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For the pdf $$f(x\mid\theta)=e^{-(x-\theta)}\exp(-e^{-(x-\theta)})\,,\qquad-\infty<x<\infty,-\infty<\theta<\infty$$, let $X_1,\cdots,X_n$ be i.i.d observations. How to find the sufficient statistic and show the complete statistic does not exit?

The answer of it is that the order statistics are minimal sufficient statistics. I want to know under what circumstances we should consider order statistics?

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    $\begingroup$ This seems like a homework question. If it is, you should add a "self study" tag. $\endgroup$ – Gordon Smyth Apr 28 '17 at 8:16
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    $\begingroup$ The premises of your question are not correct. The order statistics are not minimal sufficient and a very simple complete sufficient statistic does exist. Just apply the transformations $y=\exp(-x)$ and $\mu=\exp(\theta)$ and this will become clear. $\endgroup$ – Gordon Smyth Apr 28 '17 at 8:29
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Answer already hinted at in comments...

A complete sufficient statistic does exist for this family of distributions, a Gumbel density with scale parameter unity. The set of order statistics is not minimal sufficient here. We should consider order statistics, which are trivially sufficient for any family of distributions, when we cannot find any other (non-trivial) sufficient statistic, say for the Laplace distribution and the Cauchy distribution with unknown location parameter.

Due to independence, joint density of $(X_1,X_2,\cdots,X_n)$ is

\begin{align} f_{\theta}(x_1,x_2,\cdots,x_n)&=\prod_{i=1}^nf(x_i\mid\theta) \\&=e^{-\sum_{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_{i=1}^nx_i} \\&=g(\theta,t(\mathbf x))\,h(\mathbf x)\qquad,\text{ for all } \mathbf x=(x_1,\cdots,x_n)\in\mathbb R^n\,,\,\theta\in\mathbb R \end{align}

where $g(\theta,t(\mathbf x))=\exp\left(-e^{\theta}\sum_{i=1}^n e^{-x_i}+n\theta\right)$ depends on $\theta$ and on $x_1,x_2,\cdots,x_n$ through $t(\mathbf x)=\sum_{i=1}^n e^{-x_i}$ and $h(\mathbf x)=e^{-\sum_{i=1}^nx_i}$ is independent of $\theta$.

So by Factorisation theorem, a sufficient statistic for $\theta$ is $$T(\mathbf X)=\sum_{i=1}^n e^{-X_i}$$


Let $Y=e^{-X}$ where $X$ has the density $f(x\mid\theta)$.

Then density of $Y$ is

\begin{align}f_Y(y)&=f(-\ln y\mid\theta)\left|\frac{dx}{dy}\right|\mathbf1_{y>0} \\&=\frac{1}{y}e^{\ln y+\theta}\exp\left(-e^{\ln y+\theta}\right)\mathbf1_{y>0} \\&=e^{\theta}\exp(-ye^{\theta})\mathbf1_{y>0} \end{align}

That is, $Y\sim\text{Exp}(\lambda)$ with $\lambda=e^{\theta}(>0)$.

In other words, $e^{-X_i}\stackrel{\text{i.i.d}}\sim\text{Exp}(\lambda)$ for each $i=1,2,\cdots,n$, which implies

$$T\sim\text{Gamma}(\lambda,n)$$


To show that $T$ is indeed a complete statistic, we have to show that for any function $\psi$ of $T$, $$E_{\theta}(\psi(T))=0\implies \psi(T)=0\quad,\text{ a.e.}$$

Now, \begin{align}E_{\theta}(\psi(T))&=0 \\\implies \int_0^\infty \psi(t)e^{-\lambda t}t^{n-1}\,dt &=0 \end{align}

It can be shown using the unicity property of Mellin transform that the last implication means $\psi(t)$ itself is identically zero. There might be an easier proof for completeness but I could not find one.

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  • $\begingroup$ Since the pdf is a member of the one-parameter exponential family, it can be directly stated that $\sum e^{-X_i}$ is a complete sufficient statistic. $\endgroup$ – StubbornAtom Oct 29 '18 at 8:57

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