Sufficient Statistic Example

Example) I have no idea where I found this example, Sorry!

Let $X_{1},...,X{n}$ be a random sample from a distribution with the following density function. $f(x|\theta)= \frac {2x}{\theta}\exp (\frac{-x^2}{\theta})$, x>0 Show that $\sum_{i=1}^{n}X_{i}^2$ is a sufficient statistic for $\theta$

$\triangleright$ Think First!
We can use likelihood $L(\theta)$ is a joint density of the sample! 

$\triangleright$ Solution
$L(x_{1},...,x_{n}|\theta) = \frac {2x_{1}}{\theta}\exp (\frac{-x_{1}^2}{\theta})\cdots \frac {2x_{n}}{\theta}\exp (\frac{-x_{n}^2}{\theta})= \frac{2^n}{\theta^n} \cdot \prod x_{i}\exp \frac{-\sum x_{i}^2}{\theta}$ 

Let $g(\sum x_{i}^2, \theta)=\frac{2^n}{\theta^n} \exp \frac{-\sum x_{i}^2}{\theta}$, where g is the function of  $\sum x_{i}^2$ and $\theta$.
Also let $h(x_{1},...,x_{n}) = \prod_{x_{i}}$ 

We have $L(x_{1},...,x_{n}|\theta)$ = $g(\sum x_{i}^2, \theta)\cdot h(x_{1},...,x_{n})$ 

$\therefore \sum x_{i}^2$ is sufficient statistic for $\theta$