Inverse Matrix Definition 1 0 . . . 0 0 1 . . . 0 The Identity Matrix I n × n is . . . . . . . . . . . . . 0 0 . . . 1 It satisfies I · M = M and N · I = N for any M , N for which you can perform the multiplication. The inverse matrix to an n × n matrix A is a matrix B such that A · B = I . Alternatively B · A = I . If one holds, both hold. If the inverse exists, it is unique. The inverse is written B − 1 . Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 1 / 14
Example of Inverse 1 2 2 2 4 6 A = 0 1 3 B = − 1 − 4 − 3 . Find its inverse if it exists. 0 0 1 0 1 − 1 Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 2 / 14
Encoding Encode the message To be or not to be. In numbers, 20 , 15 , 27 , 2 , 5 , 27 , 15 , 18 , 27 , 14 , 15 , 20 , 27 , 20 , 15 , 27 , 2 , 5. A to Z are 1 to 26 and spaces are 27. You substitute, and break up into 3 × 1 column vectors. The encoded message is B · v i . To decode, you break up into groups of three and make column vectors. Then apply B − 1 . 2 4 6 B = − 1 − 4 − 3 . 0 1 − 1 ABCDEFGHIJ K L M N O P Q R S T U V W X 12345678910 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 3 / 14
Markov Chains Transition Matrix P = ( p ij ) where p ij = P ( j | i ) . Defining Properties: 0 ≤ p ij ≤ 1 , � j p ij = 1 . Regular P n for some n > 0 has positive entries only. Main Property: An equilibrium/fixed vector exists and is unique. v · P = v where v is a probability row vector. Absorbing There is an i such that p ii = 1 . If so, all OTHER entries in row i are 0; p ij = 0 for j � = i . Also there is a positive probability to go from any state to an absorbing state. Main Property: Long Term Trend. Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 4 / 14
Examples � 1 � 0 1 0 . 5 0 . 5 0 . 4 0 . 3 0 . 3 0 0 0 . 3 0 0 . 7 2 0 . 2 0 . 2 0 . 3 0 . 3 0 0 . 6 0 0 . 4 � 0 . 2 � 0 . 8 3 0 . 5 0 . 5 � 0 � 1 4 1 0 0 . 2 0 . 8 0 0 0 . 8 0 . 2 5 0 . 8 0 . 2 0 0 1 0 0 0 1 6 1 0 0 Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 5 / 14
Example Continued 0 . 68 0 . 16 0 . 16 For (5), P 2 = . So the matrix is regular. 0 . 16 0 . 68 0 . 16 0 . 16 . 16 0 . 68 Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 6 / 14
Application Example (Problem 25, Section 10.2) The probability that a complex assembly line works correctly depends only on whether the line worked the last time it was used. There is a 0 . 9 chance that the line will work correctly if it worked correctly the last time, and a 0 . 8 chance that it will work correctly if it failed last time. Set up a transition matrix, and find the long range probability that the line will work correctly. Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 7 / 14
Answer. � 0 . 9 � 0 . 1 The matrix is . This is regular. We need to solve the system 0 . 8 0 . 2 � 0 . 9 , � 0 . 1 � � � � x , y · = x , y . 0 . 8 , 0 . 2 � � � � This is 0 . 9 x + 0 . 8 y , 0 . 1 x + 0 . 2 y = x , y We can rewrite it as � − 0 . 1 x + 0 . 8 y = 0 = 0. The two equations are x = 8 y . So the equilibrium 0 . 1 x − 0 . 8 y � � vector is 8 / 9 1 / 9 . In general, the coefficient matrix of the system is P T − I , and you set every equation equal to 0: [ P T − I | 0]. Transpose means that you interchange rows and columns. Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 8 / 14
Regular Markov Chains, Summary 1 The matrix P has positive entries only. 2 There is a unique equilibrium (probability) vector V such that V · P = V . V 1 . . . V n . . . 3 lim m →∞ P m = lim m →∞ v · P m = V . . . . , . . . . . . V 1 V n 4 To solve for V , you need to solve the system [ P t − I n × n | 0] . There is a unique solution satisfying 0 ≤ V j ≤ 1 and V 1 + · · · + V n = 1 . Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 9 / 14
Previous Example − 0 . 1 0 . 8 0 0 0 0 � 0 � � 0 � − 9 − 1 1 1 / 9 �→ �→ 0 . 1 − 0 . 8 0 1 − 8 0 �→ �→ 1 1 1 1 1 1 1 1 1 1 1 1 � 0 � � 1 � 1 1 / 9 0 8 / 9 �→ �→ 1 0 8 / 9 0 1 1 / 9 Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 10 / 14
Absorbing Markov Chains 1 The matrix P has some p ii = 1 . You rearrange the states so that the � I � 0 matrix is P = . For each state there must a positive R Q probability for it to go to an equilibrium state. This means each row of F moust have a nonzero entry. � � 0 I 2 lim m →∞ P m = . The rows of ( I − Q ) − 1 R tell you ( I − Q ) − 1 R 0 the probabilities of ending up in the state corresponding to the column. The matrix F = ( I − Q ) − 1 is called the fundamental matrix. The number f ij give the number of visits to state j before being absorbed, given that the current state is i . Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 11 / 14
Gambler’s Ruin Two players A and B , play a game. They toss a coin. If heads, then A pays B $1. If tails, B pays A $1. They start out with a total of $3. The game ends whenever on of the players is broke. Analyze the game. Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 12 / 14
Classification of States I Question: How to tell if a Markov Chain is regular. 1 We say j is accessible from i if p m ( i , j ) > 0 for some m > 0 . i , j communicate, i ≃ j , if p n ( i , j ) > 0 and p m ( j , i ) > 0 for some m , n > 0 . 2 i ≃ i 3 i ≃ j implies j ≃ i 4 i ≃ j and j ≃ k implies i ≃ j 5 The sets of equivalent states are called classes 6 A Markov Chain is irreducible if all states are in a single class. 7 Regular implies irreducible. 8 Absorbing states are single classes. 9 i is called transient if f i = P ( i for some m | i ) = 1 . 10 i is called recurrent if f i = P ( i for some m | i ) < 1 . Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 13 / 14
Classification of States II To find the equivalence classes: 1 Draw the states 2 Forget the probabilities on the diagonal 3 Join two states by an edge if p ( i , j ) > 0 and p ( j , i ) > 0 . 4 If only one is > 0 , draw an arrow. Dan Barbasch Math 1105 Chapter 2 and 10, October 30 Week of October 30 14 / 14
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