Does the weibull distribution has a sufficient statistic? When using the following definition of weibull:
$f(y) = \beta \alpha y^{\alpha - 1}e^{-\beta y ^ {\alpha}} $ , 
When $\beta>0 \alpha >0$.
I could only find (using the factorization theorem) the following two possible sufficient statistics: 
$\sum ln(y_i)   $ , $\sum (y_i^\alpha)   $
The first one seems fine. However, the second one is not clear to me. From what I've learned, a statistic need to not be dependent on the parameters of the function. How then can the second statistic (depending on $\alpha$), be a statistic (let alone sufficient)? (I would guess that one can estimate $\alpha$ using the first statistic, and then use it for the second one for $\beta$, however - I did not see any nice theory explaining why such an approach would make sense).
Thanks!
 A: For a known $\alpha$ (and unknown $\beta$) it belongs to the exponential families and the sufficient (and complete) statistic can easily be derived. 
A: Does the weibull distribution has a sufficient statistic?
Yes, trivially, the data itself is always sufficient. Notice order statistic are always sufficient also - i.e. ordering the $(x_1,x_2,...x_n)$ in ascending or descending order, this is always the case when the samples are iid, i.e. order does notmatter.
You probably want a something better than order statistics, minimal sufficient? The answer seems to be: there is none. The reason is you need
$\prod_{i=1}^n f_{\alpha,\beta}(x)/f_{\alpha,\beta}(y)$ is independent iff $S(x)=S(y)$. This needs to hold for ALL $\alpha>0$. 
The only way I see that might happen is by permutation. 
http://books.google.co.uk/books?id=6wdcTfiLNS4C&pg=PA313&lpg=PA313&dq=%27sufficient+statistics%27+weibull+distribution&source=bl&ots=7C6XVI1J-x&sig=-02V4thha7rYWIbMo3FluLdSocQ&hl=en&sa=X&ei=KgPUUv2UA4mp7Qb10oGYAQ&ved=0CEAQ6AEwAg#v=onepage&q=%27sufficient%20statistics%27%20weibull%20distribution&f=false seems to confirm my answer.
