# Showing that $X_{1:n}$ is sufficient for $\eta$, by factorization

I'm asked to show that $X_{1:n}$ (the minimum order statistic) is sufficient for $\eta$, in the case of a random sample $(X_1, ... , X_n)$ where $X_i\sim EXP(1,\eta$) (this is the two-parameter exponential distribution $EXP(\theta,\eta):$ $(1/\theta)\exp(-(x-\eta)/\theta)$, $x>\eta$; in this case $\theta=1$), by using the "factorization method", that is, writing $f(x_1,...,x_n;\eta)$ as $g(s,\eta)h(x_1,...,x_n)$, where $S$ is the statistic ($X_{1:n}$, in this case), $g(s;\eta)$ does not depend on $x_1,...,x_n$ except through $s$, and $h(x_1,...,x_n)$ does not involve $\theta$.

I have $f(x_1,...,x_n;\eta)=\exp(n\eta)\exp(-\sum_{i=1}^n x_i)$. The sum in the exponential is the same as $\sum_{i=1}^n x_{i:n}$, but I need an expression that involves $x_{1:n}$ in one factor and the $x_i$'s in the other. I don't know how to do this.

I know there are formulas for joint pdf's of any set of order statistics (quite lengthy), but I really don't know how to proceed.

Thank you.

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Can you explain your use of the notation $\mathrm{Exp}(1,\eta)$? – cardinal May 22 '11 at 3:17
@cardinal: edited. – Weltschmerz May 22 '11 at 3:24

I'm going to make an educated guess that $X \sim \mathrm{Exp}(1,\eta)$ means that $X$ has density $$\newcommand{\Ind}[1]{\mathbf{1}_{(#1)}} f(x) \equiv f_X(x) = e^{-(x-\eta)}\Ind{x\geq\eta} \>,$$ where $\Ind{A}$ is the indicator function of the event $A$.
If $X_1, X_2, \ldots$ is an iid sample from this distribution, then the joint density is $$\prod_{i=1}^n f(x_i) = \prod_{i=1}^n e^{-(x_i - \eta)} \Ind{x_i \geq \eta} = e^{-\sum_{i=1}^n x_i} e^{n \eta} \prod_{i=1}^n \Ind{x_i \geq \eta} = e^{-\sum_{i=1}^n x_i} e^{n \eta} \Ind{\min_i x_i \geq \eta} \> ,$$ where the last equality follows since the the product of the indicators $\Ind{x_i \geq \eta}$ is one if and only if the indicator $\Ind{\min_i x_i \geq \eta}$ is one and otherwise is zero.
Now, take $h(x_1,\ldots,x_n) = \exp(-\sum_i x_i)$ and $g(s,\eta) = e^{n \eta} \Ind{s \geq \eta}$ with $s = \min_i x_i$ and apply the factorization theorem.
Hence $S = \min_{1 \leq i \leq n} X_i$ is sufficient for $\eta$.