For the Markov process with transition diagram shown above, I am wondering why I can expect (without any computation) the steady state to have equal probabilities for the four states A;B;C;D, no matter what the probabilities r, s and t happen to be. [As is the normal convention, missing arrows all represent transitions with probability zero.]
Is this because the four states $A,B,C,D$ have no difference between them at all? (i.e.they have equal positions?) The steady state vector will just be 1/4 1/4 1/4 1/4?
Secondly, I am also wondering how to write out the transition matrix for this process. May I ask for some help for that? Thanks so much.
The states are very similar. Every state has the same probabilities to stay, to get back to the previous state and to step forward to the next state. We have to imagine the chain as the following sequence of states $\cdots \rightleftarrows \overset {\curvearrowleft }A\rightleftarrows \overset {\curvearrowleft }B\rightleftarrows \overset {\curvearrowleft }C\rightleftarrows \overset {\curvearrowleft }D\rightleftarrows \overset {\curvearrowleft }A\rightleftarrows\cdots$
The transition matrix is
$$\begin{bmatrix} r&s&0&t\\ t&r&s&0\\ 0&t&r&s\\ s&0&t&r \end{bmatrix}$$
where a row represents the current state and a column represents the next state.
Indeed, by computation, $\frac14,\frac14,\frac14,\frac14$ are the stationary probabilities: $$\begin{bmatrix} r&s&0&t\\ t&r&s&0\\ 0&t&r&s\\ s&0&t&r \end{bmatrix}\begin{bmatrix}\frac14\\\frac14\\\frac14\\\frac14\end{bmatrix}=\begin{bmatrix}\frac14(r+s+t)\\\frac14(r+s+t)\\\frac14(r+s+t)\\\frac14(r+s+t)\end{bmatrix}$$ and $r+s+t=1$.