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I would like to ask about the statistical inference of the extreme value theorem.

Let $X_1,X_2,\dots,X_n$ be iid random variables with distribution function $F(x)$ and $M_n:=\max\{X_1,\dots,X_n\}$. By Fisher-Tippett theorem, there exist two sequences of real numbers $a_n>0$ and $b_n$ such that the following limits converge to a non-degenerate distribution function $G(x)$: $$ \lim_{n\to \infty} P(\frac{M_n-b_n}{a_n}\le x)=\lim_{n\to \infty} F(a_nx+b_n)=G(x) $$ where $G(x)$ is one of the extreme value distributions.

Assume that we only consider $$G(x)=\exp(-\exp(\frac{x-\mu}{\sigma}))$$ is the Gumbel distribution.

Question: Suppose we have a set of maximum values $\{Y_i\}_{i=1}^m$, why can the article directly (for example: here) use this set of maximum values to fit the Gumbel distribution based on the maximal likelihood method $$ l(\mu,\sigma; Y_1,\dots, Y_m)=-n\log \sigma-\sum_{i=1}^m \left(\exp(\frac{X_i-\mu}{\sigma})-\frac{X_i-\mu}{\sigma} \right) $$ ?

By extreme value theorem, we only know that $$ P(\frac{Y_n-b_n}{a_n}\le x)=G(x) $$ but not $ P(Y_n\le x)=G(x). $

So I'm a bit confused about fitting the extreme value distribution directly with $Y_i$ for $i=1,\dots, m$, don't we need to normalize it?

Assume that we only consider $$G(x)=\exp(-\exp(\frac{x-\mu}{\sigma}))$$ is the Gumbel distribution.

Question: Suppose we have a set of maximum values $\{Y_i\}_{i=1}^m$, why can the article directly (for example: here) use this set of maximum values to fit the Gumbel distribution based on the maximal likelihood method ?

By extreme value theorem, we only know that $$ P(\frac{Y_n-b_n}{a_n}\le x)=G(x) $$ but not $ P(Y_n\le x)=G(x). $

So I'm a bit confused about fitting the extreme value distribution directly with $Y_i$ for $i=1,\dots, m$, don't we need to normalize it?

I would like to ask about the statistical inference of the extreme value theorem.

Let $X_1,X_2,\dots,X_n$ be iid random variables with distribution function $F(x)$ and $M_n:=\max\{X_1,\dots,X_n\}$. By Fisher-Tippett theorem, there exist two sequences of real numbers $a_n>0$ and $b_n$ such that the following limits converge to a non-degenerate distribution function $G(x)$: $$ \lim_{n\to \infty} P(\frac{M_n-b_n}{a_n}\le x)=\lim_{n\to \infty} F(a_nx+b_n)=G(x) $$ where $G(x)$ is one of the extreme value distributions.

Assume that we only consider $$G(x)=\exp(-\exp(\frac{x-\mu}{\sigma}))$$ is the Gumbel distribution.

Question: Suppose we have a set of maximum values $\{Y_i\}_{i=1}^m$, why can the article directly use this set of maximum values to fit the Gumbel distribution based on the maximal likelihood method $$ l(\mu,\sigma; Y_1,\dots, Y_m)=-n\log \sigma-\sum_{i=1}^m \left(\exp(\frac{X_i-\mu}{\sigma})-\frac{X_i-\mu}{\sigma} \right) $$ ?

By extreme value theorem, we only know that $$ P(\frac{Y_n-b_n}{a_n}\le x)=G(x) $$ but not $ P(Y_n\le x)=G(x). $

So I'm a bit confused about fitting the extreme value distribution directly with $Y_i$ for $i=1,\dots, m$, don't we need to normalize it?

I would like to ask about the statistical inference of the extreme value theorem.

Let $X_1,X_2,\dots,X_n$ be iid random variables with distribution function $F(x)$ and $M_n:=\max\{X_1,\dots,X_n\}$. By Fisher-Tippett theorem, there exist two sequences of real numbers $a_n>0$ and $b_n$ such that the following limits converge to a non-degenerate distribution function $G(x)$: $$ \lim_{n\to \infty} P(\frac{M_n-b_n}{a_n}\le x)=\lim_{n\to \infty} F(a_nx+b_n)=G(x) $$ where $G(x)$ is one of the extreme value distributions.

Assume that we only consider $$G(x)=\exp(-\exp(\frac{x-\mu}{\sigma}))$$ is the Gumbel distribution.

Question: Suppose we have a set of maximum values $\{Y_i\}_{i=1}^m$, why can the article directly (for example: here) use this set of maximum values to fit the Gumbel distribution based on the maximal likelihood method $$ l(\mu,\sigma; Y_1,\dots, Y_m)=-n\log \sigma-\sum_{i=1}^m \left(\exp(\frac{X_i-\mu}{\sigma})-\frac{X_i-\mu}{\sigma} \right) $$ ?

By extreme value theorem, we only know that $$ P(\frac{Y_n-b_n}{a_n}\le x)=G(x) $$ but not $ P(Y_n\le x)=G(x). $

So I'm a bit confused about fitting the extreme value distribution directly with $Y_i$ for $i=1,\dots, m$, don't we need to normalize it?