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Exercise 10.12. MICROECONOMICS Principles and Analysis Frank Cowell. January 2007. Ex 10.12(1): Question. purpose : Set out a one-sided bargaining game method : Use backwards induction methods where appropriate. Ex 10.12(1): setting. Alf offers Bill a share of his cake
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Exercise 10.12 MICROECONOMICS Principles and Analysis Frank Cowell January 2007
Ex 10.12(1): Question • purpose: Set out a one-sided bargaining game • method: Use backwards induction methods where appropriate.
Ex 10.12(1): setting • Alf offers Bill a share of his cake • Bill may or may not accept the offer • if the offer is accepted game over • if rejected game continues • Two main ways of continuing • end the game after a finite number of periods • allow the offer-and-response sequence to continue indefinitely • To analyse this: • use dynamic games • find subgame-perfect equilibrium
Ex 10.12(1): payoff structure • Begin by drawing extensive form tree for this bargaining game • start with 3 periods • but tree is easily extended • Note that payoffs can accrue • either in period 1 (if Bill accepts immediately) • or in period 2 (if Bill accepts the second offer) • or in period 3 (Bill rejects both offers) • Compute payoffs at each possible stage • discount all payoffs back to period 1 the extensive form
Ex 10.12(1): extensive form Alf • Alf makes Bill an offer period 1 [offer 1 - g1] • If Bill accepts, game ends • If Bill rejects, they go to period 2 Bill • Alf makes Bill another offer [accept] [reject] • If Bill accepts, game ends • If Bill rejects, they go to period 3 (g1, 1 - g1) Alf period 2 • Values discounted to period 1 • Game is over anyway in period 3 [offer 1 - g2] Bill [accept] [reject] (dg2, d[1 - g2]) (d2g, d2[1 -g]) period 3
Ex 10.12(1): Backward induction, t=2 • Assume game has reached t = 2 • Bill decides whether to accept the offer 1 g2 made by Alf • Best-response function for Bill is • [accept] if [1 g2] ≥ d [1 g] • [reject] otherwise • Alf will not offer more than [1 g]d • wants to maximise own payoff • this offer would leave Alf with 1 − d + dg • Should Alf offer less than d[1 − g] today and get γ tomorrow? • tomorrow’s payoff is worth dg, discounted back to t = 2 • but d < 1, so 1 − d + dg > dg • So Alf would offer exactly 1 − g2 = d[1 − g] to Bill • and Bill accepts the offer
Ex 10.12(1): Backward induction, t=1 • Now, consider an offer of g1 made by Alf in period 1 • The best-response function for Bill at t = 1 is • [accept] if 1 −g1 ≥ d2 [1 − g] • [reject] otherwise • Alf will not offer more than d2g in period 1 • (same argument as before) • So Alf has choice between • receiving 1 − d2[1 − g] in period 1 • receiving 1 − d[1 − g] in period 2 • But we find 1 − d2[1 − g] > 1 − d[1 − g] • again since d < 1 • So Alf will offer 1 −g1 = d2[1 − g] to Bill today • and Bill accepts the offer
Ex 10.12(2): Question method: • Extend the backward-induction reasoning
Ex 10.12(2): 2 < T < ∞ • Consider a longer, but finite time horizon • increase from T = 2 bargaining rounds… • …to T = T' • Use the backwards induction method again • same structure of problem as before • same type of solution as before • Apply the same argument at each stage: • as the time horizon increases • the offer made by Alf reduces to 1 −g1 = δT' [1 − γ] • which is accepted by Bill
Ex 10.12(3): Question method: • Reason on the “steady state” situation
Ex 10.12(3): T = ∞ • Can we use previous part to suggest: as T→∞, 1 −g1 → 0? • this reasoning is inappropriate • there is no “last period” from which backwards induction outcome can be obtained • Instead, consider the continuation game after each period t • the game played if Bill rejects the offer made by Alf • This looks identical to the game just played • there is in both games… • …a potentially infinite number of future periods • This insight enables us to find the equilibrium outcome of this game • use a kind of “steady-state” argument
Ex 10.12(3): T = ∞ • Consider the continuation game that follows if Bill rejects at t • suppose it has a solution with allocation (γ, 1 γ) • so, in period t, Bill will accept an offer 1g1 if g1 ≥ δγ, as before • Thus, given a solution (g, 1 g), Alf would offer • 1 g1 = d[1 γ] • Now apply the “steady state” argument: • if γ is a solution to the continuation game, must also be a solution to the game at tl • so g1 = g • It follows that • g = dg • this is only true if γ = 0 • Alf will offer1 g = 0 to Bill, which is accepted
Ex 10.12: Points to remember • Use backwards induction in all finite-period cases • Take are in “thinking about infinity” • if T→∞ • there is no “last period” • so we cannot use simple backwards induction method