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

Answer 2

So $\mathbf{P}\left(\sup_{0\leq s\leq n}W_{s}<X\right)=\mathbf{P}\left(\left|W_{n}\right|<X\right)$, because it can be shown, that $$\sup_{0\leq s\leq n}W_{s}\overset{d}{=}\left|W_{n}\right|.$$ I know I should write $\sup_{0\leq s\leq n}\left|W_{s}\right|$, but I don't know any statements about the distribution of $\sup_{0\leq s\leq n}\left|W_{s}\right|$, and on the other hand the probability of this event is going to be $0$ so $\sup_{0\leq s\leq n}W_{s}$ is going to reach the $X$ level. If it reaches the $X$ exponential distributed level, then $\sup_{0\leq s\leq n}\left|W_{s}\right|$ has to do the same. $$\begin{align*} \mathbf{P}\left(\left|W_{n}\right|<X\right) & =\mathbb{E}\left(\chi_{\left\{ \left|W_{n}\right|<X\right\} }\right)=\int_{0}^{\infty}\int_{-x}^{x}\lambda e^{-\lambda x}\frac{1}{\sqrt{2\pi n}}e^{-\frac{y^{2}}{2n}}dydx=\int_{0}^{\infty}\lambda e^{-\lambda x}\int_{-x}^{x}\frac{1}{\sqrt{2\pi n}}e^{-\frac{y^{2}}{2n}}dydx=\\ & =\int_{0}^{\infty}\lambda e^{-\lambda x}\mathbf{P}\left(-x<N\left(0,n\right)<x\right)dx=\int_{0}^{\infty}\lambda e^{-\lambda x}\mathbf{P}\left(\frac{-x}{\sqrt{n}}<N\left(0,1\right)<\frac{x}{\sqrt{n}}\right)dx=\\ & =\int_{0}^{\infty}\lambda e^{-\lambda x}\left(\varPhi\left(\frac{x}{\sqrt{n}}\right)-\varPhi\left(\frac{-x}{\sqrt{n}}\right)\right)dx \end{align*}$$ Taking the $\lim_{n\rightarrow\infty}$ limit, we can use the dominated convergence theorem, and $\lim_{n\rightarrow\infty}\varPhi\left(\frac{x}{\sqrt{n}}\right)=\varPhi\left(\lim_{n\rightarrow\infty}\frac{x}{\sqrt{n}}\right)$, because $\varPhi\left(.\right)$ is bounded, so we get $$\int_{0}^{\infty}\lambda e^{-\lambda x}\varPhi\left(0\right)dx-\int_{0}^{\infty}\lambda e^{-\lambda x}\varPhi\left(-0\right)dx=\frac{1}{2}-\frac{1}{2}=0.$$ We have known the joint density function, because if $\mathcal{F}$ is the natural filtration of $W$, then $X\in\mathcal{F}_{0}\perp W$, where $\perp$ denotes the independence.

I think we can say similar for the Cauchy distributed random variable: $$\mathbf{P}\left(\sup_{0\leq s\leq n}W_{s}<X\right)=?$$ Again, I should have written $\left\{ \sup_{0\leq s\leq n}\left|W_{s}\right|<X\right\}$, but the same reason holds as before:$$\mathbf{P}\left(\sup_{0\leq s\leq n}W_{s}<X\right)=\mathbf{P}\left(\left|W_{n}\right|<X\right)=\int_{0}^{\infty}\int_{-x}^{x}\frac{1}{\pi}\frac{1}{1+x^{2}}\frac{1}{\sqrt{2\pi n}}e^{-\frac{y^{2}}{2n}}dydx=\int_{0}^{\infty}\frac{1}{\pi}\frac{1}{1+x^{2}}\left(\varPhi\left(\frac{x}{\sqrt{n}}\right)-\varPhi\left(\frac{-x}{\sqrt{n}}\right)\right)dx.$$ Taking again the limit and using the dominated convergence theorem $$\lim_{n\rightarrow\infty}\int_{0}^{\infty}\frac{1}{\pi}\frac{1}{1+x^{2}}\left(\varPhi\left(\frac{x}{\sqrt{n}}\right)-\varPhi\left(\frac{-x}{\sqrt{n}}\right)\right)dx=\int_{0}^{\infty}\frac{1}{\pi}\frac{1}{1+x^{2}}\left(\frac{1}{2}-\frac{1}{2}\right)dx=0.$$ I can't prove, but I think if a Wiener process visits every level with $1$ probability, than it doesn't matter if we examine a random or a deterministic level. We do not have any plus criterium. Correct me if I am wrong. I think $\mathbb{E}\left(\tau\right)=\infty$ in both cases: if the level is deterministic or random. Therefore we can't use the Wald identity to determine $\mathbb{E}\left(W_{\tau}\cdot\chi_{\left\{ \tau<\infty\right\} }\right)$. I wanted to use Doob's optional stopping theorem, but I can't find any $c$ where $\tau\leq c$ almost surely. I would say $$\mathbb{E}\left(W_{\tau}\cdot\chi_{\left\{ \tau<\infty\right\} }\right)=\mathbb{E}\left(X\right)=\frac{1}{\lambda}$$ using $\tau$'sdefinition at the exponential case, and $$\mathbb{E}\left(W_{\tau}\cdot\chi_{\left\{ \tau<\infty\right\} }\right)=\frac{1}{2}\cdot0+\frac{1}{2}\mathbb{E}\left(X\cdot\chi_{\left\{ X\geq0\right\} }\right)=\infty,$$ because "half of the time" $X<0$, then $\tau=0$. So $\tau=0$ with $\frac{1}{2}$ probability. $\mathbb{E}\left(\tau\right)=\frac{1}{2}\cdot0+\frac{1}{2}\cdot\infty=\infty$.