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\begin{center}
{\large {\bf Rethinking Bargaining Under Complete Information with
Outside Options }
}
7 August 2008 . Pdf'd \today.
\bigskip
David P. Myatt and Eric B. Rasmusen
{\it Abstract}
\end{center}
\begin{small}
\noindent
Myatt: Nuffield College, Room L. University of Oxford, New Road,
Oxford, England, OX1 1NF. 011-44-1865 278-578 or (01865) 278-578.
David.myatt@economics.ox.ac.uk.
\noindent
Rasmusen: Dan R. and Catherine M. Dalton Professor, Department of
Business Economics and Public Policy, Kelley School of Business,
Indiana
University, BU 438, 1309 East Tenth Street, Bloomington, Indiana,
USA 47405, (812) 855-9219 (secretary), (812) 855-3356 (direct),
Erasmuse@indiana.edu,
\url{http://www.rasmusen.org},
\url{http://www.rasmusen.org/papers/bargaining.pdf}.
{\small We would like to thank xxx for helpful comments and
Nuffield College for its hospitality. }
\end{small}
\newpage
\noindent
{\bf 1. Introduction}
The most basic problem in bargaining is what happens when two
players with perfect information about each other must both agree on
how to split a surplus or else the surplus vanishes. The ideal model
of this situation would:
1. Give plausible results (e.g., a 50-50 split between symmetric
players)
2. Be simple.
3. Allow us to parameterize bargaining strength.
For many years, the standard model was from Nash (1950). The Nash
Bargaining Solution was an idea from cooperative game theory that used
plausible axioms to construct a reduced-form solution. It gives us
the 50-50 bargaining split result and is simple, but it assumes equal
bargaining strengths. It can be seen, however, as a justification of
the even simpler Split the Difference reduced-form model: assign
Player 1 a bargaining strength of $\lambda$ and Player 2 of
$(1-\lambda)$ and assume that those are their equilibrium shares of
the surplus.
Both of these are reduced-form models, which can make it unclear
how to fit them into larger models. Also, they both evade the
question of what "the surplus" means, something which we will discuss
more below. One simple way to build a structural model is to let each
player have probability $\lambda$ of making a take-it-or-leave-it
offer. Let us call this the Probabilistic Ultimatum Game. If the
players have equal bargaining strength, this results in an expected
payoff for each of them of half the surplus. Otherwise, we can use
$\lambda$ to parameterize bargaining strength. The Probabilistic
Ultimatum Game is a structural model, but its assumptions about the
bargaining process are unrealistic, so it has not been popular.
The model most used today is that of Rubinstein
(1982). In this model, the two players alternate making offers in
discrete periods, and the game ends when one player accepts the
other's offer. Delay is costly, and in equilibrium the very first
offer is accepted. If the two players have equal discount rates, the
equilibrium split is close to 50-50, but otherwise the more patient
player has an advantage, providing a way to parameterize bargaining
strength.
The Rubinstein Model does have drawbacks. First, its feature of
alternating offers leads to the complication that it matters who gets
to offer first and that the equilibrium strategies depend on who is
making the offer in a given period. Second, the equilibrium split is
not quite 50-50 with symmetric players. It approaches 50-50 as
discount rates go to zero, but one must always add that
qualification-- and if discount rates do equal zero, the model breaks
down. Third, it does not resolve the question of what "surplus" means
any better than the Nash solultion. Fourth, it is troubling that
bargaining strength is parameterized by discounting when in most
bargaining situations the delay between offer and counter-offer is
trivial.
As Sutton (1986) explains, the availability of the outside option in
case exogenous breakdown occurs is crucial. There are two ways to run
Rubinstein (1982): time discounting, and breakdown. It is crucial
whether breakdown removes the possibility of the outside option.
The Rubinstein model has some good properties, but it doesn't have
the right feel. It is really a model of alternating monopolies. The
new model below will have that same problem.
\bigskip
\noindent
{\bf The Rubinstein Model}
\bigskip
Ann and Bob are splitting a pie of size 1. First Ann makes an offer
of a split of $x_a$ for herself and $(1-x_a)$ for Bob. If Bob accepts,
the game is over and the payoffs are
\begin{equation} \label{a1}
\pi_a= x_a \hspace{1in} \pi_b= 1- x_a.
\end{equation}
If Bob rejects, a period of time passes, at the end of which he
makes an offer of a split of $(1-x_b)$ for himself and $ x_b $ for
Ann. If Ann accepts, the game is over and the payoffs are (viewed from
the start of the game) \begin{equation} \label{a2} \pi_a= \delta_a
x_b \hspace{1in} \pi_b= \delta_b (1- x_b)
\end{equation}
because second-period payoffs are discounted by discount factors of
$\delta_a = \frac{1}{1+\rho_a}<1$ and $\delta_b = \frac{1}{1+\rho_b}
<1$.
Our main interest is in what happens when the periods are short, so
the
discount rates approach $\rho_a= \rho_b=0$ and the discount factors
approach $\delta_a = \delta_b = 1$.
There are two possible origins for discounting here: time
preference, and the possibility of the game ending exogenously. We
will
discuss both. Use time discounting for now.
If Ann rejects Bob's offer, another period of time elapses and she
makes the next offer. The two players make alternating offers until
agreement is reached, or forever if agreement is not
reached.\footnote{Unlike in some games, it makes little difference
whether the game is infinitely repeated or just repeated a large
finite
number of times.}
In any subgame perfect equilibrium, Ann's first offer will be
accepted immediately. The only reason to wait is that Ann can get a
bigger share $x_a'$ later. She would prefer to offer $x_a' +
\epsilon$ now.
It follows that the equilibrium is stationary. The players' payoffs
cannot get bigger or smaller, or they would make more generous offers
now that would equal any bigger payoffs without having to pay the
bargaining costs.
Denote Ann's equilibrium offer by $x_a^*$ and Bob's by $x_b^*$. In
making her offer, Ann realizes that Bob could reject it and, if both
players follow their equilibrium strategies, offer $x_b^*$ next
period and have it accepted. Thus, rejection gives Bob a payoff of
$\delta_b (1- x_b^*) $, as in equation \eqref{a2}. Ann can make Bob
willing to accept her offer by making it generous enough to give him
the same payoff, i.e.
\begin{equation} \label{a3}
1-x_a^* = \delta_b (1-x_b^*)
\end{equation}
In making his offer, Bob realizes that Ann could reject it and, if
both players follow their equilibrium strategies, offer $x_a^*$ next
period and have it accepted. Thus, rejection gives Ann a payoff of
$\delta_a x_a^* $. Bob can make Ann willing to accept his offer
by
making it generous enough to give her the same payoff, i.e.
\begin{equation} \label{a4}
x_b^* = \delta_a x_a^* .
\end{equation}
These two equations solve to
\begin{equation} \label{a5}
x_a^* = \frac{ 1 - \delta_b }{ 1- \delta_a \delta_b }
\end{equation}
and
\begin{equation} \label{a6}
x_b^* = \frac{ \delta_a (1- \delta_b) }{ 1- \delta_a
\delta_b }
\end{equation}
This solution is asymmetric because it gives the first offeror, Ann,
a bigger share, even if the number of possible offers is very large.
If discounting is small, though, it is not very asymmetric.
Also, it allows us to use the discount factors to parametrize
bargaining strength. We do think impatient people have less bargaining
strength, though that is not the only source.
As Rubinstein found, we cannot let $\delta_a= \delta_b = 1$ or the
solution technique fails to work.
\bigskip
\noindent
{\bf No Time Preference, but an Exogenous Risk of Breakdown }
Another version of the model, with properties almost identical to
those of the standard Rubinstein model, has an exogenous risk of
breakdown replacing time discounting. Let there be no time
preference, but suppose that with probability $1-\delta$ the game ends
suddenly and both players receive payoffs of 0.
Ann and Bob are splitting a pie of size 1. First Ann makes an offer
of a split of $x_a$ for herself and $(1-x_a)$ for Bob. If Bob accepts,
the game is over and the payoffs are
\begin{equation} \label{b1}
\pi_a= x_a \hspace{1in} \pi_b= 1- x_a.
\end{equation}
If Bob rejects, then
a period of time passes, at the end of which either
Bob makes an offer of a split of $x_b$ for Ann and $(1-x_b)$ for
Ann, which has probability $\delta$; or the game ends, which has
probability $(1-\delta)$. If Bob offers and Ann accepts, the game
is over.
The payoffs are (viewed from the start of
the game)
\begin{equation} \label{b2}
\pi_a= \delta x_b \hspace{1in} \pi_b= \delta (1-
x_b).
\end{equation}
If Ann rejects Bob's offer, another period of time elapses and she
makes the next offer. The two players make alternating offers until
agreement is reached, or forever if agreement is not reached.
In any subgame perfect equilibrium, Ann's first offer will be
accepted immediately. The only reason to wait is that Ann can get a
bigger share $x_b'$ later. She would prefer to offer $x_b' +
\epsilon$ now.
It follows that the equilibrium is stationary. The players' payoffs
cannot get bigger or smaller, or they would make more generous offers
now that would equal any bigger payoffs without having to pay the
bargaining costs.
Denote Ann's equilibrium offer by $x_a^*$ and Bob's by $x_b^*$. In
making her offer, Ann realizes that Bob could reject it and, if both
players follow their equilibrium strategies, offer $x_b^*$ next
period and have it accepted. Thus, rejection gives Bob a payoff of
$\delta (1- x_b^*) $, as in equation \eqref{b2}. Ann can make Bob
willing to accept her offer by making it generous enough to give him
the same payoff, i.e.
\begin{equation} \label{b3}
1-x_a^* = \delta (1-x_b^*)
\end{equation}
In making his offer, Bob realizes that Ann could reject it and, if
both players follow their equilibrium strategies, offer $x_a^*$ next
period and have it accepted. Thus, rejection gives Ann a payoff of
$\delta x_a^* $. Bob can make Ann willing to accept his offer by
making it generous enough to give her the same payoff, i.e.
\begin{equation} \label{b4}
x_b^* = \delta x_a^* .
\end{equation}
These two equations solve to
\begin{equation} \label{b5}
x_a^* = \frac{ 1 - \delta }{ 1- \delta^2 }
\end{equation}
and
\begin{equation} \label{b6}
x_b^* = \frac{ \delta (1- \delta ) }{ 1- \delta^2 }
\end{equation}
This solution is asymmetric because it gives the first offeror, Ann,
a
bigger share, even if the number of possible offers is very large. If
discounting is small, though, it is not very asymmetric.
\noindent
{\it Definition: The {\bf Rubinstein price} is the price generated by
equation \eqref{b5} in a bargaining game.}
If the surplus size is one, and the buyer offers first, the
Rubinstein
price is
\begin{equation} \label{b6a}
p = 1 - \frac{ 1 - \delta }{ 1- \delta^2 }
= \frac{1- \delta^2 - 1 + \delta }{ 1- \delta^2 } = \frac{
\delta - \delta^2 }{ 1- \delta^2 }.
\end{equation}
We cannot let $\delta = 1$ or the
solution technique fails to work.
This is even simpler than the time preference model, but now we have
lost any way to parameterize bargaining strength.
\bigskip
\noindent
{\bf Outside Options: Binmore-Rubinstein-Wolinsky (1986) }
Now let us add outside options of $w_a$ and $w_b$. These are
alternatives that the players have if they fail to reach agreement.
See the Sutton (1986) paper for the source of the outside option
idea.
What others have shown is that the outside options matter only if the
constraints they provide are binding.
This is puzzling. Suppose Bob has no outside option. The outside
option principle says that the bargaining split will be about .5 even
if Ann has an outside option of about .5. We would think that if Ann
could get about .5 by ceasing to bargain, she could get more by
bargaining. The Split the Difference principle would give her .75
instead. Why doesn't that happen here?
The reason is that the outside option is a threat, and the threat is
not credible. Both players expect agreement on about .5 will be
reached with certainty. Hence, Ann saying that she could withdraw and
not be hurt as much as Bob would be is irrelevant. It is not that Ann
can withdraw and do better than the Rubinstein solution; what is true
is merely that she could withdraw and not do so much worse as Bob
would do. Instead of getting about 50, she would end up with 20; but
Bob would end up with 0. That, however, is irrelevant to the
bargaining.
If this outcome is implausible, it is because something is missing
from the model. And it is indeed implausible, as the following
example will show.
\bigskip
\noindent
{\bf Competition Does Not Reduce Prices}
Let's consider an example: bargaining between a buyer and a seller.
We will start with bilateral monopoly, and then apply the model to one
buyer facing two sellers.
The buyer values the object being sold at $v>c$ and each seller has
a marginal cost of $c \geq 0$. The discount rate is $r>0$. Bargaining
with a seller costs the buyer $k \geq 0 $, so if he deals with one
seller his cost is $b$ but if he deals with two his cost is $2k$.
There is an infinite number of periods. The buyer might have an
outside option with payoff $w$ to him.
We will use a running example in which $v=100, c=0, r = .01$ (so
$\delta \approx .99$), and $k=0$.
The order of play is that the buyer chooses a seller and makes him a
price offer, which the celler accepts or rejects, all happening at the
start of the period. If it is rejected, then immediately the buyer
may switch sellers and make an offer to the second seller, or if the
buyer does not switch, after one unit of time the first seller can
make an offer to the buyer that the buyer accepts or rejects. This
continues, with either alternating offers within the same bargaining
pair or switching and a new start to bargaining with a new
bargaining pair.
\bigskip
\noindent
{\bf One Seller}
If there were just one seller, this would be the Rubinstein (1982)
model. The bargaining range is prices in the interval $[0, 100]$. The
buyer payoff range is in the range of $[0, 100]$. The equilibrium
buyer surplus from \eqref{a5} would be about $\pi_{buyer}=
\left(\frac{ 1 - \delta }{ 1- \delta^2 } \right) 100 =
\left(\frac{ 1 - .99 }{ 1- .9801 } \right) 100 \approx
50.03 $, and the equilibrium price would be the Rubinstein price,
$p \approx 49.97$.
Already, we can see two defects of the Rubinstein model. The
equilibrium price is not 50, and for a simple discount rate such as
$r=.01$ the equilibrium price cannot be represented with just two
decimals. We could instead say that the equilibrium price approaches
50 as the discount rates approaches zero, but it would be nice to be
able to avoid the verbiage of such qualifications and just say $p^*=
50$. This would be especially true if our bargaining model were
embedded in a more complex model where every step after the bargaining
process would have to include these qualifications, and the reader
would have to check whether the qualifications arise from the
bargaining submodel used or from some other assumption of the complex
model.
\bigskip
\noindent
{\bf Monopoly with an Outside Price of 80}
With equal bargaining power, discounting near zero, and a fixed
outside
option of $p=80$ from some outside seller of a substitute product
(so $w=20$), things do not
change. The equilibrium price remains the Rubinstein price of $p
\approx 49.97$.
This outcome is disconcerting because the social surplus from
bargaining would seem to be not 100, but 80, in which case the seller
is getting 3/5 of it, not 1/2, from the Rubinstein price.
\bigskip
\noindent
{\bf Two Sellers, No Switching Cost}
Now add a second seller to the model. We have two identical
sellers competing to sell to one buyer, so we would expect an
equilibrium price of $p = c =0$ and an equilibrium buyer payoff of
$\pi_{buyer} =100$. But that is not the only equilibrium.
The Rubinstein (1982) bargaining model has now become a Binmore-
Rubinstein-Wolinsky (1986) model with an outside option.
The impasse payoff, the payoff the buyer gets if bargaining breaks
down and he takes no other option, is about 0. The outside option
payoff, the payoff the buyer gets if he stops bargaining with his
current partner, is his payoff from bargaining with the other player,
which we will denote by $\pi_{buyer}^2$.
The outside option payoff $\pi_{buyer}^2$ is the buyer's payoff
from bargaining with Seller 2. The value of $\pi_{buyer}^2$ depends on
the buyer's outside option in that bargaining game, which is the
buyer's payoff, $\pi_{buyer}^1$, from quitting and bargaining with
Seller 1 instead. Thus, we have a simultaneous system to solve: we
need to find values of $\pi_{buyer}^1$ and $\pi_{buyer}^2$ that are
consistent with each other.
Consider the values $(p =0, \pi_{buyer}^1 = \pi_{buyer}^2 =100)$.
This is an equilibrium outcome. In it, the buyer starts by making an
offer of $p=0$ to Seller 1. Seller 1 knows that if he rejects it, the
buyer will make an offer of $p = 0$ to Seller 2 and Seller 2's
equilibrium strategy is to accept that offer. Thus, the buyer's
outside option has a payoff of 100, which is better than the 50.03 of
the Rubinstein solution. Knowing this, Seller 1 is willing to
accept $p =0$, and the buyer's payoff is $ 100$.
But consider instead $p = 37 $. This, too, is an equilibrium
outcome. The buyer makes an offer of $p =37$ to Seller 1. Seller 1
knows that if he rejects it, the buyer will make an offer of $p =37$
to Seller 2 and Seller 2's equilibrium strategy is to accept that
offer, so Seller 1 will end up with a payoff of 0. The buyer's
outside option has a payoff of 63, which is better than the 50.03 of
the Rubinstein solution. Thus, Seller 1 will accept $p =37$.
What if the buyer deviates by offering a lower price, $p =35$?
Seller 1 believes that if he rejects this offer as too low, the buyer
will not have a positive incentive to switch (since Seller 2 would
require a price of 37). Instead, he, Seller 1, will get to make an
offer of $ 37$ in the second round, and the buyer will accept it. So
the buyer cannot gain by deviating from $p =37$.
On the other hand, consider the price $p= 96$.
This is not an equilibrium outcome. Suppose the buyer deviates to $p=
49.97$. The seller need not fear that the buyer will go to the other
seller if he rejected $p=49.97$, because the other seller will hold
out for $p=96$ in our postulated equilibrium. The first seller does,
however, know that with the other seller irrelevant, he is in a simple
Rubinstein bargaining game with the buyer, and the equilibrium has $p
= 49.97. $
In fact, any price in $[0, 49.97]$ and any buyer payoff in $
[50.03,100]$ is an equilibrium. It all depends on expectations. The
Rubinstein model does better than the simultaneous offer game in
narrowing down the number of equilibria-- it has ruled out prices in
$[49.97, 100]$-- but it still has a continuum of them, almost all of
which yield positive profit to a seller.
At least the Bertrand equilibrium is one equilibrium, if not the
unique one. But it turns out not to be robust to a small switching
cost. Only the Rubinstein equilibrium will survive in that case.
\bigskip
\noindent
{\bf Two Sellers, Positive Switching Cost}
Now let us restore the switching cost, so $k=1$. This has a drastic
effect: now the only equilibria are near $p= 49.97$.
Consider the values $ p = 37 $ that we looked at before. The buyer
has made an offer of $p = 37$. Suppose Seller 1 rejects this and
counterproposes $p = 37.5$, which if accepted yields him a payoff in
period-1 dollars of about 37.1. The buyer's payoff from accepting is
about $.99 (62.5) \approx 61.9$. The buyer's payoff from rejecting and
bargaining with Seller 2 is about $.99(63-1) \approx 61.4$, since the
buyer would have to incur the switching cost. Thus, the buyer will
accept Seller 1's offer and the seller's deviation has been
profitable.
Let's make the argument more generally. Let $p^*$ be the equilibrium
price and let $\hat{p}$ denote the deviation offer by the seller if he
rejects the buyer's offer in the first round. The seller chooses
$\hat{p}$ to maximize it subject to two constraints: (1) $\hat{p}$
must not exceed the Rubinstein price of $\left(\frac{ \delta -
\delta^2
}{
1- \delta^2 } \right) 100 $ and (2) the buyer must
be willing to accept in the second round rather than wait and
switch
to the second seller in the third round. If the second constraint is
binding,
\begin{equation} \label{b7}
\begin{array}{lll}
\pi_{buyer} (accept \; \hat{p}) & = & \pi_{buyer}
(switch, \;and \; get \;p^*) \\
& & \\
\delta (100-\hat{p}) & = & \delta^2 (100-p^*-k) \\
& & \\
\hat{p}& = & (1-\delta) 100+ \delta p^* + \delta k \\
\end{array}
\end{equation}
For the deviation to be profitable, the seller must strictly benefit
from his deviation to $\hat{p}$. This will be true unless
\begin{equation} \label{b8}
\begin{array}{lll}
\pi_{seller} (equilibrium, p^*) & \geq & \pi_{seller}
(deviation, \hat{p}) \\
& & \\
p^* &\geq & \delta \hat{p} = \delta [Min\{\left(\frac{ \delta -
\delta }{
1- \delta^2 } \right), 100 (1-\delta) 100+ \delta p^* +
\delta k \}] \\
\end{array}
\end{equation}
If it is constraint (2) that is binding,
\begin{equation} \label{b8a}
\begin{array}{lll}
p^* (1-\delta^2) &\geq & \delta (1-\delta) 100+ \delta^2 k]
\\
& & \\
p^* &\geq & \frac{ \delta (1-\delta) 100 }{ 1-\delta^2 } +
\frac{\delta^2 k }{ 1-\delta^2 }.
\\
\end{array}
\end{equation}
We know from before that $p^*$ cannot exceed the Rubinstein solution,
so if it is constraint (2) that is binding,
\begin{equation} \label{b9}
\begin{array}{lll}
p^* \in [ \frac{ \delta (1-\delta) 100 }{ 1-\delta^2 } +
\frac{\delta^2 k }{ 1-\delta^2 } , \left(\frac{ \delta - \delta^2
}{
1- \delta^2 } \right) 100]& \approx & [ \frac{ .99 (1-.99) 100
}{
1-.99^2 } +
\frac{.99^2 (1) }{ 1-\delta^2 } , 49.97]\\
& & \\
& \approx & [ \frac{ .99}{
.0199 } +
\frac{.9801 }{ .0199 } , 49.97]\\
& & \\
& \approx& [49.7+ 49.3, 49.97 ]\\
\end{array}
\end{equation}
With the parameter values from our running example, it is constraint
(1)
that is binding, so
\begin{equation} \label{b10}
\begin{array}{lll}
p^* \in [ \delta \left(\frac{ \delta - \delta^2 }{
1- \delta^2 } \right) 100 , \left(\frac{ \delta - \delta^2
}{
1- \delta^2 } \right) 100]& \approx & [ .99 (49.97) , 49.97]\\
& & \\
& \approx & [ 49.47 , 49.97]\\
\end{array}
\end{equation}
Thus, only $p^*$ in that interval can be supported in equilibrium.
As $\delta$ gets closer to 1 (i.e. as discounting falls), the
interval of equilibrium prices shrinks
towards the Rubinstein price, $\tilde{p}$. If constraint (1) is
binding that is because the interval is $[\delta \tilde{p}, \tilde{p}]
$. If constraint (2) is binding that is because using L'Hospital's
Rule the lower bound approaches
\begin{equation} \label{b10}
\begin{array}{l}
\frac{ \frac{d}{d \delta} [\delta (1-\delta) 100 + \delta^2 k] }
{\frac{d}{d \delta} [1-\delta^2] } \\
\\
= \frac{ 100- 200 \delta + 2\delta k] }{ -2\delta } =
100- k \\
\end{array}
\end{equation}
xxx SOMETHING IS WRONG IN THE CONSTRAINT TWO MAIN FORMULA
Thus, imposing a small transactions cost eliminates the sensible
equilibrium in which price is competed down to marginal cost and
leaves an equilibrium in which the presence of a competing seller is
completely irrelevant.
\bigskip
\noindent
{\bf A New Model}
As we said earlier, it is crucial whether or not the players receive
their outside options if bargaining breaks down. If they do, then the
model does not behave at all like the model with time preference.
Rather, it concludes with the split-the-difference rule. It would
still
have the other disadvantages of (1) An advantage to the first offeror,
(2) No way to parametrize bargaining strength. So let us abandon the
alternating-offer framework and use the probabilistic-offer framework
of
Baron et al. That will solve the two problems and also be much less
confusing to work with, since we won't have to keep track of who
offers
in which period.
Let $x $ be Ann's share of the pie, so $(1-x)$ is
Ben's. There is an infinite number of periods. In each period,
Ann has probability $\alpha$ of making the offer and Ben has
probability $(1-\alpha)$. The non-offering player either accepts or
rejects. If a player accepts an offer, the game is over. If he
rejects,
the game continues to the next period with probability $\delta$ but
ends
without agreement with probability $(1-\delta)$. If it ends without
agreement, then next period Ann gets her outside option of $w_a$ and
Ben
gets his outside option of $w_b$. In any period, a player may take his
outside option instead of making or receiving an offer.
We will denote the equilibrium Ann's-share offer that Ann would
make
in period $t$ by $x_t^a$ and Ben's by $x_t^b$.
We don't need to let the periods get small to get even results, and
the offers are the same in each period rather then differing in even
and
odd periods, unlike in alternating-offer bargaining.
\bigskip
\noindent
{\bf Equilibrium}
We will just work out Ann's equilibrium offer, not Bob's, since in
equilibrium her first offer is accepted.
Think about Ann's payoff in period $t$. With probability $\alpha$,
she makes the offer and will offer $x_t^1$, which she will choose so
it is accepted by Ben. With probability $(1-\alpha)$, Ben makes the
offer, which he will make so she will just accept it. Thus,
\begin{equation} \label{e1}
\pi^a_{t} = \alpha x_t^a + (1-\alpha) x_t^b
\end{equation}
Ann will choose $x_t^a$ to make Ben indifferent about accepting. The
value of waiting for him is that with probability $(1-\delta)$ the
game
will end and he will get the outside option $w_b$ next period and with
probability $\delta$ he will get $\pi_{t+1}^b$. Ann must subtract
that
from the size of the pie (=1) to get what she retains for herself in
her
offer. Thus,
\begin{equation} \label{e2}
x_t^a = 1- [\delta \pi_{t+1}^b + (1-\delta)w_b]
\end{equation}
Of course, it may be that $w_b$ is so large that equation \eqref{e2}
cannot be solved because even by giving Bob the entire pie Ann cannot
make him willing to accept her offer, or it may be that her offer to
him must be so high that she herself would rather take the outside
offer immediately.
Similarly,
\begin{equation} \label{e3}
x_t^b = \delta\pi_{t+1}^a + (1-\delta)w_a
\end{equation}
Substituting these in to \eqref{e1} gives
\begin{equation} \label{e4}
\begin{array}{ll}
\pi^a_{t} & = \alpha [ 1- [\delta \pi_{t+1}^b + (1-\delta)
w_b]] + (1-\alpha) [\delta\pi_{t+1}^a + (1-\delta)w_a ] \\
\end{array}
\end{equation}
This is a stationary problem, so $\pi_{t+1}^a= \pi^a_{t}$. There
will
be immediate agreement in equilibrium, so the payoffs of the two
players add up to 1 and $\pi^b_{t} = 1- \pi^a_{t} $.
Thus, we can drop the time subscripts, and
\begin{equation} \label{e5}
\pi^a = \alpha [ 1- [\delta (1- \pi^a) + (1-\delta)w_b]]
+ (1-\alpha) [\delta\pi^a + (1-\delta)w_a ]
\end{equation}
We want to solve equation \eqref{e5} for $\pi^a$.
\begin{equation} \label{e6}
\begin{array}{lll}
\pi^a & = & \alpha - \alpha \delta + \alpha \delta
\pi^a - \alpha (1-\delta)w_b] + (1-\alpha) \delta\pi^a+ (1-\alpha)
(1-\delta)w_a\\
& & \\
( 1 - \alpha \delta - (1-\alpha) \delta) \pi^a & = & \alpha
- \alpha \delta - \alpha (1-\delta)w_b] + (1-\alpha) (1-\delta)
w_a\\
& & \\
\pi^a & = & \frac{\alpha - \alpha \delta - \alpha
(1-\delta)w_b] + (1-\alpha) (1-\delta)w_a }{ 1 - \alpha \delta -
(1-\alpha) \delta} \\
& & \\
\pi^a & = & \frac{(1- \delta)[ \alpha (1- w_b) +
(1-\alpha) w_a] }{ 1 - \delta} \\
& & \\
\pi^a & = & \alpha + (1-\alpha) w_a - \alpha w_b \\
\end{array}
\end{equation}
We know that $x_a = \pi^a $.
If the bargaining strengths are equal, so $\alpha=.5$, equation
\eqref{e6} boils down to
\begin{equation} \label{e9}
\begin{array}{lll}
\pi^a & = & .5 + \frac{w_a-w_b}{2}. \\
\end{array}
\end{equation}
If the bargaining strengths differ, but there are no outside options,
so $w_a=w_b=0 $, equation \eqref{e6} boils down to
\begin{equation} \label{e9}
\begin{array}{lll}
\pi^a & = & \alpha. \\
\end{array}
\end{equation}
Thus, we have a model which is simpler, more plausible, and allows
parameterization of bargaining power.
\bigskip
\noindent
{\bf The Pricing Game}
What if we use the new bargaining model for our pricing game?
In monopoly, the outcome becomes simpler. It is exactly $p=
50$.
\bigskip
\noindent
{\bf Monopoly with an Outside Price of 80}
With equal bargaining power, discounting near zero, and a fixed
outside
option of $p=80$ from some outside seller of a substitute product
(so $w=20$), things do
change. The equilibrium price changes to, from \eqref{e9},
\begin{equation} \label{e9a}
\begin{array}{lll}
\pi_{buyer}& = & (.5)(100) + \frac{w }{2} = 50+10=60, \\
\end{array}
\end{equation}
so $p^*=40. $
\bigskip
\noindent
{\bf Duopoly}
With duopoly, things change more. When there is no switching cost,
so $k=0$, the only equilibrium will be (explain more a nd prove xxxx):
\begin{equation} \label{e12}
\begin{array}{lll}
\pi_{buyer} & = & .5(100) + \frac{ w}{2}\\
& & \\
& = & 50 + \frac{w}{2}\\
& & \\
& = 100.
\end{array}
\end{equation}
The buyer gets the entire surplus and the price equals marginal cost.
Having a positive cost of switching makes only a tiny change in this
result. The change is that now the seller would refuse $p=0$ and
could make a counteroffer of $p \leq k$ which the buyer would accept
rather than switch. If $k$ is small, the change is small too.
\bigskip
\newpage
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\end{document}