Synchronous contraction of a random affine recursion ($\mathbb{E}|X_n-Y_n|={\rho }^n\mathbb{E}|X_0-Y_0|$)

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.