14.12 Game Theory — Midterm IIAnsw ers11/13/2002Prof. Muh amet YildizInstructions. This is an open book exam; you can use an y written ma terial. You ha ve onehour and 20 minutes. Each question is 35 points. Good luc k !1. Two hunte rs go h unting, whe re they will pla y a stag-hunt game, in which eac h huntersimultaneously decides whether to go after a Rabbit or a Stag and the payoffsaregivenbySRS 6,6 0,4R 4,0 4,4.Before the hunt, Hu nter 1 can give a gift to H u nter 2 which is w orth 1 u tile to H unter2 and costs 1 utile to Hunter 1. Before they start hunting, the hunters kno w whetherthe gift is given, and all these are common kno w ledge. Hen ce, the game is as follows.GiftNo Gift3,53,1R-1,55,7SRS3,53,1R-1,55,7SRS4,44,0R0,46,6SRS4,44,0R0,46,6SRS1(a) Find all subgame-perfect equilib ria in pure strategies.Eachg subgam e has two Nash equilibria in pure strategies: (S,S), (R,R ). Therefor e,there are 4 SPE in pure strategies. To describe these, let us first name thestrategies for players. Player 1 h as 8 strateg ies: GSS, GSR, GRS , GRR, NSS,NSR, NRS, and NRR, where the first letter indicates whether he giv es gift (G) ornot (N), and the second and third letters indicate the actions he tak es in case ofgift and no gift, respectively. S im ilarly, player 2 has four strategies: SS, SR , RS,and RR. Th e subgame perfect equilibria are: (NSS ,SS ), (NRR,R R ), (NRS,R S ),and (GSR,SR ).(b) Using forward induction iterativ ely eliminate all of these equilibria except for one.Firstly, GRR and GRS are dominated b y NR R. Moreo ver, GSS is not dominated.Hence, if player 1 g ives a gift, then pla yer 2 should understand that player 1 willplay S (the first forw ard-in d uction argu m ent). In that case, play er 2 must pla y S.1Now, if pla yer 1 gives a gift, then he gets 5. After these elim ination s, NR R andNSR are strictly dominated by GS R. Clearly, NSS is not domin a ted. Therefore, ifplayer 1 does not give a gift, then pla yer 2 m ust understand that pla yer 1 will playS (the second forward-induction argum ent). In tha t case, play er 2 m ust also pla yS. Therefore, (NS S, SS) is the only sub ga m e perfect equilibrium that remain s.2. Below, there are pairs of stage gam es and strategy profiles. For eac h pair, check whetherthe strategy profile is a subgame -perfect equilibrium of the game in whic h the stagegame is repeated infinitely man y times. Eac h agen t tries to maxim ize the discountedsum of his expected payoffs in the stage game, and the discount rate is δ =0.99.(a) Stage Game:1\2S RS 6,6 0,4R 4,0 4,4Strategy profile: Each player plays S in the first rou nd and in the followingrounds he pla ys what the other player played in the previous round (i.e., at eacht>0, he plays what the other player played at t − 1).This is a version of Tit — for —tat; it is not a subgame perfect equilibrium. Con-sider the case that player 1 has played S and player has 2 play ed R. Now, ifplayer 1 sticks to his strategy and plays R , then the con tin uation play will be(RS,S R ,R S ,S R ...) , which yields 4/(1 − δ2) forplayer1. IfhedeviatesandplaysS, the continuation p lay will be (SS, SS, SS , ...), which yields 6/ (1 − δ).Thedeviation is clearly beneficial.(b) Stage Game:1\2L MRT 3,1 0,0 -1,2M 0,0 0,0 0,0B -1,2 0,0 -1,2Strategy profile: Until some pla yer deviates, player 1 pla ys T and player 2 pla ysL. If anyone deviates, then each plays M thereafter.This is a subga m e perfect equilibriu m . After the deviation, th e pla yers playa N ash equilibrium forev er. Hence, we only need to check that no player ha san y incentiv e to deviate on the path of equilibrium. Player 1 has clearly noincentive to deviate. If player 2 deviates, he g ets 2 in the cur rent period and getszero thereafter. If he sticks to his equilibrium strategy, then he gets 1 forev er.The present value of this is 1/ (1 − δ) > 2. Therefore, pla y er 2 doesn’t ha ve anyincentive to deviate, either.3. We consider a game between t wo soft ware developers, who sell operating systems (OS)for personal compu ters. (We also ha ve a P C maker and the consumers, but theirstrategies are already fixed.) Eac h soft ware dev eloper i, sim ultaneously offers “bribe”bito the PC m aker. (The bribes are in the form of contracts.) Looking at the offered2bribes b1and b2, the PC maker accepts the highest bribe (and tosses a coin betweenthem if they happen to be equal), and he rejects the other. If a firm’s offer is rejected,it goes out of business, and gets 0. Let i∗denote the soft ware developer w hose bribe isaccepted. Then, i∗pa ys the bribe bi∗, and the PC maker develops its PC compatibleonly with the operating system of i∗. Theninthenextstage,i∗becomes the monopolistin the market for operating systems. In this market the in verse demand function isgiven b yP =1− Q,where P is the pri ce of OS an d Q is the demand for OS. The mar ginal cost of producingthe operating system for each softw are developer i is ci.Thecostsc1and c2areindependently and iden tically distributed with the uniform distribution on [0, 1], i.e.,Pr (ci≤ c)=0 if c<0c if c ∈ [0, 1]1 otherwise.The software developer i knows its own marginal costs, but the other firm does notkno w. Each firm tries to max im ize its own expected profit. Everything described sofar is common knowledge.(a) W hat quantity a software developer i w o uld produce if it becomes mon opolist?What w ould be its profit?Quantity isqi=1 − ci2and the profitisvi=µ1 − ci2¶2.(b) Co m pute a symmetric Bay esian Nash equilibrium in which each firm’s bribe is inthe form of bi= α + γ (1 − ci)2.We have a first price auction where the valuation of buyer i, who is the softwaredeveloper i,isvi=(1− ci)2/4.Hispayoff from paying bribe biisUi(bi; ci)=(vi− bi)Pr(bj<bi) ,wherePr (bj<bi)=Pr¡α + γ (1 − cj)2<bi¢=Pr¡(1 − cj)2< (bi− α) /γ¢=Pr³1 − cj<p(bi− α) /γ´=Pr³cj> 1 −p(bi− α) /γ´=1− Pr³cj≤ 1 −p(bi− α) /γ´=1−h1 −p(bi− α) /γi=p(bi− α) /γ.Hence,Ui(bi; ci)=(vi− bi)p(bi− α) /γ.3Bu t maximizing Ui(bi; ci) is the sa m e as maximizingγUi(bi; ci)2=(vi− bi)2(bi− α) .The first order condition yields2(bi− vi)(bi− α)+(bi− vi)2=0,i.e.,2(bi− α)+(bi− vi)=0,i.e.,bi=13vi+23α =112(1 − ci)2+23α.Therefor e,γ =112and α =23α =⇒ α =0,yieldingbi=13vi=112(1 − ci)2.(Check that the second derivativ e is 2(3bi− 2vi)=−2vi< 0.)(c) Co nsiderin g that the dem and for PCs and the deman d of O S