minimal sufficient statistic of Cauchy distribution Given $X_1, X_2,\ldots, X_n$ are i.i.d with $f(x) = \frac{1}{\pi}\frac{1}{1+x^2}$. Find the minimal sufficient statistic for $\frac{1}{\pi^{n}(\prod_{i=1}^{n} 1+(x_i-\theta)^2)}$.
My question: Using Bahadur's Factorization and Lehman-Scheffe's theorems, I obtain the sufficient statistic as $T(X) = (x_1, \ldots, x_n)$. However, the correct answer turned out to be $T(X) = (x_{(1)}, x_{(2)},\ldots, x_{(n)})$. But I cannot see why the first answer is incorrect, because if we are given the entire sample set, we could easily obtain the order statistic of that sample, so how could $T(X) = (x_1, \ldots, x_n)$ is wrong?
 A: T(X)=X is a sufficient statistics but NOT minimal. 
T(X) = X contains more information than T(X) = (x(1),x(2),…,x(n)).
T(X) = X contains RANK while the order statistic does not. The order statistic only contains the values. 
For example X = (1,2,3,4,5) and X= (1,3,2,4,5) both has the same order statistic T(X) = (1,2,3,4,5).
A: The definition of minimal sufficient statistics from Casella's book is: 
Minimal Sufficient Statistic
A sufficient statistic T(X) for $\theta$ is a minimal sufficient statistics if for any other sufficient statistics S(X), T(X) could be expressed as a function of S(X), such that S(X) should be an one-to-one transformation of T(X). 
Judge by this, I think the minimal sufficient statistic should not be unique (If $T(X) = x$ is minimal statistic, then I suspect that $S_1(X) = log(x)$ or $S_2(X) = e^x$ are sufficient statistic. Conversely, then if we take $T(X) = e^x$ be the minimal sufficient statistic, then taking one-to-one transformation $S(X) = x$ is also a sufficient statistic. So how could we say that minimal sufficient statistic is unique??
