I have a question when reading a textbook $\textbf{Introduction to stochastic integration}$ by Kuo.

Given the stochastic differential (or rather integral) equation, $$ X_{t}=\xi+\int_{a}^{t} \sigma\left(s, X_{s}\right) d B(s)+\int_{a}^{t} f\left(s, X_{s}\right) d s, \quad a \leq t \leq b, $$ where $\sigma$ and $f$ satisfy the Lipschitz and linear growth conditions.
Let $\left\{\mathcal{F}_{t} ; a \leq t \leq b\right\}$ be the filtration given by the Brownian motion $B(t)$, namely, $\mathcal{F}_{t}$ is defined by $\mathcal{F}_{t}=\sigma\{B(s) ; s \leq t\}$. Obviously, the solution $X_{t}$ is adapted to this filtration.

So, why $X_{t}$ is adapted to this filtration? I tried as following:
Given any open set $U$, we have $$X_{t}^{-1}(U)=\xi^{-1}(U)\cap[\int_{a}^{t} \sigma\left(s, X_{s}\right) d B(s)]^{-1}(U)\cap[\int_{a}^{t} f\left(s, X_{s}\right) d s]^{-1}(U)$$ which equals to either $\emptyset$ or $[\int_{a}^{t} \sigma\left(s, X_{s}\right) d B(s)]^{-1}(U)$, which are both $\mathcal{F}_t$-measurable.
Then here are two parts that I am not sure
(1)Is it correct that the way I use $\cap$ on the r.h.s. of the equation?
(2)Is my trying correct?

  • 1
    $\begingroup$ The formulation of the textbook confuses me: If $X$ is a stochastic process which is not adapted to $\mathcal F$, then (depending on how you define the Itô integral), $$\int_{a}^{t} \sigma\left(s, X_{s}\right) d B(s)$$ may simply be not well-defined, so how could $X$ then be a solution of the SDE ? $\endgroup$ Jun 26, 2022 at 10:41
  • $\begingroup$ I am indeed asking how to use definition of "adapted" to show that $X_t$ is adapted to the filtration. $\endgroup$ Jun 26, 2022 at 10:49
  • 2
    $\begingroup$ What is your definition of a solution to the SDE above? Usually, the definition of a solution already includes the condition that $X$ is adapted, cf. for example Definition 26.1 in the book Wahrscheinlichkeitstheorie by Achim Klenke, 2013. $\endgroup$ Jun 26, 2022 at 10:51
  • $\begingroup$ In my textbook$$, it does not put $X_t$ is adapted to $\mathcal{F}_t$ in the definition. $\endgroup$ Jun 26, 2022 at 11:07

1 Answer 1


It is not at all obvious that $X_t$ is adapted to that filtration, and that is actually a very powerful theorem. It is true because $\sigma$ and $f$ are Lipschitz and satisfy a linear growth condition, but it takes a significant amount of work to show that (For a proof, see for example Theorem 5.2.5 in Karatzas and Shreve's Brownian Motion and Stochastic Calculus).

Without the Lipschitz condition, it may not be the case that $X_t$ is adapted to the filtration generated by $B_t$.

For example, let $X_t$ be a Brownian motion, and define $B_t := \int_0^t \operatorname{sgn}(X_s)dX_s$. Note that $\langle B,B\rangle_t = \int_0^t \operatorname{sgn}(X_s)^2ds = \int_0^t 1ds = t$, so $B$ is a Brownian motion. Furthermore, $\int_0^t \operatorname{sgn}(X_s)dB_s = \int_0^t \operatorname{sgn}(X_s)^2 dX_s = X_t$, so $X_t$ solves the integral equation $$X_t = \int_0^t \operatorname{sgn}(X_s)dB_s.$$

However, $X_t$ is not adapted to the filtration generated by $B_t$. To see this, note that by Tanaka's formula \begin{align*} |X_t| &= \int_0^t \operatorname{sgn}(X_s)dX_s + L_t \\ &= B_t + L_t\end{align*} where $L_t$ is the local time of $X$ at $0$. Recall that, since $X$ is a Brownian motion, $L_t = \lim_{\varepsilon \rightarrow 0} \frac{1}{2\varepsilon}\int_0^t 1_{|X_s|<\varepsilon}ds$, and in particular $L_t$ is adapted to $\mathcal F^{|X|}_t = \sigma\{|X_s| ; s \le t\}$. By re-writing the Tanaka formula, we have $B_t = |X_t|-L_t$, so $B_t$ is adapted to $\mathcal F^{|X|}_t$. This implies $\mathcal F_t \subseteq \mathcal F_t^{|X|}$, so if $X$ were adapted to $\mathcal F_t$, $X$ would also be adapted to $\mathcal F_t^{|X|}$. This would imply $\mathcal F_t^{X} \subseteq \mathcal F_t^{|X|}$, and therefore (because $\mathcal F_t^{|X|}\subseteq \mathcal F_t^{X}$ trivially) $\mathcal F_t^X = \mathcal F_t^{|X|}$. This is a contradiction, though, because $X$ is a Brownian motion and therefore $\mathcal F_t^X \ne \mathcal F_t^{|X|}$.

  • $\begingroup$ I checked the bool and it seems that the theorem 5.2.5 is about how to show the (strong) uniqueness, not how to show $X_t$ is adapted to such filtration. $\endgroup$ Jun 28, 2022 at 8:23
  • $\begingroup$ @TalkingPuppet Sorry, my mistake. The condition that $X_t$ is adapted to such a filtration is called being a strong solution. There is another theorem (which I can't find at the moment) that says that strong uniqueness implies all solutions are strong, so since it looked like the book was already assuming existence, I thought this would be enough. The (strong) existence result you want is Theorem 5.2.9 in the same book. $\endgroup$ Jun 28, 2022 at 15:35
  • $\begingroup$ So, do you mean that our condition can guarantee the strong uniqueness, thus by another theorem, the strong uniqueness implies the strong solution, i.e., $X_t$ is adapted to such a filtration? $\endgroup$ Jun 29, 2022 at 8:20
  • $\begingroup$ @TalkingPuppet The Lipschitz condition implies strong uniqueness, so assuming there is a solution $X_t$, the other theorem (5.3.20-5.3.23 in Karatzas and Shreve, or 9.1.7 in Revuz and Yor) implies that solution is strong and hence adapted to the filtration. Now, we also have the linear growth condition, so Theorem 5.2.9 tells us there is a strong solution, and in particular that means the solution is adapted to the filtration. $\endgroup$ Jun 29, 2022 at 15:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .