Timeline for Graph with Poisson Clock at each Vertex
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 7, 2017 at 8:26 | vote | accept | co.sine | ||
| Apr 7, 2017 at 7:59 | comment | added | domotorp | That is why we need the n-th layer to be thicker compared to T(n), the time it takes for the root to turn -1. This way it won't be killed too fast by the (n+1)-th layer. | |
| Apr 7, 2017 at 7:47 | comment | added | co.sine | Ok, now: what makes the root change color infinitely often? I mean, in the initial configuration, the root will change its color with positive probability. But why should it repeat? In other words: say the root is $1$ and layer $n$ is $-1$, and is large and therefore the root will turn $-1$ sometime. But layer $n+1$ is also large, and is a $1$ layer, which will incline the root towards $1$ (no matter what the current value of the root is). Why the $n+1$ layer doesn't beat the $n$ layer? Just kills it before it reaches the root? What makes the sequence? Sorry if I make a silly logical mistake. | |
| Apr 7, 2017 at 6:50 | comment | added | domotorp | I'm not sure how exactly you define $P(\textit{root changes to } -1\mid A_n)$, but I suppose under any reasonable definition it should tend to $1$, as you write. | |
| Apr 7, 2017 at 6:07 | comment | added | co.sine | That is $\mathop {\lim }\limits_{n \to \infty } P(root \, changes \, to -1 | A_{n}) = 1.$ Right? | |
| Apr 7, 2017 at 5:53 | comment | added | co.sine | Fine, this can be justified using the continuity from above of the probability measure: let the root be $1$. Define $A_0$ to be the event that root will sometime become $-1$, given that all vertices at distance $d$ from the root are $1$. $A_1$ means all vertices at distances $d$, $d+1$ are $1$. $A_{\infty}$ means all vertices from distance $d$ are $-1$. $P(root \, changes \, to -1 | A_{\infty})=1$, and $\bigcap\limits_{n = 1}^\infty {{A_n} = {A_\infty }} $. Because all measures are finite, $\mathop {\lim }\limits_{x \to \infty } P({A_n}) = 1.$ Please state if you agree. | |
| Apr 7, 2017 at 5:24 | history | bounty ended | co.sine | ||
| Apr 6, 2017 at 18:23 | comment | added | domotorp | Yes, that's what I mean. | |
| Apr 6, 2017 at 15:58 | comment | added | co.sine | Tell me if I'm wrong: you state, that for the $3$-regular tree, for a given vertex named the root, and for any distance $d$ from the root, there exists a natural number $N$ such that if all vertices at distance between $d$ and $d+N$ from the root have the value $c$, and the root has value $-c$, then with probability at least $0.9$ (or any other number) sometime in the future the root will change its value to $c$. | |
| Apr 6, 2017 at 12:48 | comment | added | domotorp | 90% is any number less than and I do not know how T(n) depends on f(n). | |
| Apr 6, 2017 at 12:30 | comment | added | co.sine | Sorry for the questions, just want to fully understand your idea: where did the 90% come from? And how is f(n) comparable to T(n)? | |
| Apr 6, 2017 at 12:00 | comment | added | domotorp | @co.sine: Done! | |
| Apr 6, 2017 at 11:59 | history | edited | domotorp | CC BY-SA 3.0 |
added more details
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| Apr 6, 2017 at 11:40 | comment | added | co.sine | can you please elaborate on how the root getting a color with probability more than half is dependent only on the n+1 first layers? | |
| Apr 5, 2017 at 12:27 | history | answered | domotorp | CC BY-SA 3.0 |