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Is it true? Suppose

$$ dX_t = \nabla V_1(X_t) + \sqrt{2} dB_t \,,$$ $$ dY_t = \nabla V_2(Y_t) + \sqrt{2} dB_t \,.$$

If $V_1$ and $V_2$ are ``close", would the stationary distributions of $X_t$ and $Y_t$ be close? For example, can we bound the Wasserstein distance $W^2_2(\mu, \nu)$ where $\mu, \nu$ are the stationary distributions of $X_t$ and $Y_t$? Perhaps, by something like $\|V_1 - V_2\|^2_2$, or like $\|\nabla V_1 - \nabla V_2\|^2_2$.

user58697
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blue_egg
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  • What information do you have for $\nabla V_{i}$? Are they Lipschitz? are the Vi convex? – Thomas Kojar Aug 26 '23 at 17:54
  • Here https://arxiv.org/pdf/2207.02750.pdf "An SDE perspective on stochastic convex optimization" they study the limit for such SDEs if you have minus and convexity (take sigma to have a zero column and a ones column). – Thomas Kojar Aug 26 '23 at 17:56
  • @ThomasKojar In the particular case I'm thinking about, I'm comparing the drifts $\nabla V(z_{-i})$ and $\nabla V(z_{-j})$ where the $-i$ subscript denotes removing the $i$th component from $z \in \mathbb{R}^n$. And I suppose $V$ can be convex and smooth. – blue_egg Aug 27 '23 at 22:34

1 Answers1

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Let $$\begin{align*}dX_t&=-\nabla V_1(X_t)dt+\sqrt2 dB_t \\ dY_t&=-\nabla V_2(Y_t)dt+\sqrt2 dB_t.\end{align*}$$ It is well-known that under suitable conditions of $V_1,V_2$ (e.g. $\lambda$-convexity), the processes $X,Y$ converge to equilibrium $\mu_1,\mu_2$ resepectively, where $$\begin{align*}\mu_i(dx)&=\frac{\exp(-V_i(x))}{\int\exp(-V_i(y))dy}dx\end{align*}$$

Hence, the densities of the stationary distributions are known. However, in general it is quite difficult to bound the Wasserstein distance of measures in terms of their densities, see e.g. here.

In this particular case, i doubt that a meaningful bound can be achieved as you describe it. Consider the case where $$V_i(x)=\theta_i(x-m_i)^2$$ leading to $\mu_i$ being Gaussian with mean $m_i$ and variance $\theta_i^{-1}$.

Now note that, unless $\theta_1=\theta_2$ as well as $m_1=m_2$ (in which case the Wasserstein distance would be $0$ anyway), we have both $$\Vert V_1-V_2\Vert_2^2=\infty, \hspace{1cm} \Vert \nabla V_1-\nabla V_2\Vert_2^2=\infty.$$ Hence i doubt that one would be able to describe the "similarity" between $\mu_1$ and $\mu_2$ using such quantities.

Perhaps another approach is the following: One can view the evolution described by the SDEs as a Gradient Flow of a certain energy functional, namely the relative entropy functional $\text{Ent}_{\mu_i}$, on the Wasserstein space. Perhaps a more promising approach would be to compare $\mu_1$ and $\mu_2$ by some comparison between $\text{Ent}_{\mu_1}$ and $\text{Ent}_{\mu_2}$. For more information on viewing this problem as a Gradient Flow, see the book Lectures on Optimal Transport, specifically Chapter 18.

Small Deviation
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