This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Ec1052: Introduction to Game Theory Handout 4 Harvard University 10 March 2004 Solutions to Problem Set 2 Problem 1. Assume, that both lawyers can order the appetizer or the main course, and that they both have utility 10 and 15 over both dishes respectively (in Dollar terms). The price of the appetizer is 7 Dollars, and the price of the main course is 13 Dollars. If each lawyer would pay for herself she would choose the appetizer. However, if they share the bill the game becomes a PD. Ordering the appetizer gives utility (3,3) to both of them. If one lawyer defects and order the main dish the lawyer with the appetizer has net utility 0 and the defector has utility 5. If both order the main dish they receive utility 2 each. Hence, cooperating corresponds to Order Appetizer, and defecting corresponds to Order Main Dish. Problem 2. The deletion algorithm we defined in class is invariant to the order of deletion because all dominated strategies are removed for all players simultaneously in each step. For the exercise to make sense, we have to define an alternative algorithm with a lower speed of deletion. So lets define a sequence of strategy sets S n i with S i = S i where in each step only some or none of the dominate strategies are deleted. We continue to assume that the set S does not contain any dominated strategies any longer, i.e. if some strategy is dominated is will eventually get eliminated. We have to show that S = S . First of all it is easy to see that S k i S k i for all k and i . The claim is true by definition for k = 0. Assume that it holds for k 0. It then has to hold for k + 1 as well. Take some s i S k +1 i , for example. This strategy is a best response to some s i S k i S k i . Therefore s i S k +1 i . The above proof establishes that S S . It remains to be shown that S S . This can be done by showing that for any S k i there exists some S m i with m k such that S m i S k i . The claim is again true for k = 0. Assume it holds for k . For each i there exists some m i such that S m i i S k i . Take m = max( m 1 ,m 2 ,..,m I ). Then we have S m S k . We obtain S k +1 i by deleting all strategies which are strictly dominated when played against strategies in S k i . But these strategies are also strictly dominated when played against strategies in S m i . The new algorithm might not delete these strategies in the first step, but we assumed that they will get eventually deleted. Therefore, there exists some m > m such that S m i S k +1 i . This finishes the proof. The intuition behind the proof is simple. If a strategy s i A is strictly dom inated by s i A when the opponents strategies s i are drawn from some set B , then it is strictly dominated when the opponents strategies are drawn from some smaller set C B . Note, that this is NOT true for weakly dominated....
View
Full
Document
 Fall '10
 ding

Click to edit the document details