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:

$P(\tau < t) \geq (|W_t|>2)\\$

$=1-P(|W_t| \leq 2)\\$

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

$=1-\frac{4}{\sqrt{2 \pi t}}$

$\rightarrow 1$ as $t\rightarrow \infty$

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?


I think all they are doing is integrating and estimating

$$P(|W_t| \leq 2) = \int_{-2}^{2} \frac{d}{dr} P(W_t \leq r) dr $$

so $$ P(|W_t| \leq 2) \leq 4 \sup \limits_{r \in [-2,2]} \frac{d}{dr}P(W_t \leq r) $$ The normal density is maximal at zero and we are done.

  • $\begingroup$ It may be $$P(|W_t| \leq 2) = \int_{-2}^{2} \frac{d}{dx} P(W_t \leq x) dx. $$ $\endgroup$ – Gordon Oct 21 '15 at 0:52

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