Consider the following random variables

  • The $\{m_n\}_{n\geq 1}$ are iid and satisfy $$\mathbb{P}(m_{n}\leq x)\leq C x$$ for $x>0$ and some $C>0$.
  • The $\{L_{n,m}\}_{m\geq n\geq 1}$ satisfy $L_{n,m}\stackrel{law}{=}L_{1,m-n+1}$ and the concentration $$\mathbb{P}(L_{n,m}\geq u)\leq \frac{1}{u^{q}}\rho^{(m-n-\frac{1}{2})b},$$ where $0<\rho<1$, $q>1$ and $b>1$. The $L_{n_{1},m_{1}}$ and $L_{n_{1},m_{1}}$ are independent iff $n_{1}>m_{2}$ or $n_{2}>m_{1}$. Let $$\sigma_{n}:=(1+\sum_{m=n+1}^{\infty}L_{n,m})^{-1}.$$
  • The $\{t_{n,k}\}_{n> k \geq 0}$ satisfy $t_{n,k}\stackrel{law}{=}t_{n-k+1,1}$ and the concentration $$\mathbb{P}(t_{n,k}\geq u)\leq c \exp\left(-\frac{u^{2}}{c \rho^{n-k-\frac{1}{2}}}\right),$$ where $0\leq k< n$, $0<\rho<1$ and $c>0$. The $t_{n_{1},k_{1}}$ and $t_{n_{2},k_{2}}$ are independent iff $k_{1}\neq k_{2}$. Let $$s_{n}:=\exp\left(-\sum_{k=0}^{n-1}t_{n,k}\right).$$

Then we have the result

$$\mathbb{P}\left[\sum_{n=1}^{N}m_{n}\sigma_{n}s_{n}\leq c_{0}N\right]\leq \rho^{pN},$$ where $p>1$ and $\rho\in (0,1)$.

Q: The proof of this is a long brute force computation without the use of any theorems. However, we are hoping to find some theoretical framework that will reveal why this statement is true.

You can find the computation in section 4.3 pg 25 of the random conformal weldings paper.

  • $\begingroup$ I don't see a real question. $\endgroup$ – Jochen Wengenroth Aug 14 at 7:03
  • 2
    $\begingroup$ @JochenWengenroth : it's in the last paragraph. The poster is asking whether there is a theoretical framework which explains the statement, without the need for a long brute force computation. $\endgroup$ – James Martin Aug 14 at 8:12

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