Estimate the speed of convergence to the stationary distribution for a ergodic Markov process

I have encountered a Markov process with following transition matrix $P= \begin{bmatrix} 0.6 & 0.4 \\ 0.2 &0.8\end{bmatrix}$. This is an ergodic Markov matrix since all the elements are positive. So it has a stationary distribution, which is $\pi=(\frac{1}{3},\frac{2}{3})$. I know that no matter what the initial distribution is, it will always converge to the stationary distribution. However, it there a way to quantify how fast the convergence will be? I'm interested in the convergence speed because I'm trying to calculate the number of occurrences in state 1 during a 1000000-steps walk.

Sure. You just need to compute the eigenvalues that are not $1$. In this case your eigenvalues are the solutions to $\lambda^2-1.4\lambda+0.4=0$, which are $1$ and $0.4$. This means that $p A^n$ = $\pi + 0.4^n v_p$ for some row vector distribution $v_p$ depending on $p$, and where $\pi$ is the equilibrium distribution. Specifically $v_p$ is the component of $p$ in the direction of the (left) eigenvector with eigenvalue $0.4$. Note that $v_p$ itself is not a distribution at all.
• Thanks for the answer. Can you provide more details about why this decomposition $p A^n = \pi + 0.4^n v_p$ works? It doesn't seem so obvious to me. Dec 6 '14 at 16:39
• @andy90 It is just diagonalization. You have $A=V \begin{bmatrix} 1 & 0 \\ 0 & 0.4 \end{bmatrix} V^{-1}$ where $V$ is the eigenvector matrix. Now exponentiate $A$ and apply $p$ on the left.