Let $W_t$ be a one dimensional brownian motion and define $\tau_c=\inf(t \geq 0 : W_t=c)$. We are supposed to show that $\tau_c$ is finite a.s., find it's distribution, and show that $E \tau_c = \infty$. Here is what I have so far.
For the distribution note that $P(W_t \geq c) = P(W_t \leq c | \tau_c \leq t)P(\tau_c \leq t)+P(W_t \geq c | \tau_c > t)$. By the definition of $\tau_c$ and continuity we have $P(W_t \geq c | \tau_c > t)=0$. From symmetry we get $P(W_t \leq c | \tau_c \leq t)=\frac{1}{2}$. Hence $P(\tau_c \leq t)=2\left( 1-G(\frac{c}{\sqrt{t}}) \right)$ where $G$ is the normal gaussian distribution. Moreover, we have that $P(A_n)=P(n< \tau_c \leq n+1)=2(G(\frac{c}{\sqrt{n}})-G(\frac{c}{\sqrt{n+1}})$. Since $G\left( \frac{c}{\sqrt{t}} \right)\to 0$ as $t \to \infty$, we have that $P(\cup A_n)=\sum P(A_n) = 2(1-\lim_{t \to \infty} G(\frac{c}{\sqrt{t}})) = 1$. Hence $\tau_c$ is bounded a.s.
I am not sure how to show that the expectation of $\tau_c$ is infinite however. Any help would be appreciated.
Best Answer
Suppose $\mathbb{E}[\tau_c] < \infty$. By the optional stopping theorem, $(W_{t \wedge \tau_c})_{t \ge 0}$ is a martingale so $\mathbb{E}[W_{t \wedge \tau_c}] = \mathbb{E}[W_0] = 0$ for all $t$. Additionally, $\mathbb{E}[W_{t \wedge \tau_c}^2] = \mathbb{E}[t \wedge \tau_c] \le \mathbb{E}[\tau_c] < \infty$ so $(W_{t \wedge \tau_c})$ is bounded in $L^2$ and hence uniformly integrable. Because $\tau_c < \infty$ a.s. we have $\lim_{t \rightarrow \infty} W_{t \wedge \tau_c} = W_{\tau_c} = c$ a.s. Therefore we have $$c = \mathbb{E}[\lim_{t \rightarrow \infty} W_{t \wedge \tau_c}] = \lim_{t \rightarrow \infty} \mathbb{E}[W_{t \wedge \tau_c}] = 0,$$
a contradiction! Hence we conclude $\mathbb{E}[\tau_c] = \infty$.