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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?

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  • $\begingroup$ You should repeat the reference to "the article" that is in your other questions. $\endgroup$
    – JimB
    Commented Nov 14, 2022 at 23:17
  • $\begingroup$ @JimB Thanks. I have added one. $\endgroup$
    – Hermi
    Commented Nov 15, 2022 at 2:01

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This $M_n$ represents a sequence of variables $M_1 = max(X_1)$, $M_2 = max(X_1,X_2)$, $M_3 = max(X_1,X_2,X_3)$, etcetera. And this sequence will have a different scale and location depending on $n$ which makes that it doesn't approach a fixed distribution.

The numbers $a_n$ and $b_n$ are necessary to make a sequence of variables $M_n$ approach a distribution that is fixed. The variable $M_n$ does not approach a fixed distribution. It is a scaled and shifted variable $\frac{M_n-a_n}{b_n}$ that approaches a distribution.

The fitting of variables $Y_i$ does not have this issue. You can see these $Y_i$ as multiple instances of different $M_n$ with $n$ fixed. There is only a single $n$ and a single distribution to be found/fitted.

For example $Y_i$ could be the year maximum of daily energy use. Then $n = 365$ the number of days per year, and $i$ refers to the particular year. The distribution of $Y_{2020}$, $Y_{2021}$, $Y_{2022}$ is assumed to be constant and does not scale with some parameters like $a_n$, $b_n$ ($n$ does not refer to the index of a variable, 2020, 2021, etc., but to the sample size that is used to compute the maximum).

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    $\begingroup$ @Hermi The limiting distribution is a Gumbel distribution, but in practice we do not have this limiting distribution. Still, we can approximate it with a Gumbel distribution. The parameters $a_n$ and $b_n$ will be absorbed by the location and scale parameters of the approximate Gumbel distribution. Instead of shifting and stretching $Y_i$ to fit it to a standard Gumbel distribution, you shift and stretch the Gumbel distribution to fit the $Y_i$. $\endgroup$
    – Sextus Empiricus
    Commented Nov 17, 2022 at 15:42
  • $\begingroup$ @Hermi Yes $Y_i$ might be assumed to follow a Gumbel distribution. And the subscript $_i$ does not refer to something like the sequence $X_1, \dots X_n$ that your quote speaks about. $\endgroup$
    – Sextus Empiricus
    Commented Nov 28, 2022 at 22:15
  • $\begingroup$ Why the subscript $I$ does not refer to the sequence $X_1,\dots, X_n$? The definition $Y_{2020}=\max\{X_{2020,1},\dots, X_{2020, 365}\}$ for 2020 year. $\endgroup$
    – Hermi
    Commented Nov 28, 2022 at 22:21
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    $\begingroup$ @Hermi yes could say that you have $Y_i = max(X_{i1}, X_{i,2}, \dots, X_{i,n})$ and you have a set of such $Y_i$ which follow a Gumbel distribution. But the subscript $i$ refers to a particular year. The second subscript in the range $max(X_{i1}, X_{i,2}, \dots, X_{i,n})$ refers to a sequence where $n$ hypothetically goes to infinity. It is in the limit $n\to \infty$ that the maximum (when appropriately scaled and shifted) approaches a limit distribution. In reality you don't have this infinite range but still you got close to a Gumbel distribution. $\endgroup$
    – Sextus Empiricus
    Commented Nov 28, 2022 at 22:22

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