I have parameters of two Gumbel distributions ($\mu\_1, \beta\_1)$ and $(\mu\_2, \beta\_2)$. Since max of 2 Gumbels is a Gumbel, I'd like to compute $\mu\_m, \beta\_m$, so that:


$Gumbel(\mu\_m,\beta\_m)$ = $max(Gumbel(\mu\_1, \beta\_1), Gumbel(\mu\_2, \beta\_2))$


Any hints how to do this? 
I've found a solution for the case when $\beta\_1 = \beta\_2$ in here (point 6, page 105):


<https://books.google.de/books?id=oLC6ZYPs9UoC&pg=PA105&lpg=PA105&dq=max+of+2+gumbel&source=bl&ots=nMdvj2clIe&sig=sL3rurAqDXpBpQP2McndFbcEuxk&hl=en&sa=X&ved=0CCgQ6AEwAWoVChMI1byv4cLXyAIVgb0aCh2KQwpg#v=onepage&q=max%20of%202%20gumbel&f=false>


but I'm even not sure how this was computed.



Formula for maximum of two Gumbel distributions?
Presumably your Gumbel random variables are independent.
A Gumbel random variable $X\_i$ with location parameter $\alpha\_i$ and scale parameter $\beta\_i$ has CDF
$$ F\_i(x) = \exp(-\exp((\alpha\_i - x)/\beta\_i))$$
Then the CDF for $X = \max(X\_1,X\_2)$ is
$$F\_X(x) = \mathbb P(X\_1 \le x) \mathbb P(X\_2 \le x) = \exp(-\exp((\alpha\_1-x)/\beta\_1) - \exp((\alpha\_2-x)/\beta\_2))$$
This is **not** a Gumbel distribution unless $\beta\_1 = \beta\_2 = \beta$, in which case it is $\exp(-exp((\alpha - x)/\beta))$ where $\alpha = \beta \ln(\exp(\alpha\_1/\beta) + \exp(\alpha\_2/\beta))$.


