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I need some references(or helps) on understanding why BBM is log-correlated. As I understand it, a random field on some metric space $V$ with distance $d$ is log-correlated if $$\mathbb{E}[X_u X_v]\approx-\log d(u,v)$$ for any $u,v\in V$. But for BBM, I only read from Proposition 9 in Berestycki's notes that $$\mathbb{E}[X_u(t)X_v(t)]=\mathbb{E}\tau_{u,v}$$

where $\tau_{u,v}$ is the death time(or the branching time) of the common ancestor of $u,v$. So my questions are that

  1. What is the distance here? Is it the graph-distance on the underlying Galton-Watson tree?

  2. If yes, how to calculate the covariance to get the logarithmic relation?

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  • $\begingroup$ 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." $\endgroup$
    – Shannon Starr
    Commented Feb 2, 2022 at 11:40
  • $\begingroup$ @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 $-\log d$? $\endgroup$
    – MikeG
    Commented Feb 2, 2022 at 16:34
  • $\begingroup$ 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 $\log(d(u,v)) = (n-\log(u\wedge v))/\log(2)$. $\endgroup$
    – Shannon Starr
    Commented Feb 3, 2022 at 11:30
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    $\begingroup$ 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. $\endgroup$
    – Shannon Starr
    Commented Feb 3, 2022 at 15:26
  • $\begingroup$ @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!!! $\endgroup$
    – MikeG
    Commented Feb 4, 2022 at 0:39

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