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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.

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3 Answers 3

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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 $T(\mathbf x) = T(\mathbf y)$.

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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 $T(\textbf{X})=T(\textbf{Y})$.

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  • $\begingroup$ Matt, the thing is we're suppose to prove it for 'any' distribution belonging to the location family. Also, checking the ratio equivalence may not be an easy task. For example: the cauchy distribution. $\endgroup$ Dec 29, 2015 at 17:17
  • $\begingroup$ I'm sorry if my initial question was a bit poorly written. I've edited it since. I hope you can still help me. ;) $\endgroup$ Dec 29, 2015 at 17:47
  • $\begingroup$ 1. Is it possible to write a general form of the pdf or use the cdf? Look at example 6.2.18 in Casella & Berger for a cdf example. 2. As a one-to-one transformation of a minimal sufficient statistic is a minimal sufficient statistic, can you find another minimal sufficient statistic and then find a one-to-one transformation to the order statistics? 3. Show that the set of order statistics is a complete statistic, then use Theorem 6.2.28 to show that a complete statistic is a minimal sufficient statistic. (You must also show that a minimal sufficient statistic exists for 6.2.28 to hold.) $\endgroup$
    – Matt Brems
    Dec 29, 2015 at 19:02
  • $\begingroup$ My only other thought is: 4. Show that the order statistics is a sufficient statistic, then assume toward a contradiction that you can reduce the data further. I'm sort of throwing things at the wall, but hopefully one of these sticks and helps you out! $\endgroup$
    – Matt Brems
    Dec 29, 2015 at 19:05
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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.

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    $\begingroup$ Unless I'm missing something, this does seem like a mistake on the book. The family $N(\theta,1)$ for $\theta\in\mathbb{R}$ is a location family for which $\overline{X}$ as minimal sufficient. $\endgroup$
    – a.arfe
    Mar 7, 2016 at 16:55

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