Minimum of the Brownian motion Suppose that you have a Brownian motion $(W_t)_{t\in [0, T]}$, for some $T>0$. I'm interested in calculating:
$$\mathbb P \left(\min_{t\in [0, T]} W_t = 0\right).$$
Since $W_0=0$ p.s., it is equal to the probability that the Brownian motion is non-negative...
 A: In fact, this probability is zero, since the Brownian motion a.s. oscillates above and below zero in every neighborhood of the form $[0,\epsilon)$.
Indeed, we can easily prove a more refined statement (I am adapting the argument given in page 32 of this book).
Claim. With probability $1$, there are infinitely many positive integers $n$ such that $W_{1/n}>\tfrac{1}{\sqrt n}$ and infinitely many positive integers $n$ such that $W_{1/n}<-\tfrac{1}{\sqrt n}$.
Proof. Let $A_n$ be the event that $W_{1/n}>\tfrac{1}{\sqrt n}$. By Fatou's Lemma, the probability that infinitely many of the $A_n$ occur is at least $\limsup_{n\to\infty}\mathbb P(A_n)$. On the other hand, by Brownian scaling $\mathbb P(A_n)$ is seen to be a positive constant that does not depend on $n$. Hence there is a positive probability that infinitely many of the $A_n$ occur. Using a suitable zero-one law (Hewitt-Savage is used in the text, but this is a matter of taste), it follows that the probability is in fact $1$. Similarly for the negative event. Finally, since the intersection of two events with probability $1$ also has probability $1$, the claim follows. $\square$
