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