EC941 – Game Theory Lecture 8 Prof. Francesco Squintani Email:

EC941 - Game Theory Lecture 8 Prof. Francesco Squintani Email: [email protected] 1

Structure of the Lecture Ultimatum Game and Hold Up Problem Rubinstein Alternating Offer Bargaining Nash Axiomatic Bargaining 2

The Ultimatum Game The ultimatum game is a simple model of bargaining There are two players: person 1 offers player 2 an amount of money up to c. If 2 accepts this offer then 1 receives the remainder. If 2 rejects the offer then neither person receives any payoff. Each person cares only about the amount of money she receives, and prefers to receive as much as possible. 3

Formal Model Players: The two people. Terminal histories: The set of sequences (x, Z), where x is a number with 0 x c (the amount of money that person 1 offers to person 2) and Z is either Y (“yes, I accept”) or N (“no, I reject”). Player function: P( ) 1 and P(x) 2 for all x. Preferences: Each person’s preferences are represented by payoffs equal to the amounts of money she receives. For the terminal history (x, Y) person 1 receives c 4

Extensive Form Representation 1 Player 1 chooses first how much to give to x 0player 2. x c Then player 2 chooses x whether Yes or No. 2 Y c - x, x N 0, 0 5

Backward Induction Solution First consider each subgame Gx where player 2 either accepts or rejects the offer x by player 1. For every x 0, player 2’s optimal action is to accept (if she rejects, she gets nothing). For x 0, person 2 is indifferent between accepting and rejecting. Thus in a subgame perfect equilibrium person 2’s strategy either accepts all offers x (including 0), or accepts all offers x 0 and rejects the offer x 0. 6

1 x c x 0 x c, 0 Y N N 0, 0 Y 2 0, 0 Y c - x, x N 0, c 0, 0 7

Now consider the whole game. For each of the 2 possible subgame perfect equilibrium strategies of player 2, we find the optimal strategy of player 1. If player 2 accepts all offers x 0, then player 1’s optimal offer is x 0 (which yields her the payoff c). If player 2 accepts all offers x 0, then there is no offer x that is optimal for player 1. No offer x 0 can be optimal, because x-e is better for player 1, (as long as 0 e x). 8

The offer x 0 is not optimal, as player 2 rejects it. The only subgame perfect equilibrium is the strategy pair where player 1 offers x 0 and player 2 accepts all offers x 0. In equilibrium, player 1’s payoff is c and player 2’s payoff is zero. 9

The holdup game Before an ultimatum game in which she may accept or reject an offer of person 1, person 2 may make an investment that changes the size of the pie. She may exert little effort, resulting in a small pie, of size cL, or great effort, resulting in a large pie, of size cH. Player 2 dislikes exerting effort: her payoff is x - E if her share of the pie is x, where E 10

Extensive Form Representation 2 L H 1 1 x 0 x cL x x cH x 0 x 2 Y cL-x, x-L 2 N 0, -L Y cH-x, x-H N 0, -H 11

Backward Induction Solution Each subgame that follows 2’s choice E is an ultimatum game. It has a unique subgame perfect equilibrium, in which 1 offers x 0 and 2 accepts all offers x 0. Consider 2’s choice of effort at the start of the game. 12

So, player 1 chooses L at the beginning of the game. Thus the game has a unique subgame perfect equilibrium. In the SPE, player 2 exerts little effort and player 1 obtains all of the resulting small pie. 13

Two Period Alternating Offers Two players bargain over a pie of size 1. In period 1, player 1 makes a split proposal (x, 1-x). If player 2 accepts the proposal, it is implemented. Else, in period 2, player 2 makes proposes (y, 1-y). If player 1 accepts the proposal, it is implemented. 14

Subgame Perfect Equilibrium The subgame starting after player 2 rejects a proposal (x, 1-x) of player 1 is an ultimatum game. It has a unique subgame perfect equilibrium: player 2 proposes (0, 1) and player 1 accepts all proposals. This yields payoffs of 0 for player 1 and d2 for player 2. 15

So, player 2 rejects any proposal (x, 1-x) with 1-x d2 and accepts if 1 – x d2. Hence, player 1 initially proposes (1- d2, d2). In fact, if proposing any x 1 – d2, player 1 anticipates that player 2 will reject proposal (x, 1-x), so that player 1’s eventual payoff will be zero. Intuitively, if player 2 is more impatient (i.e., d2 is smaller), then she gets a lower payoff in the bargain. 16

Rubinstein Bargaining Model Consider the following model in which players may alternate offers indefinitely. Players. The two negotiators: 1 and 2. Terminal histories. Every sequence of the form (x0, N, x1, N, ., xt, Y) with t 0, and every (infinite) sequence of the form (x0, N, x1, N, .). To each xr corresponds a split proposal (xr, 17 1- xr).

Player Function. P(Ø) 1, P(x0, N, x1, N, ., xt) P(x0, N, x1, N, ., xt, N) 2 if t is odd, P(x0, N, x1, N, ., xt) P(x0, N, x1, N, ., xt, N) 1 if t is even. Preferences. The players’ payoffs at any terminal history (x0, N, x1, N, ., xt, Y) are dt1 xt and dt2 (1-xt), and zero at the infinite terminal history (x0, 18 N, x1, N, .).

Subgame Perfect Equilibrium All subgames starting at odd periods are isomorphic, and so are all subgames starting at even periods. So, we can introduce these stationary bounds: xiH is the largest equilibrium share of player 1 in a subgame where player i makes the initial proposal. xiL is the smallest equilibrium share of player 19 1 in a

In a SPE, player 1 rejects all proposals with a payoff lower than her minimal payoff in her next round. Hence, we obtain that in any SPE, x2L d1x1L. The smallest share of player 1 when 2 makes a proposal is no smaller than the discounted smallest share of player 1, when player 1 makes a proposal. Interchanging players, in any SPE, 1-x1H 20

Player 2 does not make any proposals with a payoff lower than her minimal payoff in her next round. I.e., player 2 never proposes a share of player 1 larger than player 1’s maximal payoff in the next round. Hence, we obtain that in any SPE, x2H d1x1H. Again, interchanging players, we obtain that in any SPE, 1-x1L d2(1-x2L). 21

Substitution yields: 1 - x1H d2(1 - d1x1H), x2L d1[1 - d2 (1 - x2L)], x2H)] x2H d1[1 - d2 (1 - Further rearranging yields: 1 - d2 d1(1 - d2) x1L 1 - d1d2 x2H 1 - x1L d2(1 - d1x1L) x1H and 1x-2Ld1 d2 1 - d2 1 - d1d2 So, the unique outcome is x 22

Nash Axiomatic Bargaining The bargaining models seen so far are “positive”: they find solutions of realistic bargaining models. A normative approach to study bargaining is to find solutions that satisfy ethically reasonable requirements. For example, if two identical players bargain over a good, we reasonably require that each gets half. 23

But what if the players’ preferences are not identical? Denote by X the set of possible agreements, and by D the failure to agree. Let ui be player i’s utility over X {D}, with di ui(D). Define expected utilities Eui over X {D}, as usual. Define the payoff set U {(v1, v2) : vi 24

Definition A bargaining problem is a pair (U, d) such that d is a member of U, U is convex, bounded, and closed, for some (v , v ) in U we have v d and v 1 2 1 1 2 d2. Definition A bargaining solution is a function from the set of bargaining problems (U, d) to U. A basic assumption of the model is that the solution depends only on the preferences on 25

We introduce the following requirements (axioms). 1 Invariance to equivalent utility representations. Let (U, d) be a bargaining problem, let ai and bi be numbers with ai 0 for i 1, 2. Let U ’ {(a1v1 b1, a2v2 b2) : (v1, v2) is in U}, and let d ’ (a1d1 b1, a2d2 b2 ). If the solution of (U, d) is (v*1, v*2), then the solution of (U’, d’) is (a1v*1 b1, a2v*2 b2) . 2 Symmetry. Let (U, d) be a bargaining 26

3 Pareto efficiency. Let (U, d) be a bargaining problem, and let (v1, v2) and (v’1, v’2) be members of U. If v1 v’1 and v2 v’2, then (v’1, v’2) is not a bargaining solution of (U, d). 4 Independence of irrelevant alternatives. Let (U, d) and (U’ , d’) be bargaining problems for which U ’ is a subset of U and d’ d . If the solution v* of (U, d) is in U’ , then bargaining solution of (U’, d’) coincides with v*. 27

Theorem (Nash bargaining solution) A unique bargaining solution satisfies the axioms INV, SYM, IIA, and PAR. This solution associates with the bargaining problem (U, d) the pair of payoffs that solves the problem: max (v1 - d1)(v2 - d2) (v1, v2) subject to (v1, v2) U and (v1, v2) (d1, d2). 28

The proof or the Theorem is as follows. First note that because U is closed and bounded, the maximization problem defined has a solution. The solution is unique because the level curves of (v1 - d1)(v2 - d2) are strictly convex, and U is convex. Hence, the Nash bargaining solution fN is well defined. N 29

Define the function H by H(v1, v2) (v1 d1)(v2 - d2) for all (v1, v2). INV: Fix a1 0, a2 0, b1 and b2. Let U’ be the set of all pairs (a1v1 b1, a2v2 b2) where (v1, v2) is in U, and let d’ (a1d1 b1, a2d2 b2) . Then for every pair (v1’, v2’) in U’ there is a pair (v1, v2) in U with vi’ aivi bi for i 1, 2. Thus the maximizer of (v ’ - d ’)(v ’ - d ’) 30

But (a1v1 b1 - d1’)(a2v2 b2 - d2’) a1a2H(v1, v2), so (v1*, v2*), equals fN(U, d), the Nash solution of (U, d). SYM: If (U, d) is symmetric, the symmetry of the level curves of (v1 - d1)(v2 - d2) implies that f1N(U, d) f2N(U, d). PAR: The function H is increasing in each of its arguments, so v does not maximize H(v) over U if there exists v’ U with v1’ vi for i 1, 2. 31

Finally, I argue that if f is a bargaining solution that satisfies the four axioms, then f fN. Let (U, d) be an arbitrary problem. I need to show that f(U, d) fN(U, d). Step 1. Let fN(U, d) z. Because there exists v U such that vi di for i 1, 2, we have zi di for i 1, 2. So, let ai 1/[2(zi - di)] and bi -di /[2(zi - 32

Define U’ {(v1’, v2’) vi’ aivi bi for i 1, 2} and d’ (a1d1 b1, a2v2 b2 ). Note that d’ (0, 0) and that aizi bi 1/2, for i 1, 2. Because the Nash solution satisfies INV, fN(U’, d’) (a1z1 b1 , a2z2 b2 ) (1/2, 1/2). By INV, f(U, d) z if and only if f(U’,d’) (a1z1 b1 , a2z2 b2 ). 33

Step 2. I claim that U’ contains no point (v1’, v2’) for which v1’ v2’ 1. Suppose to the contrary that it does, and let (t1e, t2e) (1/2 (1 – e) e v1’, 1/2 (1 – e) e v2’), for 0 e 1. The set U’ is convex, so (t1e, t2e) is in U’ for all e. But for small enough values of e we have t1et2e 1/4 , contradicting fN(U’, 0) (1/2, 1/2). 34

Step 3. The set U’is bounded, so the result of Step 2 ensures that we can find a rectangle T that is symmetric about the 45 line, that contains U’, and on the boundary of which is (1/2, 1/2). Step 4. By PAR and SYM we have f (T, 0) (1/2, 1/2). Step 5. By IIA we have f (U’, 0) f (T, 0), so that f (U’, 0) (1/2, 1/2), completing the proof. 35

This result is a very deep one within the “axiomatic normative approach” to social sciences. The approach and the result can be described as follows, in the words of John Nash: “One states as axioms several properties that it would seem natural for the solution to have and then one discovers that the axioms actually determine the solution uniquely.” (Nash,1953, pp. 129.) 36

Summary of the Lecture Ultimatum Game and Hold Up Problem Rubinstein Alternating Offer Bargaining Nash Axiomatic Bargaining 37

THE END! Thank you. 38