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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.

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    $\begingroup$ 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 $\theta$!!!) $\endgroup$
    – kjetil b halvorsen ♦
    Commented Sep 10, 2012 at 19:37
  • $\begingroup$ @kjetilbhalvorsen: for some reason, after I post, I am still seeing the LaTex code, I can't see the Greek letters. $\endgroup$
    – user13985
    Commented Sep 10, 2012 at 19:43
  • $\begingroup$ 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)? $\endgroup$
    – Zen
    Commented Sep 10, 2012 at 19:46
  • $\begingroup$ @Zen: I am asking for the sufficiency of the largest order statistic. I used X(n) to represent largest order statistic. $\endgroup$
    – user13985
    Commented Sep 10, 2012 at 19:52
  • $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Commented Sep 10, 2012 at 20:24

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – user13985
    Commented Sep 11, 2012 at 3:00
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    $\begingroup$ "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 $x_{(n)} \le \theta$. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Sep 11, 2012 at 3:18
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    $\begingroup$ 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 $\cap_{i=1}^n[x_i,\infty)=[x_{(n)},\infty)$. $\endgroup$
    – Zen
    Commented Sep 11, 2012 at 4:34

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