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Let $(B_t)$ be a standard Brownian motion and $\{ \mathcal{F}_t \}$ be the filtration generated by the Brownian motion.

For a stopping time $\tau$, we know that $\{B_{\tau + t} - B_{\tau}\}_{t \geq 0}$ is a Brownian motion independent of $\{ \mathcal{F}^{+}_{\tau} \}$.

For a fixed $a>0$, let $\tau'$ be defined by $$ \tau' := \inf \{ t \geq 0 : B_{\tau + t} - B_{\tau} = a \}.$$

I don't understand why $\tau'$ is also independent of $\{ \mathcal{F}^{+}_{\tau} \}$. Any ideas?

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    $\begingroup$ Can you show that for a fixed $s$, the event $ \tau' \lt s$ is independent of $\mathcal F_\tau^+$? $\endgroup$ – Davide Giraudo Dec 23 '14 at 16:01
  • $\begingroup$ @DavideGiraudo But the event $\{ \tau' <s \} = \bigcup_{t \in [0,s)} \{B_{\tau +s} - B_{\tau} =a \}$, which is an uncountable union. $\endgroup$ – erik Dec 23 '14 at 16:04
  • $\begingroup$ Since $\{a\}$ is a closed set, I don't know how to write this uncountable union as a countable union using rationals. $\endgroup$ – erik Dec 23 '14 at 16:05
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Can I argue as follows??

Let $\tilde{B}_t:= B_{\tau + t} - B_{\tau}$.

Let $ A = \{ \omega \in \Omega: t \mapsto \tilde{B}_t (\omega) \text{ is continuous } \}. $ Clearly, $\mathbb{P} (A) =1.$

Then,

$$ A \cap \{ \tau' \geq s \} = A \cap \bigcap_{0 < t <s } \{ \tilde{B}_t <a\} = A \cap \bigcap_{0 < q <s , q \in \mathbb{Q}} \{ \tilde{B}_q <a\}. $$

Hence, for any $S \in \mathcal{F}^{+}_{\tau}$,

$$ \mathbb{P} (\{\tau' \geq s \} \cap S) = \mathbb{P} \bigg( \bigcap_{0 < q <s , q \in \mathbb{Q}} \{ \tilde{B}_q <a\} \cap S \bigg) = \mathbb{P} \bigg( \bigcap_{0 < q <s , q \in \mathbb{Q}} \{ \tilde{B}_q <a\} \bigg) \mathbb{P}(S) = \mathbb{P} \{\tau' \geq s \} \mathbb{P}(S). $$ Hence, $\tau'$ and $\{ \mathcal{F}^{+}_{\tau} \}$ are independent.

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