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Fawen90
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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{ and } T>0 \mbox{ such that } \tau_i=\tau_j]=0$$\mathbb P[\exists i\neq j \mbox{ such that } \tau_i=\tau_j]=0$?

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

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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gmvh
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Fawen90
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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) $\mathbb P[\exists i\neq j \mbox{ and } T>0 \mbox{ such that } \tau_i=\tau_j]=0$?