Explanation Given: There are 5 rooms for the mouse to go and each room has some doors among that some doors are one-way. Definition used: The steady-state vector : If P is a stochastic matrix, then a steady state vector (or equilibrium vector or invariant probability vector) for P is a probability vector q such that P q = q . Theorem used: Consider an irreducible Markov chain with a finite state space, let n i j be the number of steps until the chain first visits state i given that the chain starts in state j , and let t i i = E [ n i i ] . Then t i i = 1 q i . Where q i is the entry in the steady state vector q corresponding to state i . Calculation: Start from room 1, mouse can be walked towards other rooms that are 2 and 4, thus the probabilities for the walk from room 1 towards 2 and 4 is 1 2 . From room 2, mouse can be walked towards another room that is 5, thus the probability for the walk from room 2 towards 5 is 1. From room 3, mouse can be walked towards other rooms that are 4 and 5, thus the probabilities for the walk from room 3 towards 4 and 5 is 1 2 . From room 4, mouse can be walked towards other rooms that are 1 and 3, thus the probabilities for the walk from room 4 towards 1 and 3 is 1 2 . From room 5, mouse can be walked towards another room that is 2, thus the probability for the walk from room 5 towards 2 and 4 is 1 2 . Based upon above computation, obtain the transition matrix for the simple random walk as follows. P = [ 0 0 0 1 2 0 1 2 0 0 0 1 2 0 0 0 1 2 0 1 2 0 1 2 0 1 2 0 1 1 2 0 0 ] . From the definition stated above, it is observed that, to obtain the steady-state vector q , solve the system P q = q , where q = [ q 1 q 2 q 3 q 4 q 5 ] . It is clear that the system P q = q is equivalent to the homogeneous system ( P − I ) q = 0 . Hence, it is enough to solve the system ( P − I ) q = 0 to obtain the steady-state vector q . The system ( P − I ) is shown as follows, [ 0 0 0 1 2 0 1 2 0 0 0 1 2 0 0 0 1 2 0 1 2 0 1 2 0 1 2 0 1 1 2 0 0 ] − [ 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 ] = [ − 1 0 0 1 2 0 1 2 − 1 0 0 1 2 0 0 − 1 1 2 0 1 2 0 1 2 − 1 1 2 0 1 1 2 0 − 1 ] . Thus, the augmented matrix is, [ − 1 0 0 1 2 0 0 1 2 − 1 0 0 1 2 0 0 0 − 1 1 2 0 0 1 2 0 1 2 − 1 1 2 0 0 1 1 2 0 − 1 0 ]

Thomas' Calculus and Linear Algebra and Its Applications Package for the Georgia Institute of Technology, 1/e

5th Edition

ISBN: 9781323132098

Author: Thomas, Lay

Publisher: PEARSON C

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Chapter 10.3, Problem 19E

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To find: The average time that the mouse will take if it starts in room 1 and return back to room 1.

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Chapter 10.3, Problem 18E

Chapter 10.3, Problem 20E

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Chapter 10 Solutions

Thomas' Calculus and Linear Algebra and Its Applications Package for the Georgia Institute of Technology, 1/e

Ch. 10.1 - Fill in the missing entries in the stochastic...Ch. 10.1 - Prob. 2PP Ch. 10.1 - In Exercises 1 and 2, determine whether P is a...Ch. 10.1 - In Exercises 1 and 2, determine whether P is a...Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - In Exercises 5 and 6, the transition matrix P for...Ch. 10.1 - Prob. 6E Ch. 10.1 - In Exercises 7 and 8, the transition matrix P for...Ch. 10.1 - In Exercises 7 and 8, the transition matrix P for...

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The average time that the mouse will take if it starts in room 1 and return back to room 1.

Solution Summary: The author calculates the average time the mouse takes to return to room 1 is 152 steps. If P is a stochastic matrix, the steady-state vector (or equilibrium vector or in

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To find: The average time that the mouse will take if it starts in room 1 and return back to room 1.

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Chapter 10 Solutions