# Sufficient statistic

$\displaystyle X_1, \cdots, X_n$ \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)$is a suff stat for (mu,sigma). I am conused in why we are taking the indicator function and why was the minmum order statistic used. Could anybody explain? Thank You. • Feb 13th 2011, 01:40 AM Moo Hello, We are taking the indicator function because$\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$. Hence the result. • Feb 13th 2011, 05:26 AM chutiya Thanks, and one more thing. If i wanted to know whether the sufficient stat above is also a minimum sufficient what shall I do? I looked at few examples in the book where two sample points x and y are considered and the ratio of their pdfs are taken, but I am confused on how to use that here. Could you tell? Thank you. • Feb 14th 2011, 09:21 AM harish21 just take the ratio of$\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.