I am wondering if there is an extreme value distribution that is closed under both the minimum and the maximum operation.
For example, for there is a Gumbel maximum distribution closed under the maximum (provided $\beta$ is the same for both distributions). Also there is the Gumbel minimum distribution closed under minimum. However, I am interested in a distribution that is closed under both such that, given two distributions $X\_1,X\_2$, I can find $Y\_1=\min(X\_1,X\_2)$ and $Y\_2=\max(X\_1,X\_2)$, which are of the same kind of distribution.
If it only holds in special cases (except the trivial case if iid) that would also be interesting.


I imagine that it is possible that no such kind of distribution exists, in which case my question is why this does not exist / if there is a proof that it cannot exist.


Thank you.



Extreme value distribution for both minimum and maximum at the same time
First, a few comments to make sure I understand the question correctly:


* I believe the correct wording is "stabie under maxima/minima" or "max/min-stable" rather than "closed". (Similarly, I think "same kind of distribution" is not widely used, a more standard term would be "distribution of the same type".)
* Max-stable and min-stable distributions are completely described; see, for example, the Wikipedia entry on [Generalized extreme value distribution](https://en.wikipedia.org/wiki/Generalized_extreme_value_distribution).


The distribution of $X$ is min-stable if and only if the distribution of $-X$ is max-stable. So the question boils down to finding a max-stable distribution with CDF $F(x)$ such that $1-F(-x)$ is a CDF of a max-stable distribution, too.


By the extreme value theorem, we necessarily have $F(x) = \exp(-t(x))$ with either
\[ t(x) = e^{-(x-\mu) / \sigma} \qquad \text{or} \qquad t(x) = (1 + \xi \tfrac{x-\mu}{\sigma})^{-1/\xi}\]
on the support of $X$. It is now a simple exercise to verify that $1 - F(-x)$ is *not* of the same form (the behaviour of $1 - F(-x)$ at $\infty$ or at the finite end of the support is wrong), so there are no non-trivial solutions.


