Finding sufficient statistic for Weibull density function I am given the follow problem and am having trouble finding the sufficient statistic.
Suppose that Y$_1$, Y$_2$, ..., Y$_n$ denote a Weibull density function, given by:
f ( y | $\theta$ ) = 

Let $Y_1, Y_2, ... , Y_n$ denote a Weibull density function, given by:
  $$
f (y | \theta ) =
\begin{cases}
\frac{2y}{\theta}e^\frac{-y^2}{\theta},  & 0 < y \\
0, & \text{elsewhere}
\end{cases}
$$
  Find the MVUE for $\theta$.

My issue here is again, in regards to finding the sufficient statistic. I begin by taking the likelihood:
L ($\theta$) = $\prod \frac{2y}{\theta}e^\frac{-y^2}{\theta}$
= $(\frac{2}{\theta})^n e^\frac{-\sum{y_i^2}}{\theta}\prod y_i$
How do I know what the sufficient statistic is? Is it:


*

*$\prod y_i$

*$-\sum{y_i^2}$


The answer is supposed to be the second one, but I'm still unclear as to how we know that. Any help will be much appreciated!
 A: By the Neyman factorization theorem it is quite clear the second option is the sufficient statistic.
Just review the theorem,

Let $X_1,X_2,...X_n $ denote a random sample from a distribution that
  has $\textit{pdf}$ or $\textit{pmf}$ of $f(x;\theta), \theta \in \Omega$. The statistic
  $Y_1 = u_1(X_1 , ... ,X_n)$ is a sufficient statistic for $\theta$ if
  and only if we can find two nonnegative functions, $k_1$ and $k_2$ such
  that $$f(x_1, x_2, ... , x_n;\theta) = k_1[ u_1(x_1, x_2, ... ,
 x_n);\theta ] k_2(x_1, x_2, ... , x_n)$$
where:
$k_1$ is a function that depends on the data $x_1, x_2, ..., x_n$ only
  through the function $u_1(x_1, x_2,..., x_n)$, and the function
  $k_2(x_1, x_2, ..., x_n)$ does not depend on the parameter $\theta$

Now look at your $L(\theta)$, the first part is $(\frac{2}{\theta})^n e^\frac{-\sum{y_i^2}}{\theta}$  this is the $k_1$ function, it depends on $ y_1, y_2,...,y_n $ only through $-\sum y_i^2$ (note, you should treat $\theta$ as a constant here, since you condition on it here). The second part is $\prod y_i$ which does not depend on $\theta$, it is the $k_2$ function.
So by the factorization theorem, you can directly say $-\sum y_i^2$ is the sufficient statistic for $\theta$.
