Show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic for $\theta$.

Question

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?

Answer 1

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: $$M = \begin{bmatrix} -1 & 1 \\ 1/2 & 1/2 \end{bmatrix}.$$