How do you calculate the unconditional probabilities from a markovian transition matrix?

Question

Consider the markovian states $s_{t}\in \{A,B\}$ and the following transition matrix $$\begin{array}{c|cc} &s_{t+1}=A & s_{t+1}=B\\ s_{t}=A & 1-\delta & \delta\\s_{t}=B&\lambda&1-\lambda\\ \end{array}$$ with $\delta+\lambda<1$. How do you calculate $prob(A)$? I'm supposed to get $\dfrac{\lambda}{\lambda+\delta}$.

Answer 1

The solution is the following.

Let $p_{j,i}^n$ be the probability of going from state $i$ to state $j$ after $n$ periods. The unconditional probabilities of a markov chain are the steady-state probabilities defined as: $$Lim_{n \rightarrow \infty}p_{i,j}^n=\pi_j$$ which exists for any irreducible Markov chain. Notice that the transition matrix in the statement, which I'll call $P$,is irreducible, that is, there exists some $n$ for which $p_{i,j}^n>0$ for all $i$ and $j$ (all states communicate).

The steady-state equations are $\mathbb{\pi}^T P = \mathbb{\pi}^T$ and $\sum_{i=0}^M \pi_i=1$: $$\begin{array}{cc} [\pi_1 & \pi_2] \end{array} \left[ \begin{array}{cc} 1-\delta & \delta\\ \lambda & 1-\lambda\\ \end{array}\right]= \begin{array}{cc} [\pi_1 & \pi_2] \end{array}$$ which is equivalent to $$\pi_1(1-\delta)+\pi_2\lambda=\pi_1$$ $$\pi_1\delta+\pi_2(1-\lambda)=\pi_2$$ $$\pi_1=1-\pi_2$$ And plugging $\pi_1=1-\pi_2$ the solution is: $$\pi_2=\dfrac{\delta}{\lambda+\delta}$$ $$\pi_1=1-\pi_2=\dfrac{\lambda}{\lambda+\delta}$$