Time-homogeneity at an arbitrary offset: the $k$-shift commutes with one-step extension

Let $P$ be a Markov kernel on $\mathsf{X}$, and write $\sigma^k$ for the shift that drops the first $k$ coordinates of a partial trajectory, $(\sigma^k v)_i = v_{k+i}$. Then for all $k, m$ and every partial trajectory $v = (v_0,\dots,v_{k+m})$,

What it says. Extending a trajectory by one step and then dropping its first $k$ coordinates is the same as dropping the first $k$ coordinates and then extending. This is the arbitrary-offset form of time-homogeneity; the case $k = 1$ is the basic statement that the shift commutes with one-step extension.

Why the offset version is needed separately. The $k=1$ case suffices to prove that the law of the whole trajectory started at a point is shift-covariant, and hence that a chain started from an invariant measure is stationary. It does not suffice for the conditional statements that quantitative mixing estimates require: bounding the dependence between $\sigma(X_0,\dots,X_j)$ and $\sigma(X_{j+n}, X_{j+n+1},\dots)$ means comparing a trajectory conditioned on its first $j{+}1$ coordinates with a chain restarted $n$ steps later, and that comparison is indexed by two independent offsets. Iterating the $k=1$ lemma cannot be done painlessly either, because the natural composition identity $\sigma^{k+1} = \sigma^k \circ \sigma$ forces the index arithmetic $(m{+}1)+k$ versus $(m{+}k)+1$, which is not definitional in $\mathbb{N}$ — addition recurses on its second argument. Writing the total length as $k+m$ from the outset, as here, makes $k+(m{+}1)$ and $(k{+}m)+1$ definitionally equal and the induction goes through without transport.

Two facts consumed, exactly as in the unit-offset case. The left side extends using the kernel $P(v_{k+m},\cdot)$ and the right side using $P\bigl((\sigma^k v)_m,\cdot\bigr)$; these agree because $(\sigma^k v)_m = v_{k+m}$ — the shift preserves the last coordinate, which is all the transition law depends on (the Markov property) — and because the kernel is the same $P$ at both times (time-homogeneity). For a genuinely time-inhomogeneous family the statement is false.

Proof. Rewrite both sides with the explicit pushforward description of a one-step extension, so each is a pushforward of the single measure $P(v_{k+m},\cdot)$ along a gluing map. After identifying the kernels, the claim reduces to a pointwise identity between the two gluing maps, settled by splitting on whether the index $i$ satisfies $i \le m$: the two branch conditions $k+i \le k+m$ and $i \le m$ are equivalent by cancellation, the first branch returns $v_{k+i}$ on both sides, and the second returns the newly drawn coordinate on both sides.