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Assume that $f_1, f_2, f_3,\ldots$ are i.i.d. random functions $[0,1]\mapsto \mathbb{R}$ such that

(1) random variables $M_k=\sup_{x\in[0,1]}f_k(x)$ have exponential tails,

(2) $f$'s are a.s. Lipschitz, and random variables $L_k=\inf\{h>0: f_k \text{ is Lipschitz with constant $h$}\}$ have exponential tails,

(3) $\mathbb{E} f_k(x) = 0$ for all $x\in[0,1]$.

My question is: what can be said about the law of r.v. $X_n=\sup_{x\in[0,1]}\sum_{k=1}^n f_k(x)$? Can one show that $\mathbb{P}[X_n>an] < e^{-bn}$ (where $a>0$ is arbitrary and $b$ depends on $a$)? And/or for any $\varepsilon>0$ there exists large enough $v=v(\varepsilon)$ such that $\mathbb{P}[X_n>v\sqrt{n}]<\varepsilon$?

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  • $\begingroup$ What do you get from the trivial bound $X_n \le \sum_{k=1}^n M_k$? $\endgroup$ Commented Nov 11, 2015 at 15:08
  • $\begingroup$ Let $Z_i$ be i.i.d. $\pm 1$ random variables. Do $f_k = c_kZ_k$ satisfy your hypotheses ? $\endgroup$
    – Michael
    Commented Nov 11, 2015 at 15:17
  • $\begingroup$ @Nate Eldredge - not much. $\sum M_k$ just grows linearly with positive speed, and we need to obtain that $\sum f_k$ somehow resembles a sum of 0-mean r.v.'s $\endgroup$ Commented Nov 11, 2015 at 15:38
  • $\begingroup$ @Michael - yes, if $c_k$ are i.i.d.r.v. with exponential tails. $\endgroup$ Commented Nov 11, 2015 at 15:39
  • $\begingroup$ I should have said $c_k$ constants, which I would say have sub-exponential tails, regardless of how big they may be. But if you accept that you can't expect $X_n$ to be order of $\sqrt n$. $\endgroup$
    – Michael
    Commented Nov 11, 2015 at 15:50

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