I have some trouble showing sufficiency for largest order statistic ${x}_{n}$.
This is from Casella's text, problem 1.6.3.
Let ${p}_{\theta}$ be a density function.
${p}_{\theta}(x)=c({\theta})f(x)$ for $0<x<\theta$.
If ${X}_{1},{X}_{2},….{X}_{n}$ are iid with density ${p}_{\theta}$, show that ${X}_{(n)}$ is sufficient for $\theta$.
I understand that by the definition of sufficiency, if the summary statistic, $T$, is independent of the parameter $\theta$, for all $t$, then it is sufficient.
How do I actually show that? It seems obvious that $c(\theta)$ and $f(x)$ will not get involved with each other. And there is not an explicit formula for me to work with, like normal or student t.
Best Answer
OK, let me do the reformulation. Let $f$ be a function defined for $x\ge 0$ such that $f(x) >0$, and define $c(\theta)^{-1} = \int_0^{\theta} f(x) \; dx$. Then we can define a probability density, parameterized by $\theta$, by $p̣_{\theta}(x) = c(\theta) f(x) I(0\le x \le \theta)$ where $I(x)$ denotes the indicator function of its argument.
Suppose $x_1, \dots, x_n$ is an iid sample from this density. Then the density of the sample can be written \begin{equation} p_{\theta}(x_1, \dots, x_n) = c(\theta)^n \prod_{i=1}^n f(x_i) \prod_{i=1}^n I(0\le x_i \le \theta) \end{equation} The last factor above can be seen to be $\begin{cases} =0 \text{ if } x_{(n)}>\theta \\ =1 \text{ if } x_{(n)} \le \theta \end{cases}$ and then the result follows from the factorization theorem.