14.12 Game Theory — Final12/10/2001Prof. Muh amet YildizInstructions. This is an open book exam; y ou can use any written material. You have 2hours 5 0 minutes. Each q ues tion is 20 points. Good luck !1. Consider the follow ing extensiv e form game.1212 2 3 1 000013IO LL L RRR(a) Find the normal form representation of this game.A:LROL 2,2 2,2OR 2,2 2,2IL 3,1 0,0IR 0,0 1,3(b) Find all rationalizable pure strategies.LROL 2,2 2,2OR 2,2 2,2IL 3,1 0,0(c) Find all pure strategy Nash equilibria.LROL 2,2 2,2OR 2,2 2,2IL 3,1 0,01(d) W hich strategies are consisten t with all of the follo w ing assum ption s?(i) 1 is rational.(ii) 2 is sequentially rational.(iii) at the node she moves, 2 kno w s (i).(iv) 1 knows (ii) and (iii).ANSWER: By (i) 1 does not play IR. Hence, b y (iii), at the node she moves, 2kno ws that 1 does not pla y IR, hence he kno ws IL. Then, b y (ii), 2 m ust play L.Therefor e, by (i) and (iv), 1 m u st pla y IL. The answ er is (IL,L).2. This question is about a milkman and a customer. At any day, with the given order,• M ilk man puts m ∈ [0, 1] liter of milk and 1 − m liter of w ater in a container andcloses the con ta iner, incurring cost cm for some c>0;• Cu stom er, without know ing m, decides on whether or not to buy the liquid atsome price p. If she buys, her pa yoff is vm− p and the milkman’s pa yo ff is p− cm.If she does not buy, she gets 0, and the milkm an gets −cm. If she buys, then shelearns m.(a) Assu m e that this is repeated for 100 days, and eac h player tries to maxim ize thesum of his or her stage pay offs. Find all subgam e-perfect equilibria of this game.ANSWER: The stage gam e has a unique Nash equilibrium, in which m =0andthe customer does not buy. Therefore, this fin itely repeated game has a uniquesubgame-perfect equilibriu m , in which the stage equilibrium is repeated.(b) N ow assume that this is repeated infinitely man y times and each play er tries tomaxim ize the discoun ted sum of his or her stage payoffs, where discount rate isδ ∈ (0, 1). What is the range of prices p for whic h there exists a subgame perfectequilibrium such that, everyda y, the milkman chooses m =1,andthecustomerbuys on the path of equilibrium pla y ?ANSWER: The milkman can guaran tee himself 0 by alw ays c hoosing m =0.Hence, h is continuation value at any history must be at least 0. Hence, in theworst equilibrium, if he deviates custom er should not buy milk forever, giving themilkman exactly 0 as the con tinuation value. Hence, the SPE w e are looking foris the milkman always chooses m=1 and the customer buys un til anyone deviates,and the milkman c hooses m=0 and the customer does not buy thereafter. If themilkman does not deviate, his con tinuation value will beV =p − c1 − δ.The best deviation for him (at any history on the path of equilibrium play ) is tochoose m =0(and not being able to sell thereafter). In that case, he will getVd= p + δ0=p.In order this to be an equilibrium, we m ust ha v e V ≥ Vd; i.e.,p − c1 − δ≥ p,2i.e.,p ≥cδ.In order that the customer buy on the equilibrium path, w e mu st also have p ≤ v.Therefor e,v ≥ p ≥cδ.3. For the game in question 2.a, assume that with probability 0.001, milkm an stronglybelieves that there is som e entit y who knows what the milkm a n does and will punishhim sev erely on the da y 101 for eac h day the milkm a n dilutes the milk (by c hoosingm<1). Call this type irrational. Assum e that this is common knowledge. For somev>p>c, find a perfect B ay esian equilibrium of this game. [If you find it ea sier,take the customers at different dates different, but assume that each customer kno w swhatever the previous customers knew.]ANSWER: [It is v ery difficult to giv e a rigorous answer to this question,so you w ould get a big partial grade for an informal answ er that show sthat y ou understan d the reputation from an incomplete-in form ation pointof view.] Irrational type alway s sets m =1. Since he will be detected whenever hesets m<1 and the custom er buys, the rational ty pe will set either m =1or m =0.We are looking for an equilibrium in which early in the relation the rational milkmanwill always set m =1and the customer will alw ays buy, but near the end of the relationthe rational milkm an will mix bet ween m =1and m =0, and the customer will mixbetweenbuyandnotbuy.In this equilibrium, if the milk m an sets m<1 or the custom er does not buy at anyt, then the rational milkman sets m =0at eac h s>t. In that case, if in addition thecostume r buys at some dates in the period {t +1,t+2,...,s− 1} and if the milkmanchooses m =1at each of those days, then the costumer will assign probability 1 tothat the milkm a n is irrational and buy the milk at s; otherwise, he will not buy themilk. On the path of such play, if the milkman sets m<1 or the customer does notbuy at an y t, then the rational milkman sets m =0and the costumer does not buy ateac h s>t. In order this to be an equilibrium, the probability µtthat the milkman isirrational at such history m u st satisfyµt(100 − t)(v − p) − (1 − µt) p ≤ 0,where the first term is the expected benefit from experimen ting (if the milkman happensto be irrational) and the second term is the cost (if he is rational). That is,µt≤pp +(100− t)(v − p).Now we determine what happens if the milkman has alw ays been setting m =1,and the customer has been buying. In the last date, the rational type will set m =0,and the rational ty pe will set m =1; hence, the buy er will buy iffµ100(v − p) − (1 − µ100) p ≥ 0,3i.e.,µ100≥pv.Since we w ant him to mix, we setµ100=pv.We derive µtfor previous d ates using the B ayes’ rule and the indifference conditionnecessary for the custom er’s m ixing. Let’s w rite αtfor the probability that the rationalmilkman sets m =1at t,andat= µt+(1− µt) αtfor the total probability that m =1at date t. Since the customer will be indi fferent between buying and not buying att +1, his expected payoff at t +1 will be 0. Hence, his expected pay off from buying att isat(v − p)+(1− at)(−p) .For indifference, this must be equal to zero, thusat=pv.On the other hand, by Ba yes’ rule,µt+1=µtat.Therefo re,µt= atµt+1=pvµt+1.That is,µ100=pvµ99=³pv´2µ98=³pv´3...Note thatat=pv= µt+(1− µt) αt⇒ αt=pv− µt1 − µt.Assum e that (p/v)100< 0.001. Then, we will ha ve a date t∗such that³pv´101−t∗< 0.001 <³pv´100−t∗.At each date t>t∗, we w ill hav e