Clarification of example for probability of Brownian motion being positive after a given point.

Question

I have this example I could use some clarification on:

Let $B_t: 1$-dimensional Brownian Motion

Find the follow probability $P(B_t > 0, \forall t > 10)$.

The example goes as follows:

Let us compute $P[B_t >0, \forall t \in (10,n)]$ first; and then let $n→ ∞$ to obtain our answer. Define $Y_s := 10^{−1/2}B_{10s}$, and note that $Y_s$ is a Brownian motion. Therefore, we apply Example 1 (p. 179) to find that $P(B_t > 0\text{ for some }10 < t < n) =P (Y_s =0 \text{ for some } 1<s<n/10) =$

$ 1 - \frac{2}{\pi} \arctan ( 1/\sqrt{n/10-1}) $

Let $n → ∞$ to find that $P(B_t >0,\forall t>10)= \lim_{n \to \infty} ( 1 - \frac{2}{\pi} \arctan ( 1/\sqrt{n/10-1}) =0$.

I understand the translation from $B_t$ to $Y_s$.

I have one question

1. How does calculating the probability the Brownian motion is positive for some time $t$ such that $10 < t < n$ and then letting $n \to \infty$ lead to the desired result? This just seems to be leading to the probability the Brownian motion is positive at some point $t > 10$ not every point $t > 10$.

Answer 1

I agree with @KaviRamaMurthy, there's a typo. On P. 181 of Lawler, he discusses the limit of that term

As $t \to \infty$ the quantity on the right-hand side tends to 1. This tells us that with probability 1 the Brownian motion eventually returns to the origin, and hence (with the help of the strong Markov property) that it returns infinitely often. This means that the Brownian motion for large $t$ has both positive and negative values.

This implies that $$P(B_t > 0, \forall t > 10) = 1 - P(Y_s = 0 \text{ for some } 1 < s < n/10) = 1 - \lim_{n \to \infty}\left(1 - \frac{2}{\pi}\arctan\frac{1}{\sqrt{n/10 - 1}}\right)\\ = 1 - 1 = 0.$$