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\noindent   9 February 2006.  Eric Rasmusen, Erasmuse@indiana.edu.\\
Http://www.rasmusen.org. 


 Overheads for Chapter  14, Pricing, of {\it Games and Information}. These do not cover the entire chapter, just enough for two lectures. 


  

\newpage
   

 

\includegraphics[width=100mm]{fig14-04.jpg}

 \begin{center}
 {\bf Figure  4: Location Models  }
\end{center}

 \begin{center}
{\bf  The Hotelling Pricing Game }\\
 (Hotelling [1929])
 \end{center}
{\bf Players}\\
Sellers Apex and Brydox, located at $x_a$ and $x_b,$ where $x_a < x_b$, and a
continuum of buyers indexed by location $x \in [0,1]$.

\noindent
{\bf The Order of Play }\\
 1 The sellers simultaneously choose  prices $p_a$ and $p_b$.\\
 2 Each buyer chooses a seller.

\newpage

\includegraphics[width=100mm]{fig14-04.jpg}

 \begin{center}
 {\bf Figure  4: Location Models  }
\end{center}

\noindent
{\bf Payoffs}\\
  Demand is uniformly distributed on the interval [0,1] with a density equal to
one (think of each consumer as buying one unit). Production costs are zero. Each
consumer always buys, so his problem is to minimize the sum of the price plus
the linear transport cost, which is $\theta$ per unit distance travelled.
 \begin{equation} \label{e13.33}
 \pi_{buyer \;at \;x} =V  -Min\{ \theta |x_a -x| + p_a, \; \theta |x_b - x| +
p_b \}.
  \end{equation}

\begin{equation} \label{e13.33aa}
\pi_a = \left\{ \begin{tabular}{llr}
 $p_a (0) =0$ & if   $p_a - p_b  >  \theta (x_b - x_a)$   & (a)\\ &    (Brydox
captures entire market) & \\
 & & \\ $ p_a (1) =p_a$  & if   $p_b - p_a  > \theta (x_b - x_a)$   & (b)\\
& (Apex captures entire market)  & \\ & & \\ $p_a ( \frac{1}{2\theta} \left[
(p_b - p_a) + \theta(x_a + x_b) \right] ) $ & otherwise (the market is divided)&
(c) \\
\end{tabular} \right.
 \end{equation}

\newpage

$$
\pi_a = \left\{ \begin{tabular}{llr}
 $p_a (0) =0$ & if   $p_a - p_b  >  \theta (x_b - x_a)$   & (a)\\ &    (Brydox
captures entire market) & \\
 & & \\ $ p_a (1) =p_a$  & if   $p_b - p_a  > \theta (x_b - x_a)$   & (b)\\
& (Apex captures entire market)  & \\ & & \\ $p_a ( \frac{1}{2\theta} \left[
(p_b - p_a) + \theta(x_a + x_b) \right] ) $ & otherwise (the market is divided)&
(c) \\
\end{tabular} \right.
$$


   A buyer's utility depends on the
price he pays and the distance he travels.  Price aside, Apex is most attractive
of the two sellers to the consumer at $x=0$ (``consumer 0'')  and least
attractive to the consumer at $x = 1 $ (``consumer 1''). Consumer   0  will buy
from Apex so long as
\begin{equation}\label{e13.34}
 V-(\theta x_a + p_a) >V- (\theta x_b + p_b),
 \end{equation} which implies that
 \begin{equation} \label{e13.35}
 p_a - p_b  < \theta (x_b - x_a), \end{equation}
 which yields payoff
(\ref{e13.33aa}a) for Apex.  Consumer    1  will buy from Brydox if
\begin{equation}\label{e13.36}
 V-[\theta(1- x_a) +p_a] < V- [\theta (1-x_b) + p_b],
 \end{equation} which implies that
 \begin{equation}\label{e13.37}
 p_b - p_a  < \theta (x_b - x_a),
 \end{equation}
 which  yields payoff (\ref{e13.33aa}b) for Apex.


\newpage


  Very likely, inequalities  (\ref{e13.35}) and (\ref{e13.37}) are both
satisfied, in which case Consumer 0 goes to Apex and Consumer 1 goes to Brydox.
Let  consumer $x^*$ be  the consumer   at the
boundary between the two markets, indifferent between  Apex and Brydox.

Notice that if Apex attracts Consumer $x_b$, he also attracts all $x > x_b$,
because beyond $x_b$ the consumers' distances from both sellers increase at the
same rate. So we know that if there is an indifferent consumer he is between
$x_a$ and $x_b$.

 Knowing this, the consumer's payoff equation,  (\ref{e13.33}),
tells us that
 \begin{equation}\label{e13.38}
 V- [ \theta(x^*- x_a) + p_a] = V-  [\theta (x_b -x^*) + p_b],
 \end{equation}
 so that
 \begin{equation}\label{e13.39}
 p_b - p_a = \theta (2x^*- x_a - x_b ),
 \end{equation} and
 \begin{equation}\label{e13.40}
 {\displaystyle x^* = \frac{1}{2\theta} \left[ (p_b - p_a) + \theta(x_a + x_b)
\right]},
\end{equation}
which generates demand curve (\ref{e13.33aa}c)-- a differentiated Bertrand
demand curve.

 \newpage

The Nash equilibrium can be
calculated   by setting up the profit
functions for each firm, differentiating with respect to the price of each, and
solving the two first-order conditions for the two prices. If there exists an
equilibrium in which the firms are willing to pick prices to satisfy
inequalities (\ref{e13.35}) and (\ref{e13.37}), then it  is
 \begin{equation}\label{e13.41}
p_a = \frac{(2 + x_a + x_b)\theta}{3}, \;\;p_b = \frac{(4 - x_a - x_b)\theta}
{3}.
 \end{equation}
 
 Apex charges a higher price if a large
$x_a$ gives it more safe consumers or a large $x_b$ makes the number of
contestable consumers greater.


Profits are positive and increasing in the transportation cost.

\newpage

  We cannot rest satisfied with the neat equilibrium of equation (\ref{e13.41}),
because the assumption that  there exists an equilibrium in which the firms
choose prices so as to split the market on each side of some boundary  consumer
$x^*$ is often violated. 


  Vickrey (1964) and  D'Aspremont,
Gabszewicz  \& Thisse (1979) have shown that if $x_a$ and $x_b$ are close
together, no pure-strategy equilibrium exists, for reasons similar to why none
exists in the Bertrand model with capacity constraints.

 If both firms charge
nonrandom prices, neither would deviate to a slightly different price, but one
might deviate to a much lower price that would capture every single consumer.


But if both firms charged that low price, each would deviate by raising his
price slightly.

  It turns out that if, for example,   Apex and Brydox are
located symmetrically around the center of the interval, $x_a \geq 0.25$, and
$x_b \leq 0.75$, no pure-strategy equilibrium exists (although a mixed-strategy
equilibrium does,  as Dasgupta \& Maskin [1986b] show).

 Hotelling should  have done some numerical examples.

  And he should have
thought about the comparative statics.   Equation (\ref{e13.41})
implies  that Apex should choose a higher price if both $x_a$ and $x_b$
increase, but it is odd that if the firms are locating closer together, say at
0.90 and  0.91, that Apex should be able to charge a higher price, rather  than
suffering from more intense competition.

\newpage

\includegraphics[width=150mm]{fig14-05.jpg}

 \begin{center}
{\bf Figure  5: Numerical Examples for Hotelling Pricing  }
 \end{center}

\noindent
 {\bf  Example 1. Everything works out simply}


 Try    $x_a = 0, x_b = 0.7$ and $\theta =0.5$.  Then equation (\ref{e13.41})
says $ p_a= (2+0+0.7)0.5/3 =0.45 $ and $p_b= (4-0-0.7)0.5/3 =  0.55$. Equation
(\ref{e13.40}) says that $  x^* = \frac{1}{2*0.5} \left[ (0.55-0.45) +
0.5(0.0+0.7)\right]= 0.45$.

 In Example 1, there is a pure strategy equilibrium and the equations generated
sensible numbers given the parameters we chose.   


\newpage


\includegraphics[width=150mm]{fig14-05.jpg}

 \begin{center}
{\bf Figure  5: Numerical Examples for Hotelling Pricing  }
 \end{center}


\noindent
 {\bf Example 2. Same location -- but different prices?} 

 Try  $x_a = 0.9, x_b = 0.9$ and $\theta =0.5$.  Then equation (\ref{e13.41})
says $ p_a= (2.0+0.9+0.9)0.5/3 \approx 0.63$ and     $p_b= (4.0-0.9-0.9) 0.5/3
\approx 0.37$.

 

   The  equations generate
numbers that seem innocuous until one realizes that   if both firms are located
at 0.9,  but  $p_a =0 .63$ and  $p_b = 0.37$, then  Brydox will    capture the
entire market!

 The result is nonsense, because  equation (\ref{e13.41})'s
derivation relied on the assumption that $x_a < x_b$, which is false in this
example.


\newpage


 (draw in Figure 5c by hand)

\noindent
{\bf  Example 3. Locations too near  each other.  } 

$x^* < x_a < x_b$.  Try  $x_a = 0.7, x_b = 0.9$ and $\theta =0.5$. Then equation
(\ref{e13.41}) says that   $ p_a= (2.0+0.7+0.9)0.5/3 =0.6$ and     $p_b=
(4-0.7-0.9)0.5/3 =0.4$.  As for the split of the market, equation (\ref{e13.40})
says that $    x^* = \frac{1}{2*0.5} \left[ (0.4-0.6) + 0.5(0.7+0.9) \right]=
0.6$.

   If the market splits at  $x^* = 0.6$ but
$x_a=0.7$ and $x_b=0.9$,  the result violates   our implicit assumption that the
players split the market. 

   Equation (\ref{e13.40}) is based on the premise
that there does exist some indifferent consumer, and when that is a false
premise, as under the parameters of Example 3,   equation  (\ref{e13.40}) will
still  spit out a value of $x^*$, but the value  will not mean anything. 


   In
fact  the consumer at  $x=0.6$ is not really indifferent between Apex and
Brydox.    He   could buy from Apex at  a total cost of  0.6 + 0.1(0.5) = 0.65
or from Brydox, at a total cost of  0.4 + 0.3 (0.5) = 0.55.     There exists no
consumer   who  strictly prefers Apex. 

  \newpage

 \begin{center} {\bf   The  Hotelling Location Game }\\
 (Hotelling [1929])
 \end{center}
 {\bf Players}\\
$n$ Sellers.

 \noindent
 {\bf The Order of Play }\\
The sellers simultaneously choose locations $x_i \in [0,1].$

 \noindent {\bf Payoffs}\\
 Consumers are distributed along the interval [0,1] with a uniform density equal
to one. The price equals one, and production costs are zero.  The sellers are
ordered by their location so $x_1 \leq x_2 \leq \ldots \leq x_n$, $x_0 \equiv 0$
and $x_{n+1} \equiv 1.$ Seller $i$ attracts half the consumers from the gaps on
each side of him,  as shown in Figure 14.6, so that his payoff is
\begin{equation}\label{e13.42}
 \pi_1 = x_1 + \frac{x_2 - x_1}{2},
\end{equation}
\begin{equation}\label{e13.43}
 \pi_n =    \frac{x_n - x_{n-1}}{2} + 1 - x_n,
 \end{equation}
or, for $i = 2, \ldots n-1$,
 \begin{equation}\label{e13.44}
\pi_i = \frac{x_i - x_{i-1}}{2} + \frac{x_{i+1} - x_i}{2}.
 \end{equation}

\includegraphics[width=150mm]{fig14-06.jpg}

 \begin{center}
  {\bf Figure 6: Payoffs in the Hotelling Location Game }
 \end{center}

\newpage

  With {\bf one seller}, the location does not matter in this model, since the
consumers are captive. If price were a choice variable and demand were elastic,
we would expect the monopolist to locate at $x=0.5$.

  With {\bf two sellers}, both firms locate at $x= 0.5$, regardless of whether
or not demand is elastic.  This is a stable Nash equilibrium, as can be seen by
inspecting Figure  4 and imagining best responses to each other's   location.
The best response is always to locate $\varepsilon$ closer to the center of the
interval than one's rival. When both firms do this, they end up splitting the
market since both of them end up exactly at the center.

\newpage

\includegraphics[width=150mm]{fig14-07.jpg}

\begin{center}
 {\bf Figure  7: Nonexistence of pure strategies with three players }
\end{center}

     With {\bf three sellers} the model does not have a Nash equilibrium in pure
strategies.  Consider any strategy profile in which each player locates at a
separate point. Such a strategy profile is not an equilibrium, because the two
players nearest the ends would edge in to squeeze the middle player's market
share. But if a strategy profile has any two players at the same point $a$, as
in Figure  7, the third player would be able to acquire a share of at least
$(0.5 - \epsilon)$ by moving next to them at $b$; and if the third player's
share is that large, one of the doubled-up players would deviate by jumping to
his other side and capturing his entire market share.  The only equilibrium is
in mixed strategies.

\newpage


\includegraphics[width=150mm]{fig14-08.jpg}

 \begin{center}
{\bf Figure  8:  The Equilibrium Mixed-Strategy Density in the Three-Player
Location Game  }
 \end{center}

Suppose all three players use the same mixing density,  with $m(x)$ the
probability density for location $x$, and positive density on  the support
$[g,h]$, as depicted in Figure  8.

 We will need the density for the distribution
of the minimum of the locations of Players 2 and 3. 

 Player 2 has location $x$
with density  $m(x)$, and Player 3's location is greater than that with
probability  $1-M(x)$, letting $M$ denote the cumulative distribution, so:

density (
  $x2=x$, $x2 < x3$  =  $m(x)[1-
M(x)]$.

 

 The  density for either Player 2 or Player 3 choosing $x$ and it being
smaller than the other firm's location is then $2m(x)[1-M(x)]$:

density( Minimum of x2 and x3 equalling x) =  $2m(x)[1-M(x)]$.





\newpage


We just found (a) and (b) 


(a) density (
  $x2=x$, $x2 < x3$  =  $m(x)[1-
M(x)]$.

 
(b)   density( Minimum of x2 and x3 equalling x) =  $2m(x)[1-M(x)]$.

If Player 1 chooses $x=g$  then his  expected payoff is
 \begin{equation}\label{e13.new1}
 \pi_1(x_1=g) =  g + \int_g^h   2m(x)[1-M(x)]   \left( \frac{  x-g } {2} \right)
dx,
 \end{equation}
 where   $g$ is  the  safe set of consumers to his left, $2m(x)[1- M(x)]$ is the
density for $x$ being the next biggest   location of a firm, and $\frac{  x-g }
{2}$ is Player 1's share of the consumers between his own location of $g$ and
the next biggest location.

If Player 1 chooses $x=h$    his  expected payoff is, similarly,
 \begin{equation}\label{e13.new1a}
  \pi_1(x_1=h) =  (1-h)  + \int_g^h   2m(x) M(x)   \left( \frac{  h-x } {2}
\right) dx,
 \end{equation} where $(1-h)$ is the set of safe consumers to his right.

   In a mixed strategy equilibrium, Player 1's payoffs from  these two pure
strategies must be equal, and they are also equal to his payoff from  a location
of 0.5, which we can plausibly guess is in the support of his mixing
distribution. Going on from this point, the algebra and calculus start  to
become fierce.    Shaked (1982) has computed the symmetric mixing probability
density $m(x)$ to be:
\begin{equation}\label{e13.45}
m(x)=  \left\{
 \begin{array}{ll}
 2 &{\rm if}\;\; \frac{1}{4} \leq x \leq \frac{3}{4}  \\
 & \\
 0 &{\rm  otherwise} \\
\end{array}
\right. \end{equation}
 Asymmetric equilibria also exist. 

  \newpage

   

  Since prices are inflexible, the competitive market does not achieve
efficiency.

 A benevolent social planner or a monopolist who could charge higher
prices if he located his outlets closer to more consumers would choose different
locations than competing firms. 

 In particular, when two competing firms both
locate in the center of the line, consumers are no better off than if there were
just one firm.

 As shown in Figure 10, the  average distance of a consumer from a
seller would be minimized by setting $x_1 = 0.25$ and $x_2 = 0.75$, the
locations that would be chosen either by the social planner or the monopolist.

\includegraphics[width=150mm]{fig14-10.jpg}

\begin{center}
 {\bf Figure  10:  Equilibrium versus Efficiency }
 \end{center}


    The Hotelling Location Model   is   well suited to politics.

 


\newpage

  


\begin{center}
{\bf   Vertical Differentiation I: Monopoly Quality Choice } 
\end{center}
{\bf Players}\\
  A seller and a continuum of buyers.

\noindent
  {\bf The Order of Play } 

 0  Nature assigns quality values to a continuum of buyers of length 1. Half  of
them  are ``weak'' buyers ($\theta=0$) who value high quality at 20 and low
quality at 10.  Half  of them  are ``strong'' buyers ($\theta=1$) who value high
quality at 50 and low quality at 15.        

1 The seller  picks    quality  $s$   to be either  $s 0$ or $ 1$.  

 2 The seller  picks price   $p$   from the interval $[0,\infty)$. 

 3 Each buyer    chooses one unit of  a good, or refrains from buying.  The
seller produces   at  constant marginal cost $c=1$, which does not vary with
quality.  

\noindent
  {\bf Payoffs}\\
 \begin{equation} \label{e100}
 \pi_{seller} =  (p -1) q.
\end{equation}
 The buyer's payoff is zero if he does not buy. If he does buy, it is
\begin{equation} \label{e101}
 \pi_{buyer}  = (10 + 5\theta)  + (10+ 25\theta)  s - p.
 \end{equation}




\newpage

\noindent
 {\bf OPTIMAL SELLER  PRICE AND QUANTITY}


\noindent
  {\bf Payoffs}\\
 \begin{equation} \label{e100}
 \pi_{seller} =  (p -1) q.
\end{equation}
 The buyer's payoff is zero if he does not buy. If he does buy, it is
\begin{equation} \label{e101}
 \pi_{buyer}  = (10 + 5\theta)  + (10+ 25\theta)  s - p.
 \end{equation}

  The seller  should clearly set the quality to be high,since then he can charge
more to the  buyer (though note that this runs contrary to a common
misimpression that a monopoly will result in lower quality than a competitive
market.)  The price should be either 50, which is the most the strong  buyers
would pay, or 20, the most the weak buyers would pay. Since $\pi (50) = 0.5
(50-1) = 24.5$ and $\pi (20) = 0.5 (20-1)+ 0.5 (20-1)=19$, the seller should
choose $p=50$.  Separation (by inducing only the strong buyer to buy) is better
for the seller than pooling.




\newpage

     \begin{center}
 {\bf   Vertical Differentiation II: Crimping the Product  }
 \end{center}
{\bf Players}\\
 A seller and a continuum of buyers.

\noindent
  {\bf The Order of Play }\\
 0  Nature assigns quality values to a continuum of buyers of length 1. Half  of
them  are ``weak'' buyers ($\theta=0$) who value high quality at 20 and low
quality at 10.  Half  of them  are ``strong'' buyers ($\theta=1$) who value high
quality at 50 and low quality at 15.       \\
1 The seller  decides to sell both qualities      $s =0$  and $s =1$ or just
one of them. \\
 2 The seller  picks prices   $p_L$   and $p_H$ from the interval $[0,\infty)
$.\\
 3 Each buyer    chooses one unit of  a good, or refrains from buying.  The
seller produces   at  constant marginal cost $c=1$, which does not vary with
quality.

\noindent
  {\bf Payoffs}\\
 \begin{equation} \label{e102}
 \pi_{seller} =  (p_L -1) q_L + (p_H -1) q_H.
\end{equation}
 and
\begin{equation} \label{e103}
 \pi_{buyer}  = (10 + 5\theta)  + (10+ 25\theta)  s - p.
 \end{equation}

  \newpage

 This is a problem of mechanism design. The seller needs to pick  $p_1$, and
$p_2$ to satisfy incentive compatibility and participation constraints if he
wants to offer two qualities with positive sales of both, and he also needs to
decide if that is more profitable than offering just one quality.

We already solved the    one-quality problem  in Vertical Differentiation I,
yielding profit of 24.5.   The monopolist cannot simply add a second, low-
quality, low-price good for the weak buyers, because the strong buyers, who
derive zero payoff from the high-quality good, would switch to the low-quality
good, which would give them a positive payoff. In equilibrium, the monopolist
will have to give the strong buyers  a positive payoff. Their participation
constraint will be non-binding, as we have found so many times before for the
``good'' type.

   Following the usual pattern, the participation constraint for the weak buyers
will be binding, so $p_L=10$. The self-selection constraint for the strong
buyers will also be binding, so
 \begin{equation} \label{e104}
 \pi_{strong} (L)  = 15 - p_L   =  50 - p_H.
 \end{equation}
  Since $p_L=10$, this results in $p_H=45$.  The price for high quality must be
at least 35 higher than the price for low quality to induce separation of the
buyer types.

\noindent
 Profits will now be:
    \begin{equation} \label{e105}
 \pi_{seller} =  (10 -1)(0.5)  + (44-1)  (0.5)  = 26.
\end{equation}
 This exceeds the one-quality profit of 24.5, so it is optimal for the seller to
sell two qualities.


  \newpage

  This result, is, of course, dependent on the parameters chosen, but it is
nonetheless a fascinating special case, and one which is perhaps no more special
than the other special case, in which the seller finds that profits  are
maximized with just one quality. The outcome of allowing price discrimination is
a pareto improvement. The seller is better off,  because profit has risen from
24.5 to 26. The strong buyers are better off, because the price they pay has
fallen from 50 to 45. And the weak buyers are no worse off. In Vertical
Differentiation I their payoff was zero because they chose not to buy; in
Vertical Differentiation I their payoffs are zero because they buy at a price
exactly equal to their value for the good.

 Indeed, we can go further. Suppose the cost for the low-quality good was
actually {\it higher} than for the high-quality good,  e.g.,  $p_L=3$ and $p_H=
1$, because the good is normally
produced as high quality and needs to be purposely damaged before it becomes low
quality.  The price-discrimination profit in
(\ref{e105}) would then be $\pi_{seller} =  (10 -3)(0.5)  + (44-1)  (0.5)  =
25$. Since that is still higher than 24.5, the seller would still price-
discriminate. The buyers' payoffs would be unaffected. Thus, allowing the seller
to damage some of the good at a cost in real resources of 2 per unit, converting
it from high to low quality,  can result in a pareto improvement!
 

    \newpage

\begin{center}
{\bf   Vertical Differentiation III: Duopoly Quality Choice }
  \end{center}
   {\bf Players}\\
 Two  sellers and a continuum of buyers.

 \noindent
  {\bf The Order of Play }\\
 0  Nature assigns quality values to a continuum of buyers of length 1. Half  of
them  are ``weak'' buyers ($\theta=0$) who value high quality at 20 and low
quality at 10.  Half  of them  are ``strong'' buyers ($\theta=1$) who value high
quality at 50 and low quality at 15.       \\
  1 Sellers 1 and 2 simultaneously choose values for   $s_1$ and $s_2$ from  the
set $\{    s_L=0, s_H=1 \}$. They may both choose the same value.  \\
 2 Sellers 1 and 2 simultaneously choose  prices   $p_1$   and $p_2$ from the
interval $[0,\infty)$.\\
 3 Each buyer    chooses one unit of  a good, or refrains from buying.  The
sellers produce    at  constant marginal cost $c=1$, which does not vary with
quality. \\

\noindent
  {\bf Payoffs}\\
 \begin{equation} \label{e106}
 \pi_{seller} =  (p  -1) q
\end{equation}
 and
\begin{equation} \label{e107}
 \pi_{buyer}  = (10 + 5 \theta)  + (10+ 25 \theta)  s - p.
 \end{equation}


  \newpage

  If both sellers both choose the same quality level, their profits will be
zero, but if they choose different quality levels, profits will be positive.
Thus, there are three possible equilibria in the quality stage of the game:
(Low, High), (High, Low), and a symmetric mixed-strategy equilibrium.   Let us
consider the pure-strategy equilibria first, and  without loss of generality
suppose that Seller 1 is the low-quality seller  and Seller 2 is the high-
quality seller.

\noindent
(1) The equilibrium prices of Vertical Differentiation II, $(p_L=10, p_H=45$),
will no longer be equilibrium prices.  The problem is that the low-quality
seller would deviate to $p_L=9$, doubling his sales for a small reduction in
price.

\noindent
 (2) Indeed, there is no pure-strategy equilibrium in prices. We have seen that
$(p_L=10, p_H=45$) is not an equilibrium, even though $p_H=45$ is the high-
quality seller's best response to $p_L=10$. $P_L >10$ will attract no buyers, so
that cannot be part of an equilibrium.   Suppose $P_L \in (1, 10)$.  The
response of the high-quality seller will be to set $p_H=p_L+35$, in which case
the low-quality seller can increase his profits by slightly reducing $p_L$ and
doubling his sales. The only   price  left  for the low-quality seller that does
not generate negative profits is $p_L=1$, but that yields zero profits, and so
is worse than $p_L=10$.  So no choice of $p_L$ is part of a pure-strategy
equilibrium.

\noindent
(3)  As always, an equilibrium does exist, so it must be in mixed strategies, as
shown below.


  \newpage

 
\noindent
  {\bf The Asymmetric  Equilibrium: Pure Strategies for Quality, Mixed for
Price}


\noindent
  The low-quality  seller picks $p_L$   on the support $[5.5, 10]$ using the
cumulative
distribution
\begin{equation} \label{e108}
  F(p_L) =     1-  \left( \frac{39.5}{  p_L +34 }\right)
\end{equation}
   with an atom of probability $  \frac{39.5}{ 44} $ at $p_L=10$.

\noindent
The  high-quality seller picks $p_H$ on the support $[40.5, 45]$ using the
cumulative
distribution
\begin{equation} \label{e109}
 G(p_H) = 2-  \left(\frac{9}{ p_H-36 }  \right)
\end{equation}

\noindent
Weak buyers    from  the low-quality seller if  $10-p_L \geq 20 - p_H$, which is
always true in equilibrium.  Strong buyers buy from the low-quality seller if
$15-p_L > 50 - p_H$, which has positive probability, and otherwise
from the
high-quality seller.

 

  This equilibrium is noteworthy because it includes a probability atom in the
mixed-strategy distribution, something not uncommon in pricing games.  

  \newpage

 To start deriving this equilibrium,
let us  conjecture  that   the low-quality seller will  not include any prices
above 10 in his mixing support but will include $p_L=10$ itself. That is
plausible  because  he
would lose all the low-quality buyers at prices above 10, but   $p_L=10$ yields
maximal
profits whenever $p_H$ is low enough that only weak consumers buy low quality.

 The low-quality seller's profit from $p_L=10$ is
$\pi_L (p=10) = 0.5 (10-1) = 4.5. $ Thus, the lower bound   of the support
of his mixing distribution (denote it by $a_L$) must also yield a profit of 4.5.
There is no point in charging a price less than the price which would capture
even  the
  strong consumers with probability one,   in which case
 \begin{equation} \label{e110}
 \pi_L (a_L) = 0.5 (a_L-1) + 0.5 (a_L-1)  =4.5,
\end{equation}
 and  $a_L=5.5$.
Thus, the   low-quality seller mixes on $[5.5, 10]$.

On that mixing support, the low-quality seller's profit must equal 4.5 for any
price. Thus,
 \begin{equation} \label{e111}
\begin{array}{ll}
 \pi_L (p_L)   =  4.5& = 0.5 (p_L-1) + 0.5 (p_L-1) Prob (15-p_L> 50 - p_H) \\
 & \\
 &= 0.5 (p_L-1) + 0.5 (p_L-1) Prob  (p_H > 35+ p_L) \\
 & \\
 & =  0.5 (p_L-1) + 0.5 (p_L-1)[ 1-G(  35+ p_L)]
\end{array}
\end{equation}


\newpage


From the previous page: 
On that mixing support, the low-quality seller's profit must equal 4.5 for any
price. Thus,
$$
 \pi_L (p_L)   =    0.5 (p_L-1) + 0.5 (p_L-1)[ 1-G(  35+ p_L)]
$$

 Thus, the $G(p_H)$ function   is such that
 \begin{equation} \label{e112}
   1-G(  35+ p_L) = \frac{4.5}{0.5(p_L-1)} -1
\end{equation}
and
 \begin{equation} \label{e113}
 G(35+ p_L) = 2- \left(\frac{4.5}{0.5(p_L-1)} \right).
\end{equation}
   We want a $G$ function with the argument $p_H$, not $(35+ p_L)$, so let's shift the argument by 35: 
 \begin{equation} \label{e114}
G(p_H)  = 2-\left( \frac{4.5}{0.5([p_H-35] -1)}\right) = 2-\left( \frac{9}{
p_H-36}\right).
\end{equation}
  As explained in Chapter 3, what we have just done is to find the strategy for
the high-quality seller  that makes the low-quality seller indifferent among all
the values of $p_L$ in his mixing support.


 

We can find the support of the high-quality seller's mixing distribution by
finding values $a_H$ and $b_H$ such that $G(a_H) =0$ and $G(b_H) =1$, so
\begin{equation} \label{e115}
 G(a_H) = 2-\left( \frac{9}{ a_H-36 }\right) =0,
\end{equation}
which yields $a_H=40.5$,  and
\begin{equation} \label{e116}
 G(b_H) = 2- \left(\frac{9}{(0) \cdot b_H-36 }\right) =1,
\end{equation}
which yields $b_H = 45$.  Thus the support of the high-quality seller's mixing
distribution is $[40.5, 45]$.



  \newpage


 Now let us find the low-quality seller's mixing distribution, $F(p_L)$. At
$p_H=40.5$, the high-quality seller  has zero probability of losing the strong
buyers to the low-quality seller, so  his profit is $0.5(40.5-1) = 19.75. $ Now
comes  the tricky step. At $p_h=45$, if  the high-quality seller had
probability one  of losing the strong buyers to the low-quality seller, his his
profit would be zero, and he would strictly prefer $p_H=40.5$. Thus, it must be
that at  $p_h=45$  there is strictly positive probability that $p_L=10$--- not
just a positive density.      So let us continue, using our finding that the
profit of the high-quality seller must be 19.75 from any price in the mixing
support. Then,
 \begin{equation} \label{e117}
\begin{array}{ll}
 \pi_H (p_H)   = 19.75& =   0.5 (p_H-1) Prob (15-p_L< 50 - p_H) \\
 & \\
 &=  0.5 (p_H-1) Prob (p_H-35< p_L  ) \\
 & \\
 & =   0.5 (p_H-1) [1-F( p_H-35   )] \\
\end{array}
\end{equation}
so
 \begin{equation} \label{e118}
 F( p_H-35) = 1- \left( \frac{19.75}{0.5(p_H-1)}\right).
\end{equation}
  Using the same substitution trick as in equation (\ref{e114}),  putting
$p_L$  instead of   $(p_H-35)$ as  the  argument for $F$, we get
\begin{equation} \label{e119}
\begin{array}{ll}
   F(p_L) &=   1- \left(\frac{19.75}{0.5(p_L+35-1)}\right)
  = 1- \left(\frac{39.5}{ p_L+34 } \right)\\
\end{array}
\end{equation}
   In particular, note that
 \begin{equation} \label{e120}
 F(5.5) =  1- \left(\frac{39.5}{ 5.5+34 }\right) =0,
\end{equation}
 confirming our earlier finding that  the minimum $p_L$ used is 5.5, and
\begin{equation} \label{e121}
 F(10) =  1- \left(\frac{39.5}{ 10+34 }\right) = 1-\frac{39.5}{44} <1.
\end{equation}

\newpage

$$
 F(10) =  1- \left(\frac{39.5}{ 10+34 }\right) = 1-\frac{39.5}{44} <1.
$$



   Equation (\ref{e121}) shows that at the upper bound of the low-quality
seller's mixing support  the
cumulative mixing  distribution   does not equal 1, an oddity we
usually do not see in mixing distributions. What it implies is that there is an
atom of probability at  $p_L=10$,  soaking up
all the remaining probability beyond what equation (\ref{e121}) yields for the
prices below 10. The atom must equal     $\frac{39.5}{44} \approx 0.9$.

  Happily, this solves our paradox of zero high-quality seller profit at $p_H=
45$. If $p_L=10$ has probability $\frac{39.5}{44} $, the profit from $p_H=45$ is
$0.5 (\frac{39.5}{44} )(45-1) = 19.75.  $   Thus,
the profit from $p_H=45$ is the same as from $p_H=40.5$, and the seller is
willing to mix between them.

  \newpage

  

 The  duopoly sellers' profits are  4.5 (for low-quality) and 19.75 (for high
quality) in the asymmetric equilibrium of
Vertical Differentiation III,  a total of 24.25 for the industry.   This is less
than either the 24.5 earned by the nondiscriminating  monopolist of Vertical
Differentiation I or  the 26 earned by the discriminating monopolist of Vertical
Differentiation II.  But what about the  mixed-strategy equilibrium for Vertical
Differentiation III?


  \newpage


 
\noindent
  {\bf The Symmetric  Equilibrium: Mixed Strategies for Both  Quality and Price}

Each player chooses low quality with probability $\alpha = 4.5/24.25$ and high
quality otherwise. If they choose the same quality, they next  both choose a
price equal to 1, marginal cost. If they choose different qualities, they choose
prices according to the mixing distributions in the asymmetric equilibrium.

    This equilibrium is easier to explain. Working back from the end, if they
choose the same qualities, the two firms are in undifferentiated  price
competition and will choose prices equal to marginal cost, with  payoffs of
zero. If they choose different qualities, they are in the same situation as they
would be in the asymmetric equilibrium, with expected  payoffs of 4.5 for the
low-quality firm and 19.75 for the high-quality firm. As for choice of product
quality, the expected payoffs from each quality must be equal in equilibrium, so
there must be a higher probability of both choosing high-quality:
\begin{equation} \label{e123}
 \pi(Low)  =   \alpha (0)    + (1-\alpha) 4.5   = \pi(High) =  \alpha (19.75)
+ (1-\alpha)  (0).
\end{equation}
Solving equation (\ref{e123}) yields $\alpha = 4.5/24.25 \approx 0.17$, in which
case each player's payoff is  about 3.75.
Thus, even if a player is stuck in the role of low-quality seller in the pure-
strategy equilibrium,  with an expected payoff of 4.5, that is better than the
expected  payoff he
would get in the ``fairer'' symmetric equilibrium.


  \newpage

We can conclude that if the players could somehow arrange what equilibrium would
be played out, they would arrange for a pure-strategy equilibrium, perhaps by
use of cheap talk and some random focal point variable.


 Or, perhaps they could change the rules of the game so that they would choose
qualities sequentially.   Suppose one  seller  gets to choose
quality first. He would of course choose high quality, for a payoff of 19.75.
The second-mover, hwoever,
choosing low-quality, would  have  a payoff of 4.5,   better than the expected
payoff
in the symmetric mixed-strategy equilibrium of the simultaneous quality-choice
game. This is the same phenomenon as the pareto superiority of a sequential
version of the Battle of the Sexes over the symmetric mixed-strategy equilibrium
of the simultaneous-move game.

What if Seller 1 chooses both quality and price first, and   Seller 2 responds
with   quality and price? If Seller 1 chooses low quality,  then his optimal
price is $p_L=10$, since the second player will choose high quality and  a price
low enough to attract the strong buyers--- $p_H=45$, in equilibrium--- so Seller
1's payoff would be 0.5(10-1) = 4.5. If Seller 1 chooses high quality,  then his
optimal price is $p_H=40.5$, since the second player will choose low  quality
and would choose a price high enough to lure away the strong buyers if
$p_H<40.5$. If, however, $p_H=40.5$, Seller 2 would give up on attracting the
strong buyers and pick $p_L=10$. Thus, if Seller 1 chooses both quality and
price first, he will choose high quality and $p_H=40.5$ while Seller 2 will
choose low quality and $p_L=10$, resulting in the same payoffs  as in the
asymmetric equilibrium of the simultaneous-move game, though no longer in mixed
strategies.

\newpage


  What Product Differentiation III shows us is that product differentiation can
take place  in oligopoly vertically as well as horizontally.

 Head-to-head
competition reduces profits, so firms will try to differentiate in any way that
they can.

 This increases  their profits, but it can also benefit consumers---
though  more obviously in the case of  horizontal differentiation than in
vertical.

 Keep in mind, though, that in our games here we have assumed that high
quality costs no more than low quality.

 Usually high quality is more expensive,
which means that having more than one quality level can be efficient.

 Often poor
people prefer  lower quality, given the cost of higher quality, and even a
social planner would provide a variety of quality levels. 

Here, we see that even
when only high quality would be provided in the first-best, it is better that a
monopolist provide two qualities than one, and a duopoly is still better for
consumers.



 

 
\end{Large}

  \end{document}










