1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber Feb 28 '18 at 14:59
2
$\begingroup$

You can use a Von Mises condition which often help for distributions in the Gumbel domain of attraction. This involves the inverse of the hazard rate used by Alecos, which turns out to be called Mills ratio for the normal distribution.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.