Let $(B_t)_{t \geq 0}$ be a Brownian motion and $T$ a stopping time such that $T < \infty$ a.s. Then the Strong Markov property (SMP) asserts that under the probability measure $\mathbb{P}( \, \cdot \, \vert \, T < \infty)$, the process $$ B_{t}^{(T)} = \mathbb{1}_{\{T < \infty\}} (B_{T+t} - B_T) $$ is a Brownian motion independent of $$ \mathcal{F}_{T} = \{A \in \mathcal{F}_{\infty} \colon A \cap \{T \leq t\} \in \mathcal{F}_t \text{ for all } t \geq 0\}, $$ where $\mathcal{F}_{\infty} = \sigma(B_s \colon s \geq 0)$ and $\mathcal{F}_t = \sigma(B_s \colon 0 \leq s \leq t)$.

Now I have a question to a remark I found in a book: By the (SMP) $$ T = \sup \{s \in [0,1] \colon B_s = 0\} $$ is not a stopping time. Why is this a consequence of the (SMP), how can we formally prove this?


1 Answer 1


Choose $r \in (0,1)$ such that $\mathbb{P}(T \leq r) \geq 1/2$. Then

$$W_t(\omega) := B_{T+t}(\omega)-\underbrace{B_T(\omega)}_{0} = B_{T+t}(\omega) \neq 0 \qquad \text{for all $t \in [0,1-r]$, $\omega \in \{T \leq r\}$}.$$

This means that $(W_t)_{t \geq 0}$ cannot be a Brownian motion (see below). By the strong Markov property, this implies that $T$ is not a stopping time.

Lemma Let $(B_t)_{t \geq 0}$ be a Brownian motion. With probability $1$ there exists a sequence $(a_n)_{n \in \mathbb{N}} = (a_n(\omega))_{n \in \mathbb{N}}$ such that $B_{a_n}(\omega)=0$ and $a_n \to 0$ as $n \to \infty$.

  • $\begingroup$ Thank you for your answer, but I have some additional questions: 1) Why do we know that such $r$ exists in $(0,1)$? $\endgroup$ Commented Aug 14, 2017 at 12:49
  • $\begingroup$ @user63841219 Ah sorry, that was stupid... I got your definition of TT wrong. I'm sorry. I've now edited my answer. $\endgroup$
    – saz
    Commented Aug 14, 2017 at 14:43
  • $\begingroup$ Thank you, now its clearer. But still I don't see why such $r$ exists and why we need it. $\endgroup$ Commented Aug 14, 2017 at 15:14
  • $\begingroup$ @user63841219 Clearly, $T \leq 1$; on the other hand, it cannot hold that $T=1$ almost surely (since this would imply $B_T=B_1 = 0$ almost surely), and therefore there has to exist some $r \in (0,1)$such that $\mathbb{P}(T \leq r) \geq 1/2$. Why we need it: Because we want that $$W_t \neq 0 \quad \text{on some deterministic interval}$$ with probability $>0$. $\endgroup$
    – saz
    Commented Aug 14, 2017 at 15:18

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