1
$\begingroup$

Well known facts in extreme value theory:

  • Let $\{X_i\}_{\forall i \in \{1,...,n\}}$ be i.i.d. random variables with cdf $F$. If there exists $\{a_n\}_{n\in \mathbb{N}}>0$, and $\{b_n\}_{n\in \mathbb{N}}\in \mathbb{R}$ such that $Z_n\equiv \frac{M_n-b_n}{a_n} \Rightarrow_n Z$, where $Z$ has distribution of the same type as Gumbel and $M_n\equiv \max_{i\in \{1,...,n\}}X_i$, then we say that $F$ is in the domain of attraction of the Gumbel.

  • The norming constants can be taken $$ b_n\equiv F^{-1}(1-\frac{1}{n}) $$ and $$ a_n\equiv A(b_n) $$ where $A:\mathbb{R}\rightarrow (0,\infty)$ is called auxiliary function of $F$ and $F^{-1}$ denotes quantile function.

  • Auxiliary functions are not unique. If the pdf $f$ of $F$ exists, an auxiliary function is $$ A(x)\equiv \frac{1-F(x)}{f(x)} $$

Question: suppose that $F$ is Gumbel; what is $A$? I found in some sources that $A(x)=1$ but I don't understand how to show it. I am also confused because when I compute $$ A(x)\equiv \frac{1-F(x)}{f(x)}= \frac{1-e^{-e^{-x}}}{e^{-(x+e^{-x})}} $$ I don't get $1$.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Is this self-study? $A(x)$ (the inverse of the hazard rate) is not equal to $1$ but its tends to $1$ for $x \to \infty$. What matters is the value of $A(b_n)$ for large $n$. Moreover, note that the Gumbel is max-stable so $Z_n$ has the same distribution as $X$ for some $a_n$ and $b_n$ that are easily found. $\endgroup$
    – Yves
    Apr 18, 2018 at 15:04
  • $\begingroup$ I'm confused: for $x$ very large the numerator of $A(x)$ is almost zero. $\endgroup$
    – Star
    Apr 18, 2018 at 15:12
  • 2
    $\begingroup$ Yes, as is the denominator. Both are $\sim e^{-x}$ so the ratio is $\sim 1$. $\endgroup$
    – Yves
    Apr 18, 2018 at 15:16
  • $\begingroup$ OK, thank you. Regarding the $b_n$ in my question, would it work to use $F^{-1}(1-\frac{1}{\sqrt{n}})$ instead of $F^{-1}(1-\frac{1}{n})$? $\endgroup$
    – Star
    Apr 18, 2018 at 15:30
  • 1
    $\begingroup$ No because $(1 - 1/ \sqrt{n})^n$ does not tend to a finite limit, which is the property ensuing that $b_n$ is in the bulk of the distribution of $M_n$. See my answer to thequestion stats.stackexchange.com/q/106068/10479 $\endgroup$
    – Yves
    Apr 18, 2018 at 16:04

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.