I have the following Markov matrix $A=\begin{bmatrix} 1-\beta & \beta \\ \alpha & 1-\alpha \\ \end{bmatrix} $
And initial row vector $X_0 = (0,1)$
I need to compute the state vector $X_3$ which means $X_0 A^3 $. I started doing it but got a very complex expression with $\alpha$ and $\beta$.
Is there any trick that I'm missing here that could make my life easier ?
The general formula for calculating the state state vector is of a Markov Chain with two states $\mathbb{X}=\{A,B\}$ (A,B are the states, not to be confused with $a,b$ the probability values in your transition matrix):
$$\begin{equation}(\pi_{n}(A),\pi_{n}(B))=(\pi_{n}(A),\pi_{n}(B))P^{n} \end{equation}$$
In order to calculate $P^{n}$ for any value of $n$ we can diagonalise matrix $P$.The characteristic equation $|\lambda I-P|=0$ of $P$ has eigenvalues $\lambda_{1}=1$ and $\lambda_{2} = 1-(b+a)$ (a,b) from your matrix.
The according eigenvectors are : $$u_{1}= \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ and $$u_{2}= \begin{pmatrix} -b \\ a \end{pmatrix}.$$ Therefore $$P = \begin{pmatrix} 1 &-b\\ 1 &a \end{pmatrix}\begin{pmatrix} 1 &0\\ 0 &\lambda_{2} \end{pmatrix}\begin{pmatrix} 1 &-b\\ 1 &a \end{pmatrix}^{-1}.$$
So we have $$P^{n} = \begin{pmatrix} 1 &-b\\ 1 &a \end{pmatrix}\begin{pmatrix} 1^n &0\\ 0 &\lambda_{2}^n \end{pmatrix}\begin{pmatrix} 1 &-b\\ 1 &a \end{pmatrix}^{-1} = \frac{1}{b+a}\begin{pmatrix} a+b \lambda_{2}^n & b- \lambda_{2}^n \\ a-a \lambda_{2}^n & b+ \lambda_{2}^n \end{pmatrix}$$
Substituting in the first equation we take : $$(\pi_{n}(A),\pi_{n}(B))=\left( \frac{a}{b+a},\frac{b}{b+a} \right) + \lambda_{2}^n \frac{b \pi_{0}(A) - a \pi_{0}(B)}{b+a}(1,-1)$$
