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
$P[\exists v \in \{2^u,...,2^{u+1} \} | \frac{1}{v} \sum\_{t=0}^v X\_t - \mu| > 
\frac{\delta}{\sqrt{v}}] \leq \exp(-\frac{\delta^2}{2 \sigma^2})$.



Understanding some Hoeffding-type martingale inequality
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 $h:=\delta/\sqrt{2n}$. 


