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Let $(X_i)_{i\in \mathbb{N}}$ iid random variables such that there exists $\alpha>0$ where $\mathbb{P}\left(X_1\in [x,x+1]\right)\leq \alpha$ for all $x\in \mathbb{R}$. Assume $\alpha$ small enough, does there exist a universal constant $C>0$ so that $$\mathbb{P}\left(\sum_{i=1}^N X_i\in [x,x+1]\right)\leq \frac{C\alpha}{\sqrt{N}}$$ for all $x\in\mathbb{R}$ and $N\in \mathbb{N}^*$?

A few remarks :

  • From the central limit theorem this is very natural to be expected. However here we do not assume $\mathbb{E}(X^2)<\infty$ and it is not so clear to me how to use the characteristic functions with the hypothesis.
  • In the case of discrete variable $X\in \mathbb{Z}$, one should just do the computation. However I can't see why should it be the "worse case scenario" and how to compare it with the more general case.
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1 Answer 1

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Yes, this is a special case of an inequality by Kesten (see Theorem 2 and Corollary 1).

In particular, letting $L=\lambda=:t$ in that Corollary 1, we get the following:

Let $X_1,\dots,X_n$ be independent identically distributed random variables, with $S_n:=\sum_1^n X_k$. Then for any positive real $t$, $$Q(S_n,t)\le\frac{CQ(X_1,t)}{\sqrt{(1-Q(X_1,t))\, n}},$$ where $C$ is a universal positive real constant and $Q(X,u):=\sup_{x\in\mathbb R}P(x\le X\le x+u)$ for positive real $u$.

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    $\begingroup$ This is a great reference. Thanks! But is it really enough ? I wish one could get $Q(X,t)$ in the numerator ? $\endgroup$
    – RaphaelB4
    Sep 24, 2021 at 12:58
  • $\begingroup$ @RaphaelB4 : This is now fixed. $\endgroup$ Sep 24, 2021 at 13:57
  • $\begingroup$ Perfect! That's exactly what I need. (and thanks Kesten again by the way.) $\endgroup$
    – RaphaelB4
    Sep 24, 2021 at 14:55

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