# Showing that the order statistic $X_{(n)}$ is sufficient

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.

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.

• You must reformulate your question, as it stands, it does not make very much sense (maybe only a TeXnical problem?). As it stands, $f(x)$ does not depend on $\theta$, so your parameter $\theta$ is not even identifiable! (If you try to normalize your density so that it always integrates to one, you will cancel $\theta$!!!) Commented Sep 10, 2012 at 19:37
• @kjetilbhalvorsen: for some reason, after I post, I am still seeing the LaTex code, I can't see the Greek letters. Commented Sep 10, 2012 at 19:43
• The title is a little misleading. Are you asking about the sufficiency of the order statistics $(X_{(1)},\dots,X_{(n)})$ of a random sample, or are you asking about the sufficiency of the maximum $X_{(n)}$ (maybe for the $\mathrm{Uniform}[0,\theta]$ model, which is easy to prove)?
– Zen
Commented Sep 10, 2012 at 19:46
• @Zen: I am asking for the sufficiency of the largest order statistic. I used X(n) to represent largest order statistic. Commented Sep 10, 2012 at 19:52
• Even with the edits this question does not make sense, for reasons pointed out in previous comments, and will need to be improved if it is to remain open.
– whuber
Commented Sep 10, 2012 at 20:24

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 $$p_{\theta}(x_1, \dots, x_n) = c(\theta)^n \prod_{i=1}^n f(x_i) \prod_{i=1}^n I(0\le x_i \le \theta)$$ 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.
• what does it mean when you said "the last factor"? Also, the ordered stat ${x}_{(n)}$ does not show up in the likelihood expression itself? Commented Sep 11, 2012 at 3:00
• "The last factor" is $\prod_{i=1}^n I(0\le x_i \le \theta)$ which clearly is the same as $\prod_{i=1}^n I(0\le x_{(i)} \le \theta)$ so depends on the order statistics. The point is that the product is one only if all the $n$ factors are one, only if $x_{(n)} \le \theta$. Commented Sep 11, 2012 at 3:18
• Good answer, kjetil. If it is still confusing, try this. Using a different notation for the indicators, write the density as $f_\theta(x_i)=c(\theta)f(x_i)I_{[0,\theta]}(x_i)$. Now, observe that $I_{[0,\theta]}(x_i)=I_{[x_i,\infty)}(\theta)$. So, the joint density is $f_\theta(x)=[c(\theta)]^n\prod_{i=1}^n \left(f(x_i)I_[x_i,\infty)(\theta)\right)$. But the products of indicators is the indicator of the intersection, so $\prod_{i=1}^n I_[x_i,\infty)(\theta)=I_{\cap_{i=1}^n[x_i,\infty)}(\theta)$. And it is clear that $\cap_{i=1}^n[x_i,\infty)=[x_{(n)},\infty)$.