How do you calculate the unconditional probabilities from a markovian transition matrix? 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 $\dfrac{\lambda}{\lambda+\delta}$.
 A: 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:
$$\pi_2=\dfrac{\delta}{\lambda+\delta}$$
$$\pi_1=1-\pi_2=\dfrac{\lambda}{\lambda+\delta}$$
