Discussion
Consider a discrete time Markov chain with state space \(S = \{1, 2\}\) and the transition probability diagram
| From | To 1 | To 2 |
|---|---|---|
| 1 | 0.25 | 0.75 |
| 2 | 0.5 | 0.5 |
If \(\pi = (\pi_{1}, \pi_{2})\) is the stationary distribution of the Markov chain, then which one of the following statements is true?
Answer: B
The stationary distribution is \((\pi_{1}, \pi_{2}) = \left(\frac{2}{5}, \frac{3}{5}\right)\).
For a two-state chain the stationary distribution balances the flow between the states: the probability flowing from 1 to 2 equals that flowing from 2 to 1.
\[ \begin{aligned} 0.75\,\pi_{1} &= 0.5\,\pi_{2} \Rightarrow \pi_{2} = 1.5\,\pi_{1} \\ \pi_{1} + 1.5\,\pi_{1} &= 1 \Rightarrow \pi_{1} = \frac{2}{5} \end{aligned} \]
so \(\pi_{2} = \frac{3}{5}\). Check with the first column: \(0.25 \times \frac{2}{5} + 0.5 \times \frac{3}{5}\) \(= 0.1 + 0.3 = 0.4 = \pi_{1}\).
\(\left(\frac{1}{2}, \frac{1}{2}\right)\) would need equal flows 0.75 and 0.5. \(\left(\frac{1}{4}, \frac{3}{4}\right)\) simply copies the first row, and \(\left(\frac{1}{3}, \frac{2}{3}\right)\) fails the balance equation (\(0.25 \ne 0.33\)).
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