# Approximation/bound to a_n and b_n in normal maxima to Gumbel

I was reading this which tries to find $a_n$ and $b_n$ such that

$$F\left(a_n x+b_n\right)^n\rightarrow^{n\rightarrow\infty} G(x) = e^{-\exp(-x)},$$

where $F$ is the cdf of a standard normal.

The answer from Alecos states that

$$a_n = \frac 1{n\phi(b_n)},\;\;\; b_n = \Phi^{-1}(1-1/n)$$

I am trying to get a sense of how fast $a_n$ converges to 0 and how fast $b_n$ diverges, it is easy to show that $b_n$ is bounded by some $O(\sqrt{log n})$. In fact, the answer from whuber provides the following approximations

$$a_n^\prime = \frac{\log \left(\left(4 \log^2(2)\right)/\left(\log^2\left(\frac{4}{3}\right)\right)\right)}{2\sqrt{2\log (n)}},\ b_n^\prime = \sqrt{2\log (n)}-\frac{\log (\log (n))+\log \left(4 \pi \log ^2(2)\right)}{2 \sqrt{2\log (n)}}$$

I do not fully understand how to obtain the above results, and it seems to involve a lot of tedious math, all I need is to show something like

$$\lim_{n\to\infty}\dfrac{a_n}{(\log n)^{-1/2}}=const$$

do anyone know an easy way to show the above? Or could anyone point me to the right reference?

• There's an extremely easy way: use Mills' Ratio to approximate $a_n$ using Alecos' formula. If you believe my answer, you're already done because it is explicitly proportional to $(\log(n))^{-1/2}$ and you can read the constant right off of the formula. – whuber Feb 28 '18 at 14:59

In the book by Embrechts, Klüppelberg and Mickosch section 3.3 and example 3.3.29, a quite simple derivation of the wanted result is given, based on the appealing fact that if two distributions in the Gumbel domain of attraction are tail equivalent, the same constants $a_n$ and $b_n$ can be used for both.