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16 votes
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Why is pre-specification of punishment order necessary to manipulate compliance?

Denote the cost of the punishment by $p$, the cost of registering by $r$. In order for the plan to work, you need to have $p > r$. If the order is known, then the person at the top will choose to ...
Giskard's user avatar
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11 votes

Why is pre-specification of punishment order necessary to manipulate compliance?

A game can have multiple Nash equilibria. You've correctly observed that, even if the punishment order isn't pre-specified, everyone registering is still a Nash equilibrium: each citizen knows that, ...
Ilmari Karonen's user avatar
9 votes
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Relaxing the notion of Nash Equilibrium

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_{...
Michael Greinecker's user avatar
9 votes
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Correlated equilibrium intution

Correlated equilibrium concepts are often related to communication, most easily implemented via a mediator. Communication can archive two things. It helps with coordination and it can transmit private ...
Michael Greinecker's user avatar
9 votes

Correlated equilibrium intution

An example to highlight the difference between correlated and uncorrelated random strategies in games of perfect information. Consider a game of Chicken with the payoff matrix Straight Chicken ...
Giskard's user avatar
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9 votes
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Prove that for every Nash equilibrium $\sigma^*$, the probability distribution $p_{\sigma^*}$ is a correlated equilibrium

A strategy profile $\sigma^*=(\sigma_i^*,\sigma_{-i}^*)$ is a Nash equilibrium if for all player $i$, \begin{equation} u_i(s_i,\sigma_{-i}^*)\ge u_i(s_i',\sigma_{-i}^*), \quad \forall s_i\in\mathrm{...
Herr K.'s user avatar
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8 votes

Is a mixed strategy ever the best response to a pure strategy?

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 ...
induction_is_a_laddah's user avatar
8 votes
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Effect of bounding action space on the set of equilibria

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&...
VARulle's user avatar
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7 votes

In a game with alternating moves and complete information, the Nash equilibrium cannot be a non-trivial mixed equilibrium?

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 (...
VARulle's user avatar
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7 votes
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SPNE of a normal form game

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 ...
Bayesian's user avatar
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7 votes
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symmetry of equilibria with heterogeneous players

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 ...
Herr K.'s user avatar
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7 votes
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MWG 8.B.7 - Any strictly dominant strategy must be a pure strategy

Fix any $\sigma_{-i}$. Assume $\sigma_i$ is strictly dominant but not a pure strategy. Let $X$ be the support of $\sigma_i$. Since $\sigma_i$ strictly dominates all pure strategies $s_i\in X$, we have ...
VARulle's user avatar
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6 votes
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Rosen's Diagonal Strict Concavity condition

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 ...
Julien V's user avatar
6 votes

Rosen's Diagonal Strict Concavity condition

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 ...
muxamilian's user avatar
6 votes

Can a game with a unique pure strategy Nash equilibrium also have a mixed strategy equilibria?

...
Amit's user avatar
  • 8,891
6 votes

Game theory software

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
caverac's user avatar
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6 votes
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Bayesian-Nash equilibrium in a first-price auction

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 ...
Giskard's user avatar
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6 votes
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Example of a game with no Nash equilibria but at least one correlated equilibrium

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 $(x,m)$,...
Michael Greinecker's user avatar
6 votes

Interpretation of Solution Concepts

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 ...
Michael Greinecker's user avatar
6 votes
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Existence of Symmetric Pure Strategy Equilibrium

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, ...
brunosalcedo's user avatar
6 votes
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In a game with alternating moves and complete information, the Nash equilibrium cannot be a non-trivial mixed equilibrium?

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 ...
Michael Greinecker's user avatar
6 votes
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Finitely repeated Prisoner’s Dilemma with switching cost

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". ...
Herr K.'s user avatar
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6 votes

Is there really a Nash equilibrium in this example?

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....
qwerty's user avatar
  • 517
6 votes

Auctions and finding nash equilibrium of a dynamic game

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 ...
Giskard's user avatar
  • 29.2k
6 votes

Auctions and finding nash equilibrium of a dynamic game

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 ...
tdm's user avatar
  • 12.2k
6 votes

Existence of Nash Equilibrium

The first step is showing that $BR^\epsilon$ is a continuous function from the compact convex set $\Sigma$ to itself, which amounts to showing $BR_n^\epsilon$ is a continuous function for each $n$. ...
Michael Greinecker's user avatar
5 votes

Game theory software

Game theory explorer (GTE) offers a web-based solver that searches for Nash equilibria of the inputted games. Its documentation highlights the main differences between GTE and Gambit: Gambit has ...
Herr K.'s user avatar
  • 15.4k
5 votes

Zero sum game, constant sum game

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 \...
Amit's user avatar
  • 8,891
5 votes
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Mixed Nash equilibrium

(i) $A$ & $B$ If player 1 play $A$ with probability $p$ and $B$ with probability $(1-p)$, where $0<p<1$, then player 2's expected payoff from playing $D$ is $4p+4(1-p) = 4$ $E$ is $6p + 2(...
Amit's user avatar
  • 8,891

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