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I am reading the proof that the unique solution of the SDE is a Markov process from René Schilling's Brownian Motion. In the below proof, I have one question.

When we define a functional $\Phi$ so that $$\Phi(X_s^{0,\xi} (\omega),s,t,\omega) = X_s^{0,\xi}(\omega) + \int_s^t b(r,X_r^{0,\xi})dr + \int_s^t \sigma(r,X_r^{0,\xi})dB_r,$$ why is this functional $\Phi$ measurable with respect to $\sigma(B_r - B_s:r \ge s)$?

I think we can see $\Phi$ as a sum of three random processes. First one is just $X_s^{0,\xi}$. Second and third are limits of sums, so we can look at each sum instead. Second sum is a Riemann sum, hence I see that it is $\mathscr{F}_t$ measurable by the progressive measurability of $X_r$.

The third sum is of the form $\sum_j \sigma(t_{j-1},X_{t_{j-1}}^{0,\xi})(B_j - B_{j-1})$. So this is a sum of multiples of $\mathscr{F}_{t_{j-1}}$ and $(B_j - B_{j-1})$ measurable processes.

But I cannot figure out from this why $\Phi$ would be $\sigma(B_r - B_s: r \ge s)$ measurable.

I could not figure out this part and I would greatly appreciate some help.

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The (A.5) mentioned in the proof is the following.

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nomadicmathematician
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1 Answers1

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I think that you may misinterpret the meaning of the flow $\Phi$. Let me try to give some details. The author defines a mapping $\Phi:\mathbb{R}^n\times\Delta\times\Omega\to\mathbb{R}^n$ where $\Delta=\{(s,t)\in[0,\infty)^2:\,s\leq t\}$ as the unique strong solution to the SDE $$ \Phi(x,s,t)=x+\int_s^t b(r,\Phi(x,s,r))\,dr+\int_s^t\sigma(r,\Phi(x,s,t))\,dB_r. $$ For each fixed $x,s,t$, $\Phi(x,s,t)$ is certainly measurable with respect to $\sigma(B_r-B_s,r\in[s,t])$. However, you are absolutely right that this is no longer the case for $\Phi(X_s^{0,\xi},s,t)$, but he doesn't use this. Instead, he conditions on $\mathcal{F}_s$, with respect to which $X_s^{0,\xi}$ is measurable and can then work with the deterministic flow $\Phi(x,s,t)\Big|_{x=X_s^{0,\xi}}$.

user427574
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  • I have a few questions: why is $\Phi(x,s,t)$ measurable with respect to $\sigma(B_r - B_s, r\in [s,t])$? I added equation (A.5) in the text. So to apply this equation in the proof, we have to show that $X_s^{0,\xi}(\cdot): \Omega \to \mathbb{R}^n$ is $\mathscr{F}_s$ measurable, and the last component function in $\Phi$ is $\sigma(B_r - B_s:r\in [s,t])$ measurable. I do not know how we can explicitly show this from the given form of $\Phi$. How are the $\Phi(x,s,r,\omega)$'s defined so that given state $x$ at time $s$ and $\omega$, we know the value at the $r$th time? – nomadicmathematician Aug 10 '20 at 15:48
  • Well, the fact that $X_s^{0,\xi}$ is $\mathcal{F}_s$-measurable follows immediately from the definition of a solution to an SDE. Similarly, we require $\Phi$ to be adapted. Hence, you only have to show that $\int_s^t f(r),dB_r$ is $\sigma(B_r-B_s,r\in[s,t])$-measurable for adapted $f$. This follows easily by a monotone class argument. – user427574 Aug 10 '20 at 18:03
  • But we also have $\int_s^t b(r,\Phi(x,s,r))dr$ which is $\mathscr{F}_t$ measurable. Why does this not affect things? – nomadicmathematician Aug 10 '20 at 20:14
  • The only randomness in $\Phi$ comes in from the stochastic integral. – user427574 Aug 11 '20 at 08:38
  • $X_t^{0,\xi}(\omega) = X_s^{0,\xi}(\omega) + \int_s^t b(r,X_r^{0,\xi}(\omega)dr + \int_s^t \sigma(r,X_r^{0,\xi}(\omega))dB_r(\omega)$, so there is randomness associated with $X_r^{0,\xi}$. – nomadicmathematician Aug 11 '20 at 08:57
  • You're still having the same error in your thinking. The flow $\Phi$ is defined as in my answer with DETERMINISTIC initial condition. – user427574 Aug 11 '20 at 11:27