Random level reaching of Wiener process

Question

The question:

$W$ denotes a Wiener process in filtration $\mathcal{F}$, where $X$ is an $\mathcal{F}_{0}$-measurable

a, Exponential

b, Cauchy

distributed random variable. Let us define $$\tau=\inf\left\{ t\geq0:\left|W_{t}\right|\geq X\right\} .$$
What is the probability, that $\tau$ is finite, so $\mathbf{P}\left(\tau<\infty\right)=?$ And what is $\mathbb{E}\left(\tau\right)=?$ and $\mathbb{E}\left(W_{\tau}\cdot\chi_{\left\{ \tau<\infty\right\} }\right)=?$, where $\chi_{\left\{ \tau<\infty\right\} }$ is the indicator variable of the $\left\{ \tau<\infty\right\}$ event?

Here are my thoughts so far…

I know a Wiener process visits every deterministic level, but I don't know if the same holds for random levels. Is there any plus boundary criterium? From the task I would say yes, there must be some criterium at least for the expected value for $X$, because why would the task give us two different random variables? I mean the exponential $X$ has finite expected value, but in the case of the Cauchy distribution we can't say the same. For $\mathbb{E}\left(\tau\right)$ I would say it is $\infty$, because a Wiener process visits every deterministic level with $1$ probability, but it costs $\infty$ time in expectation, so $\mathbb{E}\left(\tau\right)=\infty$. If the previous happens in the deterministic case, then it should holds for the random case as well, but I can't say any compelling reason.

For $\mathbb{E}\left(W_{\tau}\cdot\chi_{\left\{ \tau<\infty\right\} }\right)$ I wanted to use Wald's identity, but there $\mathbb{E}\left(\tau\right)$ must be finite, so I couldn't use it appropriately. I tried to use Doob's Optional stopping theorem from this site: https://en.wikipedia.org/wiki/Optional_stopping_theorem. Unfortunatelly, I didn't find the continuous-time version of the theorem so far, but as I know, the same holds for the continuous case. (Please, let me know, if I am wrong with this statement.) Using this theorem the answer should be $0$, but I am not so sure I can use this. Most of the times I have always used the words “bounded” and “finite” like they are synonims of each other. This example opened my eyes, that it is not necessary correct. I didn't find it on wikipedia, but as I know Doob's Optional stopping theorem is an “if and only if” statement. For example a Wiener process reaches every deterministic $a\neq0$ level, so if $\tau$ is the random level reaching time, then $\mathbf{P}\left(\tau<\infty\right)=1$, so we can say it is almost surely finite, but I can't say any upper boundary for that, therefore we can't use Doob's Optional stopping theorem because $\mathbb{E}\left(W_{\tau}\right)=a\neq W_{0}=0$. Is it a good train of thought? So if something is bounded or finite, then it doesn't necessary mean the same.

Answer 1

Hint

1. $$\mathbb P\{\tau=\infty \}=\lim_{n\to \infty }\mathbb P\{\tau>n\},$$ and $$\mathbb P\{\tau>n\}=\mathbb P\left\{\sup_{t\in [0,n]}|W_t|<X\right\}=\int_{0}^\infty \mathbb P\left\{\sup_{t\in [0,n]}|W_t|<x\right\}\mathbb P\{X\in \,\mathrm d x\}.$$

Now, I would be surprised that $\mathbb P\left\{\sup_{t\in [0,n]}|W_t|<x\right\}$ can be computed analytically.

2. $$\mathbb E[\tau]=\int_0^\infty \mathbb P\{\tau>t\}\,\mathrm d t.$$