I have the following exercice to do and I would like to know if what I did is correct please.
Consider $X_t$ a continuous stochastic process and $\tau$ the reaching time to the interval $[a,b]\subset[0,+\infty[$. If $\tau<\infty$ what can you say about $X_{\tau}(\omega)$ ? Compute $\mathbb{P}(\tau<\infty)$ when $X$ is a Brownian motion.
For the first question I would say that $X_{\tau}(\omega)\in[a,b]$ almost surely since $\tau$ is bounded !
For the second question, I will use the property of the Brownian motion that is the fact his law is uniquely determined by its covariance (in order to take a scaling Brownian motion when needed) and the $0-1$ law.
Since $X_t$ is a Brownian motion, it is continuous in $t$, thus it suffices to show that it is recurrent process : it pass through all values in $[0,+\infty[$. Thus we need to show that
$$ \mathbb{P}(\{\omega : \sup_{t\geq 0}, B_t(\omega) = +\infty\}) = 1 $$ That is for all $c>0$ $$ \mathbb{P}(\{\omega : \exists t\geq 0, B_t(\omega) > c\}) = 1 $$
Thus we have
$$ \mathbb{P}(\{\omega : \exists t\geq 0, B_t(\omega) > c\}) = \mathbb{P}(\{\omega :\forall n\in \mathbb{N}^{*}, \forall t\geq 0, \frac{1}{n} B_t(\omega) > \frac{1}{n}c\}) = \mathbb{P}(\{\omega : \forall n\in \mathbb{N}^{*}, \exists t\geq 0, \frac{1}{c n} B_{n^2c^2t}(\omega) > \frac{1}{n}\}) = \mathbb{P}(\{\omega : \forall n\in \mathbb{N}^{*}, \exists t\geq 0, Y_t > \frac{1}{n}\}) $$
Where $Y_t=\frac{1}{c n}B_{n^2c^2t}$ is also a Brownian motion. Thus we get
$$ \mathbb{P}(\{\omega : \forall n\in \mathbb{N}^{*}, \exists t\geq 0, Y_t > \frac{1}{n}\}) = \mathbb{P}(\cap_{n\in \mathbb{N}^{*}}\{\omega : \exists t\geq 0, Y_t > \frac{1}{n}\}) = \mathbb{P}(\{\omega : \exists t\geq 0, Y_t > 0\}) = 1 $$
Where the last equality follows from an application of the $0-1$ law to the set $A_n =\{ \omega : \sup_{0\leq t\leq\frac{1}{n}}B_t(\omega)>0\}$ in order to get that
$$ A = \{\omega : \forall \epsilon>0 \sup_{0\leq t\leq\epsilon}B_t(\omega)>0\} = \cap_{n\in\mathbb{N}^{*}}\{ \omega : \sup_{0\leq t\leq\frac{1}{n}}B_t(\omega)>0\}\in\mathcal{F}_{0+} $$
But for any $\epsilon>0$ we know that $\mathbb{P}(B_{\epsilon}>0)= \frac{1}{2}$. Thus $\mathbb{P}(A)=1$.
Then for all $c>0$ we have
$$ \mathbb{P}(\{\omega : \sup_{t\geq 0} B_t(\omega) > c\}) = \mathbb{P}(\{\omega : \exists t\geq 0, Y_t > 0\}) = 1 $$
And so $\mathbb{P}(\{\omega : \sup_{t\geq 0}, B_t(\omega) = +\infty\}) = 1$
From this we infer that the process will reach the interval for some $t\in[0,+\infty)$. Thus we conclude that
$$ \mathbb{P}(\tau<\infty) = 1 $$
I would like to know if this is correct and if there is a more efficient way to show this please ?
Thank you !
Snoop's suggestion in the comments is fast if you know the result for discrete-time random walks. I feel your approach can be streamlined a bit.
Clearly, it suffices to show that $P(\{\sup_t B_t= \infty\}) = 1$. To that end, consider the random variable $Z = \sup_{t \geq 0} B_t$ and notice that $Z$ and $cZ$ are identically distributed for any $c>0$, implying that the support of $Z$ is a subset of $\{0, \infty\}$. Fix a number $r > 0$ and observe that:
\begin{align*} P(Z=0) &\leq P(B_r \leq 0 \text{ and } B_u \leq 0 \text{ for all } u \geq r) \\ &= P\left( B_r \leq 0, \, \ \sup_{t\geq 0} (B_{t+r} - B_r) \leq |B_r| \right) \\ &= P\left( B_r \leq 0, \, \ \sup_{t\geq 0} (B_{t+r} - B_r) = 0 \right) \\ &= P(B_r \leq 0) P(Z=0) & (\star )\\ &= \frac{1}{2} P(Z=0) \end{align*}
where in the third line and fourth we used that we used the fact that $B_{t+r} - B_r$ is a Brownian motion independent of $B_r$, so its supremum has the same distribution as $Z$ (and is supported at a subset of $\{0, \infty\}$. This implies that $P(Z=0)=0$, as desired.