Extreme value theory: show that $ \lim_{n\rightarrow \infty}a_n $ exists and is finite

Question

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?

Answer 1

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.