Consider a strongly convex potential $U: \mathbb{R}^d \to \mathbb{R}$ and the Langevin diffusion $$dX = -\nabla U(X) dt + dW \qquad (\*)$$ where $W$ is a standard Brownian motion. If $(X\_t)\_{t \geq 0}$ and $(Y\_t)\_{t \geq 0}$ are two solutions driven by the same Brownian motion, one can check that $t \mapsto \mathbb{E} \| X\_t-Y\_t \|^2$ is decreasing since $d \| X\_t-Y\_t \|^2 = -2 \langle X\_t-Y\_t, \nabla U(X\_t) - \nabla U(Y\_t)\rangle \leq -2 \lambda \|X\_t - Y\_t\|^2$ where $\textrm{Hessian}(x) \geq \lambda I\_d$ on $\mathbb{R}^d$. In other words, given two starting positions $x\_0, y\_0 \in \mathbb{R}^d$, one can construct a coupling $(X\_t,Y\_t)$ of the diffusion $(\*)$, one starting from $x\_0$ and the other one from $y\_0$, such that $\mathbb{E} \|X\_t-Y\_t\| \leq e^{- \lambda t} \|x\_0-y\_0\|$. Indeed, if the function $U$ is not strongly convex, one can only say that $\mathbb{E} \|X\_t-Y\_t\|$ can be made non-increasing. This can also be expressed in terms of Wasserstein distance between $P\_t(x,dx)$ and $P\_t(y,dy)$ where $P$ is the transition operator of the diffusion $(\*)$.


**Question**: given $x\_-,x^+,y\_-,y^+ \in \mathbb{R}^d$, can we find a coupling $(X\_t,Y\_t)$ of the conditioned diffusion $(\*)$ with
$$X\_0=x\_-, \quad X\_T=x^+, \quad Y\_0=y\_-, \quad Y\_T=y^+,$$
and such that there is still a contraction property of the type $$\mathbb{E} \|X\_{T/2}-Y\_{T/2}\| \leq \beta \big( \|x\_- - y\_-\| + \|x\_+ - y\_+\|\big)$$
where $0<\beta<\frac12$ is some constant that does not depend on $(x\_-,x^+,y\_-,y^+)$. Is this known or have already been studied? Any reference welcome.


**PS**: one can check that this is true for a quadratic potential $U(x) = \frac12 \|x\|^2$ since the Ornstein-Uhlenbeck process is easy to study. The case of a brownian bridge (i.e. vanishing potential) is also straightforward and corresponds to the case $\beta = \frac12$.


**PPS**: it is important to exploit the structure of the SDE. Indeed, one can find counter examples of Markov processes $(X\_t)\_{t \geq 0}$ that have the contraction property $\mathbb{E} \|X\_t-Y\_t\| \leq e^{-\lambda t} \|x\_0 - y\_0\|$ but that do not verify the 'conditioned' contraction property.



contraction property for conditioned SDEs
I don't think so. In the limit, it would mean that if $U$ is merely convex, then we still must have this inequality with $\beta=\frac 12$. Now take any function $U$ on $\mathbb R$ that is $0$ on $[-1,1]$ and is strictly convex beyond the ends. Choose two distinct starting points in $(-0.5,0.5)$. Then, if what you said were true, we would have $F(t)=E\|X\_t-Y\_t\|$ concave. However, it tends to $0$ because you cannot stay on $[-1,1]$ forever and $F'(0)\ge 0$ because for small $t$, 
$$
F(t)=E\|X\_t-Y\_t\|\ge\|x-y\|-2E(|W\_t|\cdot 1\_{\max\_{s<t}|W\_s|>0.5})\ge F(0)-Ce^{-ct^{-1}}\,.
$$
Alas, concave tending to $0$ functions with non-negative derivative at the origin do not exist.


