When is a coarsening of a Markov chain Markov?

Question

Let $(X_t)_{t \in \mathbb{N}}$ be a discrete-time Markov process on the finite state space $\mathcal{X}$, with transition matrix $T$. Suppose $f: \mathcal{X} \to \mathcal{Y}$ is a (deterministic) function such that $(f(X_t))_{t \in \mathbb{N}}$ is a Markov process on the state space $\mathcal{Y}$. What can we say about $f$ in relation to the transition matrix $T$ of the original chain?

Ideally, I would like an exact characterization of when the derived process is Markov. If $\mathcal{X} = \mathcal{Y} \times \mathcal{Z}$ and $f$ is projection onto the first component, then I think it suffices for the transition probability $T((y_1,z_1),(y_2,z_2))$ to be constant in $z_1$ or constant in $z_2$. There is also the case when $f$ is injective. Are there any other cases?

Answer 1

Are there any other cases?

Sure there are. Say, $(X_t)_{t\in\mathbb{N}}$ is the same as a chain $(Y_t)_{t\in\mathbb{N}}$ except that it has two copies $a_0,a_1$ of a certain state $a$ of $(Y_t)_t$. Every time $(Y_t)_t$ has a transition to $a$, $(X_t)_t$ flips a coin to decide which of $a_0$ or $a_1$ to go. For transitions out of $a$, $(X_t)_t$ simply ignores the distinction between $a_0$ and $a_1$. The function $f$ maps $a_0$ and $a_1$ to $a$ and every other state to itself.

The general criterion on $f$ to ensure that $(f(X_t))_t$ is a Markov chain is that for every $x,x'\in\mathcal{X}$ with $f(x)=f(x')$ and every $y\in\mathcal{Y}$, the transition probabilities should satisfy $T(x,f^{-1}(y))=T(x',f^{-1}(y))$.

A related question: under what conditions the projected process is a $k$-step Markov chain?