Let's solve each part step-by-step.

(a) One-step Transition Probability Matrix

Let $P$ be the transition probability matrix where $P(i, j)$ denotes the probability of moving from position $i$ to position $j$ in one step. There are 6 positions arranged in a circle, so after position 6, it wraps back to position 1. The probabilities for each movement are as follows:

• Moving clockwise to the next position with probability $\frac{1}{4}$.
• Moving counter-clockwise to the previous position with probability $\frac{1}{4}$.
• Staying at the current position with probability $\frac{1}{2}$.

The transition probability matrix $P$ is then a 6x6 matrix where each entry $P(i, j)$ represents the probability of moving from position $i$ to position $j$.

Let's label each position as $1, 2, 3, 4, 5,$ and $6$. Using the above rules, we can fill out the matrix as follows:

\frac{1}{2} & \frac{1}{4} & 0 & 0 & 0 & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 & 0 \\ 0 & \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 \\ 0 & 0 & \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 \\ 0 & 0 & 0 & \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & 0 & 0 & 0 & \frac{1}{4} & \frac{1}{2} \end{pmatrix}$$ ### (b) State-Transition Diagram To visualize the Markov chain, we can draw a circle with six nodes representing positions $$1, 2, \ldots, 6$$. Each node has: - An arrow pointing to itself with probability $$\frac{1}{2}$$ (indicating the probability of staying in place). - An arrow to the next position clockwise with probability $$\frac{1}{4}$$. - An arrow to the previous position counterclockwise with probability $$\frac{1}{4}$$. This will form a circular diagram where each position has two links to adjacent positions and one loop back to itself. ### (c) Probability of Returning to the Same Position After Two Steps Now, we need to calculate the probability that a kid at position $$4$$ returns to position $$4$$ after two jumps. To find this probability, we compute the square of the transition matrix $$P$$, denoted by $$P^2$$, and look at the entry $$(4, 4)$$ of $$P^2$$. 1. **Identify possible two-step paths:** - Stay in place twice: $$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ - Move clockwise, then counterclockwise: $$\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$$ - Move counterclockwise, then clockwise: $$\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$$ 2. **Calculate the total probability:** $$P(X_2 = 4 \mid X_0 = 4) = \frac{1}{4} + \frac{1}{16} + \frac{1}{16} = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$$ So, the probability that the kid returns to the same position after two steps is $$\frac{3}{8}$$. --- Would you like any additional details on the transition matrix computation, the state-transition diagram, or the two-step probability calculation? --- ### Follow-up Questions 1. How would the transition probability matrix change if the probabilities of each movement were different? 2. What happens to the state-transition diagram if we increase the number of positions? 3. How would the probability calculation change if we consider three jumps instead of two? 4. Is this Markov chain periodic or aperiodic, and why? 5. What is the long-term behavior of this Markov chain? ### Tip For probability calculations in Markov chains, always examine both direct probabilities and all possible paths across multiple steps.