Let $X$ be standard normal and $a>b>0$, prove that $\lim\limits_{\epsilon\to 0}\epsilon^2\log P(|\epsilon X -a|

Question

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|

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|

I tried using L'Hopital's rule:

$$\frac{\log P(|\epsilon X -a|

Then we need to compute $\frac{d}{d\epsilon }P(|\epsilon X -a|

$$\frac{d}{d\epsilon }P(|\epsilon X -a|

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?

Answer 1

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|

Answer 2

Let $I(\epsilon)$ be the integral given by

$$I(\epsilon)=\frac{1}{\sqrt{2\pi}}\int_{(a-b)/\epsilon}^{(a+b)/\epsilon}e^{-t^2/2}\,dt$$

Then, we have the following estimates. An upper bound for $I(\epsilon)$ is

$$I(\epsilon)= \frac{1}{\sqrt{2\pi}}\int_{(a-b)/\epsilon}^{(a+b)/\epsilon}e^{-t^2/2}\,dt\le \frac{1}{\sqrt{2\pi}}e^{-(a-b)^2/(2\epsilon^2)}\frac{b}{2\epsilon}\tag 1$$

For $0<\epsilon<2b$, a lower bound for $I(\epsilon)$ is

$$\begin{align} I(\epsilon)&= \frac{1}{\sqrt{2\pi}}\int_{(a-b)/\epsilon}^{(a+b)/\epsilon}e^{-t^2/2}\,dt\\\\ &\ge \frac{1}{\sqrt{2\pi}}\int_{(a-b)/\epsilon}^{(a-b)/\epsilon+1}e^{-t^2/2}\,dt\\\\ &\ge \frac{1}{\sqrt{2\pi}} e^{-\frac12\left(\frac{a-b}{\epsilon}+1\right)^2}\\\\ &=\frac{1}{\sqrt{2\pi e}}\,e^{-(a-b)/\epsilon}\,e^{-(a-b)^2/(2\epsilon^2)}\tag 2 \end{align}$$

Using $(1)$ and $(2)$ together shows

$$-\frac{(a-b)^2}2-\epsilon(a-b)+\epsilon^2 \log\left(\frac{1}{\sqrt{2\pi e}}\right) \le \epsilon^2 \log(I(\epsilon))\le -\frac{(a-b)^2}2 +\epsilon^2 \log\left(\frac{b}{2\sqrt{2\pi}\,\epsilon}\right)$$

whence application of the squeeze theorem yields the coveted limit.