Chapter 10.1-10.3, Markov Chains I Review of Matrix Multiplication on page 484 Transition Matrix Markov Chain Probability Vector Sample Problems: page 487 1, 10, 15, 28 Regular Chains Equilibrium Vector Summary on page 495 Sample Problem page 497, #17 Long Term Behaviour Dan Barbasch Math 1105 Chapter 10-11, Review Week of November 27 1 / 7
Chapter 10.1-10.3, Markov Chains II Example (Problem 17, page 497) 0 . 9 0 . 05 0 . 05 0 . 15 0 . 75 0 . 10 0 0 1 Dan Barbasch Math 1105 Chapter 10-11, Review Week of November 27 2 / 7
Chapter 10.1-10.3, Markov Chains III Answer. The matrix is not regular; it has an absorbing state. It is absorbing, because any state has positive probability to reach the absorbing state. The vector, [0 , 0 , 1], is an equilibrium vector. To compute all equilibrium vectors, solve the system v 1 + v 2 + v 3 = 1 together with � � � � · P = : v 1 v 2 v 3 v 1 v 2 v 3 v 1 + v 2 + v 3 = 1 − 0 . 1 v 1 + 0 . 15 v 2 = 0 0 . 05 v 1 − 0 . 25 v 2 = 0 0 . 05 v 1 + 0 . 1 v 2 = 0 The only solution is [0 , 0 , 1]. Dan Barbasch Math 1105 Chapter 10-11, Review Week of November 27 3 / 7
Absorbing Markov Chains Definition on page 501 Summary on page 506 Fundamental Matrix Reviews on pages 505, 506 Sample Problem #23 Problems from Chapter Review. 1 Read the Summary. 2 Concept Check, # 1-14, 31-34. 3 Problems 16, 18, 20, 24, 27-30, 36, 38, 40, 42, 43-44, 49 Dan Barbasch Math 1105 Chapter 10-11, Review Week of November 27 4 / 7
Long Range Trend for Problem 17, page 497 In standard form the matrix is 1 0 0 0 . 05 0 . 9 0 . 05 0 . 10 0 . 15 0 . 75 � 0 . 05 � � 0 . 90 0 . 05 � So R = and Q = . The fundamental matrix and the 0 . 10 0 . 15 0 . 75 long term trend are � 100 / 7 � � 1 � 20 / 7 F = ( I − Q ) − 1 = , FR = . 60 / 7 40 / 7 1 Since there is only one absorbing state, each state will end up in the absorbing state with probability 1; obvious conclusion. The matrix Q tells you how many times (on the average) the chain will spend in states 1 and 2 before being absorbed. Dan Barbasch Math 1105 Chapter 10-11, Review Week of November 27 5 / 7
Chapter 11.1, 11.2, Game Theory I Payoff Matrix Dominated Strategies Strictly Determined Games Saddle Points (two ways of computing) Sample Problem page 525 #33 Mixed Strategies, Expected Value, Value of the Game Optimum Strategy, summary on page 532 Sample Problems, page 532 # 2, 16 Problems from the Review Section. 1 Read the Summary pertaining to 11.1-11.2. 2 Concept Check. # 1-12 3 Problems # 13-18, 20, 22, 24, 26, 29-36, 47-49, 60-62. Dan Barbasch Math 1105 Chapter 10-11, Review Week of November 27 6 / 7
Problem 33, page 525 When A , B are in the same city, Player A get 65% . The payoff to A is 65 − 50 = 15 . When A is in city 1, and B in city 2, the market shares are 1 2 3 A 0 . 8 × 0 . 3 0 . 2 × 0 . 45 0 . 6 × 0 . 25 B 0 . 2 × 0 . 3 0 . 8 × 0 . 45 0 . 4 × 0 . 25 The boxed numbers come from the text. The other ones complete the market in the given city. So the market share of A is 0 . 24 + 0 . 09 + 0 . 15 = 48 . The payoff is 48 − 50 = − 2 . Exercise Complete the payoff matrix and check it equals 15 − 2 6 7 15 9 3 − 3 15 Dan Barbasch Math 1105 Chapter 10-11, Review Week of November 27 7 / 7
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