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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1. Determine whether the following sets are subspaces of $\mathbb{R}^3$ under the operations of addition and scalar multiplication defined on $\mathbb{R}^3$. Justify your answers.(a) $W_1=\left\{\left(a_1, a_2, a_3\right) \in \mathbb{R}^3: a_1=3 a_2\right.$ and $\left.a_3=\mid a_2\right\}$(b) $W_2=\left\{\left(a_1, a_2, a_3\right) \in \mathbb{R}^3: a_1=a_3+2\right\}$(c) $W_3=\left\{\left(a_1, a_2, a_3\right) \in \mathbb{R}^3: 2 a_1-7 a_2+a_3=0\right\}$(d) $W_4=\left\{\left(a_1, a_2, a_3\right) \in \mathbb{R}^3: a_1-4 a_2-a_3=0\right\}$(e) $W_s=\left\{\left(a_1, a_2, a_3\right) \in \mathbb{R}^3: a_1+2 a_2-3 a_3=1\right\}$(f) $W_6=\left\{\left(a_1, a_2, a_3\right) \in \mathbb{R}^3: 5 a_1^2-3 a_2^2+6 a_3^2=0\right\}$

The annual aggregate claim amount of an insurer follows a compound Poisson distribution with parameter 1,000. Individual claim amounts follow a Gamma distribution with shape parameter a = 750 and rate parameter λ = 0.25. 1. Generate 20,000 simulated aggregate claim values for the insurer, using a random number generator seed of 955.Display the first five simulated claim values in your answer script using the R function head(). 2. Plot the empirical density function of the simulated aggregate claim values from Question 1, setting the x-axis range from 2,600,000 to 3,300,000 and the y-axis range from 0 to 0.0000045. 3. Suggest a suitable distribution, including its parameters, that approximates the simulated aggregate claim values from Question 1. 4. Generate 20,000 values from your suggested distribution in Question 3 using a random number generator seed of 955. Use the R function head() to display the first five generated values in your answer script. 5. Plot the empirical density…

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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

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