1. QR 38 Bargaining, 4/24/07 I. The bargaining problem and Nash solution II. Alternating offers models

2. I. The bargaining problem and Nash solution Bargaining constant in IR, but haven’t said much directly about it. What exactly is bargaining; how to represent it using game theory? Bargaining is in large part a coordination problem. Two parties need to agree on the distribution of a good. Many equilibria, but disagreement over which is preferred (distributional conflict).

3. Bargaining problems But failure to reach an agreement leaves all parties worse off. So bargaining involves: 1.Potential for mutual gains 2.Conflict over how to divide these gains Bargaining is not zero-sum: a surplus exists, compared to the situation where no bargain is reached.

4. Solutions A solution to a bargaining problem involves: Specification of situations in which a bargain will be reached How the surplus will be divided Most models, and Schelling’s discussion, focus on the second question

5. Types of bargaining models Two types of bargaining models, drawing on cooperative and non-cooperative game theory. Nash devised a cooperative solution Later the Nash bargaining solution was shown to be the equilibrium of a non- cooperative game as well

6. Two-player bargaining model Consider a two-person bargaining situation. If the parties reach an agreement, they get a total value v to split between themselves. If they don’t reach an agreement, A gets a and B gets b.

7. Payoffs a and b are called backstop payoffs, BATNA, or reservation points. Often set these equal to zero to simplify the problem. The surplus equals the total benefit from reaching an agreement: surplus=v-a-b

8. Solutions Assume that each player gets BATNA plus a fraction of the surplus: A gets the fraction h B gets k (=1-h). Let x be the total A gets: –x=a+h(v-a-b) –x-a=h(v-a-b) –This says that the additional benefit A gets from an agreement (x-a) is some fraction h of the total surplus.

9. Solutions Let y be the total B gets: –y=b+k(v-a-b) –y-b=k(v-a-b) (the benefit B gets is fraction k of the surplus) These are the Nash formulas. Think of them as dividing the surplus in the proportion h:k Can write (y-b)/(x-a)=k/h –Then think of k/h as the slope of the line specifying the solutions

10. Nash bargaining solution A’s payoff (x) B’s payoff (y) a b (a, b) Line with slope k/h v(=x+y) v Q a’ (a’, b) Q’

11. Nash bargaining solution A nice way to think about the problem, but it doesn’t tell us where h and k come from. Can think of h and k as bargaining strengths. Need more context to use this solution. –Nash assumed h=k; then get determinate solution for x and y.

12. Nash bargaining solution Note that being able to move the reversion point (a,b) in your direction provides you with a higher payoff. What would this mean in IR? –Usually making a threat that would hurt yourself if you had to implement it, like a trade war.

13. II. Alternating offers model To get more insight, need a model with more context. An important general model is an alternating offers model. A dynamic model, with some number of periods.

14. Alternating offers model In each period, one player has the opportunity to make an offer to the other. The other can accept or make a counteroffer. This process continues until an offer has been made and accepted.

15. Alternating offers model With a finite number of periods, can use rollback to find the equilibrium. But in an infinitely-repeated game, why would this process ever end? Have to assume that the surplus becomes less valuable over time; discounting. –This could result because the surplus itself is shrinking (some probability it will disappear, e.g.), or because the players are impatient. –The two are conceptually similar, although D&S present separately.

16. Solving alternating offers model Assume that two players are bargaining over the division of a dollar. A dollar tomorrow is as good as having only 95 cents today. –Remember how we used discount factors to address situations like this (repeated games). Assume BATNAs are zero.

17. Solving alternating offers The player making the offer suggests that he gets x. We want to solve for x using backward induction That is, x is the equilibrium outcome. Let A start. A knows that B will get x in the next round, because x is the equilibrium payoff to the player making the offer. So A has to offer something today that is worth the same as getting x in the next round.

18. Solving alternating offers So A has to offer B 0.95x now. Leaves A with 1-0.95x. But we called what A is offering x: So x=1-0.95x x=1/1.95 x=0.512 So, the equilibrium is for the player who gets to make the first offer to get 0.512 The player who goes second will get 1-.512=0.488

19. Solution The equilibrium is reached immediately even though an unlimited number of counteroffers are allowed That is, the outcome is efficient; the surplus does not decay. A first-mover advantage results: the player making the first offer gets more (x>1/2).

20. Solution with different discount rates This example assumed that the two players had the same discount factor (.95). What if the two players have different discount rates? For A, let a dollar tomorrow may be worth only 0.90 today. B’s discount rate is.95. –So A is willing to accept a smaller amount in order to be paid sooner. In equilibrium, the more impatient player gets less.

21. Different discount rates Let x be the amount A gets when he goes first Let y be the amount B gets when he starts A has to offer B 0.95y. –So x=1-0.95y B has to offer A 0.90x –So y=1-0.90x

22. Different discount rates We can solve these equations for x and y: x=1-.95y y=1-.9x x=1-.95(1-.9x)=1-.95+.855x=.05+.855x.145x=.05 x=0.345 y=1-.9(.345)=0.690

23. Different discount rates So if A goes first, A gets.345 and B gets 1-.345=.655 If B goes first, A gets.31 and B gets.69 So A gets less than B because of impatience, even if A goes first

24. Generalized solution A sees $1 today as worth $(1+r) tomorrow B sees $1 today as worth $(1+s) tomorrow Means that A sees $1 tomorrow as worth $1/(1+r) today x=(s+rs)/(r+s+rs) y=(r+rs)/(r+s+rs)

25. Generalized solution rs is usually very small So it is approximately true that x=s/(r+s) y=r/(r+s) Then we can see x and y as the shares that go to each player (verify that x+y=1). Write as y/x=r/s The shares that players get are inversely proportional to their rates of impatience.

26. Solution a=1/(1+r) b=1/(1+s) When A makes an offer, has to give B the equivalent of getting y the next period; this is by. So: x=1-by y=1-ax x=1-b(1-ax)=1-b+abx x-abx=1-b x(1-ab)=1-b

27. Solution x=(1-b)/(1-ab)=(1-(1/1+s))/(1-(1/1+r)(1/1+s)) =(1+s-1)/(1+s-(1/1+r))=s/(1+s-(1/1+r)) =s(1+r)/((1+r)+s(1+r)-1) =(s+rs)/(1+r+s+rs-1) =(s+rs)/(r+s+rs)  s/(r+s) y=(r+rs)/(r+s+rs)  r/(r+s)