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?
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.