Timeline for Distribution of the cover time of a finite path?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2016 at 10:23 | vote | accept | emme | ||
| Nov 4, 2016 at 18:51 | vote | accept | emme | ||
| Nov 5, 2016 at 10:23 | |||||
| Nov 4, 2016 at 16:48 | history | edited | Serguei Popov | CC BY-SA 3.0 |
added 171 characters in body
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| Nov 4, 2016 at 16:04 | comment | added | Serguei Popov | I mean, let us suppose that your vertices are rather labeled $\{0,\ldots,n-1\}$, and you start in $0$. Now, instead of considering the random walk with reflection in $0$ (and waiting until it comes to $n-1$), you may as well consider the simple random walk on $\{-(n-1),\ldots,n-1\}$ (in fact, the reflected random walk is just its absolute value). So, the hitting time of $n-1$ for the reflected random walk is the same as the hitting time of $\{-(n-1),n-1\}$ for the "normal" random walk on $\{-(n-1),\ldots,n-1\}$. | |
| Nov 4, 2016 at 15:45 | comment | added | emme | This makes sense, but I lose you towards the end of your answer: "Once you hit one of the extremes, you have to go to the other one, and the distribution of this time ..." can you clarify that part? There is no vertex 0 in the given path. There is no negative labelled vertex so what is $\{-(n-1),n-1\}$? | |
| Nov 4, 2016 at 15:07 | history | answered | Serguei Popov | CC BY-SA 3.0 |