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\title{EC943- Solutions to Problem Set N$^\circ$ 1}
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\author{Carlos Noton\\ University of Warwick}
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\date{Term 2 - 2012}

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\section{Sustaining Collusion}
\begin{itemize}
\item[a)]  The profits of collusion $V^c$ can be expressed as:
\begin{eqnarray*}
V^c&=& \sum_{t=0}^{\infty}\delta^t\left[\frac{1-P^m}{N}\right][P^m-c]=\frac{[1-P^m][P^m-c]}{N(1-\delta)}
\end{eqnarray*}
If deviates, the firm can serve the entire demand in the first period. However, the Nash-Bertrand price afterwards ($P^c=c$) implies zero profits forever afterwards. Formally, the
deviation profit $V^d$ is given by:
\begin{eqnarray*}
V^d&=&\left[1-P^m\right][P^m-c]+
\sum_{t=1}^{\infty}\delta^t\left[\frac{1-P^m}{N}\right][c-c]=\left[1-P^m\right][P^m-c]
\end{eqnarray*}
Thus, collusion can be sustainable if and only if:
\begin{eqnarray*}
V^c\geq V^d &\Leftrightarrow&
\frac{[1-P^m][P^m-c]}{N(1-\delta)} \geq
\left[1-P^m\right][P^m-c]\\
&\Leftrightarrow& \delta \geq 1-\frac{1}{N}
\end{eqnarray*}
Clearly, the required discount factor is increasing with the
number of firms. Collusion is harder to sustain as the number of firms
increase.

\item[b)] If firm observes the other firms' prices with a lag of two periods, the cooperation payoff is the same, whereas the deviation payoff is different. The deviation payoff will enjoy an extra period of profits:
\begin{eqnarray*}
V^d&=&\left[1-P^m\right][P^m-c]+\delta \left[1-P^m\right][P^m-c]
+\sum_{t=2}^{\infty}\delta^t\left[\frac{1-P^m}{N}\right][c-c]\\
&=&(1+\delta)\left[1-P^m\right][P^m-c]
\end{eqnarray*}
Consequently, the collusion can be sustainable if and only if:
\begin{eqnarray*}
V^c\geq V^d &\Leftrightarrow&
\frac{[1-P^m][P^m-c]}{N(1-\delta)} \geq
(1+\delta)\left[1-P^m\right][P^m-c]\\
&\Leftrightarrow& 1 \geq N (1-\delta^2)
\\
&\Leftrightarrow& \delta \geq \sqrt{1-\frac{1}{N}}
\end{eqnarray*}
Notice that for any natural number $N$, we have
$\sqrt{1-\frac{1}{N}}\geq 1-\frac{1}{N}$. Hence, the information
lag makes harder to collude (since it requires a higher $\delta$). Intuitively, the information lags make deviation more profitable.

\item[c)] Generalize your finding for any number of lags $\lambda$. Can you find an expression of $N$ as a function of $\delta$ and $\lambda$?
Denote $\pi^m=[1-P^m][P^m-c]$, so we have
\begin{eqnarray*}
V^c\geq V^d &\Leftrightarrow&
\frac{\pi^m}{N(1-\delta)} \geq
(1+\delta+\delta^2+...+\delta^{\lambda-1})\pi^m\\
&\Leftrightarrow& \frac{1}{N (1-\delta)} \geq \frac{ 1-\delta^\lambda }{1-\delta }
\\
&\Leftrightarrow& \delta \geq \left(1-\frac{1}{N}\right)^{\frac{1}{\lambda}}
\end{eqnarray*}

\item[d)] If $N=2$ and $\delta=0.8$. What is the number of largest number of lags, $\lambda^*$ that still allows for collusion? Compute the value of cooperation and deviation for $(\lambda^*+1)$ for a general $\pi^m$. (Notice that $\log(a)<0$ if $a\in(0,1)$.)
From previous part:
\begin{eqnarray*}
 \delta \geq \left(1-\frac{1}{N}\right)^{\frac{1}{\lambda}} &\Leftrightarrow&  \ln(\delta) \geq \frac{1}{\lambda}\ln\left( 1 - \frac{1}{N}\right)  \Leftrightarrow   \lambda  \leq \frac{ \ln\left( 1 - \frac{1}{N}\right) }{\ln(\delta)}\approx 3.11
\end{eqnarray*}
Given $\delta=0.8$, 2 firms can sustained collusion with at most  $\lambda^*=3$ lags. Evaluating the value function with $\lambda^*+1=4$ lags:
\begin{eqnarray*}
V^c &=& \frac{\pi^m}{N (1-\delta)}= 2.5 \pi^m\\
V^d &=& \frac{ 1-\delta^{(\lambda^*+1)} }{1-\delta }\pi^m= 2.952 \pi^m
\end{eqnarray*}
Deviation is more profitable than cooperation with four lags!
\end{itemize}


\section{Multimarket Fun}
Based on \citeasnoun{BernheimWhinston_1990}.
\begin{itemize}
\item[a.1)] What is the minimum $\delta$ that ensures cooperation for firm 1?

Collusion payoff in market A for low cost firm 1 is:
\begin{eqnarray*}
V_1^c&=& \sum_{t=0}^{\infty}\delta^t\left[\overline{s}Q^m\right][P^m-C_l]=\frac{[\overline{s}Q^m][P^m-C_l]}{(1-\delta)}
\end{eqnarray*}
Deviation is given by
\begin{eqnarray*}
V_1^d&=& [Q^m][P^m-C_l]+ \sum_{t=1}^{\infty}\delta^t 0 =[Q^m][P^m-C_l]
\end{eqnarray*}
Therefore, collusion is sustain if
\begin{eqnarray*}
V_1^c&=& \sum_{t=0}^{\infty}\delta^t\left[\overline{s}Q^m\right][P^m-C_l]=\frac{[\overline{s}Q^m][P^m-C_l]}{(1-\delta)}
\end{eqnarray*}
Deviation is given by
\begin{eqnarray*}
V_1^c\geq V_1^d&\Leftrightarrow& \frac{[\overline{s}Q^m][P^m-C_l]}{(1-\delta)}\geq[Q^m][P^m-C_l] \\
&\Leftrightarrow& \frac{\overline{s}}{(1-\delta)}\geq 1 \Leftrightarrow \delta \geq 1-\overline{s}
\end{eqnarray*}
\item[a.2)] What is the minimum $\delta$ that ensures cooperation for firm 2?
Redoing the same computation as in part [a.1], we have
\begin{eqnarray*}
V_2^c\geq V_2^d&\Leftrightarrow& \frac{1-\overline{s}}{(1-\delta)}\geq 1 \Leftrightarrow \delta \geq \overline{s}
\end{eqnarray*}

\item[a.3)] What is the minimum $\delta$ that ensures cooperation of both firms? (as a function of $\overline{s}$)
We need to ensure that both inequalities hold. Hence:
\begin{eqnarray*}
 \delta \geq 1-\overline{s} \text{ and } \delta \geq \overline{s} \Rightarrow \delta \geq \max\{1-\overline{s},\overline{s}\}
\end{eqnarray*}

\item[b.1)] Compute that collusive payoff for each firm $A$ and $B$ if the less efficient firm withdraws completely from its respective less efficient market.
The collusive payoff is easy. A monopoly in one market:
\begin{eqnarray*}
V_1^c =\underbrace{\frac{[Q^m][P^m-C_l]}{(1-\delta)}}_{\text{Monopoly in market A, withdraw from market B}}
\end{eqnarray*}
Similarly for firm 2:
\begin{eqnarray*}
V_2^c =\underbrace{\frac{[Q^m][P^m-C_l]}{(1-\delta)}}_{\text{Monopoly in market B, withdraw from market A}}
\end{eqnarray*}
\item[b.2)]  Compute the non-cooperative payoff for a given firm if deviates in both markets reaching the entire demand for themselves in both markets (and triggering zero profits forever).

The payoff of deviation for firm 1 is:
\begin{eqnarray*}
V_1^d \underbrace{[Q^m][P^m-C_l]+[Q^m][P^m-C_h]}_{{\text{Monopoly in both markets but zero profits afterwards in both markets}}}
\end{eqnarray*}
Similarly for firm 2:
\begin{eqnarray*}
V_2^d \underbrace{[Q^m][P^m-C_h]+[Q^m][P^m-C_l]}_{{\text{Monopoly in both markets but zero profits afterwards in both markets}}}
\end{eqnarray*}

\item[b.3)] Show that the minimum $\delta$ that can sustain collusion in this multi-market case.
We show that both players have the same payoffs to compare. In order to be collusion the most profitable option the comparison for firm $j=1,2$ is the following:
\begin{eqnarray*}
\underbrace{\frac{[Q^m][P^m-C_l]}{(1-\delta)}}_{\text{Monopoly in the most efficient market, withdraw from the other}} \geq \underbrace{[Q^m][P^m-C_l]+[Q^m][P^m-C_h]}_{{\text{Monopoly in both markets but zero profits afterwards}}}
\end{eqnarray*}
Therefore
\begin{eqnarray*}
&&V_j^c> V_j^d \\
&\Leftrightarrow& \frac{1}{(1-\delta)} \geq \frac{2P^m-C_l-C_h}{P^m-C_l} \Leftrightarrow \frac{P^m-C_l}{2P^m-C_l-C_h} \geq 1-\delta \\
&\Leftrightarrow& \delta \geq 1-\underbrace{\frac{P^m-C_l}{2P^m-C_l-C_h}}_{\geq 0.5}
\end{eqnarray*}

\item[b.4)] Is it larger or smaller than the case in [a] ?

The smallest value of $\delta$ in part [a] is $0.5$, whereas in part [b] any value is always smaller than $0.5$. Multi-market contact makes collusion easier.
\end{itemize}

\section{Green-Porter Revisited}
Based on \citeasnoun{GreenPorter_1984}. Suppose the Green and Porter framework covered in the class except that the punishment profits are not zero but equals to $\frac{\pi^m}{10}$.
  \begin{itemize}
  \item[a)] If this game is played for  a finite number of periods equals to $666$. What are the conditions over $\delta$ to reach collusion?

  \textit{If the time is known and finite, there is no possibility of collusion}. Backward induction kills any chance of collusion for all possible value of $\delta$.

  \item[b)] Do the algebra carefully to find the equations for $V^c$ and $V^d$.

  Redoing the lecture slides we have that collusion :
  \begin{eqnarray*}
V^c &=& p \left[\frac{\pi^m}{2}+ \delta V^c   \right]+ (1-p)\left[ 0 + \delta V^p   \right]
\end{eqnarray*}

If zero demand there is a punishment phase with profits $\pi^0$ for $\tau$ periods. Denote $V^p$ the discounted expected profits of
the punishment phase.
\begin{eqnarray*}
V^p =  \sum_{t=0}^{t=\tau-1} \delta^t \times \pi^0 +  \delta^{\tau} V^c = \left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0 + \delta^{\tau} V^c
\end{eqnarray*}
Solving the equation system, we can replace $V^p$ in the $V^c$ formula:


\begin{eqnarray*}
V^c &=& p \left[\frac{\pi^m}{2}+ \delta V^c   \right]+ (1-p)\left[ 0 + \delta \left\{ \left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0 + \delta^{\tau} V^c\right\}   \right]\\
\Rightarrow V^c    &=& \frac{1}{1-p\delta -(1-p)\delta^{\tau+1}} \left[p\frac{\pi^m}{2} + (1-p)\delta  \left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0 \right]
\end{eqnarray*}
Notice that if $\pi^0=0$ we recover the lecture's formula. Replacing in $V^p$ yields the other value:
\begin{eqnarray*}
V^p = \left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0 +  \frac{\delta^{\tau}}{1-p\delta -(1-p)\delta^{\tau+1}} \left[p\frac{\pi^m}{2} + (1-p)\delta  \left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0 \right]
\end{eqnarray*}

  \item[c)] What is the value of $k$ that can sustain collusion if $\delta=0.8$, the probabilities of boom, $p=0.5$ and the punishment length is $\tau=1$?


The usual comparison, where the deviation payoff is given by
\begin{eqnarray*}
V^d &=& p \left[\underbrace{\pi^m}_{deviation} + \delta\underbrace{V^p}_{punishment} \right]+ (1-p)\left[ 0 + \delta V^p\right]
\end{eqnarray*}
Thus, collusion is sustainable iif:
\begin{eqnarray*}
V^c & \geq & V^d \\
\Leftrightarrow  p \left[\frac{\pi^m}{2}+ \delta V^c   \right]+ (1-p)\left[ \delta V^p   \right] &\geq& p \left[\pi^m + \delta V^p \right]+ (1-p)\left[ \delta V^p   \right]\\
\Leftrightarrow  p \left[\frac{\pi^m}{2}+ \delta V^c   \right] &\geq& p \left[\pi^m + \delta V^p \right]\\
\delta &\geq& \frac{\pi^m}{2 (V^c-V^p)} \\
\delta  \frac{V^c-V^p}{\frac{\pi^m}{2}} &\geq& 1
\end{eqnarray*}
Computing the difference of value functions, we have:
\begin{eqnarray*}
V^c-V^p&=& V^c - \left\{\left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0 + \delta^{\tau} V^c\right\}= V^c(1 -\delta^{\tau} )-\left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0 \\
&=& \frac{(1 -\delta^{\tau} )}{1-p\delta -(1-p)\delta^{\tau+1}} \left[p\frac{\pi^m}{2} + (1-p)\delta  \left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0 \right]-\left( \frac{1-\delta^{\tau} }{1-\delta }\right)\pi^0
\end{eqnarray*}
Dividing by $\pi^m/2$ (using $\pi^0=k\pi^m/2$)
\begin{eqnarray*}
\delta \frac{V^c-V^p}{\frac{\pi^m}{2}}&=& \delta  \left\{\frac{(1 -\delta^{\tau} )}{1-p\delta -(1-p)\delta^{\tau+1}} \left[p + k(1-p)\delta \left( \frac{1-\delta^{\tau} }{1-\delta}\right) \right]-k \left( \frac{1-\delta^{\tau} }{1-\delta}\right) \right\}
\end{eqnarray*}
Replacing the value of $\tau=1$ in the collusive condition, we have:
\begin{eqnarray*}
\delta \frac{V^c-V^p}{\frac{\pi^m}{2}}&=& \delta  \left\{\frac{(1 -\delta)}{1-p\delta -(1-p)\delta^{2}} \left[p + k(1-p)\delta \left( \frac{1-\delta}{1-\delta}\right) \right]-k\left( \frac{1-\delta}{1-\delta}\right) \right\}\\
&=& \delta  \left\{\frac{(1 -\delta)}{1-p\delta -(1-p)\delta^{2}} \left[p +k(1-p)\delta  \right]-k\right\}
\end{eqnarray*}

Replacing the value $\delta=0.9$ and $p=0.5$ to ensure that $\delta \frac{V^c-V^p}{\frac{\pi^m}{2}}\geq 1$.
\begin{eqnarray*}
 0.9  \left\{\frac{(1 -0.9)}{1-0.5*0.9  -0.5*0.9^2} \left[0.5 + 0.5*0.9k  \right]-k\right\}&\geq&1\\
0.68966*(0.5 + 0.45k) - k &\geq& 1.\overline{1}\\
0.344828 -0.68966k &\geq&  1.\overline{1}\\
 -1.\overline{1} &\geq& k\\
\end{eqnarray*}
Notice that $k<0$, so $\pi^0$ should be negative to ensure cooperation. In this general case, larger monopoly profits, $\pi^m$, increase the temptation of deviation. Therefore, to keep the collusive behavior, the punishment profits, $\pi^0$ have to be more negative accordingly.
\end{itemize}

\pagebreak

\section{BibTeX and bibliography styles}

References and citations are best handled in a consistent way by using BibTeX. In this method, you supply all the relevant information about references in a ``.bib file'' without regard to ordering or style.

\bigskip

Then you let BibTeX format all citations and reference entries according to the chosen bibliographic style, and you don't have to sweat all the font and punctuation and ordering details yourself. The reference/bibliography section usually appears at the end of an publication. In your LaTeX file, these two commands insert the reference section: 
\bigskip
\begin{verbatim}
\bibliographystyle{xxx}
\bibliography{yyy}
\end{verbatim}

The ``xxx'' is a style name -- plain or abbrv or siam or alpha or any of dozens of other available styles, many listed below.

\bigskip
The ``yyy'' is the name of the bib file (yyy.bib) containing the reference database.

\bibliographystyle{econometrica}
\bibliography{REF-Example}

\pagebreak
\section{Labeling and Tables}
Equations \ref{eq: VAR} and \ref{eq: other} are important to learn \emph{labeling}.
Using $\log$s, the estimated model is the following VAR(1) system. :
\begin{eqnarray}\label{eq: VAR}
\left[\begin{array}{c }
 W_{ft} \\
 Y_{ft}
 \end{array}\right]
= \lambda_0+ \lambda_f' \left[\begin{array}{c }
 W_{ft-1} \\
 Y_{ft-1}
 \end{array}\right]
 + \left[\begin{array}{c }
 v_{1,ft} \\
 v_{2,ft}
 \end{array}\right]
\end{eqnarray}
where
\begin{eqnarray}\label{eq: other}
\mathbb{E}( v_{1,ft} v_{2,rp})=\Gamma \neq 0 \text{ if and only if } f=r \text{ and } t=p
\end{eqnarray}

We present table \ref{NameoftheTable} and table \ref{Exc Rat CORR}, which is just another table for you to learn \LaTeX
\begin{table}[!htbp]
\begin{center}
\caption{\textbf{Mean of Retailer Markups by counties}} \label{NameoftheTable}
\begin{tabular}{|c||c|c|c||c|c|c|}\hline
 & \multicolumn{3}{c||}{\textbf{EDLP}}  & \multicolumn{3}{|c|}{\textbf{HL}}   \\
Manufacturer $\backslash$ County           & \ Rich \ & \ Mid \ & \ Poor \ &  \  Rich  \  &  \    Mid   \ &  \  Poor \  \\\hline \hline
   Nestl\'{e}  &  \ \ 7.5  &   \   \                         7.4  &          \           \          7.3  &                            10.9  &                            10.0  &                       \ \        9.1  \\\hline
 Non Nestl\'{e}  & 10.8  &                            11.7  &                            11.7  &                            15.0  &                            14.9  &                            14.4  \\\hline
\end{tabular}
\end{center}
\end{table}

and also:
\begin{table}[!htbp]
\begin{center}
\caption{\textbf{Correlation Matrix of Exchange Rate Shocks}}\label{Exc Rat CORR}
\begin{tabular}{|c ||c c c c c c||}\hline
Yearly  &   Bel &   Fra &   Ger &   Ita &   UK  &   Jap \\  \hline
Bel &   1.00    &       &       &       &       &       \\
Fra &   0.93    &   1.00    &       &       &       &       \\
Ger &   0.97    &   0.91    &   1.00    &       &       &       \\
Ita &   0.81    &   0.84    &   0.79    &   1.00    &       &       \\
UK  &   0.66    &   0.65    &   0.65    &   0.70    &   1.00    &       \\
Jap &   0.62    &   0.60    &   0.62    &   0.45    &   0.46    &
1.00    \\  \hline
\end{tabular}
\end{center}
\end{table}


From STATA to \LaTeX, click \href{http://www.ats.ucla.edu/stat/stata/latex/tabstat_stata.htm}{HERE}. \bigskip

From Excel to \LaTeX, click \href{http://www.mackichan.com/index.html?techtalk/v30/30ts71.htm~mainFrame}{HERE}\bigskip

\textbf{...or just Google it! :)}

\pagebreak
\section{Figures}
Figure \ref{CB} is a Cobb-Douglas utility, $U(x,y)=x^{\lambda}y^{1-\lambda}$ function with $\lambda=0.25$.
\begin{figure}[!htbp]
\begin{center}
\caption{\textbf{Utility Function}}\label{CB}
{\scalebox{0.8}[0.8]{\includegraphics{U_CD.eps}}}
\end{center}
\end{figure}


\end{document}
