Find UMVUE in a uniform distribution setting Let $X_1, ..., X_n$ be independent and uniformly distributed on $(\theta_1-\theta_2,\theta_1+\theta_2)$ for $\theta_1 \in \mathbb{R}$, $\theta_2>0$.
Find UMVUEs for
a) $\theta_1$ and $\theta_2$,
b) $\theta_1/\theta_2$.
Naturally, I would like to use the Lehmann-Scheffé theorem that says:
If $V$ is a complete, sufficient statistic for $\theta$ and $\mathbb{E}_\theta[g(V)]=h(\theta)$ holds. Then $g(V)$ is an UMVUE for $h(\theta)$.
So first, I have to find such a statistic for any of my $\theta$s.
In the lecture we learned that $(X_{(1)}, X_{(n)})$ sufficient for $(a,b)$ if the $X_i$ are uniform on $(a,b)$ (where $X_{(1)}<...<X_{(n)}$ is the order statistic for $X_1, ..., X_n$).
So it follows that in my setting $(X_{(1)}, X_{(n)})$ is sufficient for $(\theta_1-\theta_2,\theta_1+\theta_2)$? But looking at the proof, $(X_{(1)}, X_{(n)})$ is already a sufficient statistic for $(\theta_1, \theta_2)$, too, right?
So I have to check for completeness and thus show that for all functions $g$ mapping from the range of $V$ to $\mathbb{R}$ from $\mathbb{E}_\theta[g(X_{(1)}, X_{(n)})]=0$ for all $\theta=(\theta_1, \theta_2)$ follows $\mathbb{P}_\theta(g(X_{(1)}, X_{(n)})=1)=1$ $\mathbb{P}_\theta$-almost-surely.
Here I'm stuck: How can I show that?
 A: I don't think you can. At least if X is distributed on $(\theta,\theta + 1)$, then $(X_{(1)}, X_{(n)})$ is sufficient, but not complete. 
Lets say $g(X_{(1)}, X_{(n)})$ is a function of your statistic. Then
$\mathbb{E}_{\theta}[X_{(n)} - X_{(1)}] = \mathbb{E}_{\theta}[X_{(n)}] - \mathbb{E}_{\theta}[ X_{(1)}] = (\theta + n/(n+1)) - (\theta + 1/(n+1)) = (n-1)/(n+1)$
Then $g$ defined as 
$g(X_{(1)}, X_{(n)}) := X_{(n)} - X_{(1)} - (n-1)/(n+1)$ 
would be 0 $\forall \theta$ and thus not complete.
As I said, the setting is slightly different though. Let me know if you find a definite solution - I'm looking at the same problem. 
A: That $T=(X_{(1)},X_{(n)})$ is sufficient for $\theta_1,\theta_2$ is verified easily using Factorisation theorem. And the proof of $T$ being a complete statistic is similar to the argument presented here.
Alternatively, if $g$ is any (measurable) function of $T$, then after some effort it can be shown by differentiating both sides of $E\,[g(T)]=0$ wrt $\theta_1,\theta_2$ that $g$ is identically zero with probability $1$ for all $\theta_1,\theta_2$.
Since $X_i\stackrel{\text{ i.i.d}}\sim U(\theta_1-\theta_2,\theta_1+\theta_2)$, we have $Y_i=(X_i-(\theta_1-\theta_2))/(2\theta_2)\stackrel{\text{ i.i.d}}\sim U(0,1)$ for all $i=1,\ldots,n$.
Now it is known that $Y_{(1)}\sim \text{Beta}(1,n)$ and $Y_{(n)}\sim \text{Beta}(n,1)$.
So, 
\begin{align}
E\,[X_{(1)}]&=2\theta_2 E(Y_{(1)})+\theta_1-\theta_2
\\&=\frac{2\theta_2}{n+1}+\theta_1-\theta_2
\end{align}
And 
\begin{align}
E\,[X_{(n)}]&=2\theta_2 E(Y_{(n)})+\theta_1-\theta_2
\\&=\frac{2n\theta_2}{n+1}+\theta_1-\theta_2
\end{align}
Solving for $\theta_1$ and $\theta_2$ from the above equations we get 
$$\theta_1=E\left[\frac{X_{(1)}+X_{(n)}}{2}\right]\qquad,\qquad \theta_2=E\left[\frac{n+1}{2(n-1)}\left(X_{(n)}-X_{(1)}\right)\right]$$
These unbiased estimators of $\theta_1,\theta_2$ are functions of $T$, hence they are the corresponding UMVUEs by Lehmann-Scheffe theorem.

We can guess that an unbiased estimator of $\theta_1/\theta_2$ based on $T$ is some function of $(X_{(1)}+X_{(n)})/(X_{(n)}-X_{(1)})$.
Now,
\begin{align}
E\left[\frac{X_{(1)}+X_{(n)}}{X_{(n)}-X_{(1)}}\right]&=E\left[\frac{Y_{(n)}+Y_{(1)}}{Y_{(n)}-Y_{(1)}}\right]+\frac{\theta_1-\theta_2}{\theta_2}E\left[\frac{1}{Y_{(n)}-Y_{(1)}}\right]
\end{align}
Using the joint density of $(Y_{(1)},Y_{(n)})$ the above expression is found to  be exactly equal to $\frac{n}{n-2}\left(\frac{\theta_1}{\theta_2}\right)$, suggesting that the UMVUE of $\theta_1/\theta_2$ is $$\frac{n-2}{n}\left(\frac{X_{(n)}+X_{(1)}}{X_{(n)}-X_{(1)}}\right)$$
