contraction property for conditioned SDEs

Consider a strictly 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 strictly 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.

• @ Alekk :I don't know if it helps but if you have sufficent structure added I think that you can derive the conditionned SDE with respct to another brownian motion. Here are a few references that might help, chapter 6 of Protter's book, the LNM of Mansuy and Yor about Enlargmetn of filtration (especialy the markovian case) where initial enlargement of filtration is studied. I also know this paper by Baudoin "Conditioned SDEs Theory and Applications" that might help. This idea would be then to reproduce your initial argument on the conditoned SDEs for $Y$ and $X$ in this framework. Best regards Oct 25 '12 at 9:35
• Thanks TheBridge: yes, but to get the expression for the conditioned SDE you typically need to have analytical expressions for the transition probabilities (and then do a h-transform), which is not the case in general. Will have a look at the mentioned references. Oct 25 '12 at 10:09
• @ Alekk : for the Makovian case yes I remember that the correction of the sde on the drift has to be done, so you have a modification of $\nabla U$ term by adding this correction which involves a ratio of transition kernels of the unconditional distribution (which must be the h-transform you mention). So why not examine sufficient conditions on this correction (and by extension on the Kernel) to still get the inequality of the unconditional case for some possibly new $\lambda$ ? I'm sure you tried this (simple) approach but don't see exactely where it fails. Best regards. Oct 25 '12 at 11:47
• adding this correction the drift then reads $-\nabla U(X) + c(X)$ and this leads to $d \|X-Y\|^2 = -2\langle X-Y, \nabla U(X)-\nabla U(Y) -c(X)+c(Y)\rangle$ which has no reason to be negative. Oct 25 '12 at 12:17
• @ alekk : Got it thank's. Best regards Oct 25 '12 at 15:44

1 Answer

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.

• Thank you Fedja. I afraid I do not completely understand your comments, though. Are you talking about the non-conditioned case. Your potential verifies $\lambda = \inf_x U^{′′}(x)=0$ so one can only say that $F(t)$ can be made non-increasing (and tend to $0$). Also, my question was asking if it were possible to find a coupling $(X,Y)$ such that conditioned on $(X_0,Y_0)=(x_-,y_-)$ and $(X_T,Y_T)=(x_+,y_+)$ the difference $|X_{T/2}-Y_{T/2}|$ is on average less than $\frac{1}{2}(|x_--y_-|+|x_+-y_+|)$. I am sorry if I misunderstood your comment. Oct 25 '12 at 9:02
• Alekk, if you have the conditioned coupling for ALL $x_\pm,y_\pm$, then you can get the unconditioned inequality E|X_t-Y_t|\le \frac 12(|x_-y_|+E|X_{2t}-Y_{2t}||by integrating over $x_+,y_+$, right? Also, the process starts over at any time, so I can shift this by $s$ and integrate over whatever the distribution of $(X_s,Y_s)$ is. When I wrote "concave", I meant "convex" of course, but the rest should be clear. The inequality for small $t$ merely means that you have the pure BM until you reach the endpoint and that is next to impossible to do within short time. Oct 25 '12 at 10:09
• Thank you Fedja, I now understand what is going on and can cook up simple examples where one can see why the contractive coupling cannot be constructed. Many thank!. Oct 25 '12 at 14:20