Problem Set 3 Micro, Susen Wang Try to do most problems in MWG (1995), Chapters 7—9. Question 3.1. (Mixed-Strategy Nash Equilibrium). A principal hires an agent to perform some service at a price (which is supposed to equal the cost of the service). The principal and the agent have initial wealth w p =1.3 and w a =0.5, respectively. The principal can potentially lose l =0.8. If the agent o?ers low quality, the probability of losing l is P 1 =80%;if the agent o?ers high quality, the probability of losing l is P 3 =50%. The quality is unobservable to the principal. The price of a low quality productis(paidtotheagent)isc 1 =0.08 and the price of a high quality product is c 3 =0.2; by the competitive market assumption, c 1 and c 3 are the costs of producing the products (the agent bears the costs). The agent is required by regulation to provide high quality, but he may cheat. After the bad event happens, the principal can spend F =0.32 for an investigation; if the agent is found to have provided the low quality, the agent will have to pay for the loss l totheprincipal.Thisgamecanbewritteninthe following normal form: low quality, α high quality, 1?α investigate, ρ w p ?c 3 ?P 1 F, w a + c 3 ?c 1 ?P 1 l w p ?c 3 ?P 3 (F + l),w a not to investigate, 1?ρ w p ?c 3 ?P 1 l, w a + c 3 ?c 1 w p ?c 3 ?P 3 l, w a where ρ = probability of principal investigating, α = probability of agent delivering low quality. Find mixed-strategy Nash equilibria. Question 3.2. (Pure-Strategy Nash Equilibrium). Find pure-strategy Nash equi- libria for the above exercise. 3—1 Question 3.3. For the game in Example 3.11, find all the pure strategy Nash equilibria. Obviously, among these Nash equilibria, only the one that is found by backward induction satisfies sequential rationality. Question 3.4. For Example 3.13, find the mixed strategy SPNE. Question 3.5. Find all the pure-strategy Nash equilibria of the game in Example 3.20. Question 3.6. [A revised version of Exercise 9.C.7 in MWG (1995, p.304)]. (a) For the following game, find all the pure-strategy NEs. Which one is the SPNE? o P2 ? ? ? ? ? ? 2 4 .. P1 ? ? ? ? ? ? 2 2 1 δ 2 δ 1 γ 2 γ 1 δ 2 δ B T D U D U ? ? ? ? ? ? 1 1 ? ? ? ? ? ? 1 5 P2 (b) Now suppose that P2 cannot observe P1’s move. Draw the game tree, and find all the mixed-strategy NEs. (c) Following the game in (b), now suppose that P1 may make a mistake in implementing his strategies. Specifically, after P1 has decided to play T, hemayactuallyimple- ment T with probability p and mistakenly implement B with probability 1 ? p; symmetrically, after P1 has decided to play B, he may actually implement B with probability p and mistakenly implement T with probability 1?p. 1 Draw the game tree and find all the BEs. 1 In MWG (1995), it is P2 who may make a mistake in observing P1’s strategies. In this case, there is no mistake in implementation; it is just a mistake in identifying the acutal strategy. 3—2 Answer Set 3 Econ522, Susen Wang Answer 3.1. Assume that the principal can commit ex ante to investigate or not before a loss occurs. In other words, the principal can only make up his mind on investigation before she has su?ered a loss. Before a loss occurs, the game box of surpluses is low quality, α high quality, 1?α investigate, ρ w p ?c 3 ?P 1 F, w a + c 3 ?c 1 ?P 1 l w p ?c 3 ?P 3 (F + l),w a not to investigate, 1?ρ w p ?c 3 ?P 1 l, w a + c 3 ?c 1 w p ?c 3 ?P 3 l, w a In each cell, the value on the left is the surplus of the principal and the value on the right is the surplus of the agent. The optimal choice of α is to make the principal indi?erent between investigation and no investigation: w p ?c 3 ?αP 1 F ?(1 ?α)P 3 (F + l)=w p ?c 3 ?αP 1 l?(1 ?α)P 3 l, (5) implying ?αP 1 F ?(1?α)P 3 F = ?αP 1 l, implying α(P 1 l + P 3 F ?P 1 F)=P 3 F, implying α = P 3 F P 1 l + P 3 F ?P 1 F = 0.5 ×0.32 0.8 ×0.8+0.5 ×0.32?0.8× 0.32 =0.29. The choice of ρ is to make the agent indi?erent between cheating and no cheating: w a + c 3 ?c 1 ?ρP 1 l = w a , (6) implying ρ = c 3 ?c 1 P 1 l = 0.20 ?0.08 0.8 ×0.8 =0.19. Answer 3.2. By substituting the parameter values into the game box of surpluses, we have 3—3 cheat, α not to cheat, 1?α investigate, ρ 0.84, ?0.02 0.54, 0.5 not to investigate, 1?ρ 0.46, 0.62 0.7, 0.5 By Proposition 3.2, to find pure-strategy Nash equilibria, we can restrict to pure strategies only. Thus, simply by inspecting each cell one by one, we know that there is no pure- strategy Nash equilibrium. Answer 3.3. The strategy sets for players 1 and 2 are simple: S 1 = {L,R}, S 2 = {a,b}. There are three information sets for player 3. Denote a typical strategy of player 3 as s 3 =(a 1 ,a 2 ,a 3 ), where a 1 is the action if the information set on the left is reached, a 2 is the action if the information set in the middle is reached, and a 3 is the action if the information set on the right is reached. Player 3 has eight strategies: s 13 =(l,l,l),s 23 =(r,l,l),s 33 =(l,l,r),s 43 =(r,l,r), s 53 =(l,r,l),s 63 =(r,r,l),s 73 =(l,r,r),s 83 =(r,r,r). The normal form is P1 plays L P3 s 13 s 23 s 33 s 43 s 53 s 63 s 73 s 83 P2: a 2,0,1 -1,5,6 2,0,1 -1,5,6 2,0,1 -1,5,6 2,0,1 -1,5,6 b 2,0,1 -1,5,6 2,0,1 (-1,5,6) 2,0,1 -1,5,6 2,0,1 (-1,5,6) P1 plays R P3 s 13 s 23 s 33 s 43 s 53 s 63 s 73 s 83 P2: a 3,1,2 3,1,2 3,1,2 3,1,2 (5,4,4) (5,4,4) (5,4,4) (5,4,4) b 0,-1,7 0,-1,7 -2,2,0 -2,2,0 0,-1,7 0,-1,7 -2,2,0 -2,2,0 All the pure strategy Nash equilibria are indicated in the boxes. 3—4 To find all the Nash equilibria, we can check each cell one by one. A cell cannot be a Nash equilibrium if one of the players doesn’t stick to it. In each cell, we can first check to see if player 3 will stick to his strategy, by which we can quickly eliminate many cells. A sequentially rational NE must be an outcome from backward induction. Example 3.11 shows that backward induction only leads to one outcome: s 1 = R, s 2 = a, s 3 = s 63 , which is one of the Nash equilibria. Answer 3.4. In the proper subgame with the normal form: Firm I Small, σ 2 Large, 1?σ 2 Firm E: Small, σ 1 -6, -6 (-1, 1) Large, 1?σ 1 (1, -1) -3, -3 The equilibrium σ 2 is to make firm E indi?erent between his two strategies: ?6σ 2 ?1(1 ?σ 2 )=σ 2 ?3(1 ?σ 2 ), implying σ ? 2 = 2 9 . Sincethegameissymmetric,wealsohaveσ ? 1 = 2 9 . Then, the expected payo? is π ? I = ?6σ ? 2 ? 1(1 ?σ ? 2 )=? 19 9 . We also have π ? E = ? 19 9 . Thegameisreduced to: o Out In ? ? ? ? ? ? ? ? 9 19 9 19 Firm E ? ? ? ? ? ? 2 0 Then, firm E will choose ‘out’. Thus, the SPNE is σ ? E =(out, choose small niche with probability 2 9 ), σ ? I = choose small niche with probability 2 9 . Answer 3.5. Firm I has one information set H I containing two nodes. Based on this information, firm I has two strategies: s 11 = Fight, s 21 = Accom. 3—5 Firm E has two informations H 1 and H 2 , where H 1 contains the initial node. Denote firm E’s strategies as s 2 =(a 1 ,a 2 ), where a 1 is an action at H 1 and a 2 is an action at H 2 . We can then find the normal form: Firm E (out, fight) (out, accom.) (in, fight) (in, accom.) Firm I: fight (2, 0) (2, 0) -1, -3 -1, -2 accom. 2, 0 2, 0 -2, 1 (1, 3) We can easily find the pure-strategy Nash equilibria, as indicated in the above box. Among these three NEs, there is one SPNE, which is s I = accom., s E =(in, accom.), and there is one BE, which is s I = fight, s E =(out, accom.). This example indicates that 1. BE and SPNE don’t imply each other. 2. BE eliminates two NEs, one of which is SPNE. SPNE also eliminates two NEs, one of which is BE. Answer 3.6. (a) There are two information sets for P2. Let (a 1 ,a 2 ) be a typical P2’s strategy, where a 1 is an action taken at the left information set and a 2 is an action taken at the right information set. The normal form of the game is P2 (D, D) (D, U) (U, D) (U, U) P1: B 4, 2 (4, 2) 1, 1 1, 1 T 5, 1 2, 2 5, 1 (2, 2) There are two pure-strategy NEs: σ ? =[B, (D,U)] and σ ? =[T, (U,U)]. The first one is the SPNE. 3—6 (b) Thegametreeis: o P2 ? ? ? ? ? ? 2 4 .. P1 ? ? ? ? ? ? 2 2 1 δ 2 δ 2 H 1 γ 2 γ 1 δ 2 δ B T D U D U ? ? ? ? ? ? 1 1 ? ? ? ? ? ? 1 5 The normal form is P2 DU P1: B 4, 2 1, 1 T 5, 1 (2, 2) There is a pure-strategy NE: σ ? =(T, U). Since playing T is a strictly dominant strategy for P1, this NE is the only mixed-strategy NE. (c) Thegametreeis o P2 ? ? ? ? ? ? 2 4 .. P1 21 )1( γγ pp ?+ ? ? ? ? ? ? 2 2 1 δ 2 δ 2 H 1 γ 2 γ 12 )1( γγ pp ?+ 1 δ 2 δ B T D U D U ? ? ? ? ? ? 1 1 ? ? ? ? ? ? 1 5 Figure 3.6 In this game tree, the beliefs are μ 1 = pγ 1 +(1?p)γ 2 , μ 2 = pγ 2 +(1?p)γ 1 , whicharederivedfromBayesrulebyallowingthepossibilityofanerrorinimplemen- tation, where γ i ≥ 0 and γ 1 + γ 2 =1. We have μ 1 + μ 2 =1. We can also have the 3—7 following game tree, where P2’s beliefs are also derived from Bayes rule. Since the two game trees are equivalent, we will thus use Figure 3.6 only. o P2 .. P1 1 γp 2 H 1 γ 2 γ 2 γp B T ? ? ? ? ? ? 2 4 D U ? ? ? ? ? ? 1 1 Nature p p?1 p?1p B BT T 1 )1( γp? ? ? ? ? ? ? 1 5 D U ? ? ? ? ? ? 2 2 ? ? ? ? ? ? 1 5 D U ? ? ? ? ? ? 2 2 ? ? ? ? ? ? 2 4 D U ? ? ? ? ? ? 1 1 . . . . 2 )1( γp? We now solve for BEs in the game tree of Figure 3.6. We solve by backward induction. We find D null U ? 2μ 1 + μ 2 > μ 1 +2μ 2 ? (1 ?2p)γ 2 > (1?2p)γ 1 . (7) Then, first, if D null U, P1 will choose T, i.e., γ 1 =0 and γ 2 =1. To be consistent with (7), we need p< 1 2 . Thus,wehaveoneBEwhenp< 1 2 : γ ? 1 =0 and δ ? 1 =1. Second, if D ? U, P1 will also choose T, i.e., γ 1 =0 and γ 2 =1. To be consistent with (7), we need p> 1 2 . Thus,wehaveanotherBEwhenp> 1 2 : γ ? 1 =0 and δ ? 1 =0. Third, if D ～ U, (7) implies (1 ? 2p)γ 2 =(1? 2p)γ 1 . If p null= 1 2 , we have γ 1 = γ 2 , i.e., γ i = 1 2 . P1 compares the expected profits for the two choices: π B =4δ 1 + δ 2 and π T =5δ 1 +2δ 2 . Since π T > π B , P1 chooses T, i.e., γ 1 =0and γ 2 =1, which is inconsistent with γ i = 1 2 . If p = 1 2 , we still have π B =4δ 1 + δ 2 and π T =5δ 1 +2δ 2 . Since π T > π B , P1 chooses T, i.e., γ 1 =0and γ 2 =1. Thus, we have another BE when p = 1 2 : γ ? 1 =0 and δ ? 1 can be any value in [0, 1]. In summary, we have three BEs: Error BE p< 1 2 P1 plays T, P2 plays D p> 1 2 P1 plays T, P2 plays U p = 1 2 P1 plays T, P2 plays any strategy (pure or mixed) End 3—8