solutionspset4

# solutionspset4 - Ec1052 Introduction to Game Theory Handout...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ec1052: Introduction to Game Theory Handout 11 Harvard University 28 April 2004 Solutions to Problem Set 4 Problem 1. There is no right answer to this one. We have shown in previous problem sets that the Nash equilibrium (NE) is stronger than iterated deletion of dominated strategies because we can find sets of strategies that are in S ∞ but not NE, but can not do the reverse. 1 The Nash existence theorem shows that the NE is not too strong in the sense that (in a finite game) a NE will always exist. However, you could argue that NE is not strong enough because it often fails to give a unique prediction of how the game will be played. There is a huge literature (mostly from the 1980s) trying to find acceptable stronger conditions - we will see some of these when we look at dynamic games. Also, the more recent literature on evolution is partly motivated by the problem of how to select between NE. Problem 2. 2(a) Clearly any strategy pair gives a payoff of zero to both parties. Hence, it is impossible to get the strict inequality required in the definition of ESS. 2(b) By proposition 4 from lecture 9, we know that if every ESS is a NE. Therefore, we can confine ourselves to checking NE. It is straightforward to show that there is no pure strategy NE in the game. When considering replicator dynamics or ESS, we can confine ourselves to symmetric strategy profiles (the population is playing against itself). Note that if any one of the strategies is eliminated for both the row and the column player, we are left with a 2*2 game that is solvable by IDSDS. Therefore the only NE is must include all pure strategies. In fact the unique equilibrium involves playing each strategy with probability 1 / 3. The payoff in this equilibrium is 5 / 3 (it is symmetric). Suppose we add weight ² > 0 to the proportion of players playing A , and subtract ²/ 2 from the proportion of players playing B and C . The old strategy (1 / 3 , 1 / 3 , 1 / 3) still earns an expected payoff of 5/3, but now, A earns 5 / 3+ ε/ 2 > 5 / 3. Therefore, there is no ESS in the game. 1 Remark by Markus: We have seen in experiments (for example 2/3 of the average) that subjects often do not even play profiles in S ∞ . Since Nash equilibrium is a refinement of IDSDS it seems too strong a solution concept for games where agents have to iterate very often. However, many of these games have a tendency to converge to Nash equilibria over time. For example, if the 2/3 of the average game is repeated many times typically bids go gradually down as subjects start to understand the iteration process. Hence the criticism of NE on the basis that it is too strong has more validity in the short run rather than the long run. Ec1052 Handout 11: Solutions to Problem Set 4 2 Problem 3. There are four steady states: playing any of the pure strategies with probability 1 and the NE discovered above. Once a strategy is played by none of the population, it can never be played by any of the population....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

solutionspset4 - Ec1052 Introduction to Game Theory Handout...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online