1
$\begingroup$

Assume $\rho$ is probability density defined on $\mathbb{R}^d\times\mathbb{R}^d$. I am interested in the Wasserstein gradient flow of a functional: \begin{equation*} \mathcal{E}(\rho)=\iint_{\mathbb{R}^d\times\mathbb{R}^d}c(x,y)\rho(x,y)~dxdy+\rm{KL}(\rho_a|\varrho_a)+\rm{KL}(\rho_b|\varrho_b) \end{equation*} Here $\rho_a(x)=\int\rho(x,y)~dy$, $\rho_b(y)=\int\rho(x,y)~dx$ are marginals of $\rho$ and $\varrho_a$,$\varrho_b$ are given probability densities on $\mathbb{R}^d$. I tried to compute the Wasserstein gradient flow of this functional $\mathcal{E}$: $\frac{\partial \rho}{\partial t}=-\rm{grad}_\rho \mathcal{E}(\rho)$,this gives the following Fokker Planck Equation: \begin{equation*} \frac{\partial\rho}{\partial t}=\nabla\cdot\left(\rho\nabla\left(c(x,y)+\log\left(\frac{\rho_a(x)}{\varrho_a(x)}\right)+\log\left(\frac{\rho_b(y)}{\varrho_b(y)}\right)\right)\right) \quad\quad (*) \end{equation*} Here is my question: Is it possible to give a Stochastic Differential Equation whose probability density is exactly the equation (*)? If we directly formulate the SDE like: \begin{equation} d\left[\begin{array}{c}\mathbf{X}_t\\\mathbf{Y}_t\end{array}\right]=\left[\begin{array}{c}\nabla_x(c(\mathbf{X}_t,\mathbf{Y}_t)-\log\varrho_a(\mathbf{X}_t))\\\nabla_y(c(\mathbf{X}_t,\mathbf{Y}_t)-\log\varrho_b(\mathbf{Y}_t))\end{array}\right]dt-\left[\begin{array}{c}\nabla_x\log\rho_a(\mathbf{X}_t)\\\nabla_y\log\rho_b(\mathbf{Y}_t)\end{array}\right]dt \quad\quad (\#) \end{equation} It is clear that the first term \begin{equation*} \left[\begin{array}{c}\nabla_x(c(\mathbf{X}_t,\mathbf{Y}_t)-\log\varrho_a(\mathbf{X}_t))\\\nabla_y(c(\mathbf{X}_t,\mathbf{Y}_t)-\log\varrho_b(\mathbf{Y}_t))\end{array}\right]dt \end{equation*} in (#) can be treated as the drift term, I am now quite confused about the second term in (#). What I want to do is to introduce Brownian Motion to replace the second term. If we consider the case \begin{equation*} d\mathbf{X}_t=-\nabla_x\log\rho(\mathbf{X}_t)dt \end{equation*} where $\rho$ is the law of $\mathbf{X}_t$. Then the corresponding Fokker Planck is $\frac{\partial \rho}{\partial t}=\Delta\rho$. So now if we let $\tilde{\mathbf{X}}_t$ satisfies \begin{equation*} d\tilde{\mathbf{X}}_t=\sqrt{2}~d\mathbf{B}_t \quad \tilde{\mathbf{X}}_0\stackrel{d}{=}\mathbf{X}_0 \end{equation*} Here $\mathbf{B}_t$ is the standard Brownian Motion. We will know $\mathbf{X}_t$ and $\tilde{\mathbf{X}}_t$ have the same law. (i.e.$\mathbf{X}_t\stackrel{d}{=}\tilde{\mathbf{X}}_t$ for any $t\geq 0$) But we can't simply replace the second term in (#) by $\left[\begin{array}{c} \sqrt{2}d\mathbf{B}_t^{(1)} \\ \sqrt{2}d\mathbf{B}_t^{(2)} \end{array}\right]$. Otherwise, the corresponding Fokker Planck becomes \begin{equation*} \frac{\partial\rho}{\partial t}=\nabla\cdot\left(\rho\nabla\left(c(x,y)-\log\varrho_a(x)-\log\varrho_b(y)\right)\right)+\Delta \rho \end{equation*} which is definitely different from (*). Are there any experts who can help me on this?

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.