91Ó°ÊÓ

Given that the Markov chain has four states \(1,2,3,4\). Given transition matrix to find state-2 at time \(n + 2\) and state-3 at time \(n + 3\).

Here we calculate the Markov chain for particular rows and columns. Not calculate the whole transition matrix.

Here, wee define the square matrix forthef third row and a second column for the given transition probability matrix. Then from the given matrix to find that

\(\begin{aligned}{}{p^2} &= p \times p\\ &= \left[ {\begin{aligned}{{}{}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{aligned}} \right] \times \left[ {\begin{aligned}{{}{}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{aligned}} \right]\\ &= \left[ {\frac{1}{8}} \right]\end{aligned}\)

Finally it is solved this way.

Here we calculate the cube of the matrix of first row and third column of the given transition matrix. Then it is defined as

\(\begin{aligned}{}{p^3} &= {p^2} \times p\\ &= \left[ {\begin{aligned}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{aligned}} \right] \times \left[ {\begin{aligned}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{aligned}} \right]\\ &= \left[ {\frac{1}{8}} \right]\end{aligned}\)

Hence it is solved by this process.