## Given Information A Markov chain has transition matrix: \[ P = \begin{pmatrix} & 1 & \\ \frac{1}{2} & & \frac{1}{2} \\ & 1 & \end{pmatrix} \] --- ## Solution ### 1. **Is the chain irreducible?** **Irreducible:** A Markov chain is irreducible if it's possible to get from any state to any other state (possibly in multiple steps). - From state 1: can go to state 2 in 1 step. - From state 2: can go to states 1 and 3 in 1 step. - From state 3: can go to state 2 in 1 step. **Check all pairs:** - 1 → 3: 1 → 2 → 3 (possible in 2 steps, since 1→2, 2→3) - 3 → 1: 3 → 2 → 1 (possible in 2 steps, since 3→2, 2→1) **Conclusion:** **Yes, the chain is irreducible.** --- ### 2. **Find the stationary distribution, if it exists.** Let stationary distribution be \(\pi = (\pi_1, \pi_2, \pi_3)\). We want \(\pi P = \pi\), i.e., \[ \begin{cases} \pi_1 = \cdot \pi_1 + \frac{1}{2} \pi_2 + \cdot \pi_3 = \frac{1}{2}\pi_2 \\ \pi_2 = 1 \cdot \pi_1 + \cdot \pi_2 + 1 \cdot \pi_3 = \pi_1 + \pi_3 \\ \pi_3 = \cdot \pi_1 + \frac{1}{2} \pi_2 + \cdot \pi_3 = \frac{1}{2}\pi_2 \end{cases} \] with \(\pi_1 + \pi_2 + \pi_3 = 1\). From the first and third equations: \(\pi_1 = \pi_3 = \frac{1}{2}\pi_2\) Therefore, \(\pi_1 = \pi_3\). Let \(\pi_1 = \pi_3 = x\), \(\pi_2 = y\). Then: - \(x + y + x = 1 \implies 2x + y = 1\) - From above, \(x = \frac{1}{2}y\) Substitute: - \(2\left(\frac{1}{2}y\right) + y = 1 \implies y + y = 1 \implies y = \frac{1}{2}\) - So, \(x = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\) **Stationary distribution:** \[ \boxed{ \pi = \left(\frac{1}{4}, \frac{1}{2}, \frac{1}{4}\right) } \] --- ### 3. **Is the chain ergodic?** A chain is **ergodic** if it is irreducible, aperiodic, and positive recurrent. - **Irreducible:** Yes (shown above). - **Aperiodic:** Let's check the period of each state: - State 2 can only return to itself in even steps (since 2→1/3→2), so period is 2. - All states have period 2. **Conclusion:** **The chain is not ergodic** (since it is not aperiodic; period is 2). --- ## **Summary** 1. **Irreducible:** Yes 2. **Stationary distribution:** \( \left(\frac{1}{4}, \frac{1}{2}, \frac{1}{4}\right) \) 3. **Ergodic:** No (periodic, not aperiodic)