Discussion
Consider a Markov chain on \(S = \{1, 2, 3, 4, 5\}\) with its transition probability matrix given by :
| P | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| 1 | 3/4 | 0 | 1/4 | 0 | 0 |
| 2 | 1/2 | 0 | 0 | 1/2 | 0 |
| 3 | 1/2 | 0 | 1/2 | 0 | 0 |
| 4 | 0 | 0 | 0 | 2/3 | 1/3 |
| 5 | 0 | 0 | 0 | 1/3 | 2/3 |
Then :
Answer: D
There exist infinitely many stationary distributions.
From states 1 and 3 the chain moves only between 1 and 3, so {1, 3} is a closed recurrent class. Likewise {4, 5} is closed and recurrent. State 2 leads to 1 or 4 and never returns, so it is transient.
Each closed class has its own stationary distribution: on {1, 3}, \(\pi_{1} = 2/3, \pi_{3} = 1/3\); on {4, 5}, \(\pi_{4} = \pi_{5} = 1/2\). Every convex combination of the two is also stationary, giving infinitely many.
Uniqueness (A) needs a single recurrent class. Not all states are transient (C), and with a finite chain at least one stationary distribution always exists (B fails).
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