Timeline for How long for a simple random walk to exceed $\sqrt{T}$?
Current License: CC BY-SA 2.5
3 events
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| Jan 31, 2010 at 20:40 | comment | added | Yemon Choi | Thanks! it really has been too long since I learned/practiced any of this... | |
| Jan 31, 2010 at 20:30 | comment | added | Douglas Zare | For any $k \gt 0$, the infimum of the times where a Brownian motion is more than $k \sqrt T$ is 0. This may be counterintuitive, but it's a consequence of the combination of time inversion ($t W(1/t)$ is also Brownian) and the law of the iterated logarithm. | |
| Nov 22, 2009 at 14:51 | history | answered | Yemon Choi | CC BY-SA 2.5 |