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## Hot answers tagged nash-equilibrium

17

Two areas that have been profoundly affected by game theoretic research stemming from Nash's contribution are Oligopoly theory There are actually a few examples of what would come to be known as Nash equilibrium in the industrial organization literature that predate Nash's work (for example, Cournot's 1838 analysis of oligopoly competition). However, until ...

12

You're not alone in your skepticism of the relevance of game theory. Some of the greats, including Gary Becker, were at times dismissive of the practical/empirical importance of game theory (see the introduction/preface of his Economic Theory book). No doubt it is in a way foundational to the economic sciences (see Myerson's great essay on Nash's ...

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This is only half a joke : Nash-equilibrium gives a very good prediction on the relative size of groups of foraging ducks on a pond when two food sources are established at opposite sides of the pond. A very good explanation can be found at https://headbiotech.wordpress.com/nash-equilibrium-example-on-ducks/, among other places (https://headbiotech....

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Let's review the definitions of the two concepts. Let $\sigma$ be a profile of strategies and $\mu$ a system of beliefs. A pair $(\sigma,\mu)$ is a weak perfect Bayesian equilibrium (WPBE) if $\sigma$ is sequentially rational given $\mu$, and $\mu$ is derived from $\sigma$ using Bayes rule whenever applicable. and A pair $(\sigma,\mu)... 8 Fixing the strategy of the opponent, a mixed strategy never yields a strictly higher utility if you are expected utility-maximizing. The reason is that the expected utility from a mixed strategy is at most as high as the highest utility from the pure strategies which this mixed strategy plays with positive probability. That is not to say that a mixed ... 8 No, there is not. Consider a game with two players, Ann and Bob. Both choose such vectors with entries$0$or$1$of the form$(a_1,a_2,\ldots,a_J)$or$(b_1,b_2,\ldots,b_J)$, respectively. If$\sum_{i=1}^J a_i+b_i$is odd, Ann wins and Bob loses. If the number is even, Ann loses and Bob wins. Clearly, one of them loses in every profile of pure strategies, ... 7 The following paper compares the efficiency of the Bertrand and Cournot game in the case of product differentiation. However, their utility function is more general than the Dixit/Stiglitz case. You also do not have an infinite number of firms competing in each differentiated product but just one firm per differentiated product (the number of differentiated ... 7 Yes, there is always a pure Nash equilibrium. See: I Milchtaich (1996). Congestion games with player-specific payoff functions. Games and economic behavior 13 (1), 111-124. You are interested in the special case of singleton congestion games with player-specific payoff functions. And yes, they can be computed in polynomial time. See Corollary 7 in: ... 7 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. 7 the equilibria of the game in which the strategies of users who face the same reward and costs (i.e. same type) are the same. This sounds like an ex ante (or ex interim or ex post) symmetric equilibrium, depending on the timing of the realization of the types. If both dimensions of the types are realized at the beginning of the game, then I'd go with the ... 7 Let$N=2$and for$(x,y)$and$(p,q)$in$[0,1]^2$let$d_{p,q}(x,y)$be the Euclidean distance between$(p,q)$and$(x,y)$, i.e.$d_{p,q}(x,y)=[(p-x)^2+(q-y)^2]^{1/2}$. Choose$k>0$such that$k<a-b$and$k<b$. Now define $$u(p,q):=\max\{ -d_{p,q}(a,a),-d_{p,q}(b,b),k-d_{p,q}(b,0),k-d_{p,q}(0,b) \}.$$ This function is continuous as a maximum of ... 6 Yes there absolutely is. It is generally agreed upon that utility is (at least) ordinal. That means I can compare utility levels for a single person (not necessarily across people) and the numbers have meaning in this sense. So for a single person:$U(down, right) = 4 > 1 = U(Up, Left)$. This means (Down,Right) is strictly better for each agent than (... 6 The diagonally strict concavity property is better known as the strict monotonicity property of the pseudo-gradient. An operator$\Psi:\mathbb{R}^n \to \mathbb{R}^n$is strictly monotone if the following holds true: $$\forall x_0, x_1 \in \mathbb{R}^n: \left< x_0 - x_1, \Psi(x_0) - \Psi(x_1) \right> >0.$$ In your case$\Psi$would be$g$. The ... 6 So you want to find a maximum of$\sigma(s,z)$. If$\sigma$is diagonally strictly concave you can do so by starting at any point and just following the gradient$g(s,z)$until you find the maximum and no matter where you start, you will always end up at the same point (Start at the lower black points and follow the direction of the gradient (the direction ... 6 The conventional definition of Bayesian Nash equilibrium (BNE) is as follows: A pure strategy BNE is a profile of type-contingent strategies $$(s_i(\theta_i),s_{-i}(\theta_{-i}))=(s_1(\theta_1),\dots,s_{i-1}(\theta_{i-1}),s_i(\theta_i),s_{i+1}(\theta_{i+1}),\dots,s_m(\theta_m))$$ such that for each$i\in P$, $$s_i(\theta_i)\in\arg\max_{s_i'\in S_i} ... 6 A leader-leader Stackelberg is a situation in the Stackelberg model where both firms believe they are leaders. This leads to global production being much higher than expected by both firms, as they anticipate small production by the other firm in response to their high production. 6 Player 2 +---+-------+-------+-------+ | | A | B | C | +---+-------+-------+-------+ Player 1 | A | (1,1) | (0,0) | (0,0) | | B | (0,0) | (2,1) | (1,2) | | C | (0,0) | (1,2) | (2,1) | +---+-------+-------+-------+ If we allow 3\times 3 games, then above is an example of the ... 6 I have used gambit (python) in the past and could recommend it. It also includes a small GUI which makes things very intuitive. R also has GameTheory if you prefer this platform instead 6 It is actually assumed that b_i(v_i) is of the form \alpha_i+\beta_i \cdot v_i. So it is an affine function. Linearity only works if the bottom of the uniform distribution is 0. A somewhat intuitive reasoning is that if the valuations are uniformly distributed over [c,d] and b_i(v_i) is a symmetric equilibrium, then b_i(v_i)+k should define a ... 6 There are three players, 1,2,3, and the action spaces of players 1 and 2 are both [0,1]\times\mathbb{N}, the action space of player 3 is [0,1]. The generic action of player 1 is written as (x,m), the generic action of player 2 is written as (y,n), and the generic action of player 3 is written as (x'). The payoff of player 3 is simply 1 if x'=x... 6 One reason why there exists no ultimate overview is that these issues are still under debate. A great entry point would be the survey "Foundations of Strategic Equilibrium" by John Hillas and Elon Kohlberg in Volume 3 of the Handbook of Game Theory with Economic Applications. A preprint version of the survey can be found here. 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 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. 6 A couple hints. Regarding the lower bound on$\epsilon$: What happens if deviation occurs at stage$T$? In other words, there is no opportunity for your so-called "punishment stages". Regarding the upper bound on$\epsilon$: Suppose player 2 deviates at stage$T-1$but player 1 does not. What must be true about$\epsilon$in order for player 1 to ... 6 The Nash equilibrium is indeed (down, right). Note that your chart has helpfully underlined the max value among all possible strategies each player can play, conditional on what the other player plays. The steps on the graph are useful as well: If no numbers are underlined in a cell, that strategy set is strictly dominated by something else. If one number ... 6 You are only required to get Nash equilibria and not sequentially rational/subgame perfect equilibria. Hence Player 2's actions at information sets that do not occur (that do not reflect Player 1's actual strategy) do not need to be best responses. All you have to make sure is that no one is better off by deviating. In case 4., regardless of what Player 2 ... 6 Let's first determine the sets of actions of the players. An action of player 1 is simply a bid$x_1 \in \mathbb{R}_+$. An action of player 2 is a function:$f_2: \mathbb{R}_+ \to \mathbb{R}_+$that determines for every action$x_1$of Player$1$an action$x_2 = f_2(x_1) \in \mathbb{R}_+$. Let us denote by$F_2$the set of all actions of player 2. Let's now ... 5 No. In static games, common knowledge of rationality is equivalent to rationalizability. Bernheim, in "Rationalizable Strategic Behavior" (Econometrica, July 1984) gives an example on page 1012 in Figure 1 of a normalform game in which there is a unique Nash equilibrium, yet multiple strategies are rationalizable. 5 Suppose$(\sigma_1^*, \sigma_2^*)$is the Nash Equilibrium of the game$G$consisting of players$\{1,2\}$having strategy sets$A_1$and$A_2$and the payoff functions$u_1:A_1\times A_2 \rightarrow \mathbb{R}$and$u_2:A_1\times A_2 \rightarrow \mathbb{R}$. Consider the game$G'$in which everything else is same as$G$except that the payoff functions are$...

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