Consider a random sample from a uniform distribution $X_i\sim \mathrm{UNIF}(0,\theta)$, where $\theta$ is unknown. Show that $S=X_{n:n}$ is sufficient for $\theta$ by the factorization criterion.
Factorization Criterion :
if $X_1,...,X_n$ have joint pdf $f(x_1,...,x_n;\theta)$ and if $S=(S_1,...,S_k)$, then $S_1,...,S_k$ are jointly sufficient for $\theta$ if and only if
$f(x_1,...,x_n;\theta)=g(s;\theta)h(x_1,...x_n)$
based on the questions, it is known that
$X_i\sim \mathrm{UNIF}(0,\theta)$ so the pdf is $f(x) = 1/\theta$ , $0<=x<=\theta$
and the CDF is $F(x) = x/\theta$
so, the joint pdf is
$f(x_1,...,x_n;\theta)=f(x_1;\theta)...f(x_n;\theta)$
$=(1/\theta)^n$
$=1/\theta^n$
$=g(x_{n:n};\theta)h(x_1,...,x_n)$
where $=g(x_{n:n};\theta) = g(s;\theta) =1/\theta^n$ if $s<\theta$ and zero otherwise, and $h(x_1,...,x_n) = 1$ if $0<x_{1:n}$ and zero otherwise.
So, $S=X_{n:n}$ is sufficient for $\theta$
But I don't understand why $g(x_{n:n};\theta) = g(s;\theta) =1/\theta^n$?
The issue is that the joint density needs to be specified completely. A single observation has density $$f_X(x \mid \theta) = \frac{1}{\theta} \mathbb 1(0 \le x \le \theta),$$ where $\mathbb 1$ is the indicator function $$\mathbb 1 (0 \le x \le \theta) = \begin{cases}1 , & 0 \le x \le \theta \\ 0, & \text{otherwise}. \end{cases}$$ So we have the joint density of $n$ iid observations $$f_{\boldsymbol X}(x_1, \ldots, x_n \mid \theta) = \theta^{-n} \prod_{i=1}^n \mathbb 1 (0 \le x_i \le \theta).$$ The product is equal to $1$ if and only if $$0 \le x_{1:n} \le x_{n:n} \le \theta,$$ that is to say, the smallest observation must be at least $0$ and the largest must not exceed $\theta$. Hence the full joint density is $$f_{\boldsymbol X}(x_1, \ldots, x_n \mid \theta) = \theta^{-n} \mathbb 1(x_{1:n} \ge 0) \mathbb 1(x_{n:n} \le \theta).$$ Now by the factorization theorem, $$f = h(\boldsymbol x) g(T(\boldsymbol x) \mid \theta)$$ for the choice $h(\boldsymbol x) = \mathbb 1(x_{1:n} \ge 0)$, which does not depend on $\theta$; and $$g(T\mid \theta) = \theta^{-n} \mathbb 1(T \le \theta),$$ and $$T(\boldsymbol x) = x_{n:n}.$$ This last function $T$ is the sufficient statistic.