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Let $D={\max }_{x,y}\rho (x,y)$ and let $\pi$ be the invariant distribution. Iterating the assumed Wasserstein contraction givesSince $\rho (x,y)\ge 1_{\{x\ne y\}}$, the coupling characterization of total variation distance impliesThe right side is at most $\epsilon$ whenwhich proves the claimed mixing bound.

A Markov kernel $P$ has Wasserstein contraction factor $c<1$ for a ground metric $\rho$ whenfor all probability measures for which the distances are finite. Writing $c=e^{-\alpha }$ gives a positive one-step contraction rate $\alpha$. Iteration yields $W_{\rho }(\mu P^n,\nu P^n)\le c^n{W}_{\rho }(\mu ,\nu )$. If $\pi$ is a stationary distribution with finite first moment, this proves geometric convergence in Wasserstein distance to $\pi$ and uniqueness within that moment class. A continuous-time version instead uses factor $e^{-\alpha t}$ for its transition semigroup.

For the random affine recursion $X_{n+1}=a{X}_n+{\xi }_{n+1}$ on the real line with $|a|<1$ and independent finite-first-moment noise, Synchronous coupling uses the same fresh ${\xi }_{n+1}$ for both chains. It gives $|X_{n+1}-Y_{n+1}|=|a||X_n-Y_n|$, hence the contraction factor $|a|$ for $W_1$. The stationary random series ${\sum }_{j\ge 0}{a}^j{\xi }_{-j}$ converges absolutely almost surely because its expected absolute sum is finite. Thus this example also supplies existence, which a uniqueness argument alone would not establish.