# Thread: Sufficient statistic

1. ## 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)$is a suff stat for (mu,sigma). I am conused in why we are taking the indicator function and why was the minmum order statistic used. Could anybody explain? Thank You. 2. Hello, We are taking the indicator function because$\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$. Hence the result. 3. Thanks, and one more thing. If i wanted to know whether the sufficient stat above is also a minimum sufficient what shall I do? I looked at few examples in the book where two sample points x and y are considered and the ratio of their pdfs are taken, but I am confused on how to use that here. Could you tell? Thank you. 4. just take the ratio of$\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.

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