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

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2 Answers 2

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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.

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    $\begingroup$ By the way, you have not defined what $T$ is in your question. $\endgroup$
    – Julius
    Commented Jul 12, 2023 at 13:07
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    $\begingroup$ Thanks for the answer. $T$ is a typo and I've edited $\endgroup$
    – Fawen90
    Commented Jul 12, 2023 at 13:45
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    $\begingroup$ @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. $\endgroup$
    – Julius
    Commented Jul 12, 2023 at 14:00
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    $\begingroup$ Fantastic. Many thx $\endgroup$
    – Fawen90
    Commented Jul 12, 2023 at 14:35
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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.

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  • $\begingroup$ 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/… ? $\endgroup$
    – Fawen90
    Commented Jul 12, 2023 at 17:36
  • $\begingroup$ @Fawen90 : No, no good idea for that one. $\endgroup$
    – Iosif Pinelis
    Commented Jul 12, 2023 at 18:36

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