The transition matrix for a Markov process is T = State 1 State 2 [ .3 .4 .7 .6 ] S tate 1 S tate 2 and the initial-state distribution vector is X 0 = State 1 State 2 [ .6 .4 ] Find X 2 .

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Answer 1

Textbook Question

Chapter 9.BMO, Problem 1BMO

The transition matrix for a Markov process is

$T = State 1 State 2 [ .3 .4 .7 .6 ] S tate 1 S tate 2$

and the initial-state distribution vector is

$X 0 = State 1 State 2 [ .6 .4 ]$

Find $X 2$ .

Expert Solution & Answer

To determine

To find:

The vector $X2$ .

Answer to Problem 1BMO

Solution:

The vector $X2$ is $[0.3660.634]$

Explanation of Solution

Given:

The transition matrix for a Markov process is

$T=State 1State 2 [.3.4.7.6]State 1State 2$

and the initial-state distribution vector is

$X0=State 1State 2[.6.4]$

Approach:

From the given data,

Write the expression of the probability distribution after one observation

$X1=TX0→(1)$

Write the expression of the probability distribution after two observations

$X2=TX1→(2)$

Calculation:

Substitute $[.6.4]$ for $X0$ and $[.3.4.7.6]$ for $T$ in equation (1)

$X1=[.3.4.7.6]⋅[.6.4]=[(.3×.6)+(.4×.4)(.7×.6)+(.6×.4)]=[.18+.16.42+.24]=[.34.66]$

Similarly, Substitute $[.34.66]$ for $X1$ and $[.3.4.7.6]$ for $T$ in equation (2).

$X2=[.3.4.7.6]⋅[.34.66]=[(.3×.34)+(.4×.66)(.7×.34)+(.6×.66)]=[.102+.264.238+.396]=[.366.634]$

Conclusion:

Hence, vector $X2$ is $[.366.634]$ .

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Answer 2

Textbook Question

Chapter 9.BMO, Problem 1BMO

The transition matrix for a Markov process is

$T = State 1 State 2 [ .3 .4 .7 .6 ] S tate 1 S tate 2$

and the initial-state distribution vector is

$X 0 = State 1 State 2 [ .6 .4 ]$

Find $X 2$ .

Expert Solution & Answer

To determine

To find:

The vector $X2$ .

Answer to Problem 1BMO

Solution:

The vector $X2$ is $[0.3660.634]$

Explanation of Solution

Given:

The transition matrix for a Markov process is

$T=State 1State 2 [.3.4.7.6]State 1State 2$

and the initial-state distribution vector is

$X0=State 1State 2[.6.4]$

Approach:

From the given data,

Write the expression of the probability distribution after one observation

$X1=TX0→(1)$

Write the expression of the probability distribution after two observations

$X2=TX1→(2)$

Calculation:

Substitute $[.6.4]$ for $X0$ and $[.3.4.7.6]$ for $T$ in equation (1)

$X1=[.3.4.7.6]⋅[.6.4]=[(.3×.6)+(.4×.4)(.7×.6)+(.6×.4)]=[.18+.16.42+.24]=[.34.66]$

Similarly, Substitute $[.34.66]$ for $X1$ and $[.3.4.7.6]$ for $T$ in equation (2).

$X2=[.3.4.7.6]⋅[.34.66]=[(.3×.34)+(.4×.66)(.7×.34)+(.6×.66)]=[.102+.264.238+.396]=[.366.634]$

Conclusion:

Hence, vector $X2$ is $[.366.634]$ .

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Answer 3

Textbook Question

Chapter 9.BMO, Problem 1BMO

The transition matrix for a Markov process is

$T = State 1 State 2 [ .3 .4 .7 .6 ] S tate 1 S tate 2$

and the initial-state distribution vector is

$X 0 = State 1 State 2 [ .6 .4 ]$

Find $X 2$ .

Expert Solution & Answer

To determine

To find:

The vector $X2$ .

Answer to Problem 1BMO

Solution:

The vector $X2$ is $[0.3660.634]$

Explanation of Solution

Given:

The transition matrix for a Markov process is

$T=State 1State 2 [.3.4.7.6]State 1State 2$

and the initial-state distribution vector is

$X0=State 1State 2[.6.4]$

Approach:

From the given data,

Write the expression of the probability distribution after one observation

$X1=TX0→(1)$

Write the expression of the probability distribution after two observations

$X2=TX1→(2)$

Calculation:

Substitute $[.6.4]$ for $X0$ and $[.3.4.7.6]$ for $T$ in equation (1)

$X1=[.3.4.7.6]⋅[.6.4]=[(.3×.6)+(.4×.4)(.7×.6)+(.6×.4)]=[.18+.16.42+.24]=[.34.66]$

Similarly, Substitute $[.34.66]$ for $X1$ and $[.3.4.7.6]$ for $T$ in equation (2).

$X2=[.3.4.7.6]⋅[.34.66]=[(.3×.34)+(.4×.66)(.7×.34)+(.6×.66)]=[.102+.264.238+.396]=[.366.634]$

Conclusion:

Hence, vector $X2$ is $[.366.634]$ .

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Answer 4

Textbook Question

Chapter 9.BMO, Problem 1BMO

The transition matrix for a Markov process is

$T = State 1 State 2 [ .3 .4 .7 .6 ] S tate 1 S tate 2$

and the initial-state distribution vector is

$X 0 = State 1 State 2 [ .6 .4 ]$

Find $X 2$ .

Expert Solution & Answer

To determine

To find:

The vector $X2$ .

Answer to Problem 1BMO

Solution:

The vector $X2$ is $[0.3660.634]$

Explanation of Solution

Given:

The transition matrix for a Markov process is

$T=State 1State 2 [.3.4.7.6]State 1State 2$

and the initial-state distribution vector is

$X0=State 1State 2[.6.4]$

Approach:

From the given data,

Write the expression of the probability distribution after one observation

$X1=TX0→(1)$

Write the expression of the probability distribution after two observations

$X2=TX1→(2)$

Calculation:

Substitute $[.6.4]$ for $X0$ and $[.3.4.7.6]$ for $T$ in equation (1)

$X1=[.3.4.7.6]⋅[.6.4]=[(.3×.6)+(.4×.4)(.7×.6)+(.6×.4)]=[.18+.16.42+.24]=[.34.66]$

Similarly, Substitute $[.34.66]$ for $X1$ and $[.3.4.7.6]$ for $T$ in equation (2).

$X2=[.3.4.7.6]⋅[.34.66]=[(.3×.34)+(.4×.66)(.7×.34)+(.6×.66)]=[.102+.264.238+.396]=[.366.634]$

Conclusion:

Hence, vector $X2$ is $[.366.634]$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To find:

The vector $X2$ .

Answer to Problem 1BMO

Solution:

The vector $X2$ is $[0.3660.634]$

Explanation of Solution

Given:

The transition matrix for a Markov process is

$T=State 1State 2 [.3.4.7.6]State 1State 2$

and the initial-state distribution vector is

$X0=State 1State 2[.6.4]$

Approach:

From the given data,

Write the expression of the probability distribution after one observation

$X1=TX0→(1)$

Write the expression of the probability distribution after two observations

$X2=TX1→(2)$

Calculation:

Substitute $[.6.4]$ for $X0$ and $[.3.4.7.6]$ for $T$ in equation (1)

$X1=[.3.4.7.6]⋅[.6.4]=[(.3×.6)+(.4×.4)(.7×.6)+(.6×.4)]=[.18+.16.42+.24]=[.34.66]$

Similarly, Substitute $[.34.66]$ for $X1$ and $[.3.4.7.6]$ for $T$ in equation (2).

$X2=[.3.4.7.6]⋅[.34.66]=[(.3×.34)+(.4×.66)(.7×.34)+(.6×.66)]=[.102+.264.238+.396]=[.366.634]$

Conclusion:

Hence, vector $X2$ is $[.366.634]$ .

Answer 8

Expert Solution & Answer

To determine

To find:

The vector $X2$ .

Answer to Problem 1BMO

Solution:

The vector $X2$ is $[0.3660.634]$

Explanation of Solution

Given:

The transition matrix for a Markov process is

$T=State 1State 2 [.3.4.7.6]State 1State 2$

and the initial-state distribution vector is

$X0=State 1State 2[.6.4]$

Approach:

From the given data,

Write the expression of the probability distribution after one observation

$X1=TX0→(1)$

Write the expression of the probability distribution after two observations

$X2=TX1→(2)$

Calculation:

Substitute $[.6.4]$ for $X0$ and $[.3.4.7.6]$ for $T$ in equation (1)

$X1=[.3.4.7.6]⋅[.6.4]=[(.3×.6)+(.4×.4)(.7×.6)+(.6×.4)]=[.18+.16.42+.24]=[.34.66]$

Similarly, Substitute $[.34.66]$ for $X1$ and $[.3.4.7.6]$ for $T$ in equation (2).

$X2=[.3.4.7.6]⋅[.34.66]=[(.3×.34)+(.4×.66)(.7×.34)+(.6×.66)]=[.102+.264.238+.396]=[.366.634]$

Conclusion:

Hence, vector $X2$ is $[.366.634]$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To find:

The vector $X2$ .

Answer 12

Answer to Problem 1BMO

Solution:

The vector $X2$ is $[0.3660.634]$

Answer 13

Explanation of Solution

Given:

The transition matrix for a Markov process is

$T=State 1State 2 [.3.4.7.6]State 1State 2$

and the initial-state distribution vector is

$X0=State 1State 2[.6.4]$

Approach:

From the given data,

Write the expression of the probability distribution after one observation

$X1=TX0→(1)$

Write the expression of the probability distribution after two observations

$X2=TX1→(2)$

Calculation:

Substitute $[.6.4]$ for $X0$ and $[.3.4.7.6]$ for $T$ in equation (1)

$X1=[.3.4.7.6]⋅[.6.4]=[(.3×.6)+(.4×.4)(.7×.6)+(.6×.4)]=[.18+.16.42+.24]=[.34.66]$

Similarly, Substitute $[.34.66]$ for $X1$ and $[.3.4.7.6]$ for $T$ in equation (2).

$X2=[.3.4.7.6]⋅[.34.66]=[(.3×.34)+(.4×.66)(.7×.34)+(.6×.66)]=[.102+.264.238+.396]=[.366.634]$

Conclusion:

Hence, vector $X2$ is $[.366.634]$ .

Answer 14

Ch. 9.1 - What is a finite stochastic process? What can you...Ch. 9.1 - Prob. 2CQ Ch. 9.1 - Consider a transition matrix T for a Markov chain...Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E

The transition matrix for a Markov process is T = State 1 State 2 [ .3 .4 .7 .6 ] S tate 1 S tate 2 and the initial-state distribution vector is X 0 = State 1 State 2 [ .6 .4 ] Find X 2 .

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Answer to Problem 1BMO

Explanation of Solution

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