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) $\mathbb P[\exists i\neq j \mbox{ such that } \tau\_i=\tau\_j]=0$?



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


