For $n \geq 2$, let $X_1,X_2,\ldots,X_n$ be independent samples from $P_{\theta}$, the uniform distribution $U(\theta,\theta +1),\theta \in \mathbb R$ For $n \geq 2$, let $X_1,X_2,\ldots,X_n$ be independent samples from $P_{\theta}$, the uniform distribution $U(\theta,\theta +1),\theta \in \mathbb R$. Let $X_{(1)},X_{(2)},\ldots,X_{(n)}$ be order statistics of the sample.
(a) Show that $(X_{(1)},X_{(n)})$ is a sufficient statistic for $\theta$.
Thoughts: By factorization theorem, this one is clear.
(b) Is $(X_{(1)},X_{(n)})$ complete?
Thoughts: I tried to use the definition of being complete here, but I don't know how to deal with the two dimensions here, which troubles me.
(c) Find $a_n$ and $b_{\theta}$ such that $a_n (b(\theta)-X_{(n)}) \rightarrow Z$ in distribution, where $Z$ has an exponential distribution with density $f(x)=e^{-x},x>0$.
(d) What is the MLE of $\theta$ given the sample?
Thoughts: I got the likelihood function: $\prod_{i=1}^{n} x_{i} \mathbb 1_{\theta < X_{(1)} \leq X_{(n)} < \theta +1}$ and the MLE is $X_{(n)}-1<\hat \theta <X_{(1)}$. I'm not sure whether this is right.

That $(X_{(1)},X_{(n)})$ is not complete follows from the fact that the expected value of $X_{(n)} - X_{(1)}$ does not depend on $\theta$.  Thus you can subtract its expected value from it and get a nonzero unbiased estimator of $0$.
And you should say that $X_{(1)},X_{(2)},\ldots,X_{(n)}$ is a sample, not that $X_{(1)},X_{(2)},\ldots,X_{(n)}$ "are samples".
(Maybe I'll add to this later.)
