# When does a factorization of a transition function (as in a dynamic Bayesian network) transfer to the stationary distribution?

I want to know the relation between a factorization of a transition probability function for a Markov chain and its stationary distribution (assuming it has one).

Specifically, a dynamic Bayesian network (DBN) is a Bayesian network with two layers -- one for time $t$ and one for $t+1$. When viewed as a Markov chain the DBN assumes that the transition probability can be factored according to a directed acyclic graph $\mathcal{G}$: $$P(S_{t+1}|S_t) = \prod_{j=1}^NP(S_{t+1,j}|\text{Pa}_j^{t+1},S_t).$$ Here $\text{Pa}_j^t$ refers to the parents of $S_{tj}$ at time $t$, according to $\mathcal{G}$. In this case I assume the graph $\mathcal{G}$ is defined only over nodes $S_{tj}$ at the same time, $t$, and that conditional dependence between times only occurs through self-recurrence $S_{tj}\to S_{t+1,j}$ (see example figure).

My question is: under what conditions will the stationary distribution of the Markov chain, $\rho$, also factorize according to $\mathcal{G}$? That is, as: $$\rho(S_t) = \prod_{j=1}^N\rho (S_{t,j}|\text{Pa}_j^{t}).$$

It seems that in general there is no reason to expect this to be true. At least in the case that $P$ does not imply a unique stationary distribution then some $\rho$ may factorize according to $\mathcal{G}$ but some may not.

However, intuitively, it seems like an ergodic Markov chain, whose stationary distribution can be inferred from a single long realization of the chain, will inherit the factorization of the transition dynamics throughout this long realization. But I do not know how to make this intuition more precise to test the idea.

Are there other properties besides ergodicity that might be used instead? Am I missing something obvious/misunderstanding something? Or are there no general conditions that can be specified so that the stationary distribution factors according to $\mathcal{G}$?

If the condition is $P(S_{t+1}|S_t)=\Pi_{j=1}^NP(S_{t+1,j}|Pa^{t+1}_j,S_t)$, then counterexample is everywhere.
If the condition is $P(S_{t+1}|S_t)=\Pi_{j=1}^NP(S_{t+1,j}|Pa^{t+1}_j,S_{t,j})$, then things are much better, but still not true.
Consider an example with DAG $S_3\leftarrow S_1\to S_2$. If the factorization for stationary distribution is valid, then we should have $$S_{t+1,2} \text{ is independent with } S_{t+1,3}\text{ conditioned on } S_{t+1,1}$$ Generally this is not true, since conditioned on $S_{t+1,1}$, $S_{t+1,3}$ contains some extra information of $S_{t,1}$, and $S_{t,1}$ has some extra information of $S_{t+1,2}$.
My guess is that, the above is the only problematic scheme (i.e. some nodes with common ancestor are not linked together) if the condition is $P(S_{t+1}|S_t)=\Pi_{j=1}^NP(S_{t+1,j}|Pa^{t+1}_j,S_{t,j})$.