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 corre- lated 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<p − − ≤ and if p<.5 .7(pq + (1 p)(1 q)) .7(1 p) < 1 p − − ≤ − − Thus we see that B is never a BR to any belief of p and q. Thus B is not independent rational- izable. 2 Part e. Show (by example) that there exist independent rationalizable ac- tions 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).