91Ó°ÊÓ

Construct a transition probability matrix, denoted as P, based on the given information. The matrix has five rows and five columns, representing the five states. The element P_{ij} represents the transition probability from state i to state j. P = \(\left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 & 0 \\ 1/3 & 1/3 & 1/3 & 0 & 0 \\ 1/4 & 1/4 & 1/4 & 1/4 & 0 \end{array}\right)\)

Now, we will find the stationary probabilities, denoted as \(\pi = (\pi_0, \pi_1, \pi_2, \pi_3, \pi_4)\), which is the limiting probability distribution of the chain. To do this, we need to find the eigenvector corresponding to eigenvalue 1: \(\pi P = \pi\) And also, the probabilities should sum to 1: \(\sum_{i=0}^{4} \pi_i = 1\)

Using the matrix P and the above equations, we can write the following system of linear equations: \(\begin{cases} \pi_0 + \frac{1}{2} \pi_1 + \frac{1}{3} \pi_2 + \frac{1}{4} \pi_3 + \frac{1}{4} \pi_4 = \pi_0 \\ \pi_1 + \frac{1}{2} \pi_2 + \frac{1}{3} \pi_3 + \frac{1}{4} \pi_4 = \pi_1 \\ \frac{1}{2} \pi_1 + \frac{2}{3} \pi_2 + \frac{1}{4} \pi_3 + \frac{1}{4} \pi_4 = \pi_2 \\ \frac{1}{3} \pi_1 + \frac{1}{3} \pi_2 + \frac{1}{4} \pi_3 + \frac{1}{4} \pi_4 = \pi_3 \\ \frac{1}{4} \pi_1 + \frac{1}{4} \pi_2 + \frac{1}{4} \pi_3 + \frac{1}{4} \pi_4 = \pi_4 \\ \pi_0 + \pi_1 + \pi_2 + \pi_3 + \pi_4 = 1 \end{cases}\) Solving the system of linear equations by using any computational method (such as Gaussian elimination or matrix inversion), we find that the limiting probabilities are: \(\pi_0 = 0, \pi_1 = \frac{256}{305}, \pi_2 = \frac{32}{305}, \pi_3 = \frac{12}{305}, \pi_4 = \frac{5}{305}\) Thus, the stationary probability distribution of the Markov chain is: \(\pi = \left(0, \frac{256}{305}, \frac{32}{305}, \frac{12}{305}, \frac{5}{305}\right)\)