Say I have a finite, ergodic Markov chain with states ${0,1,2,3}$ and with the following transition matrix:
$$\begin{bmatrix} \frac{7}{10} & \frac{3}{10} & 0 &0\\ \frac{1}{10} & \frac{6}{10} & \frac{3}{10} &0\\ 0 & \frac{3}{10} & \frac{6}{10} & \frac{1}{10} \\ 0& 0& \frac{3}{10} & \frac{7}{10}\\ \end{bmatrix}$$
If I define $M_n$ to be the value of the state (so either $0,1,2$ or $3$) divided by $5$ at time $n$, is $M_n$ a martingale? I am inclined to believe it is not because the expected value of $M_{n+1}$ given the filtration at $n$ is not equal to $M_n$ (at least by my calculations).
For every martingale $(M_n)$, $\mathbb{E}(M_{n+1} | M_n ) = M_n$. But here, $\mathbb{E}(M_{n+1}|M_n=0) = \frac{3}{50}$. Therefore, $(M_n)$ is not a martingale.