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Suppose you know the following for a particular three-player game: The space of strategy profiles \(S\) is finite. Also, for every \(s \in S\), it is the case that \(u_{2}(s)=3 u_{1}(s), u_{3}(s)=\left[u_{1}(s)\right]^{2}\), and \(u_{1}(s) \in[0,1] .\) (a) Must this game have a Nash equilibrium? Explain your answer. (b) Must this game have an efficient Nash equilibrium? Explain your answer. (c) Suppose that in addition to the information given above, you know that \(s^{*}\) is a Nash equilibrium of the game. Must \(s^{*}\) be an efficient strategy profile? Explain your answer; if you answer "no," then provide a counterexample.

Short Answer

Expert verified
(a) Yes, a Nash equilibrium must exist. (b) No, it need not be efficient. (c) No, a counterexample shows it might not be efficient.

Step by step solution

01

Analyze Nash Equilibrium Existence

To determine if a finite game must have a Nash equilibrium, remember that according to Nash's theorem, every finite game (finite set of strategy profiles) has at least one Nash equilibrium. Since the set of strategies is finite in this game, there must exist at least one Nash equilibrium.
02

Define Efficient Nash Equilibrium

A Nash equilibrium is efficient if it maximizes the total utility across all players. Here, total utility can be represented as \(U(s) = u_1(s) + u_2(s) + u_3(s)\). Substitute the given utility functions: \(U(s) = u_1(s) + 3u_1(s) + [u_1(s)]^2\). Simplifying, \(U(s) = 4u_1(s) + [u_1(s)]^2\). An efficient Nash equilibrium maximizes this function over possible strategy profiles.
03

Evaluate Need for Efficient Nash Equilibrium

While a Nash equilibrium must exist, it doesn't necessarily have to be efficient unless it's proven that maximizing \(4u_1(s) + [u_1(s)]^2\) corresponds to the Nash equilibrium conditions for all players. Hence, efficient Nash equilibria are not guaranteed unless specific conditions are satisfied.
04

Consider Specific Nash Equilibrium Efficiency

Given that \(s^*\) is a Nash equilibrium, it may or may not be efficient. Consider \(u_1(s) = 0\) which makes utilities all zero. It could be a Nash equilibrium, but certainly isn't efficient, as choosing \(u_1(s) = 1\) results in \(U(s) = 5\). Thus, \(s^*\) need not be efficient; a counterexample using \(u_1(s) = 0\) confirms this.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nash Equilibrium
A Nash Equilibrium represents a situation in a game where no player can benefit by changing their strategy while the other players keep theirs unchanged. This concept forms the backbone of game theory.
In a finite game, like the one described in the exercise, Nash's theorem guarantees the existence of at least one Nash equilibrium. This is because the finite nature of the strategy profiles ensures that the players will reach an outcome where no one has anything to gain by switching strategies.
In our exercise, the set of strategies is finite, and therefore, according to Nash's theorem, there is at least one Nash equilibrium. This means that the players can find a steady state where none wants to change their behavior, providing a critical insight into strategic decision-making.
Efficient Strategy
An efficient strategy, or efficient Nash equilibrium, occurs when the sum of utilities across all players is maximized. This makes the strategy beneficial for everyone involved, as it results in the highest possible overall benefit.
To determine efficiency, we calculate the total utility: \(U(s) = u_1(s) + u_2(s) + u_3(s)\). For the given exercise, substituting the utility functions gives \(U(s) = 4u_1(s) + [u_1(s)]^2\).
A strategy profile is efficient if it maximizes this utility function. However, not all Nash equilibria are efficient. Even if a Nash equilibrium is achieved, it doesn't necessarily mean total utility is maximized. Hence, an additional evaluation is required to ensure efficiency.
Utility Maximization
The concept of utility maximization is central to understanding players' decisions in a game. Each player aims to choose a strategy that maximizes their utility function, given the choices of other players.
In our example, the player's utilities are functionally linked: \(u_2(s) = 3u_1(s)\) and \(u_3(s) = [u_1(s)]^2\). This implies that as player 1's utility changes, the others follow predictably.
Utility maximization involves selecting strategy profiles where players reach an agreement that improves or maximizes their earnings or satisfaction. While a Nash equilibrium will ensure players are satisfied with their choices, utility maximization goes further by aiming for the most beneficial outcome for all, aligning with efficient strategies.
Finite Games
Finite games are types of games where the number of players, strategies, and possible outcomes is limited. This characteristic makes them particularly interesting in game theory analysis because they are generally simpler to analyze.
In the exercise, the strategy space \(S\) is finite. This simplifies the problem, as we can apply Nash's theorem to ascertain that at least one Nash equilibrium exists.
The finite nature also means that we can enumerate all possible strategies and outcomes, providing an organized view of the strategic landscape. Finite games can often be represented in matrices or trees, making them accessible and straightforward to understand, allowing deeper insights into the strategic interactions between players.

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Most popular questions from this chapter

Compute the Nash equilibria of the following location game. There are two people who simultaneously select numbers between zero and one. Suppose player 1 chooses \(s_{1}\) and player 2 chooses \(s_{2}\). If \(s_{i}

Is the following statement true or false? If it is true, explain why. If it is false, provide a game that illustrates that it is false. "If a Nash equilibrium is not strict, then it is not efficient."

Consider the following \(n\)-player game. Simultaneously and independently, the players each select either X, Y, or Z. The payoffs are defined as follows. Each player who selects X obtains a payoff equal to \(\gamma\), where \(\gamma\) is the number of players who select Z. Each player who selects Y obtains a payoff of \(2 \alpha\), where \(\alpha\) is the number of players who select X. Each player who selects \(Z\) obtains a payoff of \(3 \beta\), where \(\beta\) is the number of players who select Y. Note that \(\alpha+\beta+\gamma=n\). (a) Suppose \(n=2\). Represent this game in the normal form by drawing the appropriate matrix. (b) In the case of \(n=2\), does this game have a Nash equilibrium? If so, describe it. (c) Suppose \(n=11\). Does this game have a Nash equilibrium? If so, describe an equilibrium and explain how many Nash equilibria there are.

This exercise asks you to consider what happens when players choose their actions by a simple rule of thumb instead of by reasoning. Suppose that two players play a specific finite simultaneous-move game many times. The first time the game is played, each player selects a pure strategy at random. If player \(i\) has \(m_{i}\) strategies, then she plays each strategy \(s_{i}\) with probability \(1 / m_{i}\). At all subsequent times at which the game is played, however, each player \(i\) plays a best response to the pure strategy actually chosen by the other player the previous time the game was played. If player \(i\) has \(k\) strategies that are best responses, then she randomizes among them, playing each strategy with probability \(1 / k\). (a) Suppose that the game being played is a prisoners' dilemma. Explain what will happen over time. (b) Next suppose that the game being played is the battle of the sexes. In the long run, as the game is played over and over, does play always settle down to a Nash equilibrium? Explain. (c) What if, by chance, the players happen to play a strict Nash equilibrium the first time they play the game? What will be played in the future? Explain how the assumption of a strict Nash equilibrium, rather than a nonstrict Nash equilibrium, makes a difference here. (d) Suppose that, for the game being played, a particular strategy \(s_{i}\) is not rationalizable. Is it possible that this strategy would be played in the long run? Explain carefully.

Consider a two-player game and suppose that \(s^{*}\) and \(t^{*}\) are Nash equilibrium strategy profiles in the game. Must it be the case that \(\left\\{s_{1}^{*}, t_{1}^{*}\right\\} \times\left\\{s_{2}^{*}, t_{2}^{*}\right\\}\) is a weakly congruous strategy set? Explain why or why not.

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