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We have a collection of random intervals $\{I_{k}:=(X_{k},Y_{k})\}_{k=1}^{\infty}\subset [0,1]$ s.t.

  • For deterministic $l_{k}\to 0$ we have $0<l_{k}^{a_{1}}\leq Y_{k}-X_{k}\leq l_{k}^{a_{2}}$.
  • The endpoints $\{X_{k},Y_{k}\}_{k\geq 1}$ are all correlated ,even for different k, but their lengths $|I_{k}|=Y_{k}-X_{k}, |I_{k+m}|=Y_{k+m}-X_{k+m}$ are independent when they don't intersect by a buffer: $$Y_{k+m}+\epsilon_{k+m}<X_{k}\text{ or }Y_{k}<X_{k+m}-\epsilon_{k+m},$$ where deterministic $\epsilon_{k}>0$ are decreasing to zero: $\epsilon_{k}>\epsilon_{k+1}\to 0$.

Let $a_{k,n}:=E[X_{k}X_{n}],b_{k,n}:=E[X_{k}Y_{n}],c_{k,n}:=E[Y_{k}Y_{n}]$ be the correlations.

P: Our problem is to find the probability of arranging the intervals $I_{k}$ in [0,1] so that their lengths will be independent of each other.

We are not asking for counterexamples. All we ask is whether it reminds you of some framework where this question might be placed in. We will do the rest of the work of figuring it out if and how exactly. Just mention some reference and we will go through it.

Attempts

1)random tree framework with factorial number of children

The probability of the first two $|I_{1}|,|I_{2}|$ being independent (we mean their lengths) is the union of the events

$$v_{1}:=\{Y_{2}+\epsilon_{2}<X_{1}\}\text{ or }v_{2}:=\{Y_{1}<X_{2}-\epsilon_{2}\},$$

where $v_{1},v_{2}$ are the first two vertices. If $v_{1}$ occurs, then we ask for the probability that $|I_{3}|,|I_{1}|$ are independent and $|I_{3}|,|I_{2}|$ are independent. So now $v_{1}$ will have three children vertices $v_{1,1},v_{1,2},v_{1,3}$.

Similarly, at the kth level a vertex $v$ will have $(k+1)!$ number of children.

The problem is to find the probability of at least one infinite unary subtree i.e. at least one chain of events/vertices. That will give the probability of non-intersecting (with buffer) intervals.

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  • $\begingroup$ You seem ready to start a numbered list of attempts, but there is only one. Also, this looks to me like you are asking someone else to formulate a clear question for you, rather than asking a well focussed question yourself; but perhaps that is just because I am not an expert. $\endgroup$ – LSpice Jul 4 at 0:55

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