So I have a random sample X_{1}, X_{2}, ... , X_{n}from a Uniform distribution (-theta, theta).

T(X)=max{X_{(n)}, -X_{(1)}} where X_{(n)}=max{X_{1},...,X_{n}} and X_{(1)}=min{X_{1},...,X_{n}}.

I need to show that T(X) is a minimal sufficient statistic.

So this is how I've started:

Firstly I have to get the sampling distribution of T(X), f_{T}(t). (when I use less than or greater than, I also mean or equal to but for neatness typing I'll leave them out)

P[T(X)<u]

=P[max{X_{(n)},-X_{(1)}}<u]

As it's the max being less than u, the other one will also be less than u, and as the events are independent this can be written as:

P[X_{(n)}<u]P[-X_{(1)}<u]

=P[max{X_{1},...,X_{n}}<u]P[min{X_{1},...,X_{n}}>-u]

From here I'm a bit unsure.

So from this position I'm guessing I need the pdf's of the order statistics, which can be derived from this formula:

f_{X(i)}(u)=(n!)/((i-1)!(n-1)!)*[F(u)]^{i-1}*[1-F(u)]^{n-1}*f(u)

So I have f_{X(1)}(u)=(n!)/(n-1)!*[1-F(u)]^{n-1}*f(u)

and f_{X(n)}(u)=(n!)/(n-1)!*[F(u)]^{n-1}*f(u).

In this case f(u)=1/2theta, from the uniform distribution, and so F(u)=0 for x<-theta, x/2theta for -theta<x<theta, 1 for x>theta. I can substitute these in to the above to get equations for f_{X(1)}(u) and f_{X(n)}(u).

Is this all right? Where do I proceed from here?

Thanks