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Oct 17, 2015 at 12:08 history edited Nate River CC BY-SA 3.0
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Oct 17, 2015 at 12:06 comment added Nate River @DouglasZare You are right. Thanks for pointing this out.
Oct 17, 2015 at 11:58 comment added Douglas Zare The intersection of $E_{j,n}$ with $E_{j+1,n}$ means that the Brownian motion hits $0$ on both of these intervals, not that it hits $0$ at the intersection of the intervals. This argument is not correct. By the way, the upper limit in the question is $2^{2n}$ rather than $2^{2^n}$.
Oct 17, 2015 at 11:14 review First posts
Oct 17, 2015 at 11:47
Oct 17, 2015 at 11:11 history answered Nate River CC BY-SA 3.0