Find sufficient and complete statistic: Uniform distribution, $\theta \geq 1$ Take a random sample $X_1,X_2, \dots X_n$  from the distribution f$(x;\theta)=\frac{1}{\theta}$ for $0 \leq x \leq \theta$, where $\theta \geq 1$ (pay attention!).
I need to find an efficient estimator (in terms of MSE).
I've managed to check if $\max(1, X_{(n)})$ is sufficient and complete. But I can't transform it to get an unbiased estimator.
So, my questions are:

*

*Could you suggest a suitable transformation?

*If not, could you find another sufficient and complete estimator?

 A: I think you want to find UMVUE of $\theta$
Note That:Let $\Omega$ be a set of all unbiased estimators of 0 with finite variances and T be an unbiased estimators of $\nu$ with $E(T^2)< \infty$ and suppose that $T=h(\tilde{T})$, where $\tilde{T}$ is a sufficient statistics for $\theta \in \Theta$ and h is Borel function.Let $\Omega_{\tilde{T}}$ be the subset of $\Omega$ consisting of Borel functions of $\tilde{T}$.Then a necessary and sufficient condition for T to be a UMVUE of $\nu$ is that $E[T(X)U(X)]=0$ for any $U \in \Omega_{\tilde{T}}$ and any $\theta \in \Theta$.
So we can use this note for find the UMVUE of  $\theta$.
Let $U(X_{(n)})$ be an unbiased estimator of 0.Since $X_{(n)}$ has the Lebesgue p.d.f  $n\theta^{-n}x^{n-1}I_{(0,\theta)}(x)$ 
$0=\int_{0}^{1}U(x)x^{n-1}dx+\int_{1}^{\theta}U(x)x^{n-1}dx$
for all $\theta \geq 1$ .This implies that $U(x)=0$ a.e Lebesgue measure on $[1,\infty)$ and $\int_{0}^{1}U(x)x^{n-1}dx=0$
Consider $T=h(X_{(n)}$. To have $E(TU)=0$, we must have  $\int_{0}^{1}h(x)U(x)x^{n-1}dx=0$. Thus we may consider the following function
\begin{equation*}
 h(x)  = \left\{
        \begin{array}{ll}
            c& \quad  0\leq x \leq 1 \\
            x & \quad x > 1
        \end{array}
    \right.
\end{equation*}

Where c=1 and b are some constants.From the previous discussion, Since $E[h(X_{(n)})U(X_{(n)})]=0 \hspace{2cm} \theta \geq1$ 
Since $E[h(X_{(n)})]=\theta$, we obtain that
$\theta=cP(X_{(n)}\leq1)+bE[X_{(n)}I_{(1,\infty)}(X_{(n)})]=c\theta^{-n}+[\frac{bn}{n+1}](\theta-\theta^{-n})$.
Thus , $c=1$ and $b=\frac{n+1}{n}$. The UMVUE of $\theta$ is then 
\begin{equation*}
 h(x)  = \left\{
        \begin{array}{ll}
            1& \quad  0\leq X_{(n)} \leq 1 \\
            (1+\frac{1}{n})X_{(n)} & \quad X_{(n)} > 1
        \end{array}
    \right.
\end{equation*}
