Explicit solution to the invariant distribution of a Markov chain Let $\{X_t\}_t$ be a discrete-time Markov chain with right stochastic transition matrix $P$ and a unique invariant distribution $\pi$. Let the state space be $\{1,\dots,S\}$. Is there an explicit solution (potentially an approximation) for the element $\pi^{(s)}$ of $\pi$ in terms of the elements of $P$ for a given $S$?
For instance, if
\begin{equation} P = \begin{bmatrix} 1-\alpha, \alpha \\ \beta, 1-\beta \end{bmatrix} \end{equation}
then
\begin{equation} \pi = \begin{bmatrix} \frac{\beta}{\alpha+\beta} \\ \frac{\alpha}{\alpha+\beta} \end{bmatrix}, \end{equation}
and I am interested in the generalization for $S > 2$.
 A: For ergodic $\mathbf{P},$ you seek $S$-vector$\pi$ with $\pi\mathbf{P}=\pi.$ So $\pi$ is a left eigen-vector of $\mathbf{P}.$ You can use R to compute $\pi.$
Illustration:
P = matrix(c(.1, .2, .7,
             .2, .7, .1,
             .7, .1, .2), byrow=T, nrow=3)

P
     [,1] [,2] [,3]
[1,]  0.1  0.2  0.7
[2,]  0.2  0.7  0.1
[3,]  0.7  0.1  0.2

g = eigen(t(P))$vec[,1]
g = as.numeric(g)
pi = g/sum(g);  pi
[1] 0.3333333 0.3333333 0.3333333

pi %*% P
          [,1]      [,2]      [,3]
[1,] 0.3333333 0.3333333 0.3333333

I used a doubly-stochastic matrix so the answer would be obvious,
but the method works generally.
The invariant (stationary) vector is proportional to the left eigen vector of smallest
modulus.
Notes on R code:
(a) Use transpose t(P) because R finds right eigen-vectors.
(b) Use $vec[,1] to get the first column of the display of eigen vectors; R prints the eigen-vector of smallest modulus first.
(c) Use as.numeric to suppress (possible) complex-number notation. [For an
ergodic $\mathsf{P},$ the first eigen-vector is always real, but
other vectors in the display may be complex.]
(d) Use pi = g/sum(g) to make the elements of $\pi$ sum to $1.$
(e) The last line is to verify that $\pi$ is the desired vector.
(Unnecessary, but guards against typos in the code.)
