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This question follows from the previous post Question on the martingale representation theorem which has not been answered. I consider thus a particular case. Let $(B_t)_{t\ge 0}$ be a standard Brownian motion and $\tau:=\inf\{t\ge 0: |B_t|=1\}$. Then $(B_{t\wedge \tau})_{t\ge 0}$ is a Markov martingale. Therefore, if my claim in Question on the martingale representation theorem is correct, then there exists some Brownian motion $(W_t)_{t\ge 0}$ s.t.

$$B_{t\wedge \tau}=\int_0^t\sigma(s, B_{s\wedge \tau})dW_s,\quad \forall t\ge 0.$$

Could we identify $\sigma$ for this case?

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  • $\begingroup$ I guess one can simply set $B_t = W_t$, $\sigma(s,x)=1$ if $|x|<1$ and $\sigma(s,x)=0$ if $|x|=1$? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 31, 2021 at 16:59
  • $\begingroup$ @MateuszKwaśnicki Your observation is amazing! It's absolutely right. What I thought is to compare the quadratic variation, which yields $t\wedge \tau=\int_0^t |\sigma(s,B_{s\wedge \tau})|^2ds$. But then I don't know how to get $\sigma$... $\endgroup$
    – GJC20
    Commented Jul 31, 2021 at 18:25
  • $\begingroup$ @MateuszKwaśnicki So, for this case, the existence of $\sigma$ is ensured for this particular case. As the general case, see my previous post mathoverflow.net/questions/398025/…, do you believe it is still true? $\endgroup$
    – GJC20
    Commented Jul 31, 2021 at 18:27

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