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Suppose that $\varepsilon_i$ are independent Rademacher random variables (that is, $ \mathbb{P}(\varepsilon_i=-1) = \mathbb{P}(\varepsilon_i=1) =1/2 $. Fix an $a\in\mathbb{R}^n$ and define the random variable $X=\sum_{i=1}^n a_i\varepsilon_i$.

I am interested in lower and upper bounds on $\mathbb{E}e^{-|X|/2}$ of the form $$ L(a) \le \mathbb{E}\exp\left(-\frac12|X|\right) \le U(a) $$ that satisfy the bounded ratio property: $$ \sup_{a\in\mathbb{R}^n} \frac{U(a)}{L(a)} <\infty. $$

Note that the supremum is over $\mathbb{R}^n$ for a fixed $n$ and will almost certainly depend on $n$.

Any ideas much welcome.

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    $\begingroup$ An easy one is $U(a) = 2^{n-1} L(a) = \exp(-\frac m 2)$, where $m = \min_{\varepsilon \in \{-1,1\}^n} |\sum a_i \varepsilon_i|$. $\endgroup$
    – Mikael de la Salle
    Commented Oct 16 at 14:28
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    $\begingroup$ In the previous comment, Mikael de la Salle suggested $L(a)$ and $U(a)$ with $U(a)=2^{n-1}L(a)$. In the same spirit, here are $L(a)$ and $U(a)$ with $U(a)=L(a)$: $U(a)=L(a)=2^{-n}\sum_{\delta\in\{-1,1\}^n}\exp(-\,|\sum_1^n a_i\delta_i|/2)$. So, the question is this: In what terms do you want $L(a)$ and $U(a)$ to be expressed? $\endgroup$
    – Iosif Pinelis
    Commented Oct 16 at 21:41
  • $\begingroup$ Dear @IosifPinelis, I would be happy with any analytically tractable expression in $a$. $\endgroup$
    – Aryeh Kontorovich
    Commented Oct 17 at 17:45
  • $\begingroup$ This may be relevant: arxiv.org/abs/2006.16834 $\endgroup$
    – van der Wolf
    Commented Oct 25 at 10:28

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You can get one of the bounds using the recently proved Tomaszewski's Conjecture, which can be reformulated as that for $$ X= \sum_{i=1}^{n} a_{i} \epsilon_{i} $$ we have $$ \mathbb P\left(|X|\le \Vert a\Vert\right)\ge \frac12 \quad\text{where } \Vert a\Vert=\sqrt{a_1^2+\dots+a_n^2}. $$ Hence $$ \mathbb E e^{-|X|/2}\ge \frac12 e^{-\Vert a\Vert/2}. $$

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