Timeline for A walk on a $2D$ Poisson process in which every step goes to the nearest unvisited point: expected distance from origin after $365$ steps?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 14 at 13:19 | history | bounty ended | Dan | ||
| Feb 14 at 13:19 | vote | accept | Dan | ||
| Feb 10 at 4:18 | comment | added | achille hui | Cool, it is sort of unexpected that $\xi_k$ falls off that slow. It is too bad I can't upvote again ;-) | |
| Feb 9 at 19:13 | history | edited | mjqxxxx | CC BY-SA 4.0 |
added 1943 characters in body
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| Feb 9 at 15:32 | comment | added | mjqxxxx | Yes, that's the question... assuming $E[r_n \cdot r_{n+k}]$ depends only on $k$ (which isn't guaranteed either, but no matter) -- call it $\xi_k$ -- you have $E[R_N^2]=N \xi_0 + \sum_{k=1}^{N-1} 2(N-k)\xi_k$. Which is going to be $O(N)$ if $\xi_k$ is summable, e.g., decays exponentially or as $k^{-\beta}$ with $\beta > 1$. | |
| Feb 9 at 6:54 | comment | added | achille hui | For the third case, one thing that may help is look at correlation of the displacements in each step. Let's say $r_n$ is the $n^{th}$ displacement. If $E[r_n\cdot r_{n+k}]$ falls off fast enough (say exponentially or algebraically $k^{-\beta}$ with large enough $\beta$) as $k$ increases, we should expect $E(N) \sim \alpha\sqrt{N}$ remains to hold. | |
| Feb 9 at 6:01 | comment | added | mjqxxxx | As to the original question: after $365$ days, by my calculations, the uncle is expected to be about $31.5$ miles away. But you shouldn't be surprised if he's as close as $9$ miles or as far away as $55$ miles... each of these happens about $5\%$ of the time. | |
| Feb 9 at 5:51 | history | answered | mjqxxxx | CC BY-SA 4.0 |