Show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic for $\theta$. Let $X_{1}, X_{2}, ..., X_{n}$ be a random sample from $\text{Uniform}(\theta,\theta+1)$ population with $-\infty<\theta<\theta+1< \infty$ show that $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$. Also, show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic.
For the first part I did the following
$$f(x|\theta,\theta+1)=I_{(\theta,\theta+1)}(x_{(1)},x_{(n)})$$
then
$$\frac{f(x|\theta,\theta+1)}{f(y|\theta,\theta+1)}=\frac{I_{(\theta,\theta+1)}(x_{(1)},x_{(n)})}{I_{(\theta,\theta+1)}(y_{(1)},y_{(n)})}$$
This is a constant function in $\theta$ iff $x_{(1)}=y_{(1)}$ and $x_{(n)}=y_{(n)}$ s.t. $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$.
However, I am not sure how to proceed to show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic. Can some help me with this?
 A: Any invertible function of a minimal sufficient statistic is also minimal sufficient.  Such a function is given by $$M(x,y) = (y-x, (x+y)/2)$$ which has the inverse $$M^{-1}(x,y) = (y - x/2, y + x/2).$$  This can also be conceptualized as a $2 \times 2$ matrix:  $$M = \begin{bmatrix} -1 & 1 \\ 1/2 & 1/2 \end{bmatrix}.$$
A: If you know$X_{(1)}$ and $X_{(n)}$, then you can certainly compute $\frac{X_{(1)} + X_{(n)}}{2}$ and $X_{(n)}- X_{(1)}$ from them. Likewise given the latter two, you can calculate the first two.  So if either pair is sufficient, the other is as well, because you are losing no data switching from one to the other.  Having a sufficient statistic for a parameter means that going back to the sample for more details won't improve your parameter estimate.
Intuitively, $\theta < X_{(1)} <$(the other data)$ < X_{(n)} < \theta + 1$. If you are trying to find an upper bound for $\theta$, and you have $X_{(1)}$, then going back to the sample for more data will not improve matters.  Likewise, the lower bound for $\theta + 1$ is most restricted by $X_{(n)}$.
