Timeline for Probability that planar Brownian motion doesn't "encircle" 0
Current License: CC BY-SA 3.0
7 events
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| Sep 18, 2015 at 16:48 | comment | added | Iosif Pinelis | I have corrected the answer, in view of the comments by Nate Eldredge and Will Sawin. | |
| Sep 18, 2015 at 16:48 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
I have corrected the answer, in view of the comments by **Nate Eldredge** and **Will Sawin**.
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| Sep 18, 2015 at 15:06 | comment | added | Will Sawin | @NateEldredge However, we do have $g(ab) \leq g(a) g(b)$ and $g(a) \leq 1$, so $g$ is monotonic. This means that for each $\alpha$, either $g(a) \leq \lambda |x|^\alpha$ for some $\lambda$ or $g(a) > |x|^\alpha$. Given that we have upper and lower bounds, this means that there exists a precise asymptotic constant $\alpha$ such that $|x|^\alpha <g(a) < O( |x|^{\alpha+\epsilon})$. | |
| Sep 18, 2015 at 14:50 | comment | added | Nate Eldredge | I don't see why (1) is an equality. Let $C_r$ be the circle of radius $r$ centered at $O$. It sounds like you are saying "If the BM is going to encircle $O$ before hitting $C_r$, and it hasn't done so by the time it hits $C_u$ (at some point $y$), then the path started at $y$ must encircle $O$ before hitting $C_r$." But that's not true - what if the path from $x$ to $y$ wound halfway around the origin, and after $y$ it completed the other half? Then the combined path encircles the origin, while neither of the pieces before or after $y$ does so. | |
| Sep 18, 2015 at 12:26 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
added 77 characters in body
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| Sep 18, 2015 at 7:13 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
Added Remark 1 on the measurability of $f$.
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| Sep 18, 2015 at 6:52 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |