Disintegration of the chain law over its first $k+1$ coordinates

Let $P$ be a Markov kernel with initial distribution $\lambda$, and let $\mathbb{P}_\lambda$ be the resulting law on path space. Then for every $k$,

where $\mathrm{fr}_k(\omega) = (\omega_0,\dots,\omega_k)$ and $\mathrm{traj}_k$ is the kernel sending an initial segment to the law of the trajectory continuing from it.

What it says. The law of the chain is recovered by drawing its first $k+1$ coordinates from their own marginal and then continuing with $\mathrm{traj}_k$. This is the disintegration of the path measure over the past up to time $k$ — the precise sense in which $\mathrm{traj}_k$ is the conditional law of the trajectory given $(X_0,\dots,X_k)$, and the form in which conditioning is actually applied.

Why it is needed. Mixing estimates all have the shape: fix a past event $A$ and a future event $B$, condition on the first $k+1$ coordinates, and compare the conditional probability of $B$ with its stationary value. The conditioning step is exactly this identity. Once it is available, a past event — which by the identification of the past $\sigma$-algebra is a pullback $\mathrm{fr}_k^{-1}(A_0)$ — becomes a function of the integration variable in the outer integral, while a future event is handled inside $\mathrm{traj}_k(u)$ by the restart property. Without the disintegration the two identifications have nothing to act on.

Proof. Two observations, both at the level of kernels. First, the marginal: pushing the trajectory kernel forward along $\mathrm{fr}_k$ gives $\mathrm{partialTraj}(0,k)$ precomposed with the embedding of a point as a length-one segment, so $(\mathrm{fr}_k)_*\mathbb{P}_\lambda = Q \circ \lambda$ for that kernel $Q$. Second, the decomposition: Mathlib's $\mathrm{traj}_0 = \mathrm{traj}_k \circ \mathrm{partialTraj}(0,k)$ together with the fact that precomposition commutes with composition on the right gives $\mathrm{traj}_k \circ Q =$ the trajectory kernel of the chain. Associativity of composition with a measure then turns $\mathrm{traj}_k \circ (Q \circ \lambda)$ into $(\mathrm{traj}_k \circ Q) \circ \lambda = \mathbb{P}_\lambda$.

No hypothesis on $\lambda$ beyond being a probability measure is needed; in particular the chain need not be started from an invariant measure.