# Tag Info

## Hot answers tagged nash-equilibrium

6

There is no guarantee that symmetric games have symmetric equilibria. See this paper for concrete examples. There is also no guarantee that symmetric games have pure-strategy equilibria. For example, the following game is symmetric and has no pure-strategy equilibria \begin{array}{cccc} & \mathrm{a} & \mathrm{b} & \mathrm{c}\\ \mathrm{a} &... 6 There is a series of papers that address precisely this question. The most famous ones are probably Walker and Wooders (2001) and Chiappori, Levitt, and Groseclose (2002) that deal with penalty kicks and tennis serves. Both papers conclude that the behavior of professional athletes is consistent with them playing g a mixed strategy equilibrium. A more recent ... 6 Note :Q = \sum_{i=1}^n q_i. Thus the optimisation problem of firm i is: \begin{align} max_{x_i\in\mathbb{R}_+}\pi_i(q_i,Q) \end{align} where \pi_i(q_i,Q) = \big(Q^{-1}(q_i;q_{-i}) -c_i\big)q_i. Assuming interior solution, the first order condition is \begin{align} \frac{\partial\pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial Q}\frac{\partial Q}{\... 6 If you draw the corresponding game tree, you will see that "equivalent to simultaneous move game" implies that the game has no proper subgame and the only subgame is the whole game. This is because the information set of the second player covers every move of the first player. Therefore, every Nash equilibrium is trivially also subgame perfect. 5 Level-k reasoning in the stag hunt game is analyzed in Gracia-Lázaro, Carlos, Luis Mario Floría, and Yamir Moreno. "Cognitive hierarchy theory and two-person games." Games 8.1 (2017): 1. The idea that playing s guarantees its payoff is discussed in Aumann, Robert "Nash equilibria are not self-enforcing, in ‘‘Economic Decision-Making: Games, ... 5 The following two claims hold in the general n-shop case. Claim 1. In equilibrium a shop closest to an edge (0 or 1) cannot be alone. Proof. Such a shop could gain customers by moving slightly inward. Claim 2. In equilibrium at most two shops can be in any location. Proof. Assume there is an equilibrium where there are three or more shops in a location. ... 5 Is there any more information about why von Neumann had this attitude? There must be, but I have never seen it. Or can we infer a reasonable answer? Here is one possibility based on hearsay among the Game Theory community. Von Neumann thought that there was a sharp distinction between zero-sum games and other games. In zero-sum games, there was no ... 5 This statement is wrong. Consider Alternating Matching Pennies with imperfect information (the follower doesn't observe the leader's move). The strategic form of this game is just the classical (simultaneous-move) Matching Pennies Game and the unique NE has both players mixing. 4 The Key to BNE is that players that know something (about the state of the world or their type) can condition their strategies with their information. That is, for example, in question 3, type A could choose strategy U, while type B could choose strategy D. Therefore from the second player's perspective, there are four possible pure strategies of his ... 4 I would have thought that the if the opponent strategy \sigma_{-i} is given, even if it is a mixed strategy, thenu_i(\lambda\sigma_i+(1-\lambda)\sigma_i', \sigma_{-i})= \lambda u_i(\sigma_i, \sigma_{-i})+(1-\lambda)u_i(\sigma_i', \sigma_{-i})$$and so if u_i(\sigma_i, \sigma_{-i})=u_i(\sigma_i', \sigma_{-i}) then u_i(\lambda\sigma_i+(1-\lambda)\... 4 The Nash bargaining solution DOES maximize the Nash product. You have to separate the playing of the game from the bargaining problem. If the players negotiate a binding agreement they will realize that their maximal total payoff from playing the game is 3. This can be achieved by playing (T,L) or (T,C), or any mixture of those two profiles. The ... 4 Pure strategy Nash equilibria are a subset of mixed strategy Nash equilibria, so as long as your statement a mixed strategy Nash equilibrium always exists and all such equilibria have the same value is true, all pure strategy Nash equilibria will have the same value as well. 4 As Michael Greinecker noted, the stag hunt is the leading example of a symmetric 2x2-game with a payoff-dominated but risk-dominant NE. In symmetric 2x2 coordination games, a pure NE is risk dominant iff it is the unique best reply to the mixture (\frac12,\frac12). Since Level-0 types are usually assumed to mix uniformly over pure strategies, all higher-... 4 As is clear from the answer of VARulle, complete information is of no use. Every (finite) game in normal-form is the normal form of an extensive form game of complete information. The situation is different for games of perfect information, and one can prove a result to the effect that "Almost all finite games of perfect information have equilibria that ... 4 Check for the Nash equilibria (pure or mixed) of the one-shot game. Repetition of the strategy profile of the Nash equilibria of the one-shot version yields one set of subgame perfect equilibria: For instance, play (A,A) in the first stage and for any action profile played at the first stage, play (A,A) in the second stage. The same holds true for the ... 4 The two (pure) Nash equilibrium in this game is (Betray, Silent) and (Silent, Betray). Let us see why (Betray, Silent) is an equilibrium. Let us look at person A. Person B is playing Silent. If she plays Betray, she gets 2. If she deviates to Silent, she gets 1. So she would not play Silent. Now consider person B. Person A is playing Betray. If she ... 4 Yes. There are no proper subgames then, so all NE are trivially SP. 3 Suppose player i plays the mixed strategy \mathbb{P}_i(B)= p_i, and assume for now that the support of \mathbb{P}_i is \{B,F\} (i.e. player 1 plays a fully mixed strategy). For both B and F to be in 1's support, he must obtain the same expected payoff from either strategy (otherwise, he would put all the weight on the strategy with the higher ... 3 Since the choice of q_i can be conditioned on (s_i,s_j), strategies in this game are of the form (\hat s_i, \hat q_i(s_i,s_j)). For the given values \alpha=3 and k=1, the SPNE can be calculated as the profile where s^*_i=1/3 and q^*_i\equiv 1. Indeed, production levels q_i^*=1 are the unique NE in all subgames, independently of the chosen ... 3 Although I don't have the book here, I am quite certain that mixed strategies are already present in the book from 1944. I believe that I have read somewhere the claim that "mixed strategies" where actually introduced in an earlier version of the book (published in German in 1928, says Wikipedia). However, what I have read mostly is the claim that ... 3 Ariel Rubinstein has a fabulous paper in which he illustrates that Common Knowledge and Almost Common Knowledge are very different! http://www.cs.cornell.edu/courses/cs6764/2018fa/The_Electronic_Mail_Game.pdf 3 The gist/shortened and generalized version of the above answer: In the context where Q = \sum_i q_i the equation$$ \frac{\partial \pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial Q} = \frac{\partial \pi_i}{\partial q_i} + \frac{\partial Q}{\partial q_i}\frac{\partial \pi_i}{\partial Q} $$holds as$$ \frac{\partial Q}{\partial q_i} = 1. 

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Adding to @soslow's answer: once you have an SPE, it should be easy to construct a non-subgame-perfect NE by modifying the off-equilibrium actions in such a way that 1) the players have no incentive to deviate to those actions and 2) the action profile is not a NE in any subgame. For example, one SPE of the game is play $(A,A)$ in stage 1, and play $(A,A)$ ...

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If a sequential game can be validly represented in the 'normal form' then that means the game has only one sub-game - the whole game. In that case any NE is also SPNE.

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I'd like to understand how this infinite recursion (for lack of a better term...) of knowledge figures mathematically into say, equilibrium concepts. It figures mathematically, for example, in the simplification of strategy sets through elimination of strictly-dominated alternatives. Others can correct me, but the point isn't to write down a mathematical ...

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The idea of a rational expectations equilibrium is more general than BNE. It simply means that the belief system of agents is consistent with the model and incorporates all available information. This abstract idea can be applied for games, markets or other types of interactions. BNE is a solution concept for non-cooperative games. The expectations being ...

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First, a caveat: I'm on the job market this year in the midst of the couple weeks when calls are rolling in. Hence, this seemed like a nice way to kill some time (semi-)productively. This is also a disclaimer in case I've made an error :) Now, let's look at the one you're suggested to try for, where player $1$ chooses $0$ with $p$ and $1$ with $1-p$. Again ...

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There are several scientific papers analyzing this on topologies other than the unit interval. A 2D example is Spatial competition of firms in a two-dimensional bounded market by Aoyagi and Okabe. Unfortunately there are usually no equilibria in these cases. The intuition behind this is that the consumers arriving from different directions are usually not "...

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Mixed strategy Nash equilibrium cannot involve strictly dominated strategies. In particular, Cooperate is strictly dominated for player 1 ($6<8$ and $3<4$). Therefore, no $b\in[0,1]$ can make player 1 indifferent between Cooperate and Non-cooperate. You made a mistake by trying to solve for $b$ by equating player 1's expected payoffs from his two pure ...

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To find the SPNE you want to consider all possible scenarios that the seller might face in the second period, and what his best strategy would be. This will help you study the seller's actions in the first period since they will lead to different scenarios in the second one. In fact, saying "the subgame perfect pricing policy is set p1 = 1199 and p2 = 299" ...

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