Explicit Laplace approximation for tail of gaussian distribution

Question

I'm studying some lecture notes by S. R. Srinivasa Varadhan about Large Deviations Theory and I have some trouble understanding a simple equation right on page 2 where it says $$\frac{\sqrt{n}}{2\pi} \int_l^\infty \exp{\left(-\frac{nx^2}{2}\right)}\mathrm dx=\exp{\left(-\frac{n l^2}{2}+o(n) \right)}.$$ The left side of the equation looks to be the same as $1-\Phi(\sqrt{n}x)=\Phi(-\sqrt{n}x)$ where $\Phi$ is the standard normal CDF. But I don't see how to get there: Is this a simple (approximative) calculation or a fact about the normal distribution?

Answer 1

Regarding the left hand side, \begin{align*} \frac{\sqrt{n}}{2\pi} \int_l^\infty \exp\left\{-\frac{nx^2}{2}\right\}\,dx &=\frac{\sqrt{n}}{2\pi} \int_l^\infty \exp\left\{-\frac{1}{2}\left(\frac{x}{1/\sqrt n}\right)^2\right\}\,dx\\ &=\frac{\sqrt{n}}{2\pi}\cdot\sqrt{2\pi}(1/\sqrt n) \int_l^\infty \frac{1}{\sqrt{2\pi}(1/\sqrt n)}\exp\left\{-\frac{1}{2}\left(\frac{x}{1/\sqrt n}\right)^2\right\}\,dx\\ &=\frac{\sqrt{2\pi}}{2\pi}\int_l^\infty \frac{1}{\sqrt{2\pi}(1/\sqrt n)}\exp\left\{-\frac{1}{2}\left(\frac{x}{1/\sqrt n}\right)^2\right\}\,dx\\ &=\frac{\sqrt{2\pi}}{2\pi}\int_{l\sqrt n}^\infty \frac{1}{\sqrt{2\pi}(1/\sqrt n)}\exp\left\{-\frac{1}{2}u^2\right\}\,\left(\frac{1}{\sqrt n}\,du\right)\\ &=\frac{\sqrt{2\pi}}{2\pi}\int_{l\sqrt n}^\infty \frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}u^2\right\}\,du\\ &=\frac{1}{\sqrt{2\pi}}\left[1-\Phi\left(\sqrt n l\right)\right] \end{align*}