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For an adapted stationary process with the displayed bounds, the Itô formula and stationarity give ${\int }_0^t(\mathbb{E}[Z_s{g}_s]+\lambda )ds=0$. The Cauchy-Schwarz inequality makes each integrand nonnegative, so equality holds almost everywhere. Then $\mathbb{E}(g_s+\lambda Z_s{)}^2\le 0$, proving $g=-\lambda Z$ for almost every time and outcome. Thus the process solves the Ornstein-Uhlenbeck stochastic differential equation and its initial value, rather than stationarity alone, determines its coupling with W.

Drive two copies of a random affine recursion with the same coefficient pairs, independent of their initial coupling. Then $X_n-Y_n=A_n(X_{n-1}-Y_{n-1})$. If $\rho =\mathbb{E}|A_1|<1$ and $\mathbb{E}|X_0-Y_0|<\infty$, independence of random variables givesThis proves convergence in L1 and hence convergence in probability of the difference. If one copy starts in a stationary distribution, Slutsky's theorem shows that the other copy converges to that distribution. The argument compares paths before taking their one-time probability distributions; it does not require the paths ever to become exactly equal.

If each coordinate law satisfies the same convex transport-entropy inequality, then the product law satisfies the sum-cost version. Sequentially couple conditional coordinates, apply Jensen inequality, and use the chain rule for relative entropy.

A transport-entropy inequality controls an optimal expected transportation cost between probability distributions by their Kullback-Leibler divergence.