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Interpretation of Mixed Strategy Nash Equilibria (In Games of Complete Information), Known As the “Purification Theorem”

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Interpretation of Mixed Strategy Nash Equilibria (In Games of Complete Information), Known As the “Purification Theorem” ECON 803: MICROECONOMIC THEORY II Arthur J. Robson Fall 2016 Assignment 8 (due in class on November 15) 1. This exercise concerns Harsanyi’s (1973) interpretation of mixed strategy Nash equilibria (in games of complete information), known as the “purification theorem”. Consider the following matching-pennies game: HT H 1, −1 −1, 1 T −1, 1 1, −1 (a) Find the unique Nash equilibrium of this game. Does player 1 have a strict incentive to play her equilibrium strategy? ∗ ∗ Solution: Obviously there is no pure NE. Let s1 = (p, 1 − p) and s2 = (q, 1 − q) be ∗ ∗ the mixed NE strategies. Thus, s1 and s2 must satisfy ∗ ∗ ∗ ∗ u1(H, s2 ) = u1(T, s2 ) u2(s1 , H) = u2(s1 , T). 1 From these equations it is straightforward to derive that p = q = 2 . If player 2 commits ∗ to playing s2 , then player 1 does not have a strict incentive to play s1; in fact, any mixed strategy of player 1 yields the same payoff for him. Player 1’s payoffs are now subject to a perturbation based on a parameter t, which has the value e > 0 (assume e to be “small”) with probability 1/2, or the value −e, also with probability 1/2. When the parameter is t, the game is HT H 1 + t, −1 −1 + t, 1 T −1, 1 1, −1 Player 1 is informed of the realized value of t; player 2 knows only the distribution of t. This game is then a Bayesian game, as defined by Harsanyi, and discussed in class, where t is player 1’s “type”. (b) There is an equilibrium of this modified game in which player 2 continues to play as in matching pennies. Prove this, while also describing how player 1 then chooses as a function of t. Does each type of player 1 now have a strict incentive to choose in this fashion? Page 1 of 5 Econ 803 (Fall 2016) Assignment 8 Due: November 15 ∗ 1 1 Solution: Let P2 continue to play s2 = 2 H + 2 T. P1’s best response is 8 ∗ <H if t = e s1 (t) = , :T if t = −e since ∗ ∗ u1(H, s2 ; e) = e > 0 = u1(T, s2 ; e) ∗ ∗ u1(H, s2 ; −e) = −e < 0 = u1(T, s2 ; −e). ∗ ∗ We need to show that s2 is a best response to s1 (t). It suffices to argue that P2 is ∗ indifferent between H and T when P1 plays s1 (t): 1 1 E [u (H, s∗(t))] = u (H, H) + u (H, T) = 0 t 2 1 2 2 2 2 1 1 E [u (T, s∗(t))] = u (T, H) + u (T, T) = 0. t 2 1 2 2 2 2 Therefore, 8 ∗ <H if t ≥ 0 ∗ 1 1 s1 (t) = and s2 = H + T (1) :T if t < 0 2 2 constitute a Bayesian Nash equilibrium (BNE), in which P2 continues to play as if she were in the original unperturbed game, while P1’s equilibrium strategy varies depending on his realized type. Moreover, P1 has a strict incentive to play his (pure) ∗ equilibrium strategy s1 (t). (c) What would happen to the equilibrium of the modified game and these incentives of player 1 if the distribution of t were instead uniform on [−e, e]? Solution: There is no pure strategy equilibrium. Suppose P2 plays a mixed strategy ∗ s2 = pH + (1 − p)T with p 2 (0, 1). It follows that ∗ ∗ Et[u2(H, s1 (t))] = Et[u2(T, s1 (t))]. ∗ 1 This means s1 (t) must effectively play H and T each with probability 2 . Observe that, given s2, P1 chooses H when p(1 + t) + (1 − p)(t − 1) ≥ −p + 1 − p ) t ≥ 2 − 4p, (2) and chooses T if the inequality is reversed. Since the median of the uniform distribution 1 on [−e, e] is zero, we have 2 − 4p = 0, or p = 2 . Therefore, the same pair of strategies in (1) constitute a BNE of this game. Page 2 of 5 Econ 803 (Fall 2016) Assignment 8 Due: November 15 As in part (b), P1 has strict incentives to play his equilibrium strategy, provided that P2 ∗ plays s2 . There is one exception: when t = 0, P1 is indifferent between any of his pure or mixed strategies. This exception is insignificant though, because t = 0 is a measure zero event. (d) What would happen if the distribution of t were now an arbitrary continuous distribution on [0, e]? Solution: The analysis is similar to the one in part (c). There is still no pure strategy ∗ equilibrium. To justify P2 playing a mixed strategy s2 = pH + (1 − p)T, P1 must 1 effectively play H and T each with probability 2 . As shown in (2), P1 chooses H if t ≥ 2 − 4p and chooses T if the inequality if reversed. 1 Let F be the cdf of t. Denote by tm the median under F, i.e. F(tm) = 2 . Hence, we have 2−tm 2 − 4p = tm, or p = 4 . As a result, the BNE of this game is 8 <H if t ≥ tm 2 − t 2 + t s∗(t) = and s∗ = m H + m T. 1 2 4 4 :T if t < tm P1 has a strict incentive to play his (pure) equilibrium strategy, except for the measure zero event of t = tm. Together, parts (b) through (d) illustrate that we may interpret mixed strategy NE in a complete information game, where each player has only weak incentive to stick to their equilibrium strategy, as the limit of the pure strategy BNE in an incomplete information game as the uncertainty about the opponent’s payoff vanishes. Stephen Morris has a nice essay on this topic: http://www.princeton.edu/~smorris/pdfs/ Morris-Purification.pdf. 2. In a first-price sealed-bid auction, each bidder simultaneously submits a bid to the auctioneer for a single object. The bidder who submits the highest bid gets the object and pays a price equal to the amount of his bid. Thus, the payoff of a particular bidder i is 8 <vi − bi if bi > maxj6=i bj ui = :0 otherwise where vi is bidder i’s private valuation of the object. Suppose there are n bidders, whose valuations are independent draws from the uniform distribution on [0, 1]. Bidder i observes his own valuation vi but only knows the distribution of other bidders’ valuations. We want to Page 3 of 5 Econ 803 (Fall 2016) Assignment 8 Due: November 15 derive a symmetric Bayesian Nash equilibrium where each bidder i follows a strictly increasing and differentiable strategy b(vi). (a) Suppose bidders j 6= i bid according to the equilibrium strategy bj = b(vj). Write down bidder i’s expected payoff as a function of n, his bid bi, his valuation vi, and the inverse of the equilibrium strategy b−1(·). Solution: Bidder i’s expected payoff is −1 Pr max b(vj) ≤ bi (vi − bi) = Pr max vj ≤ b (bi) (vi − bi) j6=i j6=i n−1 −1 = b (bi) (vi − bi), (3) where the second equality follows from the assumption that each vj is independently and uniformly distributed on [0, 1]. (b) Maximize the function you get in part (a) with respect to bi. Hence derive the first-order ∗ condition that determines bidder i’s best response bi . Solution: Differentiating (3) w.r.t bi, we get − n−2 (n − 1) b 1(b ) n−1 i ( − ) − −1( ) = 0 −1 vi bi b bi 0 b (b (bi)) ∗ bi=bi ∗ (c) At a symmetric equilibrium, bi = b(vi). Use this condition and the FOC to derive a first-order differential equation involving b(·) and its derivative b0(·). ∗ Solution: With bi = b(vi), the FOC simplifies to n−2 n−1 0 (n − 1)(vi) (vi − b(vi)) − (vi) b (vi) = 0. (4) (d) To solve for the function b(·), we conjecture that it takes the form b(v) = cvk, where c and k are parameters. Verify that there exist values of c and k that satisfy the differential equation in part (c). What is the equilibrium strategy in this symmetric Bayesian Nash equilibrium? Solution: b(v) = cvk implies that b0(v) = ckvk−1. Substitute these into (4) to obtain n−2 k n−1 k−1 (n − 1)vi (vi − cvi ) − vi ckvi = 0. n−1 Henceforth, it is straightforward to verify that k = 1 and c = n . Page 4 of 5 Econ 803 (Fall 2016) Assignment 8 Due: November 15 n−1 Thus, the symmetric equilibrium strategy is b(vi) = n vi. Note that this is precisely the expectation of the highest competing bid conditional on it being smaller than vi; that is, b(vi) = E[maxj6=i vj j maxj6=i vj < vi]. Page 5 of 5.
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