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

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 $n \to \infty$.

Answer 1

Using Cramér theorem of Large Deviation Theory, you can say that: $$ \displaystyle\lim_{n\to\infty}\frac{1}{n}\log\mathbb P(\frac{1}{n}\sum_{i=1}^n X_i\geq \delta)=-I(\delta) $$ where $$ I(x)=\sup_{t\in\mathbb R}[xt-\log \phi(t)] $$ for $\phi(t)$ the moment generating function of $X_i$ which in this case is $\phi(t)=e^{t^2/2}$. Hence: $$ I(\delta)=\sup_{t\in\mathbb R}[\delta t-\frac{t^2}{2}]=\frac{\delta^2}{2}. $$ Now it is enough to see the following using the assumption that the standard normal RV is symmetric around zero: $$ \displaystyle\lim_{n\to\infty}\frac{1}{n}\log \mathbb P(\frac{1}{n}\sum_{i=1}^n X_i\geq \delta)=\displaystyle\lim_{n\to\infty}\frac{1}{n}\log\frac{1}{2}\mathbb P(|\frac{1}{n}\sum_{i=1}^n X_i|\geq \delta)\\ =\displaystyle\lim_{n\to\infty}\frac{1}{n}\log\mathbb P(|\frac{1}{n}\sum_{i=1}^n X_i|\geq \delta)+\displaystyle\lim_{n\to\infty}\frac{1}{n}\log\frac{1}{2}\\ =\displaystyle\lim_{n\to\infty}\frac{1}{n}\log\mathbb P(|\frac{1}{n}\sum_{i=1}^n X_i|\geq \delta) $$

Answer 2

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.