Timeline for Verifying if Markov process's dependency on time is not random
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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S Sep 19, 2017 at 20:26 | history | suggested | oxonianftw | CC BY-SA 3.0 |
made question more straightforward
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Sep 19, 2017 at 19:35 | review | Suggested edits | |||
S Sep 19, 2017 at 20:26 | |||||
S Sep 18, 2017 at 21:57 | history | suggested | oxonianftw | CC BY-SA 3.0 |
more explicit
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Sep 18, 2017 at 21:44 | review | Suggested edits | |||
S Sep 18, 2017 at 21:57 | |||||
Sep 18, 2017 at 8:18 | comment | added | Mateusz Kwaśnicki | @IndoUbt: This is why I wrote "roughly". I do not even claim comparability of $P(k)$ and $(1-2/n^2)^k$ for large $k$. The decay is in fact exponential, of the form $C \lambda^k$, where $\lambda$ is the largest eigenvalue of a killed Markov chain that describes both objects simultaneously (and it is killed when the objects meet). However, I am not sure if this answers the original question. | |
Sep 18, 2017 at 6:45 | comment | added | Indo Ubt | @MateuszKwaśnicki For a 3 row grid, $(1-2/(3)^2)^k$ would give a < 1 value for k < 2. Similarly for all $n\ge3$ | |
Sep 17, 2017 at 22:29 | comment | added | Mateusz Kwaśnicki | I bet it will decrease exponentially, roughly as $(1-2/n^2)^k$, where $k$ is the number of steps. This is because after a long period of time, the position of each of the objects becomes almost uniform on black/white squares of the chessboard for even/odd $k$, and roughly independent from whether the objects already met or not. | |
Sep 17, 2017 at 22:10 | review | First posts | |||
Sep 17, 2017 at 23:11 | |||||
Sep 17, 2017 at 22:05 | history | asked | oxonianftw | CC BY-SA 3.0 |