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Use the same Brownian motion for both solutions: this is a synchronous coupling. For $D_t=X_t-Y_t$, the Itô formula givesOn any fixed finite horizon the stochastic integral has mean zero. Indeed, boundedness of $\sigma$ and the assumed second-moment bounds make the expectation of its squared integrand integrable in time. With $h(t)=\mathbb{E}{D}_t^2$, the contraction condition yields, for every $s\le t$,The allowed Gronwall inequality gives the mean-square contraction of synchronously coupled diffusionsThe identical estimate can also be obtained by localizing the nonnegative local supermartingale $e^{kt}{D}_t^2$ and using Fatou's lemma, a formulation useful when the drift has linear growth.

There is a genuine compatibility issue in the printed global assumptions. If $|b|\le B$, then for $d=|x-y|$,It cannot be at most $-k{d}^2$ for all $d>4B/k$. Thus a globally bounded drift cannot satisfy the stated strict contraction on all of the real line. The stochastic estimate above is the requested conditional calculation. For a non-vacuous application, global boundedness of the drift must be relaxed while retaining suitable existence and moment hypotheses; it is not silently changed here.

The two recursions use a synchronous coupling: they receive exactly the same fresh inputs. Subtraction cancels the additive term:The variable $U_n$ is independent of the preceding difference and has expected value $1/2$. Thus expectation of a product of independent random variables givesBy Markov's inequality, for every $\epsilon >0$,This proves the requested convergence in probability; in fact the difference has L1 convergence. It is synchronous contraction of a random affine recursion, and the estimate holds for every real $c$.

A random affine recursion updates a real random variable by $X_n=A_n{X}_{n-1}+B_n$. Assume the coefficient pairs $(A_n,B_n)$ are independent and identically distributed random variables, independent of the initial value. The two coordinates within a pair may be dependent. Fresh coefficients make the recursion a time-homogeneous Markov process: its next conditional distribution depends on the current value alone.

The competition between the random multiplier and the fresh input determines whether initial conditions are forgotten. A synchronous coupling makes this particularly visible by using the same coefficient sequence for two initial values.

A Synchronous coupling drives two processes with the same fresh random inputs. For example, recursions $X_{n+1}=F(X_n,{\xi }_{n+1})$ and $Y_{n+1}=F(Y_n,{\xi }_{n+1})$ use the same input ${\xi }_{n+1}$ at each step. This can reveal synchronous contraction of a random affine recursion even when independently simulated trajectories would conceal the relationship.

A Synchronous coupling drives two solutions of a stochastic differential equation by the same Brownian motion. Their difference has noise coefficient $\sigma (X)-\sigma (Y)$. A dissipativity inequality for the drift and this coefficient gives the mean-square contraction of synchronously coupled diffusions.

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.