Suppose we have a location-scale family with pdf $$\frac{1}{\sigma}f(\frac{x - \mu}{\sigma}).$$ Let $X_1$, $X_2$, $X_3$,..., $X_n$ be a random sample from the location-scale family. Is the statistics $$T(X) = (X_{(1)}, X_{(2)}, X_{(3)},..., X_{(n)})$$ minimal sufficient for the parameters ($\mu$,$\sigma$)? ($X_{(k)}$ is the kth order statistic) I have read somewhere that $T(X)$ is minimal sufficient for the parameters ($\mu$,$\sigma$) (I know that it is sufficient).

Can someone please help? If the answer is positive, can someone please give me a proof?


This question is Example 6.10, page 36, in Lehmann and Casella Theory of Point Estimation. If the density $f$ is unknown or simply outside exponential families, like the Cauchy distribution, the order statistics $$T(X) = (X_{(1)}, X_{(2)}, X_{(3)},..., X_{(n)})$$ is minimal sufficient.

The result is based on Theorem 6.12, page 37, that, for a finite family of distributions $\mathcal{P}=\{p_0,\ldots,p_k\}$, and a sample $X$, the statistic $$T(X) = (p_1(X)/p_0(X),\ldots,p_k(X)/p_0(X))$$ is minimal sufficient. The result follows from this property by considering $n$ different values of the Cauchy parameters and by noticing that if $T$ is minimal sufficient for $\mathcal{P}_0$ and sufficient for $\mathcal{P}_1\supset\mathcal{P}_0$, then it is minimal sufficient for $\mathcal{P}_1$.

  • $\begingroup$ Thank you very much for your answer. What exactly is a finite family? What are the $p_i$ ? $\endgroup$ Feb 8 '15 at 9:13
  • $\begingroup$ Sorry for asking, but what is a 'finite family of distributions'? I have heard of things like the exponential family and the location-scale family, but not a 'finite family'. $\endgroup$ Feb 8 '15 at 9:21
  • $\begingroup$ Ps: You can access the relevant pages in Lehmann-Casella on amazon. $\endgroup$
    – Xi'an
    Feb 8 '15 at 9:31
  • $\begingroup$ @Xi'an I believe theorem 6.12 only suits a finite family of distributions. However, the Cauchy distribution family is infinite, how to use 6.12 to prove Cauchy? $\endgroup$
    – Tan
    Nov 6 '20 at 0:37
  • 1
    $\begingroup$ I got it. The lemma 6.14. Thanks! $\endgroup$
    – Tan
    Nov 6 '20 at 9:26

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