I am trying to understand the basics of Extreme Value Theory with my limited mathematical abilities.
Currently, I am stuck with the following limit:
$$ \lim_{\xi \to 0} \exp\Big(-(1 + \xi s)^{-\frac{1}{\xi}}\Big)$$
I am told this should produce the Gumbel:
$$ \exp\Big(-\exp(-s)\Big) $$
Any assistance would be much appreciated. Thanks.
I leave it to you to justify why \begin{align} \lim_{\xi \to 0}\frac{\log(1+\xi s)}{\xi} &= s. \end{align} Now, since $\exp$ is continuous, we can "pull the limit inside": \begin{align} \lim_{\xi\to 0}\exp\left( -(1+\xi s)^{-1/\xi}\right) &= \exp\left(-\lim_{\xi \to 0}(1+\xi s)^{-1/\xi}\right) \\ &= \exp\left[-\lim_{\xi \to 0} \exp\left(-\frac{\log(1+\xi s)}{\xi}\right) \right] \\ &= \exp\left[-\exp\left( -\lim_{\xi \to 0}\frac{\log(1+\xi s)}{\xi}\right) \right] \\ &= \exp\left[-\exp(-s)\right], \end{align} where in the second line I used the definition of exponentiation ($a^b := \exp(b\log(a))$) to deduce that $(1+\xi s)^{-1/\xi} = \exp\left(-\frac{\log(1+\xi s)}{\xi}\right)$.