Let $\theta \lt x_{(1)} \lt x_{(n)} \lt \theta + 1$ be ordre statistics for $X_1, \dots, X_n$ iid uniform random variables on $(\theta, \theta + 1)$.
Let $R = X_{(n)} - X_{(1)}$ and $M = (X_{(1)} + X_{(n)}) / 2$. Then we have the inverse transformation $X_{(1)} = (2M-R)/2$ and $X_{(n)} = (2M + R) /2$.
Then the joint pdf is $n(n-1)r^{n-2}$ if $0 \lt r \lt 1, \theta + (r/2) \lt m \lt \theta + 2 - (r/2)$
How are the conditions $0 \lt r \lt 1, \theta + (r/2) \lt m \lt \theta + 2 - (r/2)$ derived?
I see how $0 \lt r \lt 1$ since the the $x_{(i)}$'s could be arbitrarily close together or $(\theta + 1) - \theta = 1$ apart, but I'm having trouble figuring out a way to show the $m$ inequalities.
Anyone have any ideas?
I think you have the wrong inequality on the right. It should be $m<\theta+1-r/2.$ This can be checked to give a proper joint distribution as
$$\int_0^1\int _{\theta+\frac{r}{2}}^{\theta+1-\frac{r}{2}}n(n-1)r^{n-2}dm\, dr =\int_0^1 n(n-1)r^{n-2}(1-r)=n(n-1)\left[\dfrac{1}{n-1}-\dfrac{1}{n}\right]=1.$$
$M=\dfrac{X_{(1)}+X_{(n)}}{2}$ and $R=X_{(n)}-X_{(1)}$ implies $M-\dfrac{R}{2}=X_{(1)}> \theta,$ which gives you the inequality $\theta+r/2 < m.$
Similarly, $M+\dfrac{R}{2}=X_{(n)} < \theta+1,$ which means $m<\theta+1-r/2.$