The minimal sufficient statistic of $f(x) = e^{-(x-\theta)}e^{-e^{-(x-\theta)}}$

Question

The Casella Berger (2002) solutions manual says that the minimal sufficient statistic for

$$f(x) = e^{-(x-\theta)}e^{-e^{-(x-\theta)}}, \qquad x\in \mathbb{R}$$

are the order statistics $(X_{(1)},\dots,X_{(n)})$. This confuses me because it can be written as

$$f(x) = e^{\theta}e^{-x}e^{-e^\theta{e^{-x}}}$$

Which seems to be in the form of a full-rank exponential family, with complete sufficient statistic $T(X) = \sum_i e^{-X_i}$, and hence $T(X)$ is minimal sufficient by Bahadur's theorem. $T(X)$ seems to achieve a much greater reduction in the data aswell, so that the order statistics cannot be minimal sufficient?

I wonder if there is a property of the PDF I am missing that means it's not actually exponential family?

Answer 1

Indeed it is clear from the density of $X_1,\ldots,X_n$ that $\sum\limits_{i=1}^n e^{-X_i}$ is a minimal complete sufficient statistic for $\theta$ as the pdf is a member of a regular full-rank exponential family as you say:

\begin{align} f_{\theta}(x_1,\cdots,x_n)&=e^{-\sum\limits_{i=1}^n x_i+n\theta}\exp\left(-\sum_{i=1}^n e^{-(x_i-\theta)}\right) \\&=\exp\left(-e^{\theta}\sum_{i=1}^n e^{-x_i}+n\theta\right)e^{-\sum\limits_{i=1}^nx_i}\quad\small\text{ for all }(x_1,\ldots,x_n)\in\mathbb R^n\,,\,\theta\in\mathbb R \end{align}

The vector of order statistics is just sufficient, not minimal sufficient