Guide for solving the 2000 exam for 14.122Question 1(4See the lecture notes or FT for the formal definition. We care about continuity at infinity becauseif this property is satisfied, then checking a proposed SPE is equivalent to checking the single-deviation-property (ie that no player can achieve a higher payoff by deviating from their equilibrium strategy at asingle node, and playing according to their equilibrium strategy at all other nodes).@IYes, it has a unique NE (u,A). This is also the only outcome that survives iterated strictdominance. To see this, start by noting that for player 2, C is strictly dominated by ;A + ZB (or othercombinations of A and B will work as well), and go from there.Cc)See the notes for the definitions. The assumption that breaks down in the “name the largestnumber” game is that C, the space of strategy profiles, is not compact (because there is no limit on howlarge the numbers named by the players can be).(4This is false. If a property holds for generic games, then for any game for which the propertydoesn’t hold, there is another game, where the payoffs are arbitrarily close to the first game, for which theproperty does hold. So clearly this isn’t true in, for example, games where all the payoffs are strictly negative.(e)There are two subgames (three if you also count the whole game as a subgame). One begins atl’s node just above the terminal nodes (1,l) and (2,0). The other begins at the upper of 2’s non-singletonnodes.(BTW: According to the way FT, and I believe Glenn, defines things, all games are proper subgames ofthemselves, so by this definition the whole game is also a subgame.Gibbons, on the other hand, doesn’tcount the original game as being one of the subgames).(0The centipede game that Glenn discussed in class is an example where SPE/backward inductionrules out a reasonable outcome. If you remember, the SPE is to play down immediately, giving the players alow payoff. But it is probably more reasonable from an empirical point of view to expect players to cooperatefor a while before they play down.An example where SPE gives an unreasonable result - Glenn gives several of these in the notes. Hediscusses games in which SPE allows equilibria that involve players choosing actions at an information setthat are inconsistent with any possible set of beliefs the player could hold at that information set. Signallinggames will often have unreasonable SPE, because there are no subgames (you can’t separate parts of thegame tree without cutting information sets).(8)Firstly note that D and then B can be eliminated by iterated strict dominance. There are nopure strategy NE of the resulting 2x2 game. The mixed strategy NE is (SU + $M, $A + SC).04The NE is al = $$, a2 = E (found by solving the two best response functions simultaneously).(9Remember for the Intuitive Criterion you basically have to establish two things - heuristically (1)that the type making the speech gets a higher payoff under the deviation and some best response to thedeviation than under the PBE, and (2) that the other type does not have an incentive to make such a speech.So Irving’s argument is ok - you can verify that the B = 3 type could do better by choosing e = 1, but the0 = 2 type could not. Freddy’s argument doesn’t make sense, because neither type has an incentive to makesuch a speech.Courtesy of James Vickery. Used with permission.Question 2(4Tree not drawn. But player 1 has 3x2~2~2 = 24 pure strategies, and player 2 has 3x3x3 = 27pure strategies.64For example, can check that the strategy profile where player 1 plays: (give other player$2, D, D, D), and player 2 plays: (B if player 1 gave up 2, C otherwise) is a NE. Given the strategyof the other player, neither player can achieve a higher payoff by unilaterally adopting a different strategy.Cc)IfJ:<Othen(U,A)’1s not a NE of the subgame, and thus the only NE of the subgame is (D, B).Then by backward induction player 1 will choose to give up 0 at the first stage of the game. So the conditionfor a unique SPE is z < 0.(4We are looking for an SPE where the static NE (U, A)will be used to reward player 1 when hegives up $2 at the first stage of the game (we can’t use (U, A)as the punishment NE because x > 0 so thepunishment won’t be big enough). So along the equilibrium path, player 1 will get payoff of x, which mustbe greater than 2 + 1 (since the second stage outcome must always be a NE of the second stage game).So x > 3 will ensure that the strategy profile where player 1 plays (give other player $2, A if 1 gaveaway $2 at the first stage, D otherwise) and player 2 plays (A if player 1 gave up $2, B otherwise) is anSPE.(e)No, because the lowest-payoff NE in the second stage gives payoff of at least zero to both players(zero occurs only when x = 0). So player 1 can always keep both dollars at the first stage of the game, andin any SPE will at worst get 0 in the second stage, for a total payoff of 2.Question 3(a)01 = {type who doesn’t know 4). 02 = {1,2,3}. MIT has 2x2x2=8 pure strategies (a choicebetween two actions for each of three types), while Harvard has only 2 pure strategies.(b)The actions a2(q = 3) = don’t admit, and a2(q = 1) = admit are both conditionally dominated.In words, MIT will never let in a 4 = 1 student because even if Harvard does not make the student an offer,the payoff for MIT is less than 0. In contrast, MIT will always admit a 4 = 3 student regardless of Harvard’sst,rategy, because even if Harvard makes the student an offer, MIT’s payoff is 0.65 x 1.5 - 1 x 0.35, which isgreater than zero. No actions are conditionally dominated for MIT when q = 2, because they want to admitthe student if and only if Harvard does not admit the st,udent.Harvard has no conditionally dominated strategies.(clGiven the discussion in part (b), the only possible BNE’s are1.al = Admit; a2(q = 1) = aQ(q = 2) = Don’t admit, a2(q = 3) = Admit2.al = Don’t admit; az(q = 1) = Don’t admit, az(q = 2) = a2(q = 3) = AdmitYou can check that the first of these possible equilibria doesn’t work. In the first putative equilibrium,Harvard’s expected payoff from admitting the student isU(Admit) = -0.5 x l/3 + 0.5 x l/3 + (0.35 x 1.5 - 1 x 0.65) x l/3,which is negative. So Harvard wouldchoose to deviate and not admit the student.By similar calculations, can check that the second potential equilibrium really is a BNE.So the unique BNE will be al = Don’t admit; a2(q = 1) = D on’t