Lecture V: Bargaining Recommended Reading: Dixit & Skeath, Chapter 17 Osborne, Chapter 6.1, 16 Powell, In the Shadow of Power, Ch. 3

Lecture V: Bargaining Bargaining situation –not zero-sum –potential gains from trade / negotiation e.g., divide-the-dollar game where i & j split $1 as follows: –i & j announce bids x  [0, 1] & y  [0, 1], respectively –if x + y ≤ 1, i gets x, j gets y –but if x + y > 1, i and j each get 0 (or some small amount) –How might players resolve this situation?

Nash Cooperative Bargaining Solution i’s share = x j’s share = y Two observations: 1.Both i & j are better off splitting the entire surplus, so the bargain should fall on the Pareto frontier 2. i and j would object to bargains that returned less than their reservation value of.2 0 $1 0

Nash Cooperative Bargaining Solution i’s share = x j’s share = y If we think of j’s potential share of surplus as: y = b +  (s – a – b) (0 <  < 1) y – b =  (s – a – b) & i’s potential share of surplus as: x = a + (1-  )(s – a – b) x – a + (1-  )(s – a – b) then we can define a line, y = b +  /(1-  )(s – a – b) 0 $1 0 a b

Nash Cooperative Bargaining Solution i’s share = x j’s share = y Nash’s cooperative bargaining solution is the point where y = b +  /(1-  )(s – a – b) intersects x + y =s 0 $1 0 a b   1

Ultimatum Game i’s share = x j’s share = y Think of following (non-cooperative) game from this perspective: 1.i offers j $y, y  [0, 1] 2.if j accepts, i retains $1-y 3.if j rejects, i & j get 0 0 $1 0

Ultimatum Game i’s share = x j’s share = y 0 $1 0 i y j A R 1-y, y 0, 0 j’s strategy: y ≥ 0, so: a) accept all y b) accept y > 0, reject y = 0 i’s strategy: c)if j plays a), offer y = 0 d)if j plays b), offer ? No offer is optimal. Why?

Cons’ Share Alliance’s Share t1t1 t2t2 t3t3 t4t4 -c Modify ultimatum game in two ways: 1.Let player 2 make a counter-offer to player 1 2.Let surplus shrink by c every time an offer is rejected Alternating Offers t0t0

C ons’ Share A lliance’s Share t1t1 t2t2 t3t3 t4t4 Solve via backward induction: At t 3, i (say C) can offer A  > 0 & A accepts (something at t 3 beats nothing at t 4 )  at t 2, A must offer C at least what C could secure at t 3  at t 1, C must offer A at least what C could secure at t 2 …. Alternating Offers t0t0

C ons’ Share A lliance’s Share t1t1 t2t2 t3t3 Surplus that player i’s refusal burns goes to player j in equilibrium More rounds leads to more even split Alternating Offers t0t0

 1 y 1,  2 y 2 P1P1 P2P2 x x 1, x 2 A R P1P1 R A y 0, 0 First mover power in ultimatum game due to players’ indifference about delayed payoffs But typically delay is costly; actors prefer an agreement now to one later Bargaining & Discounting: The Rubenstein Model

 1 y 1,  2 y 2 P1P1 P2P2 x x 1, x 2 A R P1P1 R A y 0, 0 Bargaining & Discounting Consider following game: 1.P 1 proposes a division of the $ {x 1, x 2 } s.t. x 1 + x 2 = 1 2.P 2 can accept {x 1, x 2 } or counter-offer the following period with {y 1, y 2 } 3.If P 1 rejects {y 1, y 2 }, both players receive 0 4.If P 1 accepts, the players receive discounted payoffs of  1 y 1,  2 y 2, respectively. Bargaining & Discounting: The Rubenstein Model

1y1,2y21y1,2y2 P1P1 P2P2 x x 1, x 2 A R P1P1 R A y 0, 0 Bargaining & Discounting Solving the game: 1.Subgame in which P 2 proposes is an ultimatum game: P 2 proposes {y 1, y 2 } = {0, 1} and P 1 accepts. P 2 obtains  2 2.At P 1 ’s initial proposal, then: x 2 <  2 not optimal; P 2 rejects and secures  2 at next round x 2 >  2 not optimal; P 1 would be better off proposing x 2 s.t. x 2 > x 2 >  2 3.Thus, P 1 proposes {x 1, x 2 } = {1-  2,  2 } Bargaining & Discounting: The Rubenstein Model

 1 y 1,  2 y 2 P1P1 P2P2 x x 1, x 2 A R P1P1 R A y Bargaining & DiscountingBargaining & Discounting: The Rubenstein Model 0, 0 A R z  1 2 z 1,  2 2 z 2 P2P2 In last subgame P 1 proposes {z 1,z 2 } = {1,0} & P 2 accepts; P 1 gets payoff of  1 2 Previous stage: i.y 1 <  1 not optimal: P 1 rejects, P 2 gets 0 at t+1 ii.y 1 >  1 not optimal: P 2 could offer  less, keep  more

 1 y 1,  2 y 2 P1P1 P2P2 x x 1, x 2 A R P1P1 R A y Bargaining & DiscountingBargaining & Discounting: The Rubenstein Model 0, 0 A R z  1 2 z 1,  2 2 z 2 P2P2 Hence P 2 offers  1 = {  1, 1-  1 } for payoffs {  1,  2 (1-  1 )}  P 1 initially offers x 2 =  2 (1-  1 ); P 2 accepts.

Bargaining & Discounting: The Rubenstein Model Rubinstein Bargaining Model: i.Players: N = {1, 2} ii.Terminal Histories: Every sequence (x 1,N, x 2,N,... x t, Y ) iii.Players alternate making offers until one offer is accepted iv.Preferences: u i =  i x i t x x x A R A R x A R

Bargaining & Discounting: The Rubenstein Model How do we find the equilibrium? Without a final subgame, we can’t do backward induction But all subgames are identical: Stationarity Stationary strategies seem reasonable, i.e., because the game is identical after every history, sensible to assume players take same action after each history. A strategy in the game is then a ‘rule’ specifying proposals and conditions for acceptance

Bargaining & Discounting: The Rubenstein Model We can write the players’ strategies as: i.P 1 : x* & A if y 1  y 1 * (1) ii.P 2 : y* & A if x 2  x 2 * (2) Consider player 2’s actions under these strategies: If player 2 rejects, then she proposes y*, which player 1 accepts. Player 2’s payoff then equals y discounted by 1 period of delay, i.e.,  2 y 2. Thus, player 2 accepts any x   2 y 2. It follows that x 2 * =  2 y 2 (3) By symmetry, for Player 1 y 1 * =  1 x 1 (4)

Bargaining & Discounting: The Rubenstein Model So we have: x 2 * =  2 y 2 (3) y 1 * =  1 x 1 (4) Players are bargaining over a pie of size 1, thus x 1 * = 1− x 2 *; we can rewrite (4) as: 1 − y 2 * =  1 x 1 * (5) And from (3) we get: 1 − x 2 */  2 =  1 x 1 * (6)

Bargaining & Discounting: The Rubenstein Model Simplifying:  2 − (1 − x 1 * ) =  1  2 x 1 * (10) and then: x 1 * −  1  2 x 1 * = 1 −  2 (11) Dividing both sides by 1 −  1  2 we get: x 1 * = 1–  2 /(1−  1  2 ) To get player 2’s strategy, we simply use (6): y 1 * =  1 ×(1–  2 )/(1−  1  2 )

Bargaining & Discounting: The Rubenstein Model Properties of Rubenstein’s solution: Efficiency Effects of patience First mover advantage; Player 1’s payoff (for equal discount factors) is, (1-  )/(1-  2 ) = 1/(1 +  )

Bargaining in the Shadow of Power, Powell Ch. 3 Research Question: Is war more likely under 1.Preponderance of Power? State D less willing to risk conflict with strong state, S But S makes steeper demands of D 2.Balance of Power D steps more carefully in making demands of equally powerful S But S reacts more inclined to resist D’s demands Argument: war more likely when distribution of power out of step with distribution of benefits

Bargaining in the Shadow of Power, Powell Ch. 3 Set-up: 1.States S & D bargain over [x,y] 2.S or D may use force to impose settlement 3.Force is costly 4.Uncertain about other’s willingness to resort to force D S xqy

Bargaining in the Shadow of Power, Powell Ch. 3 Variation on Rubinstein Bargaining Model: 1.Normalize [x,y] to [0,1] 2.State i makes offer x  [0,1] to State j 3.State j accepts, counters, or attacks x x Accept Reject Attack SD Accept Reject Attack

Bargaining in the Shadow of Power, Powell Ch. 3 Utility Functions S & D get flows of utility from their territory Discounted over time: e.g., D’s utility from q territory: U D = q +  q +  2 q  a-1 q After an agreement that moves border to some x at time a, D’s utility from x territory: U D =  a x +  a+1 x  a+t x

Bargaining in the Shadow of Power, Powell Ch. 3 Utility Functions War gives D all territory, i.e. x = 1 with probability p War also costs d (Notation: war costs S, s) i.e., D’s utility from successful war at t=w is: U D = p(  w (1-d) +  w+1 (1-d)  w+t (1-d) whereas a defeat provides: U D = (1-p)(  w -d +  w+1 -d  w+t -d )

Bargaining in the Shadow of Power, Powell Ch. 3 Utility Functions: 1.Utility from q up to some t-1 & either: 2.Utility flow from agreement to x from t onward 3.Lottery over utility flows from victory or defeat at t onward x 1-d -d q

Bargaining in the Shadow of Power, Powell Ch. 3 Thus, D’s decision to attack hinges on several conditions: 1-d -d q 1.Cost-benefit of war hinges on lottery over flow of 1-d & flow of -d 2.Even if net flow from war, p –d > 0, war only worthwhile if also > flow from q

Bargaining in the Shadow of Power, Powell Ch. 3 Observe that the infinite sums of  s all cancel out because the comparison of flows is made from common time point, t = w. War is worthwhile to D iff: 1-d -d q t=w =

Bargaining in the Shadow of Power, Powell Ch. 3 Thus, simple algebra shows D prefers war to status quo iff p-d > q Even if p – d > s, S can alter border so that D receives flow from x > q 1-d -d q x

Bargaining in the Shadow of Power, Powell Ch. 3 Results: 1.There can be at most one dissatisfied state (i.e., with incentive to fight) Why? 2.Hence, with full information, there is no war. Why? D S x q p+s x* = p-d

Bargaining in the Shadow of Power, Powell Ch. 3 War hinges on asymmetric information 1.S does not know D’s cost to war, & vice versa 2.d ~ U[d H or d L ], s ~ U[s H or s L ] 3.Result: S averts war by x* = p + (s-d H )/2 D S x q p+s? p x d