Minimal sufficient statistics for Cauchy distribution I'm trying to find the minimal sufficient statistics for a Cauchy distributed random sample $X_1,...,X_n$, here
\begin{equation}
f(x|\theta) = \frac{1}{\pi[1+(x-\theta)^2]}
\end{equation}
I begin by guessing that the order statistics are the minimal sufficient statistics (first of all, they are sufficient). Then I try to prove that
\begin{equation}
\frac{f(X|\theta)}{f(Y|\theta)} = \prod_{i=1}^n\frac{1+(y_i-\theta)^2}{1+(x_i-\theta)^2}=C
\end{equation}
Where $C$ is a constant of $\theta$, I want to prove that the above equation holds iff $T(X)=T(Y)$, where $T(X)$ is the order statistics of $X$.
I am really stuck and don't know how to show why that if it holds then $T(X)=T(Y)$, can anyone help me on this proof? Thanks in advance!
p.s., a similar thread minimal sufficient statistic of Cauchy distribution discusses the problem but offers no proof for the minimal sufficiency.
 A: It is trivial to see the order statistics $T(X) = (X_{(1)},\dots,X_{(n)})$ are sufficient, hence we only need to prove one direction: that if the ratio is constant as a function of $\theta$, then $T(x) = T(y)$. That is, $x$ must be a permutation of $y$.
Suppose $p(x|\theta) \propto_\theta p(y|\theta)$. Since this proportion holds for all $\theta$, then divide each side by the case where $\theta=0$ so that the constant of proportionality cancels and we get
$$\frac{p(x|\theta)}{p(x|0)} = \frac{p(y|\theta)}{p(y|0)}$$
Taking reciprocal gives us
$$\prod_i \frac{1+(x_i-\theta)^2}{1+x_i^2} = \prod_i \frac{1+(y_i-\theta)^2}{1+y_i^2}$$
Note that since these polynomials are equal, they must have the same roots. Also, each polynomial is of degree $2n$ and so they can have at most $2n$ roots. But it should be clear in this form that the LHS polynomial has complex roots
$$x_i \pm i$$
Since $\big(x_i-(x_i\pm i)\big)^2 = (\pm i)^2 = -1$ While the RHS polynomial has roots
$$y_i \pm i$$
hence each side has $2n$ complex roots of that form, and since the polynomials share the same roots and cannot have more than $2n$ roots, it must be that $x$ and $y$ are permutations of one another, and so they have the same order statistics
$$T(y) = T(y)$$
