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Northwestern University ECON 410-3: Spring 2015 Problem Set 1: Solutions Prof. Jeff Ely TAs: Arjada Bardhi Egor Starkov Problem 1 8.C.4 Page 1 of 4 Northwestern University Problem 1 ECON 410-3: Spring 2015 Problem Set 1: Solutions Prof. Jeff Ely TAs: Arjada Bardhi Egor Starkov Problem 2 (a) a0i strictly dominates ai , so by definition: ui (a0i , a−i ) > ui (ai , a−i ) for any a−i ∈ A−i . Consider an arbitrary α−i ∈ ∆A−i , let S(α−i ) ⊆ ∆A−i be the support of this mixed strategy. Then, X ui (a0i , α−i ) = α−i (a−i )ui (a0i , a−i ) a−i ∈S(α−i ) ui (ai , α−i ) = X α−i (a−i )ui (ai , a−i ) a−i ∈S(α−i ) Comparing the right hand sides, it immediately follows that ui (a0i , α−i ) > ui (ai , α−i ). (b.i) As a reminder ai is strictly dominated by a0i . From αi construct a new mixed strategy αi0 such that: αi0 (a0i ) = αi (a0i ) + αi (ai ) αi0 (ai ) = 0 αi0 (ãi ) = αi (ãi ) for any ãi 6= ai , a0i So, αi0 is identical to αi except that all the probability weight given to ai in the old mixed strategy has now been transferred to a0i in the new mixed strategy. Then, X X ui (αi , α−i ) = αi (ãi )ui (ãi , α−i ) = αi (ai )ui (ai , α−i ) + αi (a0i )ui (a0i , α−i ) + αi (ãi )ui (ãi , α−i ) ãi 6=ai ,a0i ãi ∈Ai < αi0 (a0i )ui (a0i , α−i ) + X αi0 (ãi )ui (ãi , α−i ) = ui (αi0 , α−i ). ãi 6=ai ,a0i Q (b.ii) As above, αi (ai ) > 0, but now α ∈ ∆ ( i Ai ). The proof of this claim extends the idea of part b.i. Consider all action profiles ã = (ai , ã−i ) for which α(ã) > 0. Modify these profiles such that ã0 = (a0i , ã−i ) and construct a new distribution over action profiles α0 such that α0 (ã0 ) = α(ã) + α(α̃0 ) for all ã−i , and α0 (a) = α(a) for all other action profiles. From strict dominance, we know that ui (ã0 ) > ui (ã). From here, it follows that ui (α) < ui (αi0 , α−i ). Note: if we interpret the problem as finding a mixed strategy αi0 in the formal definition of the one (i.e. αi0 is not allowed to be correlated with a−i ) then the claim is not true. As a counterexample consider a two-player game where the payoffs of the row player are described by the table below: l r t 0 3 m 1 4 b 3 0 Obviously, m strictly dominates t. Consider an action distribution α = 12 (t; r) ⊕ 12 (b; l), i.e. it assigns probabilities 21 to outcomes (t; r) and (b; l) and zero to all other outcomes. The marginal on column player’s actions is then α−i = 21 l ⊕ 12 r. Any row player’s mixed strategy αi0 can be decribed as (γ, δ) such that αi0 = (1 − γ − δ) t ⊕ γm ⊕ δb and given α−i it yields u (αi0 ; α−i ) = (1 − γ − δ) 21 0 + 12 3 + γ 12 1 + 12 4 + δ 12 3 + 12 0 = 32 + γ < 3 ∀γ ∈ [0; 1] so no mixed strategy αi0 can beat α for the row player. Page 2 of 4 Northwestern University Problem 2 ECON 410-3: Spring 2015 Problem Set 1: Solutions Prof. Jeff Ely TAs: Arjada Bardhi Egor Starkov Problem 3 For the first two parts let the symmetric strategy set SM = [2, M ] ∩ N with M ≥ 3, we will show that M is not a BR. Applying the following reasoning for the more specific case of M = 100 provides an answer for part (a). The more general argument is required for understanding the set of rationalizable actions for part (b). For any belief about the opponent’s actions, p, denote K = max{i : p(s2 = i) > 0} and by p and p the belief assigned to the opponent playing K and K-1 respectively. • K = M. u(M, p) = p ∗ M + p ∗ (M − 3) + (1 − p − p)Ep [s2 − 2|s2 <M −1 ] u(M − 1, p) = p ∗ (M + 1) + p ∗ (M − 1) + (1 − p − p)Ep [s2 − 2|s2 <M −1 ] ⇒ u(M, p) < u(M − 1, p) The third term does not change, as conditional on s2 ≤ M − 1, M and M-1 will both always lose and provide a payoff of s2 − 2. As this is true for an arbitrary p, M is never a BR. • K < M. u(M, p) = Ep [s2 − 2] = pEp [s2 − 2] + (1 − p)Ep [s2 − 2] u(K, p) = p ∗ (K) + (1 − p)Ep [s2 − 2|s2 <K ] ⇒ u(K, p) − u(M, p) > 2p > 0 So M is not a BR for this case either. Because any M ≥ 3 is not a BR, it follows that the only action that survives deleted iteration of non-BR is 2, and thus a = (2, 2) is the unique rationalizable outcome. (c) Define the action σ by ( 10−3∗(99−i) P r(σ = i) = P96 1 −3i ≈ 1 − 1000 i=0 10 if i < 99 998 999 if i = 99 By playing a = 100 the payoff is 100 if opponents plays 100 and s2 − 2 for all other cases. Also note, that against and opponent playing s2 = x all strategies larger than x provide the same payoff. • (s2 = 100). The new strategy has a payoff s. greater than • (s2 = 99) The new strategy has a payoff s. greater than 998 999 998 999 ∗ 101 > 100. ∗ 99 > 97. • (3 < s2 < 99) Then a and σ have different payoffs iff the realization of σ is no greater then s2 . In these cases, σ provides a strictly greater payoff if s1 ∈ {x, x−1} and a weakly lower payoff otherwise. However, Problem 3 continued on next page. . . Page 3 of 4 Northwestern University Problem 3 (continued) ECON 410-3: Spring 2015 Problem Set 1: Solutions Prof. Jeff Ely TAs: Arjada Bardhi Egor Starkov note that the decrease in payoffs is at most 100. The increase in payoffs is 10−297+3x ∗2+10−297−3x−3 ∗3, whereas the decrease in payoffs is less than 100 ∗ 10−297+3x−6 ∗ 100 = 10−297+3x−2 . Thus the gains are greater than the losses for any x. • (s2 ≤ 3) In this case σ never gives a payoff that is worse then that of a, and with a probability of at least 10−291 it gives a higher payoff (2 instead of zero). Thus for all possible strategies of the opponent σ provides a strictly higher payoff than a. Problem 4 (a) Consider bidding a bid b ≥ 1 and let b0 denote the bid of the other player. If b0 < 1, then you win with a payoff 1 − b ≤ 0, while bidding b = 0 gives you a payoff of zero. If b0 ≥ 1 but b > b0 , you win with a payoff 1 − b < 0, while again bidding zero gives you a payoff of zero. IF b0 ≥ 1 but b0 > b you lose the auction and obtain a payoff of zero, while bidding b = 0 would have given you a payoff of zero. So, b = 0 weakly dominates any b ≥ 1. (b) Let us consider first if b = 0 is a rationalizable action. Suppose the other player plays b0 = 0 with probability β > 0. Then, if you play b = n1 instead of b = 0 you will win with probability at least β, so your payoff from b = n1 is at least β(1 − n1 ) while your payoff from b = 0 is β/2. But n > 2, so β(1 − n1 ) > β2 . If instead the other player always plays bids larger than 0, bidding an appropriately chosen b 6= 0 gives a strictly positive payoff in expectation while bidding b = 0 gives a zero payoff. So, b = 0 is never a best reply, so it not rationalizable. By induction, let’s suppose we have eliminated all bids smaller than nk for any k < n − 2. So, the surviving n−1 k set is { nk , k+1 n , ..., n }. Suppose the other player bids n with probability α and higher bids with probability k 1 − α. Then, the payoff from bidding n is: (1 − nk ) α 2 while the payoff from bidding k+1 n is at least: α(1 − Bidding k+1 n k+1 ). n is a BR as long as k < n − 2. Therefore the set of rationalizable actions is: n−2 n−1 , n n . (c) All bids b ≥ 1 remain weakly dominated, by a similar argument to part a. But, any bid bi ∈ [0, +∞) is now rationalizable. To see this, notice that an arbitrary bi is a best response to any b−i ∈ (max{bi , 1}, +∞), because if b−i is such a high bid, the best i can do is to lose the auction by bidding bi . Page 4 of 4