# Sufficient statistic

• Feb 12th 2011, 10:57 PM
chutiya
Sufficient statistic
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

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.
• Feb 13th 2011, 02:40 AM
Moo
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.
• Feb 13th 2011, 06:26 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, 10:21 AM
harish21
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.