UMVUE for $\theta$ where $X \sim Unif\{1 ,\ldots, \theta\}$ Say we have $X \sim Unif\{1, \ldots , \theta\}$ and we want to find the uniformly minimum variance unbiased estimator for $\theta$.
My first assumption was $X_{(n)}$. Which I managed to show is complete sufficient and unbiased. Can I conclude that this is UMVUE? Is there a theorem which would say that:
If a statistic is sufficient, complete and unbiased, then it is UMVUE?
Thank you for the information!
CORRECTION:
As noted in comment, $X_{(n)}$ is biased.
 A: Please see the Lehmann Scheffe theorem (Theorem 2 in http://www.stat.unc.edu/faculty/cji/lecture9.pdf), which states that if $U$ is a complete sufficient statistic, and $T$ is an unbiased estimator, than $E(T|U)$ is UMVUE. In your case, you have $ T  = U $.
Please check your claims about unbiasedness: A variant of your problem is addressed in https://math.stackexchange.com/questions/150586/expected-value-of-max-of-iid-variables?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa, where one of the answers show that the estimate is in fact biased
A: Let's take Sid's answer and work with it. We'll start out with a simple unbiased estimator of $\theta$ and work out its expectation given $X_{(n)}$.  
Our simple unbiased estimator will be based on the sample mean, whose expectation is:
$$\mathbb{E}\bar{X} = {\theta + 1 \over 2}$$
We can construct our unbiased estimator $T$ as:
$$T = 2\bar{X}-1$$
Now, given a sample of size $n$ and $X_{(n)}$, the expectation of $T$ can be calculated by noting that, conditional upon $X_{(n)}$,


*

*One of the data points is known to be $X_{(n)}$,

*The other $n-1$ data points are distributed according to a discrete uniform distribution on $\{1, \dots, X_{(n)}\}$, which has expectation $(X_{(n)} + 1) / 2$.


So, starting out by calculating $\mathbb{E}[\bar{X}|X_{(n)}]$, which is a valid approach as $T$ is linear in $\bar{X}$,
$$\mathbb{E}\bar{X} = {X_{(n)} + (n-1){X_{(n)} + 1 \over 2} \over n}$$
Rearranging terms gives us:
$$\mathbb{E}\bar{X} = {(n+1)X_{(n)}+(n-1) \over 2n}$$
and
$$\mathbb{E}T = {(n+1)X_{(n)}+(n-1) \over n} - 1$$
resulting, with some further simplification, in our UMVUE $T^*$:
$$T^* = {(n+1)X_{(n)} - 1 \over n}$$
Having done all this, I am sorry to note that this estimator will not, in general, output integer values for its estimates, even though we (I presume) know $\theta$ is an integer.  "Unbiased" and "variance" are not terms that take this into account, however, so were we to be in a real-world situation where we had to provide integer estimates of $\theta$, we'd have to do something else - if only (possibly) round off the estimates we get this way.
ETA: Finding the conditional distribution of $X|X_{(n)}$.  Given $X_{(n)}$, we know $X \in \{1, \dots, X_{(n)}\}$.  The conditional distribution $p(x|x \leq X_{(n)})$ is:
$$ p(x|x \leq X_{(n)}) = {p(x) \over P(X_{(n)})}$$  
where $P(X_{(n)})$ is the probability that $x \leq X_{(n)}$. 
Now, $P(X_{(n)}) = X_{(n)}/\theta$.  Substituting gives:
$$ p(x|x \leq X_{(n)}) = {1/\theta \over (X_{(n)}/\theta)} = {1 \over X_{(n)}}$$
This, combined with the fact that $X \in \{1, \dots, X_{(n)}\}$, is sufficient for us to conclude that $x|x\leq X_{(n)}$ is distributed according to a discrete uniform distribution on $\{1, \dots, X_{(n)}\}$.
