
Sufficient statistic
Hi, I am preparing for my test. I have a question about sufficient statistics. More to follow later.
$\displaystyle X_1, \cdots, X_n$ \mbox{is a random sample with pdf:}$
$\displaystyle f(x\mu, \sigma) = \dfrac{1}{\sigma} e^{(x\mu)/\sigma} \;,\;\mu<x<\infty\;,\;0<\sigma<\infty. \mbox{Find a two dimesional sufficient statistic for}(\mu, \sigma)$
The problem has been worked this way:
Let $\displaystyle x_{(1)}=min_i\;x_i. \mbox{Then the joint pdf is}$
$\displaystyle f(x_1,....., x_n)= \prod_{i=1}^n \dfrac{1}{\sigma} e^{(x\mu)/\sigma} \;I_{(\mu,\sigma)}(x_i)\;=\;\left(\dfrac{e^{\mu/\sigma}}{\sigma}\right)^n\cdot e^{\sum_i x_i/\sigma}\;I_{(\mu,\sigma)}(x_1)$
and thus factorization theorem concludes that $\displaystyle (X_{(1)}, \sum_i X_i)$ is a suff stat for (mu,sigma).
I am conused in why we are taking the indicator function and why was the minmum order statistic used. Could anybody explain? Thank You.

Hello,
We are taking the indicator function because $\displaystyle \forall i, x_i\in(\mu,\infty)$ (this is a consequence of the pdf of X)
For the product (in the joint pdf) to be different from 0, we need $\displaystyle \forall i,I_{(\mu,\infty)}(x_i)\neq 0 \Leftrightarrow I_{(\mu,\infty)}(x_i)=1$
So $\displaystyle \forall i,x_i\geq \mu$, which means that the minimum among the $\displaystyle x_i$'s should be $\displaystyle \geq \mu$. Hence the result.

Thanks, and one more thing. If i wanted to know whether the sufficient stat above is also a minimum sufficient what shall I do?
I looked at few examples in the book where two sample points x and y are considered and the ratio of their pdfs are taken, but I am confused on how to use that here. Could you tell?
Thank you.

just take the ratio of $\displaystyle \dfrac{f(x\mu, \sigma)}{f(y\mu, \sigma)}$, and find the conditions that would make this ratio a constant as a function of $\displaystyle \mu \;and\; \sigma^2$, which will give you the minimum sufficient statistic.