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Chapter 9 9.4 (iv)

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Xiaoran Gao

Member
Hi,

Can someone please explain why a standard Brownian motion B_t takes a negative value at some point close to t=0?

As per my understanding, at time 0, B_t is equally likely to fluctuate upwards or downwards, then the prob of B_t < 0 would be 1/2. What is faulty in my statement?

Thanks!
 
Hi
Your understanding is correct; P(B_{t}<0)=1/2. But if you keep zooming in and magnifying the Brownian motion around the origin, we can also say that P(B_{t/2}<0)=1/2, and P(B_{t/3}<0)=1/2 etc. So by looking at smaller and smaller intervals, the probability of B having a negative value stays at 1/2. This means we have complete freedom to keep searching for negative values... and we'll find one with probability 1.
 
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