3 -2 3 I2N 120 180 90

Algebra and Trigonometry (6th Edition)
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Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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let P be the transition matrix for a Markov chain with three states.Let x0 be the initial state vector for the population.Compute Find the steady state vector

3
-2
3
I2N
Transcribed Image Text:3 -2 3 I2N
120
180
90
Transcribed Image Text:120 180 90
Expert Solution
Step 1

a ] the state vector X_1 and X_2 is given by ,

X_1=PX_0

X1 = [1/2 1/3 1/3] [1201 0 1/3 2/3 180 (1/2 1/ 30 120(1/2) + 180(1/3) +90(1/3) 120(0) + 180(1/3) +90(2/3) 120(1/2) + 180(1/3)

so we get ,

X1 = 150 120 120

Step 2

now X_2 is given by

X_2=P^2X_0

P^2=\begin{bmatrix} 1/2 &1/3 &1/3 \\ 0 & 1/3 & 2/3\\ 1/2 &1/3 &0 \end{bmatrix}\begin{bmatrix} 1/2 &1/3 &1/3 \\ 0 & 1/3 & 2/3\\ 1/2 &1/3 &0 \end{bmatrix}=\begin{bmatrix} 5/12 &7/18 & 7/18\\ 1/3 & 1/3 &2/9 \\ 1/4 & 5/18 &7/18 \end{bmatrix}

X_2=P^2X_0=\begin{bmatrix} 5/12 &7/18 & 7/18\\ 1/3 & 1/3 &2/9 \\ 1/4 & 5/18 &7/18 \end{bmatrix}\begin{bmatrix} 120\\ 180\\ 90 \end{bmatrix}=\begin{bmatrix} 120(5/12)+180(7/18)+90(7/18)\\ 120(1/3)+180(1/3)+90(2/9)\\ 120(1/4)+180(5/18)+90(7/18) \end{bmatrix}

we get X_2 as ,

{\color{DarkBlue} \mathbf{X_2=\begin{bmatrix} 155\\ 120\\ 115 \end{bmatrix}}}

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