Discussion
Consider a Markov chain on S = {1, 2, 3, 4} with the transition probability matrix given by :
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| 1 | 0 | 0 | 0 | 1 |
| 2 | 0 | 0 | 0 | 1 |
| 3 | 1/2 | 1/2 | 0 | 0 |
| 4 | 0 | 0 | 1 | 0 |
Then :
Answer: B
The limits do not exist, and there is a unique stationary distribution.
The moves are \(1 \to 4\), \(2 \to 4\), \(4 \to 3\) and \(3 \to 1\) or 2. Every state reaches every other, so the chain is irreducible, and every return path has length 3 (\(4 \to 3 \to 1 \to 4\) or \(4 \to 3 \to 2 \to 4\)), so it has period 3.
A finite irreducible chain has a unique stationary distribution. Solving \(\pi = \pi P\): \(\pi_{3} = \pi_{4} = \pi_{1} + \pi_{2}\) and \(\pi_{1} = \pi_{2} = \pi_{3}/2\), so \(\pi = (1/6, 1/6, 1/3, 1/3)\).
Because of the period, \(p_{ii}^{(n)}\) is 0 unless 3 divides \(n\), so the n-step probabilities oscillate and have no limit. That rules out A, and C and D are wrong since there is exactly one stationary distribution.
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