3.18 Use first-step analysis to find the expected return time to state b for the Markov chain with transition matrix a b c a (1/2 1/2 P = b 1/4 3/4 c (1/2 1/2

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please answer 3.18 use the example 3.17 method

**3.18** Use first-step analysis to find the expected return time to state \( b \) for the Markov chain with transition matrix

\[
P = \begin{pmatrix}
1/2 & 1/2 & 0 \\
1/4 & 0 & 3/4 \\
1/2 & 1/2 & 0 
\end{pmatrix}
\]

The matrix \( P \) is a transition matrix for a Markov chain with states \( a \), \( b \), and \( c \). The elements of the matrix represent the probabilities of moving from one state to another in a single step:

- From state \( a \):
  - Probability of moving to \( a \): \( 1/2 \)
  - Probability of moving to \( b \): \( 1/2 \)
  - Probability of moving to \( c \): \( 0 \)

- From state \( b \):
  - Probability of moving to \( a \): \( 1/4 \)
  - Probability of moving to \( b \): \( 0 \)
  - Probability of moving to \( c \): \( 3/4 \)

- From state \( c \):
  - Probability of moving to \( a \): \( 1/2 \)
  - Probability of moving to \( b \): \( 1/2 \)
  - Probability of moving to \( c \): \( 0 \)

The goal is to calculate the expected return time to state \( b \) using first-step analysis.
Transcribed Image Text:**3.18** Use first-step analysis to find the expected return time to state \( b \) for the Markov chain with transition matrix \[ P = \begin{pmatrix} 1/2 & 1/2 & 0 \\ 1/4 & 0 & 3/4 \\ 1/2 & 1/2 & 0 \end{pmatrix} \] The matrix \( P \) is a transition matrix for a Markov chain with states \( a \), \( b \), and \( c \). The elements of the matrix represent the probabilities of moving from one state to another in a single step: - From state \( a \): - Probability of moving to \( a \): \( 1/2 \) - Probability of moving to \( b \): \( 1/2 \) - Probability of moving to \( c \): \( 0 \) - From state \( b \): - Probability of moving to \( a \): \( 1/4 \) - Probability of moving to \( b \): \( 0 \) - Probability of moving to \( c \): \( 3/4 \) - From state \( c \): - Probability of moving to \( a \): \( 1/2 \) - Probability of moving to \( b \): \( 1/2 \) - Probability of moving to \( c \): \( 0 \) The goal is to calculate the expected return time to state \( b \) using first-step analysis.
**Example 3.17: Consider a Markov chain with transition matrix**

\[ P = \begin{bmatrix} 
0 & 1 & 0 \\ 
1/2 & 0 & 1/2 \\ 
1/3 & 1/3 & 1/3 
\end{bmatrix} \]

From state \( a \), find the expected return time \( E(T_a|X_0 = a) \) using first-step analysis.

**Solution:**

Let \( e_x = E(T_a|X_0 = x) \), for \( x = a, b, c \). Thus, \( e_a \) is the desired expected return time, and \( e_b \) and \( e_c \) are the expected first passage times to \( a \) for the chain started in \( b \) and \( c \), respectively.

For the chain started in \( a \), the next state is \( b \), with probability 1. From \( b \), the further evolution of the chain behaves as if the original chain started at \( b \). Thus,

\[ e_a = 1 + e_b. \]

From \( b \), the chain either hits \( a \), with probability 1/2, or moves to \( c \), where the chain behaves as if the original chain started at \( c \). It follows that

\[ e_b = \frac{1}{2} + \frac{1}{2}(1 + e_c). \]

Similarly, from \( c \), we have

\[ e_c = \frac{1}{3} + \frac{1}{3}(1 + e_b) + \frac{1}{3}(1 + e_c). \]

Solving the three equations gives

\[ e_c = \frac{8}{3}, \quad e_b = \frac{7}{3}, \quad \text{and} \quad e_a = \frac{10}{3}. \]

The desired expected return time is \( 10/3 \).
Transcribed Image Text:**Example 3.17: Consider a Markov chain with transition matrix** \[ P = \begin{bmatrix} 0 & 1 & 0 \\ 1/2 & 0 & 1/2 \\ 1/3 & 1/3 & 1/3 \end{bmatrix} \] From state \( a \), find the expected return time \( E(T_a|X_0 = a) \) using first-step analysis. **Solution:** Let \( e_x = E(T_a|X_0 = x) \), for \( x = a, b, c \). Thus, \( e_a \) is the desired expected return time, and \( e_b \) and \( e_c \) are the expected first passage times to \( a \) for the chain started in \( b \) and \( c \), respectively. For the chain started in \( a \), the next state is \( b \), with probability 1. From \( b \), the further evolution of the chain behaves as if the original chain started at \( b \). Thus, \[ e_a = 1 + e_b. \] From \( b \), the chain either hits \( a \), with probability 1/2, or moves to \( c \), where the chain behaves as if the original chain started at \( c \). It follows that \[ e_b = \frac{1}{2} + \frac{1}{2}(1 + e_c). \] Similarly, from \( c \), we have \[ e_c = \frac{1}{3} + \frac{1}{3}(1 + e_b) + \frac{1}{3}(1 + e_c). \] Solving the three equations gives \[ e_c = \frac{8}{3}, \quad e_b = \frac{7}{3}, \quad \text{and} \quad e_a = \frac{10}{3}. \] The desired expected return time is \( 10/3 \).
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