# Using the Strong Markov property to show that a Brownian motion depending on a stopping time is identically distributed with a shifted BM

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 $$\omega \in \{\tau<\infty\}.$$

• Maybe I have misunderstood something, but don't you have $\tau=\sigma+\tau'$ a.s.? Commented Dec 20, 2022 at 13:15
• @jakobdt I agree but can't think of a rigorous argument. Commented Dec 20, 2022 at 19:34

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). $$\tag*{\Box} \quad$$

• Can you explain why $\tau = \tau' + \sigma$ ? Consider for $n \in \mathbb{N}$ the constant stopping times $n = \tau \geq \sigma = n$. Then isn't $\tau' = \tau = n$? Commented Dec 22, 2022 at 13:11
• Good point. See edit. Commented Dec 22, 2022 at 15:40
• @KurtG. You're using a different edition to mine but they are essentially the same. I've been thinking of a rigorous way to show the obvious identity $\tau=\tau' + \sigma$. Do you have any ideas? Commented Dec 22, 2022 at 21:42
• @KurtG. I don't think they are almost surely equal I feel that the author would have stated otherwise if so. The proof may involve the shift operator but I have no more clue at the moment. Commented Dec 24, 2022 at 6:59
• @nomadicmathematician If they are not a.s. equal what are they in your opinion? Please find the flaw -if any- in my line of reasoning. The fact that some prefer a shift operator should be irrelevant. And BTW, who are "they" exactly? The stopping times or the "functionals"? Commented Dec 24, 2022 at 8:19