[Math] Does the weibull distribution has a sufficient statistic

probability distributions

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!

Best Answer

For a known $\alpha$ (and unknown $\beta$) it belongs to the exponential families and the sufficient (and complete) statistic can easily be derived.