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.

  • 1
    $\begingroup$ I will post an answer to this when I get home. Basically you need to use complex roots and the fact that a polynomial of degree n can have at most n Roots $\endgroup$
    – Xiaomi
    Oct 29, 2018 at 8:16
  • $\begingroup$ @Xiaomi thanks in advance! $\endgroup$ Oct 29, 2018 at 8:17
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    $\begingroup$ You’re welcome - though i suspect you is in my class and the assignment is due today ? So here is the outline . Assume they are proportional and divide each side by the case when theta is 0 so that we have strict equality . Then each side is a polynomial of degree n wth roots $x_i \pm i$ and $y_i \pm i$. Since each side is equal they share these roots. So y must be a permutation or x, so the order statistics are identical $\endgroup$
    – Xiaomi
    Oct 29, 2018 at 8:25
  • $\begingroup$ @Xiaomi Haha you are right that this is part of the assignment and is due today. I think I'm not in your class though, you should be teaching a class right now. I will work on your hint first. Thank you! $\endgroup$ Oct 29, 2018 at 8:26

1 Answer 1


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

$$T(y) = T(y)$$

  • $\begingroup$ Thank you! I understand now. What I was missing was how to reform it into a problem of studying the polynomials instead. The explanation is very clear. One small typo occurred when taking the reciprocal (The denominator should be $1+x_iˆ2$, but it's no big deal). Many thanks again! $\endgroup$ Oct 29, 2018 at 16:17
  • $\begingroup$ Both side of the equation is a $2n$ degree polynomial in $\theta$ . I was having trouble understanding this and so I decided to put it in the comments. +1 for a nice answer. $\endgroup$ Aug 12, 2019 at 14:41
  • $\begingroup$ what's the meaning of "$p(x|\theta) \propto_\theta p(y|\theta)$" $\endgroup$
    – christine
    Jan 19, 2021 at 18:21
  • $\begingroup$ "Since this proportion holds for all $\theta$, then divide each side by the case where $\theta=0$ so that the constant of proportionality cancels" Does this mean the fraction on both sides contains nothing about $\theta$? $\endgroup$
    – Jackie
    Feb 4, 2023 at 0:12

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