# Can independent Brownian motions hit zero at the same time?

Consider for $$i=1,\ldots, N\ge2$$

$$X^i_t=x_i+W^i_t,\quad \forall t\ge 0,$$

where $$x_1,\ldots, x_N\in (0,\infty)$$ and $$W^1,\ldots, W^N$$ are independent Brownian motions. Denote by $$\tau_i$$ the first hitting time of $$X^i$$ at zero, i.e.

$$\tau_i:=\inf\big\{t\ge 0: X^i_t\le 0 \big\}.$$

How to prove (rigorously) $$\mathbb P[\exists i\neq j \mbox{ such that } \tau_i=\tau_j]=0$$?

Your question is asking whether two Brownian motions can both first hit zero simultaneously. In fact we can say something stronger; for $$N$$ independent Brownian motions, the set of times where each Brownian motion hits 0 are pairwise disjoint.

When $$N=2$$, this is equivalent to asking whether a standard two-dimensional Brownian motion starting at $$(x_1,x_2)$$ ever hits $$(0,0)$$; it is known that, almost-surely, it will not.

For $$N>2$$, by the union bound, the probability is less than the sum over $$i \neq j$$ of the probability that a BM starting at $$(x_i,x_j)$$ ever hits $$(0,0)$$, so this probability is also 0.

• By the way, you have not defined what $T$ is in your question. Jul 12, 2023 at 13:07
• Thanks for the answer. $T$ is a typo and I've edited Jul 12, 2023 at 13:45
• @Fawen90 I've added a sentence making reference to the fact that your question is about the first time each Brownian motion hits 0. What I have explained is a stronger result. Jul 12, 2023 at 14:00
• Fantastic. Many thx Jul 12, 2023 at 14:35

It is easy to see that the distribution of each $$\tau_i$$ is non-atomic (actually, it is absolutely continuous and can be found explicitly using the reflection principle). The desired result now follows from the Tonelli theorem, because the $$\tau_i$$'s are independent.

• Thanks Iosif. Indeed I don't know why I was not aware of the independence of $\tau_1,\ldots,\tau_N$. Btw, do you have any idea of this question mathoverflow.net/questions/450626/… ? Jul 12, 2023 at 17:36
• @Fawen90 : No, no good idea for that one. Jul 12, 2023 at 18:36