Brownian motion and hitting time

Question

I need some help with the geometrical aspect of a Brownian motion and his hitting time. If $W$ a Brownian motion and $\tau=inf { t\ge 0 \text{ st } W_t > a } $ with $a \ge 0 $. Can someone please draw the process $W_{\tau}$. Is this a stopped process meanings that after $a$ the process is stopped for ever or the process is stopped (ie: $W_s=a$ for $s<t$ such that $W_s< a$) but then if $W_t > a $ the process is "back to normal".

Can we say that $W_t=W_{min(t,\tau)}+W_{min(t,\sigma)}$ with $\sigma= inf { t\ge 0, W_t \ge a}$ ???

I don't know if it's clear.

Thanks

Answer 1

Note that $W_{0}(\omega)=0$ a.s., so $W_{\min\{t,\sigma(\omega)\}}(\omega)=W_{\sigma(\omega)}(\omega)=W_{0}(\omega)=0$, since

$$\sigma(\omega)=\inf\{t\geq 0\colon W_{t}(\omega)\leq a\},$$

for $a\geq 0$, therefore we have $\sigma(\omega)=0$ a.s..

Note that $W_{t}(\omega)\neq W_{\min\{t,\tau(\omega)\}}(\omega)$, since

$$\tau(\omega)=\inf\{t\geq 0\colon W_{t}(\omega)> a\},$$

for $a\geq 0$ and $\tau(\omega)$ is a random variable.

It may happen that $\tau(\omega)=3$, so for example $W_{10}(\omega)\neq W_{\min\{10,3\}}(\omega)$.

Answer 2

I think you still misunderstand how stochastic processes, stopping times and random variables are defined, and I urge you to go back to the definitions using an underlying sample space $\Omega$ and a mapping $X : [0, T] \times \Omega \rightarrow \mathbb{R}$.

That being said, I think I understand what you are asking (in terms of what it "looks like"), because you have a decomposition in mind that might look like this: The decomposition corresponds to $$W_t = \color{blue}{W_{t\wedge \tau}} + \color{red}{1_{\{t \geq \tau\}}(W_t - a)},$$ where $a \wedge b = \min(a,b)$ by definition, and $$\tau = \inf\{t \geq 0: W_t \geq a\}$$ and $1_A$ is the indicator function of the event $A$.

(Note: in the picture, the red curve is shifted to "continue" the process $W_t$ after time $t$, though $G_t = \color{red}{1_{\{t \geq \tau\}}(W_t - a)}$ is continuous at $t = \tau$.)

Yes, it is true that $W_{t\wedge \tau(\omega)}(\omega) \geq a$ for all $t \geq \tau(\omega)$, for each $\omega \in \Omega$. However, you asked:

Can someone please draw the process $W_\tau$

No, because $W_\tau$ is not a process. For each fixed $\omega \in \Omega$, it is a single point.