In Casella & Berger Statistical Inference,in exercise 6.8, we're asked to prove that, for any distribution belonging to the location family, the order statistics are minimal sufficient.
How does one prove this? I have no idea on how to tackle this exercise.
Any help would be appreciated.
A quote of the problem statement in Casella & Berger is the following:
Let $X_1, \dots, X_n$ be a sample from a population with location pdf $f(x - \theta)$. Show that the order statistics ... are a sufficient statistic for $\theta$ and no further reduction is possible.
Let $\mathbf{x} = (x_1, \dots, x_n)$. Notice that
$$ p_\theta(\mathbf{x})=\prod_if(x_i-\theta)=\prod_if(x_{(i)}-\theta), $$
which shows that $T=(x_{(1)}, \dots, x_{(n)})$ is a sufficient statistic by factorization theorem.
It is unclear what it means to show that no further reduction is possible. On the one hand, $T$ is not minimal in general; as mentioned in a comment by @a.arfe, $\bar{x}$ is minimal sufficient when $f(t)=e^{-t^2/2}/\sqrt{2\pi}$ and $p_\theta(x)=f(x - \theta)$. Indeed, this is the very next problem in the referenced textbook.
On the other hand, one could argue that the problem statement should be taken to mean that no further reduction is possible without further restrictions on $f$. In that case, it suffices (as pointed out in Matt Brems' answer) to notice that the ratio
$$ \frac{\prod_if(x_{(i)}-\theta)}{\prod_if(y_{(i)}-\theta)}, $$
is in general (functionally) independent of $\theta$ only when $T(\mathbf x) = T(\mathbf y)$.
In order to show that a statistic $T(X)$ is minimal sufficient for $\theta$, show that the ratio of pdfs $\frac{f(\textbf{x}|\theta)}{f(\textbf{y}|\theta)}$ is constant with respect to $\theta$ for any two samples $\textbf{x}$ and $\textbf{y}$ if and only if $T(\textbf{X})=T(\textbf{Y})$.
I'd start by finding the pdf of a location family distribution, then evaluating the ratio of the two pdfs as I wrote above. Once you've done that, you should be able to assess if the ratio is constant w.r.t. $\theta$ iff $T(\textbf{X})=T(\textbf{Y})$.
I'm might be wrong though, but here's what I found.
The normal distribution belongs to the localization-scale family. It also belongs to the exponential distribution family. Using some theorem directly related to exponential family distributions, one can prove that $T(\mathbf{X})=(\bar{X},S^2)$ are sufficient and complete for $(\mu,\sigma^2)$, hence they are also minimum sufficient for the same parameters. Because there's no bijective function from the $T(\mathbf{X})$ to the order statistics, the order statistics cannot be minimum sufficient.
I guess this is probably a mistake from Casella&Berger book.
