Strong restart: given the past, the future is a fresh chain started at the last state

Let $P$ be a Markov kernel on $\mathsf{X}$ and let $\mathrm{traj}_j$ denote the Ionescu–Tulcea kernel that, given an initial segment $u = (u_0,\dots,u_j)$, returns the law of the whole trajectory continuing from it. Let $\sigma^j$ be the shift $(\sigma^j\omega)_n = \omega_{j+n}$. Then

i.e. as an identity of kernels, pushing the conditional trajectory kernel forward along the $j$-shift gives the trajectory kernel of the chain, precomposed with "read off the last coordinate".

The strong restart property. In words: conditionally on the first $j+1$ states $(X_0,\dots,X_j) = u$, the future $(X_j, X_{j+1}, \dots)$ is a copy of the chain started afresh at $u_j$, and in particular depends on $u$ only through $u_j$. This is the Markov property in the form that quantitative arguments actually consume, and it is strictly stronger than the unconditional shift identity $\sigma_*\mathbb{P}_x = \int \mathbb{P}_y P(x,\mathrm{d}y)$: the latter describes the law of a shifted trajectory averaged over everything, while this describes it given the entire past, and asserts that the past is forgotten apart from its final state.

Why it is the gateway to mixing estimates. All three classical mixing coefficients compare the past $\sigma(X_0,\dots,X_k)$ with the future $\sigma(X_{k+n}, X_{k+n+1},\dots)$. Bounding them requires expressing the conditional law of the future given the past, and this theorem is exactly that expression: combined with the decomposition $\mathrm{traj}_0 = \mathrm{traj}_j \circ \mathrm{partialTraj}(0,j)$, it says that conditioning on the first $j+1$ coordinates and then looking $n$ steps ahead yields the chain started from $P^n(u_j,\cdot)$. A total-variation rate $\|P^n(x,\cdot)-\pi\| \le C$ then bounds the mixing coefficient at lag $n$ by $C$, uniformly in the split point.

Proof. By the fact that a path measure is determined by its finite-dimensional marginals, it suffices to check agreement after restricting to each initial segment $\mathrm{Iic}\,m$. On the left, $\mathrm{fr}_m \circ \sigma^j = \sigma^j_{\mathrm{fin}} \circ \mathrm{fr}_{j+m}$ — an identity that holds definitionally with the length written as $j+m$ — so the left marginal is the $j$-shift pushforward of $\mathrm{partialTraj}(j, j{+}m)$. On the right it is $\mathrm{partialTraj}(0,m)$ precomposed with $u \mapsto u_j$. The two are identified by induction on $m$: the base case is the observation that shifting a length-$(j{+}1)$ segment down to length one just reads off $u_j$, and the inductive step splits $\mathrm{partialTraj}(j, j{+}m{+}1)$ as one step after $\mathrm{partialTraj}(j, j{+}m)$ and applies the arbitrary-offset time-homogeneity lemma to move the shift past that step.