} \;\;\;\;Q_a(25-Q_a - 1),
\end{equation}
which has the first order condition $25 - 2Q_a -1 = 0$, so
that $Q_a =
12$
and $\pi_a =144 \; (=12(25-12-1)) $.

Apex's Cournot duopoly profit is found by solving the
problem
\begin{equation}  \label{e0d1}
\stackrel{Maximize}{Q_a} \;\;\;\; Q_a(25-[Q_a + Q_b] - 1),
\end{equation}
which has the first order condition $25 - 2Q_a -Q_b-1 = 0$,
so that if
the
equilibrium is symmetric and $Q_b = Q_a$, then $Q_a = 8$ and
$\pi_a =
64 \;
(=8(25- [8 + 8] -1) ) $.

Brydox will bid up to 64, since that is its gain from being
a
duopolist
rather than out of the industry altogether. Apex will bid up
to $80
(=144-64) $, and so \textrm{will win the auction at a price
of 64. }

\item[(13.3c) ]  What happens under Cournot behavior if Apex
can
commit to
giving away its patent freely to everyone in the world if
the entry
bill
passes? How much will Apex bid?

\underline{\textit{Answer}.} \textrm{Apex will bid some
small ${\
\epsilon} $
and win. } It will commit to giving away its patent if the
bill
succeeds,
which means that if the bill succeeds, the industry will
have zero
profits
and Brydox has no incentive to bid a positive amount to
secure entry.
\end{enumerate}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 14: Pricing}
\end{center}

\noindent \textbf{14.1: Differentiated Bertrand with
Advertising.} Two
firms
that produce substitutes are competing with demand curves
\begin{equation}  \label{e14.78}
q_1= 10 - \alpha p_1 + \beta p_2
\end{equation}
and
\begin{equation}  \label{e14.79}
q_2= 10 - \alpha p_2 + \beta p_1.
\end{equation}
Marginal cost is constant at $c=3$. A player's strategy is
his price.
Assume
that $\alpha > \beta/2.$

\begin{enumerate}
\item[(14.1a)]  What is the reaction function for Firm 1?
Draw the
reaction
curves for both firms.

\underline{\textit{Answer.}} Firm 1's profit function is
\begin{equation}  \label{e68}
\pi_1 = (p_1-c) q_1 = (p_1-3) (10 - \alpha p_1 + \beta p_2).
\end{equation}
Differentiating with respect to $p_1$ and solving the first
order
condition
gives the reaction function
\begin{equation}  \label{e69}
{\ p_1 = \frac{10 + \beta p_2 + 3\alpha}{2 \alpha}. }
\end{equation}
This is shown in Figure A.10.

\textbf{Figure A.10 The Reaction Curves in a Bertrand Game
with
Advertising}

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\item[(14.1b)]  What is the equilibrium? What is the
equilibrium
quantity
for Firm 1?

\underline{\textit{Answer.}} Using the symmetry of the
problem, set $%
p_{1}=p_{2}$ in the reaction function for Firm 1 and solve,
to give
${%
p_{1}^{\ast }=p_{2}^{\ast }=\frac{10+3\alpha }{2\alpha -
\beta }}$.
Using the
demand function for Firm 1, ${q_{1}=\frac{10\alpha +3\alpha
(\beta -
\alpha )%
}{2\alpha -\beta }}$.

\item[(14.1c)]  Show how Firm 2's reaction function changes
when
$\beta $
increases. What happens to the reaction curves in the
diagram?

\underline{\textit{Answer.}} The slope of Firm 2's reaction
curve is
$\frac{%
\partial p_2 }{\partial p_1} = \frac{ \beta}{2 \alpha}$. The
change in
this
when $\beta$ changes is $\frac{\partial^2 p_2 } {\partial
p_1 \partial
\beta}
= \frac{ 1}{2 \alpha}>0$. Thus, {\ \textrm{Firm 2's reaction
curve
becomes
steeper}}, as shown in Figure A.11.

\textbf{Figure A.11 How Reaction Curves Change When ${\
\beta}$
Increases }

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"XNPEU";}}

\item[(14.1d)]  Suppose that an advertising campaign could
increase
the
value of $\beta $ by one, and that this would increase the
profits of
each
firm by more than the cost of the campaign. What does this
mean? If
either
firm could pay for this campaign, what game would result
between them?

\underline{\textit{Answer.}} The meaning of an increase in
$\beta $ is
that
a firm's quantity demanded becomes more responsive to the
other firm's
price, if it charges a high price. The meaning is really
mixed:
partly, the
goods become closer substitutes, and partly, total demand
for the two
goods
increases. \newline
\hspace*{16pt} If either firm could pay, then a game of
``Chicken''
results,
with payoffs something like in Table A.14, where the ad
campaign costs
1 and
yields extra profits of $B$ to each firm.

\textbf{Table A.14 An Advertising ``Chicken'' Game}

\begin{tabular}{lllccc}
&  &  & \multicolumn{3}{c}{\textbf{Firm 2}} \\
&  &  & \textit{Advertise} &  & \textit{Do not advertise} \\
&  & $Advertise$ & B-1,B-1 & $\rightarrow $ & \textbf{B-1,B}
\\
& \textbf{Firm 1:} &  & $\downarrow $ &  & $\uparrow $ \\
&  & \textit{Do not advertise} & \textbf{B,B-1} &
$\leftarrow $ & 0,0 \\
\multicolumn{6}{l}{\textit{Payoffs to: (Firm 1, Firm 2).}}
\end{tabular}
\end{enumerate}

\bigskip \noindent \textbf{14.3: Differentiated Bertrand.}
Two firms
that
produce substitutes have the demand curves
\begin{equation}  \label{e14.80}
q_1= 1 - \alpha p_1 + \beta (p_2-p_1)
\end{equation}
and
\begin{equation}  \label{e14.81}
q_2= 1 - \alpha p_2 + \beta (p_1-p_2),
\end{equation}
where $\alpha > \beta$. Marginal cost is constant at $c$,
where $c <
1/\alpha $. A player's strategy is his price.

\begin{enumerate}
\item[14.3a) ]  (What are the equations for the reaction
curves
$p_1(p_2)$
and $p_2(p_1)$? Draw them.

\underline{\textit{Answer.}} Firm 1 solves the problem of
maximizing
$%
\pi_1=(p_1-c)q_1 = (p_1-c)(1-\alpha p_1+ \beta [p_2-p_1])$
by choice
of $p_1$%
. The first order condition is $1- 2(\alpha+\beta) p_1 +
\beta p_2 +
(\alpha+ \beta) c = 0$, which gives the reaction function
${p_1=
\frac{%
1+\beta p_2 + (\alpha+\beta) c}{2(\alpha+\beta)}}$. For
$p_2$: ${p_2=
\frac{%
1+\beta p_1 + (\alpha+ \beta) c}{2(\alpha+\beta)}}$. Figure
A.12 shows
the
reaction curves. Note that $\beta>0$, because the goods are
substitutes.

\textbf{Figure A.12 Reaction Curves for the Differentiated
Bertrand
Game }

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\item[(14.3b)]  What is the pure-strategy equilibrium for
this game?

\underline{\textit{Answer.}} This game is symmetric, so we
can guess
that $%
p_{1}^{\ast }=p_{2}^{\ast }$. In that case, using the
reaction curves,
${%
p_{1}^{\ast }=p_{2}^{\ast }=\frac{1+(\alpha +\beta )c}
{2\alpha +\beta
}}$.

\item[(14.3c)]  What happens to prices if $\alpha $, $\beta
$, or $c$
increase?

\underline{\textit{Answer.}} The response of $p^{\ast }$ to
an
increase in $%
\alpha $ is:
\begin{equation}
\frac{\partial p^{\ast }}{\partial \alpha }=\frac{c}{2\alpha
+\beta }-
\frac{%
2[1+(\alpha +\beta )c]}{(2\alpha +\beta )^{2}}=\left(
\frac{1}
{(2\alpha
+\beta )^{2}}\right) \left( 2\alpha c+\beta c-2-2\alpha c-
2\beta
c\right) <0.
\label{e63}
\end{equation}
The derivative has the same sign as $-\beta c-2<0$, so,
since $\beta >
0$,
\textrm{the price falls as ${\alpha }$ rises.} This makes
sense---
$\alpha $
represents the responsiveness of the quantity demanded to
the firm's
own
price.\newline
\hspace*{16pt} The increase in $p^{\ast }$ when $\beta $
increases is:
\begin{equation}
\frac{\partial p^{\ast }}{\partial \beta }=\frac{c}{(2\alpha
+\beta
)}-\frac{%
1+(\alpha +\beta )c}{(2\alpha +\beta )^{2}}=\left( \frac{1}
{(2\alpha
+\beta
)^{2}}\right) \left( 2\alpha c+\beta c-1-\alpha c-\beta
c\right) <0.
\label{e64}
\end{equation}
\textrm{The price falls with} ${\beta }$, because
$c<1/\alpha $.
\newline
\hspace*{16pt} The increase in $p^{\ast }$ when $c$
increases is:
\begin{equation}
\frac{\partial p^{\ast }}{\partial c}=\frac{\alpha +\beta }
{2\alpha
+\beta }%
>0.  \label{e65}
\end{equation}
\textrm{When the marginal cost rises, so does the price. }

\item[(14.3d) ]  What happens to each firm's price if
$\alpha$
increases,
but only Firm 2 realizes it (and Firm 2 knows that Firm 1 is
uninformed)?
Would Firm 2 reveal the change to Firm 1?

\underline{\textit{Answer.}} From the equation for the
reaction curve
of
Firm 1, it can be seen that the reaction curve will shift
and swivel
as in
Figure A.13. This is because $\frac{ \partial p_2}{\partial
p_1} =
\frac{%
\beta}{2(\alpha+ \beta}$, so $\frac{\partial^2 p_2}{\partial
p_1
\partial
\beta} = -\frac{\beta}{2(\alpha+ \beta)^2}<0$. Firm 2's
reaction curve
does
not change, and it believes that Firm 1's reaction curve has
not
changed
either, so Firm 2 has no reason to change its price. The
equilibrium
changes
from $E_0$ to $E_1$: \textrm{Firm 1 maintains its price, but
Firm 2
reduces
its price.} \textrm{Firm 2 would not want to reveal the
change to Firm
1},
because then Firm 1 would also reduce its price (and Firm 2
would
reduce its
price still further), and the new equilibrium would be
$E_2$.

\textbf{Figure A.13 Changes in the Reaction Curves }

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\end{enumerate}

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

\begin{center}
{\bf PROBLEMS FOR CHAPTER 15: Entry}
\end{center}

\noindent \textbf{15.1: Crazy Predators.} (adapted from
Gintis [2000],
Problem 12.10.) Apex has a monopoly in the market for
widgets, earning
profits of $m$ per period, but Brydox has just entered the
market.
There are
two periods and no discounting. Apex can either $Prey$ on
Brydox with
a low
price or accept $Duopoly$ with a high price, resulting in
profits to
Apex of
$-p_a$ or $d_a$ and to Brydox of $-p_b$ or $d_b$. Brydox
must then
decide
whether to stay in the market for the second period, when
Brydox will
make
the same choices. If, however, Professor Apex, who owns 60
percent of
the
company's stock, is crazy, he thinks he will earn an amount
$p^*> d_a$
from
preying on Brydox (and he does not learn from experience).
Brydox
initially
assesses the probability that Apex is crazy at $\theta$.

\begin{enumerate}
\item[(15.1a)]  Show that under the following condition, the
equilibrium
will be separating, i.e., Apex will behave differently in
the first
period
depending on whether the Professor is crazy or not:
\begin{equation}  \label{e15.a1}
-p_a +m < 2d_a
\end{equation}

\underline{\textit{Answer.}} In any equilibrium, Apex will
choose
$Prey$
both periods if the Professor is crazy. In any equilibrium,
Apex will
choose
$Duopoly$ in the second period if the Professor is not
crazy, by
subgame
perfectness.

If the equilibrium is separating, Apex will choose $Duopoly$
in the
first
period if the Professor is not crazy, and Brydox will
respond by
staying in
for the second period. This will yield Apex an equilibrium
payoff of
$2d_a$.
The alternative is to deviate to $Prey$. The best this can
do is to
induce
Brydox to exit, leaving Apex an overall payoff of $-p_a +m$
for the
two
periods, but if $- p_a +m < 2d_a,$ deviation is not
profitable. (And
if
Brydox would \textit{not} exit in response to $Prey$, $Prey$
is even
less
profitable.)

\item[(15.1b)]  Show that under the following condition, the
equilibrium can
be pooling, i.e., Apex will behave the same in the first
period
whether the
Professor is crazy or not:
\begin{equation}  \label{e15.a2}
\theta \geq \frac{d_b}{p_b+d_b}
\end{equation}

\item[\protect\underline{\textit{Answer.}}]  The only reason
for Apex
to
choose $Prey$ in the first period if the Professor is not
crazy is to
induce
Brydox to choose $Exit$. Thus, we should focus on Brydox's
decision.
Brydox's payoff from $Exit$ is 0. Its payoff from staying in
is
\[
\theta (-p_b) + (1-\theta) d_b.
\]
Exiting is as profitable as staying in if
\[
0 \geq \theta (-p_b) + (1-\theta) d_b,
\]
which implies that
\[
(p_b + d_b) \theta \geq d_b,\;\mathrm{and\; thus} \; \theta
\geq
\frac{ d_b}{%
p_b + d_b}.
\]

\item[(15.1c)]  If neither two condition (\ref{e15.a1}) nor
(\ref{e15.a2})
apply, the equilibrium is hybrid, i.e., Apex will use a
mixed strategy
and
Brydox may or may not be able to tell whether the Professor
is crazy
at the
end of the first period. Let $\alpha$ be the probability
that a sane
Apex
preys on Brydox in the first period, and let $\beta$ be the
probability that
Brydox stays in the market in the second period after
observing that
Apex
chose $Prey$ in the first period. Show that the equilibrium
values of
$%
\alpha $ and $\beta$ are:
\begin{equation}  \label{e15.a3}
\alpha = \frac{ \theta p_b }{(1-\theta)d_b}
\end{equation}
\begin{equation}  \label{e15.a4}
\beta =\frac{-p_a +m -2d_a}{m-d_a}
\end{equation}

\item[\protect\underline{\textit{Answer.}}]  An equilibrium
mixing
probability equates the payoffs from its two pure strategy
components.
First, consider Apex. Apex's two pure-strategy payoffs are:
\begin{equation}  \label{e15.a3a}
\pi_a(Prey) = -p_a + \beta d_a + (1-\beta)m = d_a + d_a =
\pi_a(Duopoly),
\end{equation}
so $\beta (d_a -m) = -m + p_a + 2d_a$ and we reach equation
(\ref{e15.a4}).

Note that we know the numerator of equation (\ref{e15.a4})
is
positive,
because we have ruled out a separating equilibrium by not
having the
inequality in part 15.1a hold. Also, the mixing probability
is less
than one
because the numerator is less than the denominator.

Now consider Brydox. Brydox's prior that Apex is crazy is
$\theta$,
but on
observing $Prey$, it must modify its beliefs. There was some
chance
that
Apex, if sane (which has probability $(1-\theta)$, would
have chosen
$%
Duopoly $, but that didn't happen. That had probability $(1-
\alpha)
(1-\theta)$ ex ante. Using Bayes' Rule, the posterior
probability that
Apex
is crazy is
\begin{equation}  \label{e15.a3c}
\frac{ \theta} {1- (1-\alpha) (1-\theta)},
\end{equation}
and the probability that Apex is sane is
\begin{equation}  \label{e15.a3d}
\frac{ ( \alpha)(1-\theta)} {1-(1-\alpha) (1-\theta)},
\end{equation}
.

Brydox's two pure-strategy payoffs after observing $Prey$
are
therefore
\begin{equation}  \label{e15.a3b}
\pi_b(Exit) = -p_b = -p_b + \frac{ \theta} {1-(1-\alpha) (1-
\theta)}
(-p_b)
+ \frac{ ( \alpha)(1-\theta)} {1-(1-\alpha) (1-\theta)}d_b =
\pi_b(Stay \;
in),
\end{equation}
so $0 = \theta (- p_b) + ( \alpha)(1-\theta) d_b$ and
\begin{equation}  \label{e15.a3e}
\alpha = \frac{ \theta p_b }{(1-\theta)d_b}
\end{equation}
If condition (\ref{e15.a2}) is false, then expression
({e15.a3e}) is
less
than 1, a nice check that we have calculated the mixing
probability
correctly (and it is clearly greater than zero).

\item[(15.1d)]  Is this behavior related to any of the
following
phenomenon?-- Signalling, Signal Jamming, Reputation,
Efficiency
Wages.

\underline{\textit{Answer.}} This is an example of signal
jamming.
Apex
alters its behavior in the first period so as to avoid
conveying
information
to Brydox. It is not signalling,because Apex is not trying
to signal
its
type. It is not reputation, because this is just a two-
period model,
not an
infinite-period one. In loose language, one might call it
reputation,
because Apex is trying to avoid acquiring a reputation for
sanity, but
it
has nothing in common with Klein- Leffler reputation models.
It is not
efficiency wages because no agent is being paid more than
his
reservation
utility so as to maintain incentives, nor is even any firm
being
rewarded
highly under the threat of losing the reward if it behaves
badly.
\end{enumerate}

\bigskip

\noindent \textbf{15.3: A Patent Race. } See what happens in
Patent
Race for
an Old Market when specific functional forms and parameters
are
assumed. Set
$f(x) = log(x)$, and for $w = f(x_i) -f(x_e)$, $g(w) = .5[1+
w/(1+w)]$
if $w
\geq 0$, $g(w) = .5[1+ w/(1-w)]$ if $w \leq 0$, $y= 2$, and
$z = 1$.
Figure
out the research spending by each firm for the three cases
of (a) $v=
10$,
(b) $v=4$, and (c) $v = 2$.

\underline{\textit{Answer.}} Under these parameters,
equation (15.13)
from
the book becomes
\begin{equation}  \label{e15.3a}
\frac{ 1/x_i}{1/x_e} = \frac{v-1}{Max(v-1 ,2)},
\end{equation}
or
\begin{equation}  \label{e15.3b}
\frac{ x_e}{x_i} = \frac{v-1}{Max(v-1,2)}.
\end{equation}

\noindent (15.3a) If $v=10$, then equation (\ref{e15.3b})
tells us
that
\begin{equation}  \label{e15.3c}
\frac{ x_e}{x_i} = \frac{9}{Max(9 ,2)} =1.
\end{equation}
Thus, $x_i =x_e$. But what do they equal? Use equation
(15.10) from
the
book, the first-order condition for the entrant. Equation
(15.10) is
\[
\frac{ d\pi_e}{dx_e} =-(1 - g ) + g^{\prime}f^{\prime}_e(v-
x_e- Z) -
g +
g^{\prime}f^{\prime}_e x_e = 0.
\]
The equation includes $g^{\prime}$, so let us first figure
that out
for the
functional form of this problem. It will be, for $w \geq 0$,
\begin{equation}  \label{e15.3d}
g^{\prime}= \frac{.5}{1+w} - \frac{.5w}{(1+w)^2}.
\end{equation}
Since $f(x_e) = log(x_e)$, it follows that $f^{\prime}(x_e)
= 1/x_e.$
Thus,
equation (15.10) becomes, once we use our knowledge that
$x_e= x_i$,
\begin{equation}  \label{e15.3e}
\begin{array}{l}
-(1-.5) + .5 (1/x_e)(v-x_e -1) - .5 + \left(\frac{.5}{x_e}
\right) x_e
=0,
\\
-.5 + \frac{.5(v-x_e-1}{x_e} -.5 +.5 =0 \\
x_e = v-x_e -1
\end{array}
\end{equation}
Solving this for $v=10$ yields $x_e =4.5$. $x_i$ must take
the same
value.

\noindent (15.3b) If $v=4$, nothing in the analysis changes
except the
last
line. Subsituting in $v$ now yields $x_e =x_i = 1.5.$

\noindent (15.3c)If $v=2$, the analysis changes, because now
in
equation (%
\ref{e15.3b}), $Max(v-1 ,2) = 2$. Thus,
\begin{equation}  \label{e15.3f}
\frac{ x_e}{x_i} = \frac{v-1}{Max(v-1,2)} = \frac{1}{2},
\end{equation}
so $x_i = 2 x_e$.

That is where I would be content for my students to stop,
but we can
go
further, if at the cost of messy algebra and some numerical
calculation.
Since it is no longer true that $x_i=x_e$, we cannot use the
simplification
of equation (15.10) in parts (a) and (b). Since $x_i > x_e$,
the $g$
function is $g(w) = .5[1+ w/(1+w)]$, and $g^{\prime}=
\frac{.5}{1+w} -
\frac{%
.5w}{(1+w)^2}$, as in part (a). Equation (15.10) becomes,
for $w=
log(x_i)
-log(x_e)$,
\begin{equation}  \label{e15.3g}
\begin{array}{lll}
   \frac{ d\pi_e}{dx_e} &=&-(1 - .5[1+ \frac{ w}{1+w}] ) +
\left(
\frac{.5}{1+w}
- \frac{.5w}{(1+w)^2} \right) (1/x_e)(v- x_e- Z)\\
 & &  - .5[1+ \frac{ w}{1+w}] +
\left( \frac{.5}{1+w} - \frac{.5w}{(1+w)^2} \right) (1/x_e)
x_e = 0.
\end{array}
\end{equation}
or
\begin{equation}  \label{e15.3h}
\begin{array}{l}
 -(1 - .5[1+ \frac{ w}{1+w}] ) + \left( \frac{.5}{1+w} -
\frac{.5w}
{(1+w)^2}
\right) (1/x_e)(v- x_e- Z) \\
 - .5[1+ \frac{ w} {1+w}] + \left( \frac{.5}{1+w}
- \frac{.5w}{(1+w)^2} \right)\\
  = 0.
\end{array}
\end{equation}
This can be rewritten, substituting for $V$ and $Z$, as
\begin{equation}  \label{e15.3i}
- 1 +.5 +.5( \frac{ w}{1+w}) ) \left( \frac{.5}{1+w} -
\frac{.5w}
{(1+w)^2}
\right) \frac{(2- x_e- 1)}{x_e} - .5 - \frac{ .5w}{1+w}] +
\left(
\frac{.5}{%
1+w} - \frac{.5w}{(1+w)^2} \right) = 0.
\end{equation}
so
\begin{equation}  \label{e15.3j}
-(1+w)^2 + .5( (1+w )-.5w) \frac{ 1 }{x_e} = 0,
\end{equation}
and
\begin{equation}  \label{e15.3k}
-(1+w)^2 + \frac{ .5}{x_e} = 0.
\end{equation}
Substituting $w= log (x_i) - log(x_e) $ and for $x_i=2x_e $
yields
\begin{equation}  \label{e15.3l}
-(1+ log(2x_e) - log(x_e))^2 + \frac{ .5}{x_e} = 0.
\end{equation}
This can be solved numerically. I used the Excel
spreadsheet, setting
up the
left-hand-side of equation (\ref{e15.3l}) as a formula and
trying half
a
dozen $x_e$ values till I found one that made the formula's
value
approximately zero (I was content with 0.007). That value
was $x_e=
.174$, in
which case $x_i=.348$.

\end{document}