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.


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.


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.

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

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