Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq 0},\mathbb{P})$ be a complete filtered probability space, where $(\mathcal{F}_t)_{t\geq 0}$ is the completed Brownian filtrate. Suppose that $\Phi(t,x)$ is the strong solution for the Stratonovich SDE, say on $\mathbb{T}^2$,

$$\Phi(t,x) = \Phi(0,x) + \int_0^t u(\Phi(s,x))ds + \int_0^t \sigma(\Phi(s,x))\circ dW(s), \tag{1}$$

where $u$ and $\sigma$ are respectively log-Lipschitz and smooth vector fields on $\mathbb{T}^2$ and $W$ is Brownian motion. Now suppose that $c(t)$ is a Hölder continuous stochastic process with state space $\mathbb{T}^2$, and consider the stochastic process

$$R(t) := \sup_{x\in \bar{B}(0,r)} |\Phi(t,x)-c(t)|^2. \tag{2}$$

Since $\Phi$ is spatially continuous (in fact, it has some Hölder regularity due to $u$ being log-Lipschitz) and $\bar{B}(0,r)$ is compact, we know that for each $(t,\omega)$ fixed, there exists $x(t,\omega)\in \bar{B}(0,r)$ such that

$$R(t,\omega) = |\Phi(t,x(t,\omega),\omega)-c(t,\omega)|^2. \tag{3}$$

Evidently, there may be more than one point $x(t,\omega)$ satisfying (3); so we have the multi-valued function

$$(t,\omega) \mapsto \{x\in \bar{B}(0;r) : |\Phi(t,x,\omega)-c(t,\omega)|^2 - R(t,\omega)=0\}, \tag{4}$$

where the values of this function are compact.

My question is the following.

Question. Is there a find a selection $x(t,\omega)$ in (3) so that the stochastic process $$t \mapsto \Phi(t,x(t))$$ is measurable and adapted?


2 Answers 2


Write $$K(t,\omega) := \{x\in \bar{B}(0;r) : |\Phi(t,x,\omega)-c(t,\omega)|^2 - R(t,\omega)=0\} \,. $$ Since this set is compact, choosing the lexicographically smallest point $x^*(t,\omega)$ in it will yield a measurable selection. For example, if this compact set is in ${\bf R}^2$, and $P_i$ is projection to the $i$'th coordinate, let $x^*_1(t,\omega):=\min P_1(K(t,\omega))$ and $$x^*_2(t,\omega):=\min P_2(\{x \in K(t,\omega) : P_1(x)=x^*_1(t,\omega)\}) \,.$$

Let me add some detail in response to the comment below. For every real $\alpha$, the event $\{\omega: x^*_1(t,\omega) \le \alpha\}$ means that for every rational $\epsilon>0$ there exists a a rational point $q=(q_1,q_2)$ in $B(0,r)$ such that $q_1<\alpha+\epsilon$ and $|\Phi(t,q,\omega)-c(t,\omega)|^2 >R(t,\omega)-\epsilon \,. $ A similar but a bit more involved formula applies to $\{\omega: x^*_2(t,\omega) \le \beta \}$.

  • $\begingroup$ Thank you for your comment, Professor Peres, but I am still not quite seeing why it follows that the map $(t,\omega) \mapsto x^*(t,\omega)$ is measurable. It seems like to show this, we want to show that the map $(t,\omega) \mapsto K(t,\omega)$ is measurable where the co-domain is the space $\mathcal{K}(\mathbb{T}^2)$ of non-empty compact subsets of $\mathbb{T}^2$ equipped with the Hausdorff distance and the induced Borel $\sigma$-algebra. However, showing this last point isn't obvious to me. $\endgroup$ Commented Jun 22, 2019 at 15:26

This answer is an expansion on the answer of Yuval Peres which addresses the concerns raised in my comment.

We will show the existence of a progressively measurable stochastic process $t\mapsto x^*(t)\in\mathbb{T}^2$ satisfying $$|\Phi(t,x^*(t,\omega),\omega)-c(t,\omega)|^2 - R(t,\omega) =0 \tag{1}$$ in several steps.

Let $\mathcal{K}(\mathbb{T}^2)$ denote the set of non-empty compact subsets of $\mathbb{T}^2$ equipped with the Hausdorff metric $d_H$. It is well-known that $(\mathcal{K}(\mathbb{T}^2), d_H)$ is a compact metric space. We turn $\mathcal{K}(\mathbb{T}^2)$ into a measurable space by endowing it with the Borel $\sigma$-algebra $\mathcal{B}(\mathcal{K}(\mathbb{T}^2))$.

Lemma 1. The $\mathcal{K}(\mathbb{T}^2)$-valued stochastic process $t \mapsto \bar{B}(c(t), R(t))$ is progressively measurable.

Proof. Since $c(\cdot)$ and $R(\cdot)$ are both $\mathbb{T}^2$- and $\mathbb{R}$-valued progressively measurable processes respectively, it follows that that $(c(\cdot),R(\cdot))$ is also progressively measurable. It follows from the triangle inequality that the map $$(\mathbb{T}^2, [0,\infty))\rightarrow (\mathcal{K}(\mathbb{T}^2),d_H), \quad (c,R)\mapsto \partial B(c,R)$$ is continuous, which in turn implies the desired conclusion. $\Box$

Although I omitted this detail in my original post, in application the stochastic flow $\Phi$ has the following property: for almost every $\omega$, the map $x\mapsto \Phi(t,x,\omega)$ is a homeomorphism of $\mathbb{T}^2$ for all $t\geq 0$. Moreover, the $C(\mathbb{T}^2)$-valued stochastic process $$t \mapsto \Phi^{-1}(t,\cdot)\in C(\mathbb{T}^2)$$ is progressively measurable. This fact leads us to the next lemma, the proof of which is an easy exercise.

Lemma 2. The map $$\Psi_1: (C(\mathbb{T}^2),\|\cdot\|_\infty) \times (\mathcal{K}(\mathbb{T}^2),d_H) \rightarrow (\mathcal{K}(\mathbb{T}^2),d_H), \qquad (f,A)\mapsto f(A)$$ is continuous.

We claim that the $\mathcal{K}(\mathbb{T}^2)$-valued process $$\tilde{K}(t) := \Phi^{-1}(t,\cdot)(\partial B(c(t),R(t))):= \{y\in\mathbb{T}^2: y=\Phi^{-1}(t,x) \text{ for some } x\in \partial B(c(t),R(t))\}$$ is progressively measurable. Indeed, fix $t\geq 0$. Evidently, the map $$\Psi_2:[0,t]\times \Omega \rightarrow C(\mathbb{T}^2)\times \mathcal{K}(\mathbb{T}^2), \quad (s,\omega) \mapsto (\Phi^{-1}(s,\cdot,\omega),\partial B(c(s,\omega), R(s,\omega)))$$ is $\mathcal{B}([0,t]) \otimes \mathcal{B}(\mathcal{K}(\mathbb{T}^2))$-measurable. So the claim follows from writing $$\tilde{K}(s,\omega) = \Psi_1\circ \Psi_2(s,\omega)$$ and using Lemma 2.

Now $\tilde{K}(t,\omega)$ isn't quite the set $K(t,\omega)$ from Yuval Peres's answer (it's larger); however, this doesn't matter for the following reason. If $x\in\tilde{K}(t,\omega)$, then I claim that $|x|\geq r$. Otherwise, since $\Phi(t,\cdot,\omega)$ is a homeomorphism, we could find an open ball $B(x,\delta)\subset B(0,r)$ such that $\Phi(t,\cdot,\omega)(B(x,\delta))$ is open. But by definition of $\tilde{K}(t,\omega)$, we have that $|\Phi(t,x,\omega)-c(t,\omega)|=R(t,\omega)$, which implies that there exists $y\in B(x,\delta)$ such that $$|\Phi(t,y,\omega)-c(t,\omega)|>R(t,\omega).$$ This last inequality contradicts the definition of $R(t,\omega)$. Consequently, the lexicographic minimum of $K(t,\omega)$, denoted by $x^*(t,\omega)$, equals the lexicographic minimum of $\tilde{K}(t,\omega)$.

To see that the process $t\mapsto x^*(t)$ is progressively measurable, we rely on the following lemma.

Lemma 3. Let $K$ be a non-empty compact set. We denote the lexicographic minimum of $K$ by $x_K^*$. Then the map $$(\mathcal{K}(\mathbb{T}^2), d_H) \rightarrow (\mathbb{T}^2,|\cdot|), \quad K\mapsto x_K^*$$ is continuous.

Proof. Suppose that $d_H(K_n,K)\rightarrow K$. Given $\epsilon>0$, let $N\in\mathbb{N}$ be sufficiently so that $d_H(K_n,K)\leq \epsilon$ for all $n\geq N$. For each $n\geq N$, there exists $y_{n,K}\in K$ such that $$|x_{K_n}^*-y_{n,K}| \leq \epsilon,$$ which implies by definition of $d_H$ that $$x_{K,j}^* \leq x_{K_n,j}^* + \epsilon, \qquad j=1,2.$$ Similarly, there exists $z_{K_n}\in K_n$ such that $$|x_{K}^* - z_{K_n}| \leq \epsilon,$$ which implies that $$x_{K_n,j}^* \leq x_{K,j}^* + \epsilon, \qquad j=1,2.$$ Hence, $|x_{K_n}^*-x_{K}^*|\leq 2\epsilon$, and since $\epsilon>0$ was arbitrary, the proof of the lemma is complete. $\Box$

Applying Lemma 3, we conclude that the stochastic process $t\mapsto x_{\tilde{K}(t)}^*$ is progressively measurable.


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