 Solution

Use the convention $\mathbb{E}{e}^{sX}\le e^{{\sigma }^2{s}^2/2}$ for the centered sub-Gaussian random variable. Applying the two-sided Chernoff bound to $X$ gives $\mathbb{P}(|X|>x)\le 2e^{-x^2/(2{\sigma }^2)}$. Since $\mathbb{E}{X}^2\ge 0$, this already supplies the requested bound:For averages of independent random variables, the stronger useful statement is that the centered square of a sub-Gaussian random variable is a sub-exponential random variable. The Bernstein bound for independent sub-exponential variables then give $\mathbb{P}(n^{-1}{\sum }_i(X_i^2-\mathbb{E}{X}_i^2)>t)\le 2\exp [-cn\min \{t^2/{\sigma }^4,t/{\sigma }^2\}]$ for a universal positive $c$. These are the concentration theorems used for quadratic scan statistics.