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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.

0xbadf00d
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  • 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 02 '19 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 '19 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 '19 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 '19 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 '19 at 20:43
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    Yes, I figured out why I was thinking that the support plays a role... but your reasoning should be fine. – saz Feb 18 '19 at 14:27
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    @0xbadf00d Your claim that (5) still holds if we go to $1_{\xi=\eta}(X-Y)$ is not obvious to me. Do you mean that the proof of (5) still works if we change up the integrand in the expectation on both sides of the inequality or do you mean that we can show your claim directly from (5)? – mortenmcfish May 15 '20 at 07:56

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