Explanation Given: The transition matrix A = [ 0.80 0.15 0.05 0.20 0.70 0.10 0.20 0.20 0.60 ] . Explanation: A vector v is said to be an equilibrium vector for a transition matrix A , if the following condition holds: v A = v Let the vector v be defined as follows, so that the multiplication holds. [ v 1 v 2 v 3 ] Then v A = v [ v 1 v 2 v 3 ] [ 0.80 0.15 0.05 0.20 0.70 0.10 0.20 0.20 0.60 ] = [ v 1 v 2 v 3 ] Simplify the left hand side to obtain [ 0.8 v 1 + 0.2 v 2 + 0.2 v 3 0.15 v 1 + 0.70 v 2 + 0.2 v 3 0.05 v 1 + 0.1 v 2 + 0.6 v 3 ] = [ v 1 v 2 v 3 ] This gives ride to the following system of equations 0.8 v 1 + 0.2 v 2 + 0.2 v 3 = v 1 0.15 v 1 + 0.7 v 2 + 0.2 v 3 = v 2 0.05 v 1 + 0.1 v 2 + 0.6 v 3 = v 3 This can be simplified to − 0.2 v 1 + 0.2 v 2 + 0.2 v 3 = 0 0.15 v 1 − 0.3 v 2 + 0.2 v 3 = 0 0.05 v 1 + 0.1 v 2 − 0.4 v 3 = 0 Also, use the property that v 1 , v 2 , and v 3 are entries of a probability vector so they should sum to 1. This gives v 1 + v 2 + v 3 = 1 The system involving these 4 equations can be solved with the help of Gauss Jordan method. The augmented matrix for the system is [ − 0.2 0.2 0.2 0 0.15 − 0.3 0.2 0 0.05 0.1 − 0.4 0 1 1 1 1 ] − 1 0.2 R 1 → R 1 gives [ 1 − 1 − 1 0 0.15 − 0.3 0.2 0 0.05 0.1 − 0.4 0 1 1 1 1 ] R 2 − 0.15 R 1 → R 2 , R 3 − 0

11th Edition

ISBN: 9780321979438

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 10.2, Problem 19E

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The equilibrium vector corresponding to the transition matrix provided.

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The equilibrium vector corresponding to the transition matrix provided.

Solution Summary: The author explains that a vector v is said to be an equilibrium vector for the transition matrix A if the following condition holds.

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