Let $X1, . . . , Xn$ be independent and identically distributed with density $P_θ(x) = $ \begin{cases} 2x/θ^2 & \text{for }0\le x<θ\\ 0 & \text{else} \end{cases} Demonstrate that $m_n = max(x_1,...,x_n)$ is a sufficient statistic .
This is what i did so far, but i'm a bit stuck. I found the likelihood to be:
$L(θ) = $ \begin{cases} (2/θ^2)^n \prod x_i & \text{if}max (x_1,...,x_n) ≤ θ\\ 0 & \text{else} \end{cases}
I know that $L(θ) = g(x)h(T(x),θ)$. But im not sure which is which. Do i use the RHS ($max(x1...)$) or LHS (($2/θ^2)^n$....)?
Would really appreciate the help.
Edit: is $g(x) = 1$ here?
You need to express all constraints inside your function, resulting in a density of the form
$$f_{\bf X} (x) = h(x)g(\theta, T({x})),$$
where $T$ is the sufficient statistic, in our case
$$T(x) = \max (x_1,\dots,x_n).$$
To do this, rewrite the expression for the density as follows:
\begin{align*} f_{\bf X} (x) &= \left(\prod_{j=1}^n \frac{2x_j{\bf 1}_{\{x_j>0\}}}{\theta^2}\right){\bf 1}_{\{\max(x_1,\dots,x_n)\le \theta\}}\\ &= \underset{=h(x)}{\underbrace{\left(\prod_{n=1}^n 2x_j{\bf 1}_{\{x_j>0\}}\right)}}\times \underset{=g(\theta,T(x))}{\underbrace{\theta^{-2n} \times {\bf 1}_{\{\max(x_1,\dots,x_n)\le \theta\}}}} \end{align*}