(Finite-Time Handout
Bargaining Horizon)
Econ 409 Fall 2007
Daisuke Nakajima
Consider the following bargaining process:
There is a pie with the size of 1, to be divided between players 1 and 2.
There are T rounds t = T; T 1; : : : ; 2; 1, where T is the initial round and
1 is the nal round. We index the times backward.
In odd rounds, player 1 makes an oer and in even rounds player 2 makes
an oer. (so it is player 1 who makes the nal oer, but the initial proposer
depends on if T is odd or even.)
In each round, if the oer is accepted, they divide the pie accordingly.
If it is rejected, then the player who rejects makes a counter oer in the
next round. This process continues until they reach an agreement or they
cannot even at period 1.
Let (xt ; 1 xt ) denote the oer at period t, where xt is the share of player
1 and 1 xt is the share of player 2.
If they reach an agreement of (xt ; 1 xt )
Ttt
x and player 2 payo is T t (1
s
factor representing the cost of the delay.
the next round is indierent to getting
at period t, player 1 payo is
s
xt ).
2 (0; 1) is the discount
You can see that getting x in
x now.
Let us solve this game backward.
(T = 1)
In this game, player 1 makes an oer and if player 2 declines it, they end up
with getting nothing. Hence, player 2 accepts any oer. Given that, player 1
oers (1; 0), and player 2 accepts it. Hence x1 = 1.
(T = 2)
In this case, it is player 2 who makes the initial oer. If player 1 declines
player 2 oer, he can get 1 in the next round, so player 1 accepts any oer
s
greater than 1 = . Given this, player 2 oers ( ; 1
) and player 1 accepts
it. Hence x2 =
(T = 3)
Now, player 1 makes the initial oer. If player 2 declines the oer, he gets
1
in the next round so he accepts any oer greater than (1 ). Given that,
player 1 oers (1
(1
); (1
)), which is accepted by player 2. Therefore,
x3 = 1
(1
).
1
What happens for general T ? Let us work for the case when T is odd (so
T = 2k + 1 where k = 0; 1; : : :)
Consider the rst period t = 2k + 1, it is player 1 who makes the initial oer.
Player 2 can obtain 1 x2k in the next round if he declines the oer now, so
he accepts any oer greater (1 than x2k ). Given that, player 1 should oer
(1
(1 x2k ); (1 x2k )). Therefore,
x2k+1 = 1
x2k )
(1
(1)
Now consider the second period t = 2k , now player 2 makes an oer. Again,
player 1 can obtain x2k 1 = x2(k 1)+1 in the next round by rejecting the current oer, so in order to have him accept the oer, player 2 should propose
( x2(k 1)+1 ; 1 x2(k 1)+1 ). Hence,
x2k = x2(k
1)+1
(2)
Combining (1) and (2), we have
x2k+1
=
1
x2(k
1
2 2(k 1)+1
=
x
1)+1
+ (1
(3)
)
To make it simple, dene
X k = x2k+1 ;
2
=
, and
= (1
)
then (3) becomes
Xk = Xk
1
+
(4)
Applying (4) recursively, we have
Xk
=
=
Xk 1 +
Xk 2 +
=
=
2
=
3
Xk
k
X0 +
2
X
+
k2
+ (1 + )
X
+
+ (1 + )
k3
3
+ (1 +
+
2
)
.
.
.
=
1+
2
+
+
k1
Since we have already seen x0 = 1, and X 0 = x1 by denition, it becomes
x2k+1
=
k
+
=
k
+
k
(1
=
2k
=
=
=
1+
+
2
(1
k1
+
k
1
1
) + (1
1
k
) + (1
)
)1
2k
(5)
2
1
2k
(1 + ) + 1
1+
1 + 2k+1
1+
1
2k
(6)
(7)
Therefore, when T = 2k + 1 (so T is odd), player 1 proposes
!
!
2k
1 + 2k+1
1 + 2k+1 (1
)
1 + 2k+1
;1
=
;
1+
1+
1+
1+
and player 2 accepts it2 .
How about when T is odd (so T = 2k )? Do we have to make long and messy
algebras again? No. Because (2) implies
x2k = x2(k
1)+1
and according to (7)
x2(k
1)+1
=
1+
2(k 1)+1
=
1+
1 + 2k
1+
1
1 Note
for the second equality:
How can we nd the value of 1 +
+
k 1?
+
S =1+
+
Let
+
k1
k1
+
then
S=
+
+
k
so
(1
k
)S = 1
Therefore,
S=
1
1
k
2 In order to check if the algebras are correct, it is a good idea to substitute the value such
as k = 0 or k = 1. If k = 0 (corresponding to T = 1), our formula says player 1 oers
(1; 0), which coincide with our previous analysis. If k = 1, (corresponding to T = 3), the
2
formula implies (1
+ 2;
), again matching with our previous analysis. Based on my
experiences, if the formula is correct for two values of k, it is almost sure that the formula is
correct in this kind of algebras.
3
so
2k
x
1+
2k 1
=
1+
Hence, when T = 2k is even, player 2 makes a proposal
!
(1 + 2k
(1 + 2k 1 )
(1 + 2k 1 )
=
;1
1+
1+
1+
4
1
2k
)1
;
1+
!