Bounding Brownian Exit-Time Probabilities with the Normal Density

Summary

The document examines a proof that a standard Brownian motion eventually exits the interval bounded by an upper and a lower target. The question concerns a step that bounds the probability of remaining inside the interval by a quantity involving the derivative of the normal cumulative distribution function. The response explains the estimate by expressing the probability of lying inside the interval as an integral of the normal density over that interval.

Since the standard normal density is largest at zero, the integral can be bounded by the interval’s width times the maximum density. As time grows, the distribution of Brownian motion spreads out, so this bound on the probability of remaining between the fixed targets tends to zero; consequently, the probability of having exited by then tends to one. The exchange clarifies the density bound behind the proof rather than presenting a full alternative proof. In the question’s displayed argument, the derivative variable and evaluation point are written unclearly; the response’s integral makes clear that the relevant derivative is with respect to the spatial threshold.

Key ideas

• The probability that Brownian motion lies within fixed bounds can be written as an integral of its density over those bounds.
• That integral is at most the interval width multiplied by the maximum density on the interval.
• For a normal distribution, the density is maximal at its mean, which is zero here.
• The resulting bound on remaining inside fixed targets shrinks as the Brownian distribution spreads over time.
• The response explains the estimation step but does not restate every detail of the complete exit-time proof.

Tags

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


# 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:

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

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

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

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

$\rightarrow 1$ 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?

## Answer by Mark Joshi (score 5)

https://quant.stackexchange.com/a/21337

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.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.