博弈論基礎(chǔ)(羅伯特.吉本斯著)中國社會科學(xué)出版社課后答案

1、FOC: kUU + B)-V(p 一B) = 0*=0FOC: kUU + B)-V(p 一B) = 0*2.1 Proof: we should show that A maximizes/ (A) + / JA) 9 this is equivalent to show /(A) + /(A) = 0 Solving the game by backward induction:At stage 2. given I. and lp, parent chooses B to maximize his payoff:max BFOC: kUU +B)-V/p 一F) = 0*At stag

2、e 1, child chooses A to maximize his payoff:FOC: J(M + B)(/3 +零)=0From (*), we have:OB dB . dB 礦莎W)+可W)SD_S)sckUIc + B) + Vlp-B) p=kn廠 _n)skUVc + B) + Vlr-B) c+p= f/z(/r(A) + ).(/；(A)+-kU"(I廣 B)< /kuu +B)+y_B)八Wr-B)VVp-B)=> r(A) + /(A) = 0( Since '()> 0, V'(-) < 0 and () <

3、; 0)Another approach:Solving the game by backward induction:At stage 2、given /t and Ip. parent chooses B to maximize his payoff:max V(Zp-B)+ /:(/(/ +B) BAt stage L child chooses A to maximize his payoff: max(7(/ (A) + B)FOC: "") + 3)(廠(4) + 務(wù)=0Because U is increasing and strictly concave,

4、sow A) + B) > 0 => (Z；(A)+獸)=0(l)(*)對A求偏導(dǎo)：HT(/") + B)a：(A)+舅)-廣匕-啾/；-尊)=0 n f A) - B)(/；( A)-獸)=WT(/") + B)(/；( A) + 獸)=0 (Since(£ 伽気=o) Because V is strictly concave, so 廠(人,(力)一 3) >0 => (/；(/!)-尊)=0(2)(1)+(2) =>/：(A) + /；(A) = 02.2 Proof:At. first, solving the game b

5、y backward induction:At stage 2, given S. parent chooses B to maximize his payoff: max V(人一B) + R(qa _S) + ",(S + 3)B卩FOC: -VVp - B) + kUAS + F) = 0(*)At stage L child maximizes his payoff: max(/j(/ -5) + (/2(S + )sFOC: 一：(/ _S) + ：(S + 3)(l+|) = 0From (*). we havedB _kU；(S + B)OSV"(Ip-B)

6、+ kU；(S + B)-殲-S) + ：(S + B)(lU(S + ?)'VIp-B) + kU2 (S + B),VI B)So =-(/, (/c-S) + U. (S + B)'十 kU；(S + B)Hence, in the game, child chooses S' such that:=0"V"(Ip_B) + kU；(S、B)On the other hand, if child chooses S' to maximize (7,(/ -S) + f/2(S+ B), where B is exogenous, the

7、n S' satisfies:_Ua_S') + U；(S'+B) = 0We need to show5 < 5Z.Denote /(5) = -/(/ -S) + US + B fS) = U(lc - S) + (/(5 + B) < 0 /(5*) = -U；(IC - 5*)+ B)>0 = /(5,)So5* < S，. If child save more, i.e.S', both the parent's and child's payoffs could be increase d2.4 Solving

8、 the game by backward induction:In Period two: given q. partner 2 choose c2. the minimize of c2 to completethe project is R_cIf V > (7?-c,)2,c2 = /?-, both receive V.If V < (/? - q )2,c2 = 0. both receive 0.In Period one: partner 1 choose q Consider four cases:(1) if R < y/(-3)V , that is V

9、 -R > 8V, q = R, partner 1 will complete theproject himself. So q = 0 (2) If y/(-3)V <R<yfv,hat is V-R2 <8V and V>R cI=0,c2=/?. If (1 + V)>/V >/?>Vv , that is /?2 >V , and (R-y/v)2<3V , the minimize of q to complete the project satisfies: V = (R-c)2.cl < R. So 5=R_H,

10、 C2=y/V.(4) If 7? > (1 + 4S)V , that is (R-JV)2 > <5V , it is not worth completing the project, hence c, =05c2 = 0 2.7 In the subgame, the equilibrium is qi =厶=(a - w).+lAt stage 1, the union chooses w to maximize its utility:max (a - w)(w - w(i)* +1FOC: ci + vt; -2iv = 0 => vv = "

11、+2j (i wThen payoffs of the union are ( )2, which is increasing with n. IF nz? + l 2increases, the total output increases, so does the demand for labor, so the union' utility increases ci + c2.13 Proof: The monopoly price is p =If the firms use trigger strategies, then if there is no firm deviat

12、e, both get ,l.(£z£)2()2 on every stage game, and the total discounted profit is - . cThe payoff from deviating on an stage is ()2. For the trigger strategies to丄(叫1 be SPNE, we must have > ( )2, that is 3> .-3222.14 The monopoly price is pH =, pL =-If the firms use trigger strategie

13、s, and there is no firm deviate, in period with1 5(冷(蔦與+(1一吋(蔦與) demand , the total payoff is ( )' +,“ H 7 2 2 -8in period with demand aL , the total discounted payoff is5(加_(y +(i _)_()1 嚴(yán)-22222 2-3a e 、In period with demand aH , payoff from deviating is ()2; in period withCl C* demand aL, payo

14、ff from deviating is ().For the trigger strategies to be SPNE, we must have丄(他二£)2 +2_22_222-SAnd丄（魚二£）2 +比龍運（七2 2-81_宀 2>嚀)2 +(1蟲忤門+ di2 2 2 2 2 2The lowest of 8 such that the finns can use trigger strategies to sustain these monopoly price levels in a SPNE is:1 2 i 2=> >(宀()2(1

15、4“一Cj Z1 、1 "一c、“ . 1 4“一c、(嘗(十)+(1 一巧()巧(十)For each value of 8 between 1/2 and * suppose the highest price is p(8)when the demand is highIf the firms use trigger strategies, and there is no firm deviate, in perioddemandwiththetotalpayoffisi5（足（。一 p（5）（p（6c） + （l-龍）?。▽懮希?#39;）-8in于（伽一卩（6）（6-0）+

16、22isperiod with demand athe total discounted payoff一 P(®)(“(6 - c) + (1 -龍)1-5in)2 + 2 2 2In period with demand , payoff from deviating is (aH -;period with demand(a CjiL, payoff from deviating is ()"For the trigger strategies to be SPNE, we must havei- p(S)(p(- c) + (1 - )->(-L)-一p(6)(

17、"(3)-c) +二；3 >(a-p(3)(p(8)-c)1 一力And+十響;)+("護尹),22-S2Then we can get the highest price p(8) from these two inequalities.a-c2.15 The monopoly output isInIf the finns use trigger strategies, when there is no firm deviate, total discounted(中payoff of each firm is :.Given other firms' e

18、quilibrium strategies, payoff(H+ l)(_rfrom deviating: + 丄辛一.nl-oFor the trigger strategies to be SPNE, we must have4 一c “£<一5(F)n + l(7)_>(«-c)2 + 4/1-5(” + l) +4”(H + l)2=- . increase with n.nun (“ + 1)2+4“If S is too small suppose the output of each finn in the most-profitable symm

19、etric SPNE is q, then the total output is Q=nq.If there is no firm deviate, payoff of each firm is : a n(l c Q-8-/()r+ +12 1-6 giaCSet (a_“g_c)ga_c_(”_l)q 2十(齊T丿z z -8Payoff from deviating: (“ C 。'亠-8ci-c ( + l)2 _力( + 3)(_1)=> ac >q>、 +1/? + !(/ +1)2 一 3(n-)2a-c (n +1)2 -/( + 3)(“-1) ”

20、 + 1(/? + l)2-(/-l)22.17 The strategy of the finn: in each period, if observe output y, pays c to the worker, otherwise, pays nothingThe strategy of the worker: if the firm has paid in period t-1, expend effort c and produce output y. otherwise, expend no efforty cFor the firm, payoff from no deviat

21、ion is:-7-oy c2_- > y, that is c< oy -oPayoff from deviating is: ySo a necessary and sufficient condition is3.2 The strategy spaces for firm 1 is q?仁qfOw)The strategy spaces for finn 2 is g 0、8).Solving for the Bayesian Nash equilibrium: h = H、 cq：For firm 1: 11w人=L，兀;=(a. q： 一一cq；For firm 2:心

22、=&心” 一 4“ 一 %皿 +(1 - &)(業(yè)一力一鼻)4 一冷FOC:-rL. = aH-q2-c-2ql =04兀L品=代_綣y_2/=U%弊= &(°h -q") + (l-&)(伐-qf)-c-2偽=0十6”6t 2 + & &q'pap& 1 一&1一& 11C31 j c L 3Assume ° =4, = 2,& = ?,c = l , then = ,q! =-.ck = are all 3939positive. Then the Bayesian Nas

23、h equilibrium is(7/ =罟,q：=g.3.4 The strategy spaces for player 1 is TT.TB.BT.BB The strategy spaces for player 2 is L.R The nonnal form of the game isLRTT1/2,1/20,0TB1/2,1/21,1BT0,00,0BB0,01,1There are three pure-strategy Bayesian Nash equilibria: (TT丄)，(TB,R), (BB, R)4.1 (a) The normal-form game is

24、 shown below:/RR2,22,2L4,10,0M3.00,1The pure-strategy Nash equilibria: (R,(L, L!). These are also the SPNE, becausethere is no proper subgameIn the extensive-form game:So L. L p = 1 and R. R p < 72 are all the perfect Bayesian equilibrium(b) The normal-form game is shown belowrR'L1.31,24,0M4.

25、00.23,3R2.42,42,4The pure-strategy Nash equilibrium is This is also the SPNE.M t p = 0 t R 一> LRpelOA(Lf:3pGiven p, payoff of player 2 is: < M':2pb：3(l_p)1 . .233Givenp:= i4.2: Proof: If there is pures(nUegy perfect Bayesian equilibrium.u2(Rp) = pIf p>l/2, ap = RfaK =M>p = O. Contrad

26、ictionIf pvl/2,(ip= 1 t 也=厶=> =1 Contradiction(Another approach:厶 =M tp = OtZ/t厶RTpGlO,l,p5*TLJL)Hence there is not pure-strategy perfect Bayesian equilibrium.Now, consider mixed strategy:Suppose player 1 plays CT = prp2,l-pt- p29 that is plays L with probability p,plays M with probability p2, pl

27、ays R with probability 1 - p, - /;2.P1 + P2p< L! i L2p > 丄 T X T M2p = -a2=(qA-q) for all gwO,lL:3qGiven a2, payoffs of player 1 is M : 3(1 - q)R:2For p =丄，we need p= p20 or p】+ p： =0. For player 1 to be willing toassign positive probability to both L and M, we need 3q = 3(l-q), which implies2

28、13(7 = . Then < M : , which implies R is the unique optimal choice by player 1,R:2contradicting p】=p嚴(yán)0.12For H + /人=0, we need 2>3q and 2 n 3(1 - q)、that is - <q< .3 3121Therefore, R(J. =(q一q) with <(/</; = are all the mixed-strategy332perfect Bayesian equilibria.u :O.5xl + O.5xO = O.5R: tf=dd:05x0 + 05x2 = l *4.3(a) (R,R)t pw 0M = 0.5tu:2p“ if 斗d i