You are given a transition matrix $P$. Find the steady-state distribution vector. HINT [See Example 4.] \[ P=\left[\begin{array}{ccc} 0.7 & 0 & 0.3 \\ 1 & 0 & 0 \\ 0 & 0.2 & 0.8 \end{array}\right] \]
Solution
Step 1 ：We are given a transition matrix \(P = \left[\begin{array}{ccc} 0.7 & 0 & 0.3 \ 1 & 0 & 0 \ 0 & 0.2 & 0.8 \end{array}\right]\).
Step 2 ：The steady-state distribution vector is a probability distribution that remains unchanged in the Markov chain process. It is the eigenvector of the transition matrix corresponding to the eigenvalue 1.
Step 3 ：We can find it by solving the linear system \((P-I)v=0\), where \(P\) is the transition matrix, \(I\) is the identity matrix, and \(v\) is the steady-state vector.
Step 4 ：Subtracting the identity matrix from the transition matrix, we get \(A = \left[\begin{array}{ccc} -0.3 & 0 & 0.3 \ 1 & -1 & 0 \ 0 & 0.2 & -0.2 \end{array}\right]\).
Step 5 ：Solving the system of linear equations, we find that the steady-state vector \(v = \left[\begin{array}{c} 0.33333333 \ 0.33333333 \ 0.33333333 \end{array}\right]\).
Step 6 ：Final Answer: The steady-state distribution vector is \(\boxed{\left[\begin{array}{c}0.33333333 \ 0.33333333 \ 0.33333333\end{array}\right]}\).
