Can Markov chain represent a non-stationary random process? I guess I will just talk about the discrete case and below is my current understanding of things.
Markov chain: A discrete random variable A has several states $A_{0}, A_{1}, A_{2} ... $ over a discrete time t measured in time steps. The probability $T_{i\rightarrow j}$ of reaching a particular state $A_{j}$ in next time step depends only on the identity of the current state $A_{i}$. For example whenever the current state is $A_{0}$, the probabilities for reaching $A_{0}, A_{1}, A_{2}$ ... are always the same regardless of any past states of A.
Non-stationary process: The probability distribution of states of a discrete random variable A (without knowing any information of current/past states of A) depends on discrete time t. For example, temperature is usually higher in summer than winter. Therefore, the probability distribution of possible temperature over time is a non-stationary random process.
My question is: Can a Markov chain accurately represent a non-stationary process? 
Does a Markov chain always represent a stationary random process?
 A: (2017-09-01 21:00: This answer just received a revenge downvote. Oh well...)

Summary: Homogenous transitions $\ne$ Stationary distribution.

Consider a homogenous Markov chain: this is a Markov chain $(X_n)$ such that the transition probabilities $q(x,y)=P(X_{n+1}=y\mid X_n=x)$ do not depend on $n$. In general, such a condition does not imply that the process $(X_n)$ is stationary, that is, that $\nu_n(x)=P(X_n=x)$ does not depend on $n$. Actually, the only case when this happens is if the initial distribution of the Markov chain is stationary, that is, if $\nu_0q=\nu_0$.
A: Yes you can for example let the state number encode the previously visited modes. For example a two letter alphabet with memory of 2 slots has $2^2 = 4$ possible states that the last two seen symbols were:


*

*AA

*AB

*BA

*BB


Now you can decide if you want the chain to map states of 2 old symbols to 2 new ones or just to one new symbol at a time. This will decide how many degrees of freedom your transition matrix (or conditional probability space) will have. If you choose just one new then you can figure out that states 1 and 3 can only end up in 1 or 2 since one of the symbols will "remain in the pipeline", reducing the possibilities for transitions.
