3
$\begingroup$

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.

  • A necessary and sufficient condition for being in the domain of attraction of Gumbel is $$ \exists \text{ }A:\mathbb{R}\rightarrow (0,\infty) \text{ s.t. } \lim_{s\rightarrow w(F)}\frac{1-F(s+v A(s))}{1-F(s)} = e^{-v}\text{ }\forall v \in \mathbb{R} $$ where $w(F)$ is the right end point of $F$ and $A$ is called auxiliary function of $F$.

Below I will focus on distributions for which this necessary and sufficient condition holds. In this case:

  • The norming constants can be taken $$ b_n\equiv F^{-1}(1-\frac{1}{n}) $$ and $$ a_n\equiv A(b_n) $$ where $F^{-1}$ 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:

Assume that $F$ is continuous and $X_i$ has unbounded support. Then, $$ \lim_{n\rightarrow \infty}b_n\equiv \lim_{n\rightarrow \infty}F^{-1}(1-\frac{1}{n})=F^{-1}(1-\lim_{n\rightarrow \infty} \frac{1}{n})=F^{-1}(1)=\infty $$

A proof that I'm considering uses also that $$ \lim_{n\rightarrow \infty}a_n $$ exists and is finite, in order to prove, in turn, that $$ \lim_{n\rightarrow \infty} F(a_nt+b_n)=1 \text{ }\forall t \in \mathbb{R} $$

Could you help me to show that $$ \lim_{n\rightarrow \infty}a_n $$ exists and is finite?

$\endgroup$
  • 1
    $\begingroup$ For any $p\gt 0,$ there exist distributions with $1-F(x) = x^{-p}$ when $x$ is sufficiently large. What happens to $A(x)$ for these distributions? $\endgroup$ – whuber Apr 19 '18 at 13:35
  • $\begingroup$ Extreme value theory allows for three distribution type, not just the Gumbel. Also it is not true that norming constants exist for any distribution. $\endgroup$ – Michael Chernick Apr 25 '18 at 20:05
  • 1
    $\begingroup$ @MichaelChernick Thank you. My question is focused on distributions in the domain of the Gumbel and for which the norming constants exist and can be defined as suggested above. $\endgroup$ – user3285148 Apr 26 '18 at 7:22
2
$\begingroup$

If I understood your question correctly, I am not sure that you have to implicate whether $\lim_{n \rightarrow \infty}a_n$ is finite or not in proving that $\lim_{n\rightarrow \infty} F(a_nt+b_n)=1$. If you know that $F$ satisfies the condition you mention and $w(F) = \infty$ then $$\lim_{s\rightarrow \infty}\frac{1-F[s+v A(s)]}{1-F(s)} = e^{-v}\text{ }\forall v \in \mathbb{R}$$

you also know that

$$\lim_{s\rightarrow \infty} [1-F(s)] = 0$$

Thus, $\forall v \in \mathbb{R}$

$$\lim_{s\rightarrow \infty}\left([1-F(s)]\frac{1-F[s+v A(s)]}{1-F(s)} \right) = 0e^{-v} = 0$$

But

$$\lim_{s\rightarrow \infty}\left([1-F(s)]\frac{1-F[s+v A(s)]}{1-F(s)} \right) = \lim_{s\rightarrow \infty}\Big(1 - F[s+v A(s)]\Big)$$

From these two, you get $$\lim_{s\rightarrow \infty}F[s+v A(s)] = 1$$

That means that for every sequence of $s$ going to $\infty$ the limit will be the same. Setting $s=a_n$ and taking into account that $b_n = A(a_n)$, you get

$$\lim_{n\rightarrow \infty}F(b_n + v a_n) = 1$$

which is what you are trying to prove. This further gives that $\lim_{n\rightarrow \infty}(b_n + v a_n) = \infty$ which is not affected by whether $\lim_{n\rightarrow \infty}a_n$ is finite or not.

If, on the other hand, you try to prove for an arbitrary continuous distribution $F$ with unbounded support that $\lim_{n\rightarrow \infty}a_n$ is finite, then $f(x) = \frac{1}{2} e^{-\sqrt x}$ gives you a counterexample.

$\endgroup$
  • $\begingroup$ Thank you. Why is that $\lim_{s\rightarrow \infty} \frac{1-F(s+vA(s))}{1-F(s)}=e^{-v}$ implies $\lim_{s\rightarrow \infty}F(s+vA(s))=1$? $\endgroup$ – user3285148 May 1 '18 at 10:48
  • $\begingroup$ Hi @user3285148! Note that $\forall v \in \mathbb{R}$, $e^{-v}$ is finite and thus $\lim_{s\rightarrow \infty} \frac{1-F(s+vA(s))}{1-F(s)}=e^{-v}$ should also be finite. But $\lim_{s\rightarrow \infty} (1-F(s)) = 0$. Now, if $\lim_{s\rightarrow \infty}(1-F(s+vA(s)))$ was not $0$, $\lim_{s\rightarrow \infty} \frac{1-F(s+vA(s))}{1-F(s)}$ should be $\infty$. Thus $\lim_{s\rightarrow \infty}(1-F(s+vA(s))) = 0$ meaning $\lim_{s\rightarrow \infty}F(s+vA(s)) = 1$. $\endgroup$ – Sotiris May 1 '18 at 16:07
  • $\begingroup$ Of course, the above assumes the existence of $\lim_{s\rightarrow \infty}(1-F(s+vA(s)))$ but is somewhat intuitive. If you want the existence proven as well, you can start instead from $\lim_{s\rightarrow \infty} \frac{1-F(s+vA(s))}{1-F(s)} (1 - F(s))$ which exists (because the individual limits of the product terms exist) and, on the one hand is $e^{-v} 0$ while on the other hand is $\lim_{s\rightarrow \infty} (1-F(s+vA(s)))$ showing both the existence and the exact value. $\endgroup$ – Sotiris May 3 '18 at 13:14
  • $\begingroup$ I have added the above details in the answer! :) $\endgroup$ – Sotiris May 3 '18 at 13:14

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.