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\begin{flushright}
Andrew Iannaccone\\
Diego Ubfal\\
Econ 201B - Zame
\end{flushright}
{\sc Game Theory - HW \#3}
\begin{enumerate}
\item \begin{enumerate}
\item The seller types are $T_S=\{g,b\}$. There are no buyer types.
The strategies for the buyer are $\{AA,AR,RA,RR\}$, where $AR$ means
accept high, reject low, etc. For the seller, the strategies are
$\{HH,HL,LH,LL\}$, where $HL$ means high price for good car, low
for bad, etc.
\item We compute the payoff matric for the pure strategies:
\begin{center}
\begin{tabular}[t]{|r | c | c | c | c |}
\hline
& AA & AR & RA & RR \\
\hline
HH & {\bf 4.5,3} & {\bf 4.5,3} & 0,0 & 0,0 \\
HL & 3.5,4 & 3,3 & 0.5,1 & 0,0 \\
LH & 1.5,6 & 1.5,0 & 0,6 & 0,0 \\
LL & 0.5,7 & 0.5,0 & {\bf 0.5,7} & 0,0 \\
\hline
\end{tabular}
\end{center}
The bold numbers represent weakly dominant strategies. So the pure
NE strategies are:
\begin{enumerate}
\item (HH,AA)
\item (HH,AR)
\item (LL,RA)
\end{enumerate}
Since the first two strategies yield the same utility, we also have
mixed strategy NE by combining them. In behavioral form, the
strategies are
\begin{enumerate}
\item $\sigma_S(H|g)=\sigma_S(H|b)=1;\\
\sigma_B(A|H)=1, \\ \sigma_B(A|L)=\alpha \quad \text{for }\alpha \in [0,1]$
\item $\sigma_S(H|g)=\sigma_S(H|b)=0;\\
\sigma_B(A|H)=0, \\\sigma_B(A|L)=0$
\end{enumerate}
\end{enumerate}
\item
The above analysis holds for any fraction of good cars,
and in each NE the car is always sold.
\item
\begin{enumerate}
\item We have both ignorant and knowledgable buyers. Additionally,
the knowledgable buyer gets a signal about the quality of the
car. We therefore have three types, denoted
$T_B = \left\{ i, kg, kb \right\}$. The strategies and seller
types are left unchanged.
\item
Knowledgable buyers cannot increase utility by deviating from
the strategies in Problem 1. Thus, the NE we found above are
still NE.\\
\\
Note, however, that knowledgable buyers who observe a bad car are
indifferent between the strategies $AA,AR,$ and $RA$. If they
choose $RA$, then sellers gain by deviating from $HH$ to $HL$.
In this problem, then gain (.5) from deviating with the
knowledgable buyers playing $RA$ is not sufficient to offset
the cost of deviating against the ignorant buyers
playing $AA$ (-1). If the percentage of knowledgable buyers were
large enough, however, an additional NE would arise.
\item
No. The equilibria are unchanged, so the outcome are also
unchanged.
\end{enumerate}
\item
\begin{enumerate}
\item In a symmetric equilibrium, each resident has the same
probability $q$ of reporting the crime. In equilibrium, $q$
is chosen so as to make each resident indifferent between
reporting and not.
\begin{align*}
u(R) &= 1-\varepsilon \\
u(N) &= 1 - (1-q)^{N-1}
\end{align*}
We set $u(R) = u(N)$ and solve for $q$ to obtain
\begin{align*}
q = 1- \varepsilon^{\frac{1}{N-1}}
\end{align*}
\item $P(R) = 1-(1-q)^N = 1-\varepsilon^{\frac{N}{N-1}}$
\item
Taking limits in (a) and (b), we find $q\to0$ and $P(R)\to1-\varepsilon$. That is, the chance of any individual reporting falls to zero, but the chance of someone reporting converges.
\end{enumerate}
\item
\begin{enumerate}
\item The probability of another resident reporting now changes
from $q$ to $pq$, so
\begin{align*}
u(N) &= 1-(1-pq)^{\frac{1}{N-1}}
\end{align*}
Again setting $u(R) = r(N)$, we get
\begin{align*}
q = \frac{1}{p} \left[ 1-\varepsilon^{\frac{1}{N-1}} \right]
\end{align*}
\item
Each resident now has a $pq$ chance of reporting, so the
cumulative probability is as in the previous problem,
\begin{align*}
P(R) = 1-(1-pq)^N = 1-\varepsilon^{\frac{N}{N-1}}
\end{align*}
\item
As $p \to 1$, our expressions for $q$ and $P(R)$ approach their
values from Problem 4. As $p\to0$, $q\to1$ so as long as anyone
sees the vandalism, they are certain to report it.
\item
Taking limits with $p$ fixed, we obtain $q\to0$ and $P(R)\to1-\varepsilon$.
\item
The limit as $p\to0$ and $N\to\infty$ is not well-defined. The
order in which we take the limits changes the result. If we
take the $p$-limit first, we get guaranteed reporting as in (c),
but if we take the $N$-limit first, we get the result from (d).
\end{enumerate}
\item
\begin{enumerate}
\item Each player has the same actions and types:
\begin{align*}
A_i &= \left\{ B,P \right\}\\
T_i &= \left\{ \text{AA,AJ,JJ} \right\}
\end{align*}
We may specify the behavioral strategies by a triple
$(p_1,p_2,p_3)$ indicating the probability of betting with
$AA,AJ,JJ$ respectively. Formally, the strategies are all the
mappings $\sigma_i$ from $T_i$ onto $[0,1]$.
\item
We can find the (unique) NE strategy by considering the best
responses to a generic strategy $(p_1,p_2,p_3)$. When a player
holds $AA$, betting stongly dominates passing, regardless of the
opponent's strategy. Thus, any NE strategy must have $p_1=1$.\\
\\
Suppose then, that our opponent plays $(1,p_2,p_3)$. Then
if we hold $AJ$, then we can calculate our expected utility from
betting or passing,
\begin{align*}
u(P) &= P(AA|AJ)(-1) + P(AJ|AJ)(-b_2) \\
&\quad + P(JJ|AJ)(-b_3) \\
u(B) &= P(AA|AJ)(-2)+ P(AJ|AJ)(1-b_2) \\
&\quad + P(JJ|AJ)(4b_3 + (1-b_3))
\end{align*}
Since our opponent is more likely to hold AJ than AA -- that is,
$P(AJ|AJ)>P(AA|AJ)$ -- a little arithmatic reveals that
$u(B) > u(P)$ regardless of our opponent's choices for
$p_2,p_3$. It follows that $p_2=1$.\\
\\
Finally, assume our opponent plays $(1,1,p_3)$ and that we hold
JJ. Here our utilities are
\begin{align*}
u(P) &= P(AA|JJ)(-1) + P(AJ|JJ)(-1) \\
&\quad+ P(JJ|JJ)(-b_3) \\
&= \frac{6}{15}(-1) + \frac{8}{15}(-1) + \frac{1}{15}(-b_3)\\
&= \frac{-14-b_3}{15}\\
u(B) &= P(AA|JJ)(-2)+ P(AJ|JJ)(-2) \\
&\quad+ P(JJ|JJ)((1-b_3)+8b_3)\\
&=\frac{6}{15}(-2) + \frac{8}{15}(-2) + \frac{1}{15}(1+7b_3)\\
&= \frac{-27+7b_3}{15}
\end{align*}
For any $b_3 \in [0,1]$, passing is less costly than betting, so
our best response is to always pass on $JJ$.\\
\\
Thus, the only possible NE strategy is (1,1,0) -- always betting
$AA,AJ$; always folding $JJ$. Since (1,1,0) is a best response to
itself, it is indeed a symmetric Bayesian NE.
\end{enumerate}
\item
\begin{enumerate}
\item
Passing with any signal above $C$ is strongly dominated. If the
opponent holds a lower signal, we gain by fighting; if he holds a
higher signal, he will fight anyway. We can therefore assume the
best response cutoff $X