How does one find the Inverse Tranform of the Gumbel distribution?
Let $X\sim \text{Gumbel}(\mu,\beta)$ with scale parameter $\beta>0$.
The CDF is then $F_X(x)=\text{e}^{-\text{e}^{-(x-\mu)/\beta}}$.
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Sign up to join this communityHow does one find the Inverse Tranform of the Gumbel distribution?
Let $X\sim \text{Gumbel}(\mu,\beta)$ with scale parameter $\beta>0$.
The CDF is then $F_X(x)=\text{e}^{-\text{e}^{-(x-\mu)/\beta}}$.
Finding the inverse transform follows a typical pattern. Start with the CDF, $F_X(x)$ and set equal to $U$.
Solve $F_X(x) = U$ for $x$.
$$\begin{align} \text{e}^{-\text{e}^{-(X-\mu)/\beta}} &= U\\ -\text{e}^{-\left(\frac{X-\mu}{\beta} \right)} &= \text{ln}(U) \\ -\left(\frac{X-\mu}{\beta} \right) &= \text{ln}(-\text{ln}(U)) \\ X-\mu &= -\beta \,\text{ln}(-\text{ln}(U)) \\ X &= \mu -\beta \,\text{ln}(-\text{ln}(U)) \quad \quad \square \end{align}$$
When sampling with this method, $U\sim \text{Uniform}(0,1)$ and $X\sim \text{Gumbel}(\mu,\beta)$.
More on the Inverse Transform here and here.
% MATLAB 2017a
% Code to generate X ~ Gumbel(mu,beta) with inverse transform method
% Parameters
mu = 1;
beta = 2;
n = 1000; % number of samples to generate
% Generation
U = rand(n,1); % U ~ Uniform(0,1)
X = mu - beta*log(-log(U)); % X ~ Gumbel(mu,beta)