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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.

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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 $\alpha = \beta \ln(\exp(\alpha_1/\beta) + \exp(\alpha_2/\beta))$.

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