Problem Set 3 - Solutions

March 2011

Question 1 Part a. Show that every action in the support of a (mixed strategy) Nash equilibrium is independent rationalizable. Let σ∗ :=<σ∗ ...σ∗ > be a Nash Equilibrium. Take a supp(σ∗). We know: 1 N i ∈ i ui(ai,σ∗ i) ui(a, σ∗ i) a Ai. − ≥ − ∀ ∈ (Using Sofia’s solution as a guide) Proof by contradiction: Suppose that there exists a a i ∈ supp(σi∗) that is not rationalizable. Let k denote the smallest natural such that there is a player i k such that ai / IRi (where ai supp(σi∗)). ∈ ∈ k 1 This implies that ai is never an independent BR against σ i for any σ i j=i∆IRi − . Yet − − ∈×# since no strategy in the support of the (mixed) NE has yet been eliminated by stage k 1 of iterative k 1 − deletion, we know σ∗ i j=i∆IRi − and by definition ai is a BR to σ∗ i. − ∈× # − Part b. Show that every independent rationalizable action is (correlated) rationalizable. (Using Dongkyu solution as a guide) Proof by induction. If at m =0we have IR0 = R0. n n n+1 n Assume that IR R . If ai IRi we then have that σ i j=i∆IRj such that: ⊆ ∈ ∃ − ∈× # ui(σi,σ i) ui(ai$ ,σ i) ai$ . − ≥ − ∀ n n n Since we assume IR R we then have that σ i j=i∆Rj . By the above condition, it ⊆ − ∈×# follows that a Rn+1. Thus if a IRn = a Rn. This holds for all i. Thus IRn Rn i ∈ i i ∈ i ⇒ i ∈ i ⊆ which implies that IR R. ⊆ Part c. Show that every action in the support of a correlated equilibrium is (correlated) rationalizable. (Using Sabyasachi’s solution as a guide) Proof by induction. Let a be in the support of the CE, N 0 k k µ. Since a A, where A := i=1Ai, a R = A. Assume a R . Then µ(a i ai) ∆(R i). By ∈ × ∈ ∈ − | ∈ − the definition of a CE we have:

! ai arg max µ(a i ai)ui(ai,a i). ∈ a! − | − i a i A i −!∈ −

1 k k+1 n+1 Thus ai is a BR to µ(a i ai) ∆(R i). This implies ai Ri i. Thus a R . This implies − | ∈ − ∈ ∀ ∈ a R. ∈ Part d. Show (by example) that not every action in the support of a correlated equilibrium is independent rationalizable. (Example by Sabyasachi) Let player 1 be the row player, 2 the column player and 3 the matrix player.

LR (A) U 1, 1, 11, 0, 1 D 0, 1, 00, 0, 0

LR (B) U 2, 2,.70, 0, 0 D 0, 0, 02, 2,.7 LR (C) U 1, 1, 01, 0, 0 D 0, 1, 10, 0, 1 The following is a correlated equilibrium: 1 µ(U, L, B)=µ(D,R,B)= . 2 Now assume players 1 and 2 randomize independently. Assume player 1 plays U with probability p and 2 plays L with probablity q. The payoﬀs of playing A, B or C are then:

A : p

B : .7(pq + (1 p)(1 q)) − − C :1 p − Note that if p .5 ≥ .7(pq + (1 p)(1 q)) .7p

2 Part e. Show (by example) that there exist independent rationalizable actions that are not in the support of any correlated equilibrium. (Example by Sabyasachi) LMR U 3, 32, 30, 2 M 2, 13, 13, 3 D 0, 33, 0 5, 5 − − All actions are independent rationalizable. Yet it must be the case that P (D) = 0 in any CE. Suppose not. If P (D,L) > 0 or P (D,R) > 0, player 2 has a strictly proﬁtable deviation to play M if told to play D. Yet if we have P (D,M) > 0 and P (D,L)=P (D,R)=0then player 2 can proﬁtably deviate to play L when told to play M.

Question 2 Part a. n Is R ∞ an iterative delete-non-best-response sequence? If yes, verify. If no, explain which { }n=1 property it fails to satisfy. (Answer given by Vitor) n Notice that R ∞ is (IDBR). The first requirement is simple, given by definition. The second { }n=1 requirement is also true, with equality. Just notice that the left side of the requirement is just the n+1 definition of Ri . The third requirement is also easy to verify. If there is some player i and some n n action ai Ri that is not a best response to R i then (since this is the requirement that defines ∈ − Rn+1) we know that a / Rn+1 and then Rn = Rn = Rn+1 = Rn+1. i i ∈ i ×i i * ×i i Part b. n Is IR ∞ an iterative delete-non-best-response sequence? If yes, verify. If no, explain which { }n=1 property it fails to satisfy. No. Refer to the example given in the solution to problem 1 d. B is not independent rational- 1 izable. In other words B/IR3. Yet we see from the correlated equilibrium that B is a BR to the ∈ 1 1 n correlated strategies µ 3 where µ 3(U, L)=µ 3(D,R)= . Thus we see B X3 . Thus IR n∞=1 − − − 2 ∈ { } fails to satisfy property 2.

Part c. Do iterative delete-non-best-response sequences converge? If yes, what can you say about their limits? (Answer given by Vitor) By the deﬁnition of (IDBR), the sequence not necessarily converges. I will provide one example. Consider the following game 1, 11, 0 . 0, 10, 0 " #

3 Name the strategies as A = U, D and A = L, R (usual meaning). Now define the following 1 { } 2 { } sequence of sets of action profiles: for the first 2 periods

0 Xi = Ai for all i; X1 = U and X1 = L , 1 { } 2 { }

and for periods n 2 ≥

n U , if n even; X1 = { } U, D if n>1 odd ${ } and Xn = L . 2 { } In the first 2 periods it clearly satisfies the requirements since they are equal to the rationalizability steps. After the second period I have to check the requirements. First notice that the set of best responses for player 1 is always U and for player 2 it is always L . Then they are always { } { } contained in Xn. Also it is only the case that in odd periods n there is a strategy (namely, D) that is not a best response to Xn, and indeed we have that Xn+1 = Xn. (In fact, subsequent sets are 2 1 * 1 always different by construction). Then the 3 requirements are satisfied and Xn is (IDBR). n { } Notice now that X1 n∞=1 oscillates each period, and thus is not convergent. Hence the same is n { } true for X ∞ even though this is a pretty simple case. { }n=1 The correction necessary to rule out such a trivial example (that turns the definition of (IDBR) uninteresting) is to require that the sequence of sets Xn be a decreasing one. This is in better { } accordance with the word “delete” in the question, since we require that the set be reduced whenever possible (instead of just requiring that it is different). n If we include in the definition that the sequence X ∞ has to be a decreasing one, then it is { }n=1 always the case that it converges after a finite number of steps (we can only eliminate something finitely many times). n n n We can also say that, if X n∞=1 is (IDBR), then Ri Xi for all n N. Notice that 0 0 { } ⊆ ∈ Xi = Ri = Ai. From property 2 of (IDBR) we have that

0 0 1 1 ai Xi ai is a best response to X i = Ri Xi . ∈ | − ⊆ Now, for some arbitrary% n, assume that Rn Xn for all i&. Then we have that, for each i i ⊆ i n+1 n n n n n+1 Ri ai Ri ai is a best response to R i ai Xi ai is a best response to X i Xi , ≡ ∈ | − ⊆ ∈ | − ⊆ % & %n+1 & the middle is justified since in the definition of Ri we are considering player i’s action from a ⊆ n n smaller set (since Ri Xi ) and the requirement we are imposing is stronger (it has to be a best ⊆ n n n response to something also in a smaller set, since R i X i). We conclude that Xi∗ = n 1Xi − ⊆ − ∩ ≥ ⊆ Ri∗. But notice that for any i, if ai Xi∗, this means that ai is a best response to some a i X∗ i. ∈ n − ∈ − If this was not the case, then for n sufficiently large X i = X∗ i and then ai is not a best response − −

4 n to X i, which means (by 3) that for n sufficiently large we must always eliminate something from n − Xi , which is impossible since this is a finite set. This already implies that Xi∗ = Ri∗ (this is an I alternative definition of the sets (Ri∗)i=1), but I will prove that they are equal below using the sequential defintion from class. Now suppose, by way of a contradiction, that Xi∗ = Ri∗ for some i. Then take some action * n1 ai Xi∗ Ri∗. Since ai / Ri∗ we know that there exists some step n1 such that ai / Ri , but since ∈ \ ∈ n1 ∈ ai Xi∗ (and it’s a decreasing sequence of sets) we know that ai Xi , which means (from the ∈ n∈1 1 previous paragraph) that ai is a best response to some a i ∆ X i− which must not have been − n1 1 ∈ − in ∆ R i− (otherwise ai would be a BR). Then there exists at least some other player j and − n 1 n 1 n 1 ' ( a X 1− R 1− a R 1− n

Question 3

(Answer given by Vitor) In this exercise we consider an alternative formal way to describe the email information system discussed in class. Instead of using the type space structure, we can describe the situation as a probability space, in which an element would describe totally the relevant information for this model, and we can describe the diﬀerential information of each player through “coarse” partitions of the sample space. This means that no agent knows everything that happens, but can distinguish over a set of situations. Consider the sample space Ω= 0, 1, 2, ... , in which an element ω Ω represents how many { } ∈ messages have been received in total (summing up player 1 and player 2). I should emphasize that we consider as a message the fact that player 1 learns that the state is θg. So ω =0means that we are in state θb and so no one receives any messages. In the situation ω =1, the state is θg, so player 1 receives the initial message and sends his ﬁrst message to player 2, which does not reach its destiny. Notice that an element describes completely all randomness in the model (from the possible states and the possible errors in the messages). So this description does not throw away any relevant dimension of the problem. Using the same probabilities described in class we have that ω P (ω) = (1 ε) ε. − Now we describe the information of each agent in this model. No one know exactly everything that happens. Player one’s information is given by the sets = 0 , 1, 2 , 3, 4 , ... , P1 {{ } { } { } } since player 1 knows when the state is bad, i.e. ω =0, and then he knows that he has received n emails, but the other agent might have received n 1 messages (and ω =2n 1) or he might have − − received n (and then ω =2n). So his information sets are of the form (2n 1, 2n), for n integer. − Player 2’s information is given by = 0, 1 , 2, 3 , ... . P2 {{ } { } }

5 If player 2 has received 0 messages, it might be the case that player 1 also did not receive any (ω = 0) or it might be the case that ht received one, then sent a message to player 2 and it did not reach (ω = 1). Then whenever player 1 has received n messages, he consider as possible that player 2 has also received n messages or that player 1 might have received n +1messages. Then his information sets are of the form (2n, 2n + 1) for n integer. For each player i denote as P (ω) as the set in the partition that contains ω. i Pi The interpretation is that whenever ω happens, player i can only know that something in Pi (ω) happened, but cannot differentiate between anything inside this set. Then the equivalent of the type of an agent in the original model, which describes all information available to him, we know have the element of the partition that he recognizes as realized. Then the equivalent of a type profile is a double (P ,P ) . Remember that, as in the previous case, most of this elements have 1 2 ∈P1 ×P2 zero probability. Now an event is any set E Ω. So the event “state is good” can be written as E = 1, 2, .... . ⊆ g { } Similarly to the case described before, we can say that player i knows event E at partition Pi if P E. And the knowledge operator can be defined as K (E)= ω Ω P (ω) E . So we have i ⊆ i { ∈ | i ⊆ } that K1 (Eg)=Eg and K2 (Eg)= 2, 3, ... . { } n n n 1 Similarly we can define K = iKi (E) and C (E)= n 0K (E) where K (E)=K K − (E) . ∗ ∩ ∩ ≥ ∗ ∗ ∗ ∗ Notice that it is also the case that C (E )= . g ∅ ' ( Question 4

Example (OR Ex. 56.3). Find the rationalizable actions of each player in the following game. Since this is a 2 player game, we know that the set of actions surviving iterative strict dominance is equivalent to the set of actions which are rationalizable. 1 1 Note that 2 b1, 2 b3 strictly dominates b4. Thus we may truncate the game and eliminate b4. Then in the truncated game, a4 is strictly dominated by a2. Since we can check there are no more dominated actions, we see R∗ = a ,a ,a and R∗ = b ,b ,b . 1 { 1 2 3} 2 { 1 2 3} Question 5 Part a.

By solving the agents ﬁrst order conditions, we ﬁnd the unique NE is given by a = where: 1 a = i. i I +1 ∀

I=2 Note that A [0, 1], and g (a)=a (1 a a ). We may rewrite g as: i ∈ i i − 1 − 2 g (a)=a (1 a ) a a . i i − i − i j We then see that: 1 1 1 g ( ,a )= a >g(a ,a ) a (.5, 1],a [0, 1] i 2 j 4 − 2 j i j ∀ i ∈ j ∈

6 1 1 since 4 >ai(1 ai) ai (.5, 1],aj [0, 1] and 2 aj > aiaj ai (.5, 1],aj [0, 1]. Thus 1 − ∀ ∈ ∈1 − − ∀ ∈ ∈ all ai > 2 are strictly dominated by ai = 2 . 1 1 Thus Ri = [0, 2 ]. n n n Assume Ri =[low , high ] (note by symmetry we need not index this by i). We already know 1 n+1 it must be the case that 0 low high 2 . We know want to ﬁnd Ri . If agent i believes ≤ ≤ ≤ n agents j plays the strategy f(aj) with support over Rj we have that i’s best response is given by:

1 E[aj] a = − . i 2

n n n+1 1 highn 1 lown n Since we know E[aj] [low , high ] we get that Ri =[ − 2 , − 2 ] Ri . Noting that 1 1 1 ∈ ∩ low =0, high = 2 we see: 1 highn lown+1 = − 2 and 1 lown highn+1 = − . 2 This implies: 1 highn 1 − 1 highn highn+2 = − 2 = + 2 4 2 and 1 lown lown+2 = + . 4 2 1 1 1 Thus using the starting points high = 2 and low =0we see: 1 lim highn = lim lown = . n n →∞ →∞ 3 Thus we see the NE outcome is the only rationalizable, and thus independent rationalizable outcome.

I 3 ≥ 1 1 2 1 Using the same argument as above, we can show that Ri = [0, 2 ]. We can then see Ri = [0, 2 ]. 1 The reason behind this is is that x [0, 2 ] is a best reply to a i = where 1 2x 1∈ 1 − 1− 2 1 aj = I− 1 i = j. Since x [0, 2 ] we know a i R i. This gives us Ri = Ri = [0, 2 ], which − ∀ * 1 ∈ − ∈ − implies Ri∗ = [0, 2 ] i. ∀ 1 Precisely the same argument can be used to show that IR∗ = [0, ] i. i 2 ∀ Part b. (Answer provided by Vitor)

7 Nash Equilibria: Now let’s characterize an arbitrary Nash equilibria a¯. Suppose (WLOG) that firm 1 produces 0, this means that 1 γ a 0. − j ≤ j 2 !≥ Then we have that a 1 > 0 and so at least some firm is producing positively. Take a firm j>2 j ≥ i that has a¯i > 0, then )

g (¯a) =a ¯ 1 a¯ γ a¯ i i  − i − j j=i !#   =¯a 1 γ a¯ (1 γ)¯a < 0, i  − j − − i j 2 !≥ but this is impossible since each firm has to have payoffat least zero! Then all firms produce positively, which means that 1 γ j=i a¯j a¯i = − # , )2 ¯ then summing up over i (and using Q = i a¯i), 1 Q¯ =) I γ (I 1) Q¯ 2 − − ' ( ⇓ I Q¯ = . 2+γ (I 1) − And we also have

1 γ j=i a¯j 1 γQ¯ + γa¯i a¯i = − # = − )2 2 1 γQ¯ a¯ = − i 2 γ − I 1 γ 2+γ(I 1) 1 a¯ = − − = . i 2 γ 2+γ (I 1) − − So this is the only Nash Equilibria of this game.

Rationalizable strategies: 0 we start with Ri = [0, 1]. Once again we can look at best responses of only pure strategies by players i since everyone only cares about the expected joint production of the opponents (and − the pure strategy set is convex). Since the best response function is continuous, the set of best k responses to Ri is going to also be an interval, with limits 1 γ (I 1) q1 = max 0, − − ; 2 . / 1 q¯1 = . 2

8 ¯1 1 1 So we have Ri = q , q¯ for all i. Notice that q1 > 0 if 0 1 1 >I 1. γ − And q1 =0if 1 I 1. γ ≤ − So now we have two possibilities: 1) If q1 =0, then we have that 1 γ0 1 q¯2 = − = , 2 2 and (I 1) 1 γ − q2 = max 0, − 2 , $ 2 2 Once again there are 2 possibilities

q2 =0

/ 1 I 1 − . γ ≤ 2

Or we have that q2 > 0 if 1 I 1 > − . γ 2 1 I 1 2 2 2 1 1 1 1 So in case − we have that R = q , q¯ = R = q , q¯ and so we have that R∗ = 0, . γ ≤ 2 i i i 2 1 > I 1 q2 > 0 q¯2 = 1 Now in the case where γ −2 , we0 have1 that 0 and1 2 , so from this period0 one1 we have that 1 γ (I 1)q ¯k qk+1 = − − ; 2 1 γ (I 1) qk q¯k+1 = − − . 2 This means that best responses will always be strictly positive from then on. Given the assumption on parameters, we have that this mapping is a contraction. We also have 1 1 γ(I 1)q∗ that, using as q∗ = 2+γ(I 1) = − 2− (given the solution of Nash equilibria) we have that −

k+1 γ(I 1) k q∗ q = − q¯ q∗ ; − 2 − k+1 γ(I 1) k 'q¯ q∗( = − 'q∗ q ( . − 2 − ' ( ' γ((I 1) And remember that the case we are focusing is equivalent to 2− < 1, then both this distances converge to zero and then 1 R∗ = q∗ = . i { } 2+γ (I 1) . − / And the only rationalizable strategy is the NE one.

9 Part b) So the requirement needed so that Rationalizable strategies and Nash Equilibrium concepts are the same is that γ (I 1) − < 1. 2 which means that either the number of firms is not very large (so that for reasonable production levels every firm still considers producing positively) or it is the case that substitution between firms is very small (and each firm does not care so much about what others are producing). In the case γ (I 1) − 1, 2 ≥ 1 we have that Ri∗ = 0, 2 and rationalizability has considerably less predictive restrictions than Nash. 0 1 Question 6

(Answers provided by Vitor)

Part a. Characterize the set of correlated equilibria. First we solve ﬁr the set of correlated equilibria, I will name the probabilities as

p p 11 12 . p p " 21 22 # So we have to make sure that, whenever an agent plays some action with positive probability, this action is optimal given the conditional expected play for the other agents. For player 1 we have 2 optimality conditions. The condition for him to play “Hawk” is

p 0+p 5 p 1+p 4 p p , (1) 11 12 ≥ 11 12 ⇔ 12 ≥ 11 and the condition for him wanting to play “Dove” is

p 1+p 4 p 0+p 5 p p . (2) 12 22 ≥ 21 22 ⇐⇒ 21 ≥ 22 For player 2. The condition under which he wants to play “hawk” is

p 0+p 5 p 1+p 4 p p , (3) 11 21 ≥ 11 21 ⇐⇒ 21 ≥ 11 and the condition for him to want to play “Dove” is

p + p 4 p 0+p 5 p p . (4) 12 22 ≥ 12 22 ⇐⇒ 12 ≥ 22 Notice that each of this conditions has to hold whenever an agent plays with positive probability an action. And in the case where an agent does not play an action with positive probability, the conditions are also satisﬁed trivially (since both sides are zero). Therefore these have to be true in any Correlated Equilibria.

10 So we have a characterization of the set of correlated equilibria, CE: CE = (p ,p ,p ,p ) ∆(A) (1) , (2) , (3) and (4) hold . { 11 12 21 22 ∈ | } Basically the requirement for a distribution to be a CE is that the diagonal entries are not played with large probabilities. This is so because whenever an agent believes that a diagonal play is occurring with large probability, he has an incentive to deviate (people do not want to “hawk” each other and to not want to “dove” each other).

Part b. Characterize the set of expected payoﬀvectors that players can achieve in correlated equilibrium. See graph in answer to f.

Part c. Characterize the set of Nash equilibria. In this game we have 3 possible NE (2 pure and one mixed) 1 1 1 1 NE = (hawk, dove) , (dove, hawk) , , , , . 2 2 2 2 . 33 4 3 44/ Part d. Characterize the set of expected payoffvectors that players can achieve in Nash equilibrium. 5 5 The payoffs vectors achieved with these 3 equilibria respectively are (5, 1) , (1, 5) and 2 , 2 . ' ( Part e. Find the correlated equilibrium that maximizes the sum of the players payoffs and minimizes the sum of the players payoffs. Since the set of correlated equilibria is defined by a set of linear inequalities, characterizing the symmetric optimal allocation is a simple linear programming problem. we want to find max p = arg max 8p22 + 6 (p12 + p22) s.t. p CE. ∈ Clearly if we were not constrained by equilibrium conditions, the optimal would be to just play (p22 = 1) and get payoffs (4, 4) (which have maximal sum =8). Clearly this is not in CE since each player would want to deviate do “Hawk”. Then binding constraints will be the ones for each player playing dove optimally. So the solution (given the linearity of the problem) is basically to reduce p22 and increase p12 and p21 so that the binding constraint holds with equality, i.e.,

p12 = p21 = p22.

Using p11 =0is clearly optimal since this is the worst possible realization in terms of payoﬀand p11 =0makes the other incentive constraints hold always. So the optimal is 1 p∗ = p∗ = p∗ = ; 12 21 11 3 p11∗ =0.

11 10 10 20 And the payoffvector achieved is 3 , 3 which has joint payoffof 3 (notice that it is bigger than the joint payoffof any NE). ' ( Part f. Plot your answers to (b) and (d) on a graph.