I check the Gumbel distribution article on Wikipedia, it says it is useful to represent the distribution of maxima. But it is not easy to understand how it works? A detailed explanation or examples may be more helpful.

I notice the pdf of the distribution is very close to the exponential distribution, $e^{-(z + e^{-z})}$ compared with $e^{-z}$? I believe the extra term could bring some useful properties, but I do know its concrete advantages, a close-form cdf?

  • $\begingroup$ The right tail of a Gumbel distribution is similar to that of an exponential distribution, but the left side is very different, with a positive density for all negative reals. Another point is that the location can easily be shifted (which is why you are seeing $z$ rather than $x$) $\endgroup$
    – Henry
    Commented Jan 22, 2021 at 1:08

1 Answer 1


Given a sample of iid $\{ X_1, ..., X_n \}$, we can compute its maximum $Y = \max \{ X_1, ..., X_n \}$. The question is, if the values of the sample follow a distribution $P(X)$, what is the distribution for the maximum value of the sample, $P(Y)$?

It can be shown that depending on the form of $P(X)$ in the limit $X\to\inf$, the form of $P(Y)$ changes. If $P(X)$ decays as a power-law for $X\to\inf$, then $P(Y)$ is a Frechet distribution. If there's a finite end-point $x_0$ which is approached as $(x_0-x)^\alpha$, then it follows a Weibull distribution, whose parameters can be easily written as a function of the sample size $n$. If $P(X)$ decays faster than a power-law, then $P(Y)$ is a Gumbel distribution (I'm not sure one can write its parameters as a function of $n$ as in the case of the Weibull). This is important because it shows universal behavior, i.e. details don't matter so it's the same for $e^{-x}$, $e^{-x^2}$ or any other form. Only matters that it decays faster than a power-law. You can see it in this book, page 24. Also, note that the discussion here for maximums also applies to minimums if we talk about the properties of the left tail of $P(X)$ instead of its right tail.

However, these results are only valid in the limit of an infinite sample size. If you want to have the exact result for finite $n$, you will need to compute by yourself the form of $P(Y)$ for the specific distribution $P(X)$, as shown e.g. here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.