Timeline for Why is Branching Brownian Motion log-correlated?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 4, 2022 at 11:48 | comment | added | Shannon Starr | Sorry! Yes, of course you are right! What you said is the correct answer, not what I said. | |
| Feb 4, 2022 at 0:39 | comment | added | MikeG | @ShannonStarr: But your $f_n$ is non-positive? I think the idea is that, if our binary tree has depth $n$, then we give edge-weights $2^k$ if the edge links the nodes on level $(n-k)$ and level $(n-k-1)$(that is, increase from leaf to root). This will give $d(u,v)=2^{n-(u\wedge v)}-1$, which is $-f$, and this works as ultrametric.Then I think my problem is solved, thank you!!! | |
| Feb 3, 2022 at 15:26 | comment | added | Shannon Starr | Sorry, what I wrote is not a metric. Take $f_n(k) = 1-2^{n-k}$ and let $d(u,v)=f_n(u\wedge v)$. The point is that the graph distance on a tree is a particular example of an ultrametric (or a metric that satisfies the ultrametric condition). Not only do you have the usual triangle inequality $d(u,w) \leq d(u,v)+d(v,w)$ for every $u,v,w$ in the metric space, but actually you have the even stronger ultrametric condition $d(u,w) \leq \max(\{d(u,v),d(v,w)\})$. For a tree, one can see that $u\wedge w \geq \min(u\wedge v,v\wedge w)$. Therefore, since $f_n$ is increasing, $d$ is an ultrametric. | |
| Feb 3, 2022 at 11:30 | comment | added | Shannon Starr | It also looks like it is proportional to the probability to be at that distance, since it is a binary tree, so that probability decays exponentially. If you say $d(u,v)$ equals $C_n \mathbb{P}(\mathsf{U}\wedge \mathsf{V}=u\wedge v)$ for two independent uniformly randomly chosen points $\mathsf{U}$ and $\mathsf{V}$ then that is $C_n 2^{-u\wedge v}$. So if you choose $C_n=2^n$ you have $\log(d(u,v)) = (n-\log(u\wedge v))/\log(2)$. | |
| Feb 2, 2022 at 16:34 | comment | added | MikeG | @ShannonStarr: Yes that's not from Berestycki's note but from talks mentioning "oh, so some examples are ...BBM..." kind of stuff. I just found on Arguin's review you mentioned they explained this beautifully with another criteria, the number of particles with correlation larger than constant times variance decays exponentially. This indeed kind of solved my question; but I am now thinking that can we find a metric $d$ here to make the correlation appearing exactly in the form $-\log d$? | |
| Feb 2, 2022 at 11:40 | comment | added | Shannon Starr | I do not see anywhere in Berestycki's notes a reference to "log correlated," otherwise you should be able to find the answer there. The expectation $\mathbf{E}[\tau_{u,v}]=u\wedge v$ the branching time. So to get what you want, you must make $d(u,v)=exp(n-(u\wedge v))$ where $u\wedge v$ is the branching time of $u$ and $v$ (which is between $0$ and $n$). This definition should be okay (because of the ``$\approx$'') and ultrametric. I also recommend the review article of Arguin, Bovier and Kistler, "Extrema of Log-Correlated Random Variables: Principles and Examples." | |
| Jan 31, 2022 at 22:04 | history | asked | MikeG | CC BY-SA 4.0 |