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Let $W_t$ be a Brownian motion. I wish to show that the stopping time $\tau \equiv \inf\left\{t \ge 0 : W_t >0\right\} = 0$ almost surely.

We have $$\{\tau = 0\} = \bigcap_{k=1}^\infty \quad\bigcup_{0 \leq t < \frac{1}{k}, t \in \mathbb{Q}} \{W_t > 0\} = \bigcap_{k=m}^\infty \quad \underbrace{\bigcup_{0 \leq t < \frac{1}{k}, t \in \mathbb{Q}} \{W_t > 0\}}_{\in \mathcal{F}_{1/m}^0 \forall m \in \mathbb{N}} \in \bigcap_{m=1}^\infty \mathcal{F}_{1/m}^0 = \mathcal{F}_0^+ $$

Thus by Blumenthal's zero one law, we have $P(\tau = 0) \in \{0, 1\}$ so it suffices to show that $P(\tau = 0) > 0$ but I find this impossible. Please help if you can.

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2 Answers 2

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Suppose that $\mathbb{P}(\tau=0)=0$, then $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0\::\: W_t \leq 0)=1.$$ Since $(-W_t)_{t \geq 0}$ is also a Brownian motion, this implies $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0\::\: W_t \geq 0)=1.$$ Hence, $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0, t \in \mathbb{Q}\::\: W_t \geq 0)=1.$$ As $\mathbb{P}(W_t=0)=0$ for each $t \geq 0$, this gives $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0, t \in \mathbb{Q}\::\: W_t > 0)=1,$$

i.e. $\mathbb{P}(\tau=0)=1$, which clearly contradicts our assumption.

Hence, $\mathbb{P}(\tau=0)>0$, and by Blumenthal's 0-1-law we conclude that $\mathbb{P}(\tau=0)=1$.

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  • $\begingroup$ I like this solution so I'll mark it as correct but I'm curious if the following also works: $P(\tau = 0) = P(\cap_{n \in \mathbb{N}}\{\tau \leq 1/n\}) = \lim_{n \rightarrow \infty} P(\{\tau \leq 1/n\}) \ge \limsup P(W_{1/n} \ge 1/n)$ $= P(N(0,1) \ge 0) = 1/2$, where the second equality holds due to nested decreasing subsets. Thanks for your solution! $\endgroup$
    – qp212223
    Apr 6, 2020 at 7:10
  • $\begingroup$ @qp212223 Should be fine. In any case, it's really the symmetry which makes this work. $\endgroup$
    – saz
    Apr 6, 2020 at 8:18
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Because $\{W_t>0\}\subset\{\tau\le t\}$, you have $1/2\le\Bbb P[\tau\le t]$, for each $t>0$. It follows that $\Bbb P[\tau=0]\ge 1/2$.

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