# Fitting Gumbel distribution based the maximal observation

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?

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

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).
• @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 $Y_i$. Nov 17, 2022 at 15:42
• @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. Nov 28, 2022 at 22:15
• 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. Nov 28, 2022 at 22:21
• @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. Nov 28, 2022 at 22:22