Understand this probability argument

Question

I've got a pretty specific question here. I've included an image of a probabilistic expression relating Brownian Motion and the stopping times for when the Brownian crosses each integer value. This text is Steele: Stochastic Calculus and Financial Applications. The question is, how is this last inequality derived?

I am not seeing exactly how we get the connection between $|B_{\tau_n}-B_n| \ge \epsilon\sqrt{n}$ and $\sup_{s:|s-n| \le \delta n} |B_s - B_n| \ge \epsilon\sqrt{n}$. There isn't really any more explanation in the book about this. I guess it should be obvious but I've spent a long time on it and can't figure it out. Thanks!

Answer 1

Define $S:=\{s:|s-n|\le \delta n\}$ for simplicity. We want to show that $$\left \{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\right \}\subset\left \{\sup_{s\in S}|B_s-B_n|\ge \epsilon\sqrt{n}\right \},$$ namely that if $$\omega\in\left\{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\right\},$$ then $$\omega\in \left\{\sup_{s\in S}|B_s-B_n|\ge \epsilon\sqrt{n}\right\}.$$

Now, let us fix an $\omega\in\{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\}.$ Then $|\tau_n(\omega)-n|<\delta n$, which implies that $\tau_n(\omega)\in S.$ Moreover, $|B_{\tau_n}(\omega)-B_n(\omega)|\ge \epsilon \sqrt{n},$ but $\tau_n\in S$ and hence clearly $\sup_{s\in S}|B_s(\omega)-B_n(\omega)|\ge \epsilon\sqrt{n}.$ Consequently, $$\left \{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\right \}\subset\left \{\sup_{s\in S}|B_s-B_n|\ge \epsilon\sqrt{n}\right \}$$ and we have that $$P\left \{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\right \}\le P\left \{\sup_{s\in S}|B_s-B_n|\ge \epsilon\sqrt{n}\right \}.$$

Answer 2

I am not an expert, but it looks to me that the event:

$$|B_{\tau_n}-B_n|\ge \epsilon\sqrt{n}\text{ and }|\tau_n-n|<\delta n$$

is contained in (i.e. implies) the event:

$$\sup_{s:|s-n|\le\delta n}|B_s-B_n|\ge \epsilon\sqrt{n}$$

whereas the event:

$$|B_{\tau_n}-B_n|\ge \epsilon\sqrt{n}\text{ and }|\tau_n-n|\ge\delta n$$

is contained in (i.e. implies) the event:

$$|\tau_n-n|\ge\delta n$$

so really this is the consequence of monotonicity of probability.