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Would anyone know how to prove the following, coming from the proof of theorem 2 in this paper (https://arxiv.org/pdf/1605.08671.pdf)?

Consider i.i.d. Sub Gaussian random variables $(X_t)_{t\geq 1}$ with parameter $\sigma$ and mean $\mu$. Let $u \geq 1$. Then $P[\exists v \in \{2^u,...,2^{u+1} \} | \frac{1}{v} \sum_{t=0}^v X_t - \mu| > \frac{\delta}{\sqrt{v}}] \leq \exp(-\frac{\delta^2}{2 \sigma^2})$.

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1 Answer 1

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In the mentioned paper, they have $\sum_{t=1}^v$ rather than $\sum_{t=0}^v$. Anyhow, the inequality is incorrect in general. Indeed, without loss of generality, $\mu=0$ and $\sigma=1$. Suppose that (say) the $X_i$'s are just standard normal. Then, letting $n:=2^{u+1}[\ge4]$ and $S_v:=\sum_{t=1}^v X_t$, we see that $S_{n/2},\dots,S_n$ are jointly normal, and so, they have a bounded joint density. So, the lhs of the inequality is $1-O(\delta^{n/2+1})=1-o(\delta^2)$ as $\delta\downarrow0$, whereas the rhs is $1-\delta^2/(2+o(1))$.

However, the inequality can be easily fixed as follows, likely without any serious damage to the proof of that theorem in that paper (again assuming $\mu=0$ and $\sigma=1$): In view of Doob's maximal inequality for submartingales and the sub-Gaussian property, the lhs of your inequality (with $\sum_{t=1}^v$ rather than $\sum_{t=0}^v$) is no greater than \begin{align*} P(\max_{0\le v\le n}|S_v|>\delta\sqrt{n/2}) &\le e^{-h\delta\sqrt{n/2}}Ee^{h|S_n|} \\ & \le e^{-h\delta\sqrt{n/2}}(Ee^{hS_n}+Ee^{-hS_n}) \\ & \le 2\exp\{-h\delta\sqrt{n/2}+nh^2/2\} =2\exp\{-\delta^2/4\}, \end{align*} where $h:=\delta/\sqrt{2n}$.

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