# Brownian motion remains nonnegative for some interval with length $1$ almost surely

Let $$B_t$$ be a continuous Brownian motion. I'm having a really difficult time to prove that the Brownian motion stays nonnegative for some interval with length $$1$$ almost surely.

The reason for this is to show that $$\int^t_0 e^{B_s}\,ds$$ using the property I have mentioned. As user Nate Eldredge suggests in his answer here, what I want to show might be a way to prove it. So the problem is

Problem. Show that $$B_t\geq 0$$ for all $$t\in [a,a+1]$$ for some $$a\geq 0$$ almost surely. Mathematically $$\mathbb P(\exists_a: B_t\geq 0 \text{ for all }t\in [a,a+1]) =1$$

Once I have solved this then the claim follows easily by strong Markov Property.

Attempt.

I have honestly no idea how to tackle this. I tried using Borel-Cantelli with events like \begin{align} A_n:=\{B_t-B_n\geq 0 \text{ for all } t\in [n,n+1], B_n\geq 0\} \end{align} and then show that $$\sum_n \mathbb P(A_n)=\infty$$ but the troubles induced by this approach is that $$A_n$$ are not independent to begin with so....

I do not need full answers, I would like some guidance to solve it myself.

## 1 Answer

A good first good guess is that $$B_t$$ is unlikely to become negative on $$[a,a+1]$$ if $$B_a$$ is already a large positive number. Since your $$a$$ is allowed to depend on the realisation of the Brownian path, this means it's reasonable to consider the stopping times $$T_n = \inf\{t: B_t = n\}$$ for $$n \in \mathbb{N}$$.

Now you should try to compute $$\mathbb{P}(B_t = 0 \text{ for some } t \in [T_n,T_{n} + 1])$$ (use the strong markov property and a standard result about Brownian hitting times). Once you've done this, it should be easy to show $$\mathbb{P}(\forall n, B_t < 0 \text{ for some } t \in [T_n, T_n + 1]) = 0$$ which implies the desired result.

• Many thanks! I will try to follow the steps! – Shashi Mar 27 at 11:27
• I love this, it was a nice guidance! – Shashi Mar 27 at 13:43