solutionspset3 - Ec1052 Introduction to Game Theory Handout...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
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 6 Harvard University 20 March 2004 Solutions to Problem Set 3 Problem 1. The following game matrix shows a generic way to capture the gist of the story in a 2 by 2 game. The teacher gets utility l from the students learning the material and has disutility E from writing questions. He also derives utility a > 0 from failing lazy students and we assume a- E > 0. The students get utility g from a good grade but have disutility e from learning. A bad grade has disutility b and we assume that g- e > b , i.e. students rather learn than get a bad grade. New Old Study Memorize l,g-e 0,g l-E,g-e a-E,b The game has no pure NE. We can calculate the mixed equilibrium (which MUST exist) in the usual manner and get: σ * 1 = g- e- b g- b , e g- b ¶ σ * 2 = a- E a , E a ¶ The mixed equilibrium looks sensible: • If the teacher has greater utility from failing lazy students then students are more likely to study. Higher cost of effort on the part of teacher has the opposite effect. • If students care more about good grades (larger g ), or if they are unhappier about bad grades (smaller b ) then the teacher is more likely to give old exams. If students have a higher cost of effort then the teacher is more likely to give a new exam. Ec1052 Handout 6: Solutions to Problem Set 3 2 Problem 2. This exercise requires a lot of brute force calculations. Let’s start with the first problem. There are two pure NE, ( T,A ) and ( D,D ). It is easy to see that there cannot be mixed equilibria where only one player mixes and the other one plays a pure strategy (the mixing player would have a unique BR in each case). Therefore, both players mix in any mixed equilibrium. Player 2’s mixed strategy can have lots of possible supports. We have to check each of them separately: • Player 2 has support { A,B } . Let’s assume that player 1 plays T with probability α and D with probability 1- α . Player 2 has to be indifferent between A and B such that 6 α- 10000(1- α ) = 5 α- 1000(1- α ). This gives us α = 9000 9001 . Also assume, that player 2 plays A with probability β and B with probability 1- β . Then player 1 has to be indifferent between T and D and we get 200 β + 3(1- β ) = 5(1- β ). This gives us β = 2 202 . However, we don’t know yet if this mixed equilibrium of the subgame is in fact a NE. We have to make sure that player 2 cannot profitably deviate to strategies C or D . The equilibrium payoff from the mixed equilibrium is the same as the payoff from A or B (by construction) and is approximately 4 . 9. Playing C gives payoff 3 and playing D has payoff approximately has negative payoff. Hence the strategies σ * 1 = ( 9000 9001 , 1 9001 ) and σ * 2 = ( 2 202 , 200 202 , , ) are a mixed Nash equilibrium....
View Full Document

Page1 / 9

solutionspset3 - Ec1052 Introduction to Game Theory Handout...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online