# Independence of Brownian motion-related stopping times

Let $(B_t,\mathcal{F}_t)_{t \geq 0}$ a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{P})$. For $a \in \mathbb{R}$ define a stopping time $\tau_a$ by $$\tau_a := \tau(a) := \inf\{t \geq 0; B_t = a \} \qquad \qquad (\inf \emptyset := \infty)$$ For $w \in \Omega$ and $b>a$ such that $\tau_b(\omega) = \tau_a(\omega)=\infty$ we define $\tau_b(\omega)-\tau_a(\omega) := 0$ and $B_{\tau_a}(\omega)=B_{\tau_b}(\omega):=0$.

One can show that the random variables $\tau(a_n)-\tau(a_{n-1}),\ldots,\tau(a_2)-\tau(a_1)$ are independent for $0<a_1 < \ldots < a_n$. The idea is to prove that $\tau(a_n)-\tau(a_{n-1})$ is $\mathcal{F}_{\geq \tau(a_{n-1})}$-measurable where $$\mathcal{F}_{\geq \tau(a_{n-1})} := \sigma(B_{\tau(a_{n-1})+s}-B_{\tau(a_{n-1})}; s \geq 0)$$

So here is the proof: \begin{align} [\tau(a_n)-\tau(a_{n-1})>t] &= [\tau(a_n)-\tau(a_{n-1})>t, \tau(a_{n-1}) < \infty] \\ &= \left[ \sup_{s \in \mathbb{Q}, 0 \leq s \leq t} (B_{\tau(a_{n-1})+s}-B_{\tau(a_{n-1})}) < a_n - a_{n-1}, \tau(a_{n-1})< \infty \right] \\ &= \left[0< \sup_{s \in \mathbb{Q}, 0 \leq s \leq t} (B_{\tau(a_{n-1})+s}-B_{\tau(a_{n-1})}) < a_n - a_{n-1}\right] \tag{1} \end{align} for all $t \geq 0$.

I have some problems with the last equality.

• There is no doubt that "$\subseteq$" holds. But what about "$\supseteq$"? For $\omega \in [\tau(a_{n-1})=\infty]$ I don't understand how the supremum $$\sup_{s \in \mathbb{Q}, 0 \leq s \leq t} (B_{\tau(a_{n-1})+s}-B_{\tau(a_{n-1})})$$ is defined. Is it by definition equal to zero?

Edit: I already figured out this one. Since they define $B_{\tau(a_{n-1})}(\omega) := 0$ for $\omega \in [\tau(a_{n-1})=\infty]$ (i.e. "$B_{\infty}=0$"), we also have $$B_{\tau(a_{n-1})+s}(\omega)=B_{\infty+s}(\omega)=B_{\infty}(\omega)=0 \qquad (0 \leq s \leq t)$$ for $\omega \in [\tau(a_{n-1})=\infty]$. And this means, that the supremum is equal to zero for all $\omega \in [\tau(a_{n-1})=\infty]$.

• For $t=0$: Since $\tau(a_n)>\tau(a_{n-1})$ almost surely, the left-hand is an event of probability one, whereas the right-hand side in $(1)$ is equal to the empty set since $$\sup_{s \in \mathbb{Q}, 0 \leq s \leq 0} (B_{\tau(a_{n-1})+s}-B_{\tau(a_{n-1})})=B_{\tau(a_{n-1})}-B_{\tau(a_{n-1})}=0$$

(Literature: "Introduction to the Theory of Random Processes" - N. V. Krylov)

Notation: $$[X \in B] := X^{-1}(B) := \{w \in \Omega; X(\omega) \in B\}$$

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