Let $\displaystyle Y_1, \ Y_2, \ \dotso,\ Y_n$ denote a random sample from the prob. density function:

$\displaystyle f(y|\theta) = \left\{ \begin{array}{rcl}

e^{-(y-\theta)} & \mbox{for} & y \geq \theta \\

0 & \mbox{for} & \mbox{other}

\end{array}\right.$

show that $\displaystyle Y_{(1)} =\min(Y_1, \ Y_2, \ \dotso,\ Y_n)$ is sufficient for $\displaystyle \theta$

I asked my prof, and all he told me is that in order for our definition of sufficiency to be valid in this case we simply need to considering the smallest y, where it should be greater then some fixed $\displaystyle \theta$. If I look at what the book says and go according to

http://www.mathhelpforum.com/math-he...-question.html I get something a little closer to what's in the book. Therefore what I get is

$\displaystyle F(y) = \int_{\theta}^u e^{-(y-\theta)} \ dy = 1-e^{\theta-u} $

so the density function for a minimum stat is $\displaystyle n[1-F(y)]^{n-1} \cdot f(y) = n[1-(1-e^{\theta-u})]^{n-1} \cdot e^{-(u-\theta)} $

$\displaystyle =n[e^{\theta-u}]^{n-1} \cdot e^{\theta-u} = n[e^{\theta-u}]^n$

following the same method as in the link I get:

$\displaystyle L(y_1, \ y_2, \ \dotso, \ y_n|\theta) = f(y_1, \ y_2, \ \dotso, \ y_n|\theta) = f(y_1|\theta) \cdot f(y_2|\theta) \cdot \dotso \cdot f(y_n|\theta)$

$\displaystyle =\prod_{i=1}^k n[e^{\theta-u_i}]^n$

$\displaystyle =n[e^{\theta-u_1}]^n \times n[e^{\theta-u_2}]^n \times \ \dotso \times n[e^{\theta-u_k}]^n$

$\displaystyle =n^k \cdot \exp \bigg{[} \sum_{i=1}^k (\theta -u_i)^n \bigg{]}$

factorizing it out would give

$\displaystyle g(\min (Y_1, \ Y_2, \ \dotso,\ Y_n)|\theta) = \exp \bigg{[} \sum_{i=1}^k (\theta -u_i)^n \bigg{]} \ \mbox{and} \ \ h(y_1,\ y_2,\ \dotso, \ y_n) = n^k$

I'm not quite sure that this is correct.

Also can it be done using these terms:

Let $\displaystyle X = \sum^n_{i=1} (\theta-u_i)$

$\displaystyle =\prod_{i=1}^k n[e^{\theta-u_i}]^n$ as $\displaystyle \prod_{i=1}^k n\cdot \exp\bigg{[}\sum^n_{i=1} (\theta-u_i) \bigg{]} = \prod_{i=1}^k n\cdot e^X$

expanding it out gives:

$\displaystyle =ne^X \times ne^X \times \ \dotso \times ne^X$

$\displaystyle =n^k e^{kX}$