Extended (or augmented) stopping times

Question

I am trying to prove that $\tau$, defined as: $$ \tau = inf\{t > 0 \mbox{ }|\mbox{ } B_t < t-1 \} $$ is a stopping time with respect to the filtration $(\mathscr{F}_{t+}^B)_{t\geq 0}$ where $\mathscr{F}_{t+}^B = \bigcap_{s > t} \mathscr{F}_{s}^B$, and $B$ is a standard Brownian Motion started at $0$.

What I have done so far is: $$ {\{\tau \leq t\}} = \bigcup_{s\in[0,t]\cap\mathbb Q } \{B_s < s - 1\} = \bigcap_{n\geq 1} \bigcup_{s\in[0,t]\cap\mathbb Q} \{B_s \leq s - 1 + \frac{1}{n} \} $$

Now I was about to conclude that the set $\{B_s \leq s - 1 + \frac{1}{n} \} \in \mathscr{F}_{s-1+\frac{1}{n}}$, and hence $$ \mathscr{F}_{s-1+\frac{1}{n}} \subseteq \mathscr{F}_{(s-1)+} $$

But from here I'm certain that I cannot say that $\mathscr{F}_{(s-1)+} \subseteq \mathscr{F}_{t+}$ to prove the claim.

Disturbingly, it seems like I am getting to the same result I obtain when proving that $\tau ' = inf\{t > 0 \mbox{ }|\mbox{ } B_t \leq t-1 \}$ is $\mathscr{F}_{t}$-measurable, so clearly I must be doing something wrong. Hope you can tell me the correct approach.

Thanks in advance.

Answer 1

Thanks for the well thought-out post.

A few general remarks:

• $\tau=\inf\left\{t\ge 0\ :\ X_t<0\right\}$, where $(X_t)$ is the stochastic process defined by $X_t=B_t-t+1$.
• By definition, $\mathscr{F}^B_t=\sigma(B_u,0\le u\le t)$. Since $t$ and $1$ are deterministic, it is clear that $\mathscr{F}^B_t=\sigma(X_u,0\le u\le t)$.
• $\tau$ is therefore the hitting time of the continuous process $(X_t)$ of the open set $]-\infty,0[$. This is known to be a $\mathscr{F}^B_{t+}$-stopping time (look up Début theorem for more details).

Other than that, you make a mistake when you write $\{\tau\le t\}=\bigcup_{s\in[0,t]}\{B_s<s-1\}$. Indeed, $\tau=t$ does not imply $\{B_t<t-1\}$ (if it did, then by continuity of $B$, you could find $t'<t$ such that $\{B_{t'}<t'-1\}$). Also note that if what you had written down were true, then you would be done since $\{B_s<s-1\}\in\mathscr F_s^B\subset\mathscr F_t^B$.

Also note that if what you had written were correct, then $\mathscr F_{(s-1)+}^B\subset\mathscr F_{(t-1)+}^B\subset\mathscr F_{t+}^B$ by the fact that the filtration is increasing, and again you would be done.

In fact, in order to avoid the problems caused by the fact that you are looking at the hitting time of an open set instead of a closed set, you proceed as follows:

$$ \{\tau<t\}=\bigcup_{s\in[0,t[\cap\mathbb Q}\underbrace{\{B_s< s-1\}}_{\in\mathscr F_s^B\subset\mathscr F_t^B}\in\mathscr F_t^B, $$

and you conclude by noting that

$$ \{\tau\le t\}=\bigcap_{n\in\mathbb N}\underbrace{\left\{\tau<t+\frac1n\right\}}_{\in\mathscr F_{t+1/n}^B}\in\mathscr F_{t+}^B. $$