Bargaining
Cake sharing or finite horizon bargaining as an extensive form
game.
Two players are trying to agree on a division of a unit size cake.
Time proceeds in discrete periods until T when the world ends.
In the first period (between 0 and 1) player 1 makes a proposal
(x1 , 1 − x1 ) where the first co-ordinate is the share intended to
player 1.
If player 2 accepts then the division is implemented.
If s/he rejects then it is his/her turn to propose a division
(y2 , 1 − y2 ) where the first co-ordinate is the share intended to
player 1.
Player 1 either accepts or rejects, and in the latter case the
game proceeds as in the first period.
If there is no agreement before the last period (period T )
ends both players get zero.
The players have linear utilities over the cake.
Bargaining
If agreement is reached in period t the pay-offs are discounted
by δ1t−1 and δ2t−1 .
Player 1 makes proposals in odd periods (periods are named by
the time index of the end of period).
There is a continuum of Nash-equilibria in this game
(construct some).
Let us assume that T is odd and solve the SPE of the game.
To do this we make the assumption that when a responder is
indifferent between accepting and rejecting s/he accepts.
In period T player 1 makes an offer (xT , 1 − xT ) (notice that
we have assumed that the offers are efficient; if any offers are
allowed one has to prove this and it is easy for SPE but more
difficult for Nash-equilibria).
Player 2 accepts whenever s/he gets more than zero, and thus,
in the SPE, xT = 1.
Bargaining
Knowing this player 2 in period T − 1 understands that if s/he
offers anything less than what player 1 can achieve by rejecting
his/her offer player 2 will end up with zero.
By rejecting player 1 can guarantee xT = 1 but that is only
next period.
Evaluating things in period T − 1this is of worth δ1 xT = δ1 .
Thus, player 2 proposes (xT −1 , 1 − xT −1 ) = (δ1 , 1 − δ1 ).
Knowing this, player 1 in period T − 2 understands that if s/he
offers anything less than what player 2 can achieve by rejecting
his/her offer player 1 will end up with δ1 .
By rejecting player 2 can guarantee 1 − δ1 but that is only next
period.
Evaluating things in period T − 2 this is worth δ2 (1 − δ1 ).
Thus, player 1 proposes
(xT −2 , 1 − xT −2 ) = (1 − δ2 (1 − δ1 ) , δ2 (1 − δ1 )).
Bargaining
Is there already a pattern to be distinguished (easily)?
Knowing how the game will be player in the last three periods
player 2 in period T − 3 reasons that player 1 can guarantee
him/herself (by rejecting) δ1 (1 − δ2 (1 − δ1 )).
Thus, player 2 proposes
(δ1 (1 − δ2 (1 − δ1 )) , 1 − δ1 (1 − δ2 (1 − δ1 )))
. Knowing all the above player 1 in period T − 4 makes
proposal
(1 − δ2 (1 − δ1 (1 − δ2 (1 − δ1 ))) , δ2 (1 − δ1 (1 − δ2 (1 − δ1 )))).
Bargaining
At this point it is clear that whenever the players are playing
the SPE found by backward induction the proposals are
immediately accepted.
Thus, to find the pattern it is enough to study period T − 4
and then extend the ’result’ to period T − (T − 1).
Let us write the share of player 1 in the proposal of period
T − 4 in the following form: 1 − δ2 + δ2 δ1 − δ2 δ1 δ2 + δ2 δ1 δ2 δ1 .
This looks nasty so let us assume that the discount factors are
equal δ1 = δ2 = δ .
Then we have 1 − δ + δ 2 − δ 3 + δ 4 .
Guess that in period 1 player 1’s share is
T
x1 = 1 − δ + δ 2 − δ 3 + ... + δ T −1 = 1−δ
1+δ .
T
Consequently, player 2’s share is 1 − x1 = δ 1−δ
1+δ .
Bargaining
When T goes to infinity player 1 gets
δ
1+δ .
1
1+δ
and player 2 gets
In this game there is a first-mover advantage.
If the time horizon is infinite to start with more complicated
techniques are needed to show that the above division is the
outcome of the unique subgame perfect equilibrium.
Bargaining
The same problem with T = ∞.
Theorem
In an infinite horizon alternating offers bargaining game the unique
subgame perfect equilibrium player 1 always proposes
x=
1 − δ2
1 − δ1 δ2
and player 2 always proposes
y=
δ1 (1 − δ2 )
1 − δ1 δ2
Bargaining
Proof.
(Just the basic idea). Denote by x i the lowest pay-off player i gets
in any subgame perfect equilibrium in any subgame where i makes
an offer. Let x̄i denote the highest pay-off player i gets in any
subgame where i makes an offer.
Consider a subgame where player i makes an offer. Player j does
not accept the offer if it gives j less than he can guarantee by
rejecting the offer or δj x j . Consequently it has to be the case that
x̄i ≤ 1 − δj x j
(1)
Player j accepts any offer that gives j at least δj x̄j . Consequently, it
has to be the case that
x i ≥ 1 − δj x̄j
(2)
Bargaining
Proof.
(Continued) Similar reasoning establishes
x̄j ≤ 1 − δi x i
(3)
x j ≥ 1 − δi x̄i
(4)
Subtracting (1) and (2) as well as (3) and (4) one gets
Bargaining
Proof.
(Continued)
(5)
x̄j − x j ≤ δi (x̄i − x i )
(6)
x̄i − x i ≤ δj x̄j − x j
Multiply (6) sideways by δj to get
δj x̄j − x j ≤ δj δi (x̄i − x i )
From (5) and (7) one can derive
(7)
Bargaining
Proof.
(Continued)
x̄i − x i ≤ δj δi (x̄i − x i )
But this can hold only if x̄i = x i . Similarly, one can show that
x̄j = x j .
Now expressions (1) and (2) are equivalent to
xi = 1 − δj xj
xj = 1 − δi xi
Solving the pair of linear equations yields
(8)
Bargaining
Proof.
(Continued)
x1 =
1 − δ2
1 − δ1 δ2
x2 =
1 − δ1
1 − δ1 δ2
1 (1−δ2 )
Note that x1 = x and 1 − x2 = δ1−δ
= y in the statement of the
1 δ2
theorem. QED
One should also show that subgame perfect equilibria exist, and one
should show that there are no profitable deviations.
Incomplete information
Consider a seller with an object which s/he valueas at zero.
A buyer has private information about his/her valuation v .
Let it be a draw from a uniform distribution on [0, 1].
Now there are gains from trade but it is unclear how big they
are.
Incompelete information
Assume that there is no complicated bargaining but that the
seller makes a take-it-or-leave-it offer.
The seller’s problem is
maxx xPr (v ≥ x)
This is equivalent to
maxx x(1 − x)
Incomplete information
The first order condition is given by
x=
1
2
This means that with probability one half gains from trade are
not realised.