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

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

\noindent    .  26 March 2005. 11 November 2005. Erasmuse@indiana.edu. Http:
//www.rasmusen.org.

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

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

\newpage

\begin{center} {\bf PROBLEMS FOR CHAPTER 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[(a)] If $c=0$, what is the equilibrium outcome?

 \noindent \underline{\textit{Answer}.} ${S_1=100}$  {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[(b)] If $c=80$, what is the equilibrium outcome?

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

 \noindent \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[(d)]  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.

 \noindent \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[(a)]  What is the threat point?

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

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

 \noindent \underline{\textit{Answer}.} See Figure A12.1.

\includegraphics[width=150mm]{a12-1.jpg}

 
\textbf{Figure A12.1: The Threat Point and Pareto Frontier }

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

 \noindent \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[(e)]  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.

 \noindent \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[(a)] If $c=0$, what is the equilibrium outcome?

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

 \noindent \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[(c)] If Jones' priors are that $c=0$ and $c=80$ are equally likely, but
only Smith knows the true value, what are the players' equilibrium strategies in
rounds 2 and 3?  (that is: what are $S_2$ and $S_3$, and what acceptance rules
will each player use?)

\noindent \underline{\textit{Answer}.}  Jones  proposes $S_2 = 20$ and accepts
$S_3   \leq 100$. Smith accepts $S_2 \geq 20 $ if $c=80$ and $S_2 \geq 100$ if
$c= 0$, and proposes $S_3= 100$ regardless of $c$.

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 $S_3=100$.

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

\noindent \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 $c=80$. Jones accepts $S_1 = 100$
with probability $\frac{1}{8} $, rejects $S_1 \in (30, 100) $, and accepts $S_1
\leq 30$.  Out of equilibrium, a supporting belief is for Jones to believe that
if $S_1$ equals neither 30 nor 100, then $Prob(c= 80)=1$.

 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 $v_s=0.2$ and $v_b= 0.8$. What is one equilibrium
outcome given these valuations and this bargaining procedure? Explain why this
can happen.

 \noindent \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.

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