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Julius
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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.

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.

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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Julius
  • 311
  • 1
  • 5

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.