This is a claim I found from Rene Schilling's Brownian Motion and Stochastic Calculus, but I don't know how to prove it.
Let $\tau= \tau(B_\bullet)$ be a stopping time which can be expressed as a functional of a Brownian path, e.g. a first hitting time, and assume that $\sigma$ is a further stopping time such that $\sigma \le \tau $ a.s. Denote by $\tau' = \tau(B_{\bullet + \sigma})$ the stopping time $\tau$ for the shifted process $B_{\bullet+ \sigma}$, which is the remaining time, counting from $\sigma$, until the event described by $\tau $ happens.
Set $W_\bullet:= B_{\bullet + \sigma} – B_\sigma$; then $\tau' = \tau(W_\bullet+ B_\sigma),$ and the functionals $u(B_\tau)$ and $u(W_{\tau'} + B_\sigma)$ have the same distribution, where $u : \mathbf{C}[0,\infty) \to \mathbb{R}$ be a bounded $\mathcal{B}(\mathbf{C})/\mathcal{B}(\mathbb{R})$ measurable functional.
The only property I have are the strong Markov properties in the following forms: $\bullet \;W_t:= B_{\sigma +t} – B_\sigma$, is again a Brownian motion which is independent of $\mathcal{F}_{\sigma +}$ for an a.s. finite stopping time $\sigma$.
$\bullet \;$ Let $\sigma$ be a stopping time. For all bounded Borel measurable $u \in \mathscr{B}_b(\mathbb{R}^d)$ and $P$ almost all $\omega \in \{\sigma < \infty\}$
$$E[u(B_{t+\sigma})|\mathscr{F}_{\sigma+}](\omega)=E[u(B_t+x)]|_{x = B_\sigma(\omega)}=E^{B_\sigma(\omega)}u(B_t).$$
$\bullet$ For all bounded $\mathscr{B}(C)/\mathscr{B}(\mathbb{R})$ measurable functionals $\Psi : C[0,\infty) \to \mathbb{R}$ which may depend on a whole Brownian path and $P$ almost all $\omega \in \{\sigma < \infty\}$ this becomes
$$E[\Psi(B_{\bullet + \sigma})|\mathscr{F}_{\sigma +}]=E[\Psi(B_\bullet +x)]|_{x=B_\sigma}=E^{B_\sigma}[\Psi(B_\bullet)].$$
How can I use these properties to show that $u(B_\tau)$ and $u(W_{\tau'} + B_\sigma)$ have the same distribution (conditional on $\mathcal{F}_{\sigma +}$ is I think what the author means)? I have been stuck with this for some time and would greatly appreciate some help. Moreover, how do these facts imply the corollary that
$$E[u(B_{\tau}) | \mathscr{F}_{\sigma +}](\omega)=E[u(W_{\tau'}+x)]|_{x=B_\sigma (\omega)}=E^{B_{\sigma}(\omega)} u(W_{\tau'})$$ holds for all $u\in \mathscr{B}_b(\mathbb{R}^d$) and P almost all $\omega \in \{\tau<\infty\}.$
Best Answer
I only found the book R.L. Schilling, L. Partzsch Brownian Motion which has the claim that $u(B_\tau)$ and $u(W_{\tau'}+B_\sigma)$ have the same distribution at the bottom of p. 63. They don't say that $u$ is defined on $\mathbf{C}[0,+\infty)$. It is obviously rather a measurable function on $\mathbb R^d\,.$
Claim. The functionals $u(B_\tau)$ and $u(W_{\tau'}+B_\sigma)$ do not only have the same distribution, they are almost surely equal.
Proof. By definition of $W_\bullet$ we have $W_{\tau'}+B_\sigma=B_{\tau'+\sigma}\,.$ By definition of $\tau'$ we have, as jacobdt already pointed out, $\tau=\tau'+\sigma\,\,$ a.s. and therefore $B_\tau=B_{\tau'+\sigma}=W_{\tau'}+B_\sigma\,\,$ a.s. and in particular $u(B_\tau)=u(W_{\tau'}+B_\sigma)\,\,$ a.s. $$\tag*{$\Box$} \quad $$
Edit.
In their Corollary 6.7 the authors formulate more clearly that $\tau=\tau(B_\bullet)$ should be a first hitting time and $\sigma$ a stopping time with $\sigma\le\tau\,\,$ a.s. This makes the crucial relationship $$\tag{1} \tau'=\tau-\sigma\quad\text{ a.s. } $$ straightforward to show. See below.
On p. 63 Schilling and Partzsch only wrote that $\tau=\tau(B_\bullet)$ needs to be a functional of the Brownian path, e.g. a first hitting time. I think this is not enough but they don't need that full generality for the Corollary as stated.
Proof of (1).
Since $\tau=\tau(B_\bullet)$ is a first hitting time there is a Borel set $A$ in $\mathbb R^d$ s.t. $$\tag{2} \tau=\inf\{t>0:B_t\in A\}\,. $$ Since $\sigma\le\tau\,\,$ a.s. it follows that $$\tag{3} B_\sigma\notin A\quad\text{ a.s. on }\quad\{\sigma<\tau\}\,. $$ By definition, $\tau'=\tau(B_{\bullet+\sigma})\,,$ i.e., $$\tag{4} \tau'=\inf\{t>0:B_{t+\sigma}\in A\}\,. $$ By a variable transformation this can clearly be written as $$\tag{5} \tau'=\underbrace{\inf\{t'>\sigma:B_{t'}\in A\}}_{\textstyle=:\,\tau^*}-\sigma\,. $$ It is clear that $\tau^*\ge\tau$ a.s. because the infimum in $\tau^*$ is taken over a smaller range of $t'$ values. However, from (3) it follows that $$\tag{6} B_t\notin A\quad\text{ a.s. on }\quad\{t\le \sigma<\tau\}\,. $$ This means that in the definition (2) of $\tau$ we are allowed to take the range of $t$ values over the smaller $t>\sigma$ as well, at least for almost all $\omega$ in $\{\sigma<\tau\}\,.$ In other words, $$\tag{7} \tau^*=\tau\quad\text{ a.s. on }\quad\{\sigma<\tau\}\,. $$ On $\{\sigma=\tau\}$ it follows from (4) that $\tau'=0\,.$ This finishes the proof of (1). $$\tag*{$\Box$} \quad $$