2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – saz Jan 2 at 10:09
  • $\begingroup$ @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 $(\omega,t)\in\Omega\times[0,\infty)$. $\endgroup$ – 0xbadf00d Jan 23 at 12:29
  • $\begingroup$ 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. $\endgroup$ – saz Jan 23 at 12:41
  • $\begingroup$ @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 $(14)$. $\endgroup$ – 0xbadf00d Jan 23 at 14:14
  • $\begingroup$ @saz Do you remember if you remembered why you were thinking that the support of $\xi$ plays a role? $\endgroup$ – 0xbadf00d Feb 17 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.