TITLE: Let $X$ be standard normal and $a>b>0$, prove that $\lim\limits_{\epsilon\to 0}\epsilon^2\log P(|\epsilon X -a|<b)=-\frac{(a-b)^2}{2}$
QUESTION [6 upvotes]: Let $X$ be a standard normal random variable, with $a,b>0$ and $a-b>0$,  prove that $$\lim_{\epsilon\to 0}\epsilon^2\log P(|\epsilon X -a|<b)=-\frac{(a-b)^2}{2}$$

I'm studying for a qual and this was a previous problem (problem 6b). From part a, we know $\lim_{\epsilon\to 0} P(|\epsilon X -a|<b)=0$
I tried using L'Hopital's rule:
$$\frac{\log P(|\epsilon X -a|<b)}{1/\epsilon^2}=\frac{1}{-2/\epsilon^3}\frac{1}{P(|\epsilon X -a|<b)}\frac{d}{d\epsilon }P(|\epsilon X -a|<b)$$
Then we need to compute $\frac{d}{d\epsilon }P(|\epsilon X -a|<b)$:
$$\frac{d}{d\epsilon }P(|\epsilon X -a|<b)=\frac{d}{d\epsilon} \int_{(a-b)/\epsilon}^{(a+b)/\epsilon} \frac{1}{\sqrt{2\pi}}e^{-t^2/2}~dt$$
Here is where I am stuck. Of course I can use fundamental theorem of calculus but it turns into a huge mess. How do I proceed from here?

REPLY [8 votes]: Laplace-type methods are the classical approach to this kind of large deviations results but it might be worth to detail the computations in the present case. We already know that
$$
P(|\epsilon X -a|<b)=\frac1{\sqrt{2\pi}}I(\epsilon)\quad \text{with}\quad  I(\epsilon)=\int_{(a-b)/\epsilon}^{(a+b)/\epsilon} e^{-t^2/2}~dt
$$
The change of variable 
$$
t=(a-b)/\epsilon+s\epsilon
$$
yields
$$
I(\epsilon)=e^{-(a-b)^2/2\epsilon^2}\epsilon J(\epsilon)\quad \text{with}\quad   J(\epsilon)=\int_{0}^{2b/\epsilon^2}e^{-s^2\epsilon^2/2-s(a-b)}~ds
$$
Now, for every $\epsilon$,
$$
J(\epsilon)\leqslant\int_{0}^{\infty}e^{-s(a-b)}~ds=\frac1{a-b}
$$
and, for every $\epsilon$ in $(0,1)$, $2b/\epsilon^2\geqslant2b/\epsilon$ hence
$$
J(\epsilon)\geqslant\int_{0}^{2b/\epsilon}e^{-2b^2-s(a-b)}~ds=\frac{e^{-2b^2}}{a-b}-o(1)
$$
Finally, $J(\epsilon)=\Theta(1)$ hence $\log J(\epsilon)=\Theta(1)$ and, as desired,
$$
\epsilon^2\log P(|\epsilon X -a|<b)=\epsilon^2\log I(\epsilon)+\Theta(\epsilon^2\log\epsilon)=\color{red}{-\tfrac12(a-b)^2}+\Theta(\epsilon^2\log\epsilon)
$$ 
More generally, for every interval $B$ (and this case can be still further extended to other Borel sets), 
$$
\epsilon^2\log P(\epsilon X \in B)=-\tfrac12\inf_{x\in B}x^2
$$