# If $(X^x_t)$ is the stochastic flow generated by a SDE and $(X_t)$ is the strong solution with $X_0=ξ$, is $X_t=X^ξ_t$ for all $t$ a.s.?

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ be a complete probability space
• $$(\mathcal F_t)_{t\ge0}$$ be a complete and right-continuous filtration on $$(\Omega,\mathcal A,\operatorname P)$$
• $$\xi$$ be an $$\mathcal F_0$$-measurable square-integrable random variable on $$(\Omega,\mathcal A,\operatorname P)$$
• $$W$$ be an $$\mathcal F$$-Brownian motion on $$(\Omega,\mathcal A,\operatorname P)$$
• $$b,\sigma:[0,\infty)\times\mathbb R\to\mathbb R^d$$ be Borel measurable with $$|b(t,x)|^2+|\sigma(t,x)|^2\le C_1(1+|x|^2)\;\;\;\text{for all }t\ge0\text{ and }x\in\mathbb R\tag1$$ for some $$C_1\ge0$$ and $$|b(t,x)-b(t,y)|^2+|\sigma(t,x)-\sigma(t,y)|^2\le C_2|x-y|^2\;\;\;\text{for all }t\ge0\text{ and }x,y\in\mathbb R\tag2$$ for some $$C_2\ge0$$

We know that there is a unique (up to indistinguishability) continuous $$\mathcal F$$-adapted process $$(X_t)_{t\ge0}$$ with $$X_t=\xi+\int_0^tb(s,X_s)\:{\rm d}s+\int_0^t\sigma(s,X_s)\:{\rm d}W_s\;\;\;\text{for all }t\ge0\text{ almost surely}\tag3.$$ We say that $$X$$ is the pathwise unique strong solution of $${\rm d}X_t=b(t,X_t){\rm d}t+\sigma(t,X_t){\rm d}W_t\tag4$$ with initial condition $$X_0=\xi$$. We observe that, if $$Y$$ is the pathwise unique strong solution of $$(4)$$ with initial condition $$Y_0=\eta$$ (for some $$\mathcal F_0$$-measurable square-integrable random variable $$\eta$$ on $$(\Omega,\mathcal A,\operatorname P)$$), then $$\operatorname E\left[\sup_{s\in[0,\:t]}\left|X_s-Y_s\right|^2\right]\le\Lambda(t)\operatorname E\left[\left|\xi-\eta\right|^2\right]\;\;\;\text{for all }t\ge0\tag5$$ for some continuous nondecreasing $$\Lambda:[0,\infty)\to[0,\infty)$$ (which only depends on $$C_2$$). Thus, $$X_t=Y_t\;\;\;\text{for all }t\ge0\text{ almost surely on }\left\{\xi=\eta\right\}.\tag6$$ Now, let $$(X^x_t)_{t\ge0}$$ denote pathwise unique strong solution of $$(4)$$ with initial condition $$X^x_0=x\in\mathbb R^d$$. We're able to assume that $$\Omega\times[0,t]\times\mathbb R\ni(\omega,s,x)\mapsto X_s^x(\omega)\tag7$$ is $$\mathcal F_t\otimes\mathcal B([0,t])\times\mathcal B(\mathbb R)$$-measurable for all $$t\ge0$$ and $$(t,x)\mapsto X_t^x(\omega)\tag8$$ is (jointly) continuous for all $$\omega\in\Omega$$. We easily obtain that $$\left(X^\xi_t\right)_{t\ge0}$$ is $$\mathcal F$$-progressive.

I want to conclude that $$X_t=X^\xi_t\;\;\;\text{for all }t\ge0\text{ almost surely .}\tag9$$

From $$(6)$$ we see that the claim is true as long as $$|\xi(\Omega)|\le|\mathbb N|$$. In general, there is a $$(\xi_n)_{n\in\mathbb N}$$ with $$\xi_n$$ being an $$\mathcal F_0$$-measurable random variable on $$(\Omega,\mathcal A,\operatorname P)$$ with $$|\xi_n(\Omega)|\in\mathbb N$$ for all $$n\in\mathbb N$$, $$|\xi_n|\le|\xi|\;\;\;\text{for all }n\in\mathbb N\tag{10}$$ and $$|\xi_n-\xi|\xrightarrow{n\to\infty}0\tag{11}.$$ From $$(10)$$, $$(11)$$ and the square-integrability of $$\xi$$, we obtain $$\left\|\xi_n-\xi\right\|_{L^2(\operatorname P)}\xrightarrow{n\to\infty}0\tag{12}$$ and hence $$\operatorname E\left[\sup_{s\in[0,\:t]}\left|X^{\xi_n}_s-X_s\right|^2\right]\xrightarrow{n\to\infty}0\tag{13}\;\;\;\text{for all }t\ge0$$ from $$(5)$$. On the other hand, by continuity of $$(8)$$, we should have $$\sup_{s\in[0,\:t]}|X^{\xi_n}_s-X^\xi_s|\xrightarrow{n\to\infty}0\;\;\;\text{for all }t\ge0\tag{14}$$ and hence obtain $$(9)$$ by uniqueness (up to equality almost surely) of the limit in probability.

• How exactly do you get (6) from (5)? I think that for (14) you need another truncation argument; it is not obvious (to me) why the continuity gives $(14)$... everything should be fine if the support of $\xi$ is contained in a ball... so it boils down to another truncation argument. – saz Jan 2 at 10:09
• @saz Get $(6)$ from $(5)$: If $\xi=\eta$ a.s., then $\operatorname E\left[\sup_{s\in[0,\:t]}\left|X_s-Y_s\right|^2\right]=0$ by $(5)$ for all $t\ge0$. So, $X$ and $Y$ are indistinguishable. And $(14)$: By continuity, $\left|X^{\xi_n(\omega)}_t(\omega)-X^{\xi(\omega)}_t(\omega)\right|\xrightarrow{n\to\infty}0$ for all $(\omega,t)\in\Omega\times[0,\infty)$. – 0xbadf00d Jan 23 at 12:29
• In (6) you are claiming something stronger than what you just proved; in (6) you are saying that the solutions coincide on $\{\xi=\eta\}$ whereas you were just now assuming that $\xi=\eta$ a.s. Re (14): It's been a while since I wrote the comment; honestly, I currently don't remember why I was thinking that the support of $\xi$ plays a role... I will think about it once more. – saz Jan 23 at 12:41
• @saz $(5)$ still holds if you replace $X-Y$ by $1_{\left\{\:\xi\:=\:\eta\:\right\}}(X-Y)$. Please let me know, if you remember your worries with $(14)$. – 0xbadf00d Jan 23 at 14:14
• @saz Do you remember if you remembered why you were thinking that the support of $\xi$ plays a role? – 0xbadf00d Feb 17 at 20:43