# ps3sol - 14.12 Game Theory Muhamet Yildiz Fall 2005...

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

14.12 Game Theory Muhamet Yildiz Fall 2005 Solution to Homework 3 Problem One a) This game has two subgames. One subgame is the game as a whole and the other is the subgame after both players say yes. In this last subgame we have 3 Nash equilibria (NE) which are I , I , C , C and a mixed strategy. The mixed strategy NE is computed as follows: Let’s say player 1 plays I with probability µ and player 2 plays I with probability . Then the mixed equilibrium should satisfy the following equations: Player 2 is indifferent between I and C µ 100 1 µ 0 µ 60 1 µ 60 u 0.6 Player 1 is indifferent between I and C 100 1 0 60 1 60 0.6 Therefore the solution is µ 0.6. This implies that the mixed Nash equilibrium is 0.6 I 0.4 C ,0.6 I 0.4 C . To compute the subgame-perfect equilibrium (SPE) we have to check if the strategy profile is a NE in every subgame. Thus, to obtain the SPE we have to assume that outcome of the smaller subgame will be one of those NE. When both players choose I in the smaller subgame, the reduced form of the game is: Y N Y 100, 100 80, 80 N 80, 80 80, 80 There are 2 types of NE in this case Y , Y and N , N . Therefore the SPE are YI , YI and NI , NI When both players choose C in the smaller subgame. The reduced form of the game is

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

View Full Document
Y N Y 60,60 80,80 N 80,80 80,80 There are 3 pure NE in this case Y , N , NY and N , N and 2 mixed NE N , pY 1 p N and pY 1 p N , N , 0 p 1. Therefore the SPE are YC , NC , NC , YC , NC , NC , NC , pY 1 p N C and  pY 1 p N C , NC . When both players choose to play the mixed NE in the smaller subgame the reduced form is the same as in the former case and therefore the NE are also the same. Thus, SPE of the game are YC , NC . NC , YC , NC , NC , NC , pY 1 p N C and  pY 1 p N C , NC . Problem Two Denote V t i the continuation value to player i when the game goes to the period t . Clearly, in the last period, t 3 n , player 2 will offer 0,1 thus V 3 n 2 1 . 1 0 and V 3 n In period t 3 n 3 k 3, k 1,2,. .. n 1 , player 2 makes an offer. He will offer 1 1 V 3 n 3 k 4 ,1 V 3 n 3 k 4 that will be accepted by player 1. In period t 3 n 3 k 2 , player 1 is the one to make an offer and will offer 2 2 1 V 3 n 3 k 3 , V 3 n 3 k 3 . 2
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/28/2011 for the course ECONOMICS 114.126 taught by Professor Muhammadyildiz during the Spring '05 term at University of Massachusetts Boston.

### Page1 / 7

ps3sol - 14.12 Game Theory Muhamet Yildiz Fall 2005...

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

View Full Document
Ask a homework question - tutors are online