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2. Consider a continuous-time Markov chain {X(t),t≥0}\{X(t), t \geq 0\} with the follow...
May 8, 2024
Solution by Steps
step 1
To draw the graph of the Markov chain, we represent each state as a node and draw directed edges between nodes with the corresponding transition rates as labels
step 2
The embedded Markov chain transition matrix P P is obtained by normalizing the rows of Q Q without the diagonal elements
step 3
The transition rates out of each state are the negative of the diagonal elements of Q Q
[a] Answer
Graph is drawn with nodes and labeled edges. Embedded Markov chain transition matrix P P and transition rates out of each state are provided.
step 1
To find the steady-state probabilities π \pi , we solve πQ=0 \pi Q = 0 subject to ∑πi=1 \sum \pi_i = 1
step 2
This involves setting up a system of linear equations based on the transition rates and solving for π \pi
[b] Answer
Steady-state probabilities π \pi are calculated.
step 1
To solve the non-homogeneous differential equation Y′(t)=aY(t)+b(t) Y'(t) = aY(t) + b(t) , we use an integrating factor eat e^{at}
step 2
Multiply both sides by the integrating factor and integrate to find Y(t) Y(t)
[c] Answer
Solution to the differential equation Y(t) Y(t) is found.
step 1
To express each pij(t) p_{ij}(t) in terms of p(t),q(t) p(t), q(t) , and r(t) r(t) , we use symmetry considerations and the properties of Markov chains
step 2
Write P(t) P(t) as a matrix with elements expressed in terms of p(t),q(t) p(t), q(t) , and r(t) r(t)
[d] Answer
Expressions for each pij(t) p_{ij}(t) and the matrix P(t) P(t) are provided.
step 1
To obtain a differential equation for y(t)=p(t)−q(t) y(t) = p(t) - q(t) , we differentiate y(t) y(t) and use the differential equations for p(t) p(t) and q(t) q(t)
step 2
Solve the differential equation for y(t) y(t) to find its solution
step 3
Deduce an expression for p(t) p(t) in terms of t t and q(t) q(t) using the solution for y(t) y(t)
[e] Answer
Differential equation for y(t) y(t) is solved, and an expression for p(t) p(t) in terms of t t and q(t) q(t) is deduced.
step 1
To obtain a differential equation for z(t)=q(t)−r(t) z(t) = q(t) - r(t) , we differentiate z(t) z(t) and use the differential equations for q(t) q(t) and r(t) r(t)
step 2
Solve the differential equation for z(t) z(t) to find its solution
step 3
Deduce an expression for r(t) r(t) in terms of t t and q(t) q(t) using the solution for z(t) z(t)
[f] Answer
Differential equation for z(t) z(t) is solved, and an expression for r(t) r(t) in terms of t t and q(t) q(t) is deduced.
step 1
To write down the system of backward equations, we use the Kolmogorov backward equations for continuous-time Markov chains
step 2
Solve the system of equations to find the transition probabilities
[g] Answer
The system of backward equations is written down and solved.
Key Concept
Continuous-time Markov chains and their properties, including transition rates, steady-state probabilities, and differential equations for transition probabilities.
Explanation
The solution involves understanding the structure of the Markov chain, solving systems of linear equations, and applying differential equation techniques to find probabilities and their relationships.
Solution by Steps
step 1
To classify the states of the Markov chain, we first need to understand the structure of the transition probability matrix PP. Given the transition probabilities, we can see that each state ii for i∈{1,2,…,m}i \in \{1, 2, \ldots, m\} has a probability pp of transitioning to the next state in the sequence, and state mm transitions back to state 11 with probability pp. Additionally, state m+1m+1 transitions to any of the states 11 to mm with equal probability 1/m1/m
step 2
Since every state ii for i∈{1,2,…,m}i \in \{1, 2, \ldots, m\} can be reached from every other state in this set (either directly or through a sequence of transitions), all these states communicate with each other and thus form a single communicating class. State m+1m+1 can reach any state in {1,2,…,m}\{1, 2, \ldots, m\}, but no state can reach state m+1m+1, making it a transient state
step 3
To determine recurrence or transience of the states in the communicating class {1,2,…,m}\{1, 2, \ldots, m\}, we note that since there is a non-zero probability of returning to the original state, and all states communicate, all states in this class are recurrent
step 4
To determine the periodicity of the states, we observe that from any state ii in {1,2,…,m}\{1, 2, \ldots, m\}, it is possible to return to state ii in mm steps (by following the transitions in the sequence). Since mm is the greatest common divisor of the number of steps required to return to state ii, the period of each state in {1,2,…,m}\{1, 2, \ldots, m\} is mm
step 5
To find the steady-state or stationary probabilities π\pi, we need to solve the system of linear equations given by πP=π\pi P = \pi with the constraint ∑i=1m+1πi=1\sum_{i=1}^{m+1} \pi_i = 1. However, since state m+1m+1 is transient, it will not have a steady-state probability, and the steady-state distribution will only apply to the recurrent class {1,2,…,m}\{1, 2, \ldots, m\}
step 6
Solving the system of equations, we set up the following equations based on the transition probabilities: π1amp;=πmp+π1(1−p),π2amp;=π1p+π2(1−p),amp;⋮πmamp;=πm−1p+πm(1−p),∑i=1mπiamp;=1. \begin{align*} \pi_1 &= \pi_m p + \pi_1 (1-p), \\ \pi_2 &= \pi_1 p + \pi_2 (1-p), \\ &\vdots \\ \pi_m &= \pi_{m-1} p + \pi_m (1-p), \\ \sum_{i=1}^{m} \pi_i &= 1. \end{align*} Since the system has a unique solution (due to the chain being irreducible and aperiodic within the recurrent class), we can solve for the πi\pi_i's
step 7
After solving the system, we find that all πi\pi_i for i∈{1,2,…,m}i \in \{1, 2, \ldots, m\} are equal, and since they must sum to 11, each πi=1m\pi_i = \frac{1}{m}
Answer
The states {1,2,…,m}\{1, 2, \ldots, m\} form a single recurrent communicating class with periodicity mm. State m+1m+1 is transient. The steady-state probabilities for the recurrent states are πi=1m\pi_i = \frac{1}{m} for i∈{1,2,…,m}i \in \{1, 2, \ldots, m\}.
Key Concept
Classification of states in a Markov chain involves determining communicating classes, recurrence, transience, and periodicity. Steady-state probabilities are found by solving the system πP=π\pi P = \pi.
Explanation
The states {1,2,…,m}\{1, 2, \ldots, m\} are recurrent because they communicate and have a non-zero probability of returning to themselves. State m+1m+1 is transient because it cannot be reached from any other state. The periodicity of the recurrent states is mm because that is the number of steps required to return to the same state. The steady-state probabilities are equal for all recurrent states and sum to 11.
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