Discussion
The following are the transition probability matrices of the Markov chains with state space \(S = \{1, 2\}\) :
\[ P_{1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad P_{2} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \]
\[ P_{3} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{2}{3} & \frac{1}{3} \end{bmatrix}, \quad P_{4} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ 0 & 1 \end{bmatrix} \]
(rows and columns are indexed by the states 1, 2)
Which Markov chains have a stationary distribution given by \(\left(\frac{1}{2}, \frac{1}{2}\right)\) ?
Markov chains with transition probability matrices.
Answer: B
The chains with stationary distribution \(\left(\frac{1}{2}, \frac{1}{2}\right)\) are \(P_{1}\) and \(P_{2}\).
Check \(\pi P = \pi\) with \(\pi = \left(\frac{1}{2}, \frac{1}{2}\right)\). For the identity \(P_{1}\), every distribution is stationary. \(P_{2}\) swaps the two states, so it maps \(\left(\frac{1}{2}, \frac{1}{2}\right)\) to itself; in general a doubly stochastic matrix has the uniform distribution stationary.
\[ \begin{aligned} \pi P_{3} &= \left(\tfrac{1}{4} + \tfrac{1}{3}, \tfrac{1}{4} + \tfrac{1}{6}\right) = \left(\tfrac{7}{12}, \tfrac{5}{12}\right) \\ \pi P_{4} &= \left(\tfrac{1}{4}, \tfrac{1}{4} + \tfrac{1}{2}\right) = \left(\tfrac{1}{4}, \tfrac{3}{4}\right) \end{aligned} \]
Neither is \(\left(\frac{1}{2}, \frac{1}{2}\right)\), so every option containing \(P_{3}\) or \(P_{4}\) is wrong. (\(P_{4}\)'s stationary distribution is (0, 1), since state 2 is absorbing.)
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