# 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 $$$$f(x|\theta) = \frac{1}{\pi[1+(x-\theta)^2]}$$$$ I begin by guessing that the order statistics are the minimal sufficient statistics (first of all, they are sufficient). Then I try to prove that $$$$\frac{f(X|\theta)}{f(Y|\theta)} = \prod_{i=1}^n\frac{1+(y_i-\theta)^2}{1+(x_i-\theta)^2}=C$$$$ Where $$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.

• 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 – Xiaomi Oct 29 '18 at 8:16
• @Xiaomi thanks in advance! – dogthepeter Oct 29 '18 at 8:17
• 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 – Xiaomi Oct 29 '18 at 8:25
• @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! – dogthepeter Oct 29 '18 at 8:26

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

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

• 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! – dogthepeter Oct 29 '18 at 16:17
• 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. – Prof.Shanku Aug 12 '19 at 14:41