TITLE: Large deviations: showing $n^{-1}\log \mathbb P(|n^{-1} \sum_{i=1}^n X_i| \ge \delta) \to -\delta^2/2$ for $X_i$ i.i.d. Gaussian
QUESTION [5 upvotes]: I am having a problem with the first example of Amir Dembo and Ofer Zeitouni book Large Deviations Techniques and Applications.
Could someone please help me confirm the following statement
if $X_i$ are i.i.d standard normal random variables,
$n^{-1}\log \mathbb P(|n^{-1} \sum_{i=1}^n X_i| \ge \delta) \to -\delta^2/2$, as $n \to \infty$.

REPLY [3 votes]: DZ put this example at the very beginning of their book because everything can be made explicit in this case, without requiring the theory they develop in the rest of the book. 
To wit, let $A_n=[|X_1+\cdot+X_n|\geqslant n\delta]$, then $X_1+\cdot+X_n$ is centered normal with variance $n$ hence $P[A_n]=P[|X_1|\geqslant\delta\sqrt{n}]$. As such,
$$
\frac2{\delta\sqrt{n}\sqrt{2\pi}}\mathrm e^{-n\delta^2/2}\frac{n\delta^2}{1+n\delta^2}\leqslant P[A_n]\leqslant\frac2{\delta\sqrt{n}\sqrt{2\pi}}\mathrm e^{-n\delta^2/2},
$$
hence
$$
-\frac{\delta^2}2-\frac{\varepsilon_n}{2n}-\frac{\eta_n}n\leqslant\frac{\log P[A_n]}n\leqslant-\frac{\delta^2}2-\frac{\varepsilon_n}{2n},
$$
where 
$$
\varepsilon_n=\log(n)+\log(\pi\delta^2/2),
\qquad
\eta_n=\log\left(1+\frac1{n\delta^2}\right).
$$
Since $\varepsilon_n/n\to0$ and $\eta_n/n\to0$, the result follows.