Conference 7Intermediate Microeconomics - Fall 201815/10/2018Problem 1: Rock, Paper, Scissors. Consider the following game, inspired from the classicRock, Paper, Scissors game:R P SR (0, 0) (−1, 1) (1, −1)P (1, −1) (0, 0) (−1, 1)S (−1, −1) (1, −1) (0, 0)1. Show there is no pure Nash equilibrium.Cycling argument:If player 1 plays R, player 2’s best response is to play PBut if player 2 plays P, player 1’s best response is to play SIf player 1 plays S, player 2’s BR is to play RIf player 2 plays R, player 2’s BR is to play PThus we get a loop in which each player has tries one of her strategies, always has aprofitable deviation in front of the other player’s best response.2. Find the mixed Nash equilibriumAssume player 1 plays R with probability x, P with probability y and S with probabilityz = 1 − y − x. Player 2 is indifferent if:−y + (1 − x − y) = x − (1 − x − y) (1)x − (1 − x − y) = −x + y (2)The first equation yields: y =13(2−3x). Plugging this value into the second equation yieldsx’s valuex − (1 − x −13(2 − 3x)) = x +13(2 − 3x)⇔ x =13Therefore y =13(2 − 313) =13and z = 1 −13−13=13.The same reasoning is true for player 2, and the mixed Nash equilibrium is (13R,13P,13S), (13R,13P,13S)1In the next question, we are going to use the following definitions and theorem:Zero sum game: players’ payoff in any outcome is equal to zeroSymmetric game: if player 1 plays A and player 2 plays B, player 1’s payoff is the sameas player 2’s payoff if player 1 were to instead play B and player 2 were to instead play Aand vice versaTheorem: In finite, two players, symmetric, zero sum game, each player’s equilibrium expectedutility must be zero.3. How does the theorem above help us find easily the mixed Nash equilibrium?First observe that the Rock, Paper, Scissors game is a finite, symmetric, zero sum gameand therefore we can apply the theorem. It actually provides a shortcut to what we didin the previous question. Because players’ expected utility at equilibrium must be zero,we know that player 1’s equilibrium probabilities must verify:−y + z = 0 (3)x − z = 0 (4)−x + y = 0 (5)Wich is equivalent to x = y = z. Since we know that x + y + z = 1, we deduce easily thatx = y = z =13.Problem 2: Revenue Equivalence Theorem. The problem’s goal is to show that theauctioneer’s revenues are the same in a first (FPA) and a second (SPA) price auction. Thisresult is an instance of the more general Revenue Equivalence Theorem.The set up is the following: there are two bidders 1 and 2, and each has private valuationvi∼ U[0, 1], i = 1, 2. The strategy of a bidder is her bidding function bi(vi).1. Assume player −i uses strategy b−i= αv−i. What is the expected payoff of player i inFPA?Ui= (vi− bi) Prbi> b(v−i)= (vi− bi) Prbi> αv−i= (vi− bi) Prv−i< bi/α= (vi− bi)biα2. Find a symmetric Nash equilibrium of the FPA.Players maximize their payoffs at equilibrium, i.e. player i solves:maxbi(vi− bi)biα2FOC gives us:−biα+vi− biα= 0vi− 2biα= 0vi− 2bi= 0bi=vi2Therefore in a symmetric Nash equilibrium, players’ strategy is to bid half their valuation.3. Show that it is a dominant strategy for player i to bid viin SPA.We are going to show that bi(vi) = viweakly dominates all other strategies.Suppose bi> vi.If b−i> biplayer i gets 0 and therefore is indifferent between bidding bi> viandbi= viIf b−i< viplayer i wins and gets vi− b−i> 0 whether she bids bi> vior bi= vi.Hence she is indifferent.If vi< b−i< biplayer i wins and gets vi− b−i< 0. But she would rather lose byplaying bi= viand get 0.Now suppose bi< vi.If b−i< biplayer i wins and gets vi− b−i> 0 whether she bids bi> vior bi= viIf b−i> viplayer i loses and is indifferent between bidding bi> viand bi= viIf bi< b−i< viplayer i loses and gets 0. Had she chosen bi= vishe would have wonand gotten vi− b−i> 0Therefore bidding bi(vi) = viis a dominant strategy for both players and there exists anequilibrium in which they both bid vi.4. Compute auctioneer’s revenue in both FPA and SPA and show that it is equal in bothcases. Use that E(max{X1, X2}) =23and E(min{X1, X2}) =13when X1, X2∼ U[0, 1]FPA: the auctioneer receives the maximum of the two bids: max{v12,v22} =12×23=13SPA: the auctioneer receives the minimum of the two bids: max{v1, v2} =13The expected auctioneer’s revenue is the same in both auction type. This result can gen-eralized as the Revenue Equivalence Theorem.Problem 3: Sequential auctionThere are two bidders 1 and 2, and each has private valuation vi∼ U[0, 1], i = 1, 2. Thestrategy of a bidder is her bidding function bi(vi). The auction is spread over two periods.At t = 1 bidder 1 chooses his bid b1. This bid is observed by bidder 2, who responds with abid b2. The player with the highest bid wins the auction and pays their own bid. If players tie,player 2 wins.We are going to find this auction’s equilibrium by backward induction.31. What is player 2’s Best Response at t = 2?Player 2 knows her value v2and observes b1. There are two possible cases:v2> b1. Then player 2 should bid b1, tie and win and earn v2− b1.v2< b1. Then player 2 should bid any value b2below b1, lose and get 0.Therefore b2(v2) =(b1if v2> b1b2< b1if v2< b12. What is player 1 Best Response at time t = 1?Player 1’s expected payoff is:U1= Pr(v2> b1) × 0 + Pr(v2< b1) × (v1− b1)= b1(v1− b1)Maximizing this payoff gives FOC:(v1− b1) − b1= 0b1=v12Therefore b1(v1) =v12Problem 4: All pay auction. Consider an auction where bidders have values v1, v2drawnindependently from a uniform distribution on [0, 1] and submit b1, b2. Suppose the bidder whosubmits the highest bid wins (with ties broken randomly), but that both bidders must pay anamount equal to their bids.1. Show that there is no pure Nash equilibrium in this game.Assume players respectively bid b1, b2∈ [0, 1] with probability 1. Assume also without lossof generality that b1< b2, and player 2 wins.Note that there is no equilibrium in which 0 < vi< bibecause it player i loses, his payoff if−bi< 0 and if he wins, it is vi− bi< 0. Therefore a profitable deviation for player i is tobid bi= 0Then candidates (b1, b2) for equilibrium satisfy either:0 ≤ b1< b2≤ v2≤ v1≤ 1. Player 1 has a profitable deviation: to play b1such thatb2< b1≤ v1and get v1− b1instead of −b10 ≤ b1< b2≤ v1≤ v2≤