Let ${\bf X}=(X_1,\dots,X_n)$ be a random sample from the pdf
$$f(x\mid\mu,\sigma)=\frac{1}{\sigma}e^{\frac{-(x-\mu)}{\sigma}}\;\;
,\;\; \mu<x<\infty\;,\;0<\sigma<\infty.$$Find a two-dimensional sufficient statistic for $(\mu,\sigma)$.
The Factorization Theorem says that I should be able to find $T_1({\bf x})$ and $T_2({\bf x})$ such that we have the decomposition
$$\hat{f}({\bf x}\mid\mu,\sigma)=g(T_1({\bf x}),T_2({\bf x})\mid\mu,\sigma)\cdot h({\bf x}).$$
Where $\hat{f}$ is the joint distribution for $\bf X$.
I just want to make sure I'm not missing something obvious, since by independence I can write $\hat{f}$ as
$$\hat{f}({\bf x}\mid\mu,\sigma)=\hat{f}(x_1,\dots,x_n\mid\mu,\sigma)=\left(\frac{e^{\mu/\sigma}}{\sigma}\right)^ne^{(-1/\sigma)\sum_{k=1}^nx_k}\cdot 1.$$
It appears that just setting $T_1({\bf x})=\sum_{k=1}^nx_k$ is a statistic sufficient to determine both $\mu$ and $\sigma$. Which worries me since why would the book ask specifically for a two dimensional statistic if a one dimensional one was all that was necessary to obtain sufficiency with regard to both parameters.
Best Answer
You're missing the piecewise aspect of the problem: Your first equality applies if $x>\mu$ and the density is $0$ if $x<\mu$.
The density that you give later is expressed as an equality that is true if $x_1,\ldots,x_n$ are all greater than $\mu$. That is the same as saying $\min\{x_1,\dots,x_n\}>\mu$. The joint density is therefore $$ \begin{cases} \text{something depending on }x_1,\dots,x_n\text{ only through }\sum_{k=1}^n x_k & \text{if }\min\{x_1,\dots,x_n\}>\mu, \\ 0 & \text{if }\min\{x_1,\dots,x_n\} < \mu. \end{cases} $$
This depends on $x_1,\dots,x_n$ through two numbers: $\sum_{k=1}^n x_k$ and $\min\{x_1,\dots,x_n\}$.