Let $(\Omega,\mathcal{F},P)$ be a probability space where $\Omega$ is the set of cadlag trajectories from $\mathbb{R}$ to a countable state space $S$. Let $X$ be the coordinate process $X_t(\omega) = \omega(t)$.
If $X$ is a homogeneous Markov chain, then does the condition that the point process which counts the jumps of $X$ (or any thinning thereof) be a stationary point process entail the condition that $X_0$ is distributed according to its steady state (and in particular that it has a steady state)?
The literature I've encountered seems to imply that the answer is no, but the condition that $P = P \circ \theta_t$, where $\theta_t : \Omega \to \Omega$ is the canonical flow given by $\theta_t(\omega) = \omega(t + \cdot)$, seems to imply that the answer is yes: if $0 < t_1 < \cdots < t_k$ then \begin{align} P\circ\theta_t[X_{t_1}=i_1,...,X_{t_k}=i_k] &= P[X_{t_1-t}=i_1,...,X_{t_k-t}=i_k] \\ &= \prod_{j=2}^k P[X_{t_j-t}=i_j|X_{t_{j-1}-t}=i_{j-1}] P[X_{t_1-t}=i_1] \\ &= \prod_{j=2}^k P[X_{t_j}=i_j|X_{t_{j-1}}=i_{j-1}] P[X_{t_1-t}=i_1] \\ &= P[X_{t_1}=i_1,...,X_{t_k}=i_k] \left( \frac{P[X_{t_1-t}=i_1]}{P[X_{t_1}=i_1]} \right); \end{align}
in order for $P$ to be compatible with the flow, this nuisance term $P[X_{t_1-t}=i_1]/P[X_{t_1}=i_1]$ must equal $1$, which certainly occurs if the chain is initially distributed according to its steady state.
But is such a strong requirement necessary?
Thanks, and sorry for the elementary nature of the issue.