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Let $(B_t)$ a Brownian motion and $\sigma $ a stopping time finite a.s.. I want to prove that $W_t=B_{\sigma +t}-B_\sigma $ is a Brownian motion.


The way to prove it is first to prove that for all $0\leq s<t$, $$\mathbb E[e^{i\xi\cdot (B_{\sigma +t}-B_{\sigma +s}})]=\mathbb E[e^{i\xi\cdot B_{t-s}}].$$

And what they do is they prove that for all $F\in \mathcal F_{\sigma ^+}$, $$\mathbb E[e^{i\xi\cdot (B_{\sigma +t}-B_{\sigma +s}}\boldsymbol 1_F]=\mathbb E[e^{i\xi \cdot B_{t-s}}]\mathbb P(F),$$ but I don't understand why the fact to introduce $F$ is relevant (I have the impression that if we directly take $F=\Omega $, and thus don't necessary introduce $F\in \mathcal F_{\sigma ^+}$, the proof would be the same). I put the proof here :

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  • $\begingroup$ You can choose $F=\Omega$ if you only want to prove that $(B_{t+\sigma}-B_{\sigma})_{t \geq 0}$ is a Brownian motion. The authors consider general $F \in \mathcal{F}_{\sigma+}$ to deduce that $(B_{t+\sigma}-B_{\sigma})_{t \geq 0}$ is independent from $\mathcal{F}_{\sigma+}$. $\endgroup$ – saz May 1 at 10:52
  • $\begingroup$ @saz: Thank a lot for your answer. Could you tell me which result says that $\mathbb E[e^{iX}\boldsymbol 1_{F}]=\mathbb E[e^{iY}]\mathbb P(F)$ for all $F\in \mathcal F$ implies that $X$ is independent of $\mathcal F$ ? $\endgroup$ – user657324 May 1 at 10:59
  • $\begingroup$ See Problem 9.8 in the book (the solution can be found on the author's webpage) or this closely related question $\endgroup$ – saz May 1 at 11:27
  • $\begingroup$ Problem 9.8 looks to be something else... @saz $\endgroup$ – user657324 May 1 at 12:05
  • $\begingroup$ Do you have 2nd edition or first? I have 2nd. In any case you find the proof in the question which I linked. $\endgroup$ – saz May 1 at 14:45

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