Sufficient statistic

**chutiya** · February 12th 2011, 09:57 PM

Hi, I am preparing for my test. I have a question about sufficient statistics. More to follow later.

$X_1, \cdots, X_n$ \mbox{is a random sample with pdf:}$

$f(x|\mu, \sigma) = \dfrac{1}{\sigma} e^{-(x-\mu)/\sigma} \;,\;\mu<x<\infty\;,\;0<\sigma<\infty. \mbox{Find a two dimesional sufficient statistic for}(\mu, \sigma)$

The problem has been worked this way:

Let $x_{(1)}=min_i\;x_i. \mbox{Then the joint pdf is}$

$f(x_1,....., x_n)= \prod_{i=1}^n \dfrac{1}{\sigma} e^{-(x-\mu)/\sigma} \;I_{(\mu,\sigma)}(x_i)\;=\;\left(\dfrac{e^{\mu/\sigma}}{\sigma}\right)^n\cdot e^{-\sum_i x_i/\sigma}\;I_{(\mu,\sigma)}(x_1)$

and thus factorization theorem concludes that $(X_{(1)}, \sum_i X_i)$ 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.

**Moo** · February 13th 2011, 01:40 AM

Hello,

We are taking the indicator function because $\forall i, x_i\in(\mu,\infty)$ (this is a consequence of the pdf of X)

For the product (in the joint pdf) to be different from 0, we need $\forall i,I_{(\mu,\infty)}(x_i)\neq 0 \Leftrightarrow I_{(\mu,\infty)}(x_i)=1$

So $\forall i,x_i\geq \mu$ , which means that the minimum among the $x_i$ 's should be $\geq \mu$ . Hence the result.

**chutiya** · February 13th 2011, 05:26 AM

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.

**harish21** · February 14th 2011, 09:21 AM

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

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