After a burn-in of $m$ steps, any initial distribution is within $C$ of stationarity on path space

Let $P$ be a Markov kernel with invariant probability $\pi$, let $\lambda$ be an arbitrary initial distribution, and suppose the $m$-step kernel satisfies the uniform bound $\|P^m(x,\cdot)-\pi\| \le C$ for every $x$. Then, on path space,

where $\sigma^m$ is the shift $(\sigma^m\omega)_n = \omega_{n+m}$.

What it says. A chain started from an arbitrary distribution, observed from time $m$ onwards, is within $C$ of the stationary chain — not merely at time $m$, but as an entire trajectory. This is the precise sense in which a chain "forgets its initial distribution", and it is the form the statement must take to be useful for limit theorems, which are statements about the whole path rather than about a single marginal.

Why it is the bridge in the Markov chain CLT. The central limit theorem asserts convergence for every initial distribution, while every proof of it establishes the limit for the stationary chain. For a uniformly ergodic chain, $C = R\,t^m$ decays geometrically, so this bound says the two path laws are geometrically close after a burn-in of $m$ steps. Since the normalized partial sums $\sqrt n(\bar f_n - \mathbb E_\pi f)$ are asymptotically unaffected by discarding a fixed number of initial terms, testing against a bounded continuous $g$ gives

and letting $n \to \infty$ and then $m \to \infty$ transfers the stationary limit law to $\lambda$.

Two distinct total-variation facts combine here, and neither alone suffices. First, a bound holding from every deterministic start transfers to any random start with the same constant, giving $\|\lambda P^m - \pi\| \le C$. Second, applying the trajectory kernel to both sides cannot increase total variation, lifting that state-space bound to path space. The first preserves the constant; the second is a contraction. Their composition is what turns a hypothesis about the $m$-step kernel into a statement about entire trajectories.

Proof. Shifting a trajectory by $m$ steps is the same as taking $m$ steps of $P$ first and then running the chain, so $(\sigma^m)_*\mathbb P_\lambda = \mathbb P_{\lambda P^m}$ — formally, the trajectory kernel composed with $P^m \circ \lambda$. Data processing bounds the distance between $\mathbb P_{\lambda P^m}$ and $\mathbb P_\pi$ by $\|\lambda P^m - \pi\|$, and the uniform hypothesis bounds that by $C$.