# Proof that the stopping time for a Brownian Motion is finite for given target levels

Given a standard brownian motion $W_t$ and defining $\tau$ as:

$\tau :=\inf\{t\geq0:W_t=1$ or $W_t=-2\}$

The proof below shows that the stopping time is finite:

\begin{align*} P(\tau < t) &\geq P(|W_t|>2) \\ &= 1-P(|W_t| \leq 2)\\ &\geq1-4\frac{d}{dt}P(W_t \leq t)|_{t=0} \\ &=1-\frac{4}{\sqrt{2 \pi t}}\\ &\rightarrow 1 \qquad \text{as t\rightarrow \infty} \end{align*}

It's all staighforward except the line were the derivative is used:

$\geq1-4\frac{d}{dt}P(W_t \leq t)|_{t=0}$

How does this line relate to the line above?

• It doesn't even make sense: if the derivative in that line is to be evaluated at $t=0$ then the value of that line should not depend on $t$, and it's nonsense to compare it to the preceding and succeeding lines which do. – Nate Eldredge Oct 20 '15 at 14:14
• It also seems pointless: it's very easy to compute explicitly that $P(|W_t| \le 2) \to 0$. For instance, by scaling we have $P(|W_t| \le 2) = P(|W_1| \le 2/\sqrt{t})$. As $t \to \infty$, by "continuity from above" (a consequence of countable additivity), this converges to $P(|W_1| \le 0) = 0$. – Nate Eldredge Oct 20 '15 at 14:16
• @Nate well it has me pretty confused too! – Bazman Oct 20 '15 at 14:18
• Where did you find this "proof"? – Nate Eldredge Oct 20 '15 at 14:19
• Well, assuming you've transcribed it correctly, I think it's at best a typo (though for what I don't know) and at worst it's just nonsense. If the book has more stuff like that I'd stop reading it. (Incidentally, I looked it up on Google Books to try to read the passage for myself, but you can't view any of the pages. But I did note that the book's subject is listed there as "Fiction / Romance / Historical / General" :-) – Nate Eldredge Oct 20 '15 at 14:28

As @NateEldredge has aleady pointed out, this particular line doesn't make sense and, moreover, the convergence

$$\mathbb{P}(|W_t| \leq 2) \xrightarrow[]{t \to \infty} 0$$

can be proved much easier using the scaling property.

However, here is a way to fix the proof: Obviously,

$$\mathbb{P}(|W_t| \leq 2) = \mathbb{P}(W_t \leq 2)-\mathbb{P}(W_t \leq -2) = \int_{-2}^2 \frac{d}{dx} \mathbb{P}(W_t \leq x) \, dx.$$

Using $W_t \sim N(0,t)$, we get

$$\mathbb{P}(|W_t| \leq 2) \leq \frac{4}{\sqrt{2\pi t}} \sup_{x \in [-2,2]} \exp(-x^2/2t) \leq \frac{4}{\sqrt{2\pi t}}.$$

This gives

$$\mathbb{P}(\tau<t) \geq 1- \mathbb{P}(|W_t| \leq 2) \geq 1- \frac{4}{\sqrt{2\pi t}}.$$