Given a random variable $X \sim f(x) = \frac{1}{x\log (\theta)}$ when $1 < x < \theta$, I am trying to find an unbiased estimator of a simple random sample of size $n$ for $d(\theta) = \log (\theta)$. The pdf of the simple random sample is $f(x_1, \dots, x_n) = \frac{\operatorname{I}_{(-\infty, \theta)}(X_{(n)})\operatorname{I}_{(1, \infty)}(X_{(1)})}{\log ^n(\theta)\prod_{i=0}^n x_i}$
I have tried every single candidate I can think of, but cannot find any suitable function. The candidates I have tried so far are:
1) The maximum, $T(X_1, \dots, X_n) = X_{(n)}$, whose pdf is given by: \begin{equation} P_T(X_{(n)} \leq Y) = P_X(X_i \leq Y\quad\forall i\in\{1,\dots,n\}) = (P_X(X\leq Y))^n \implies\\ f_T(x_1,\dots,x_n) = n(P_X(X\leq Y))^{n-1}f_X(x_1,\dots,x_n) = \frac{n\log ^{n-1}(x)}{x\log ^n(\theta)} \end{equation} The expected value of $T$ is thus: \begin{equation} E[X_{(n)}] =\int_1^\theta x\frac{n\log ^{n-1}(x)}{x\log ^n(\theta)} dx = \frac{n}{\log ^n(\theta)}\int_1^\theta \log ^{n-1}(x) \end{equation} which is not $\log (\theta)$.
2) $T(X_1, \dots, X_n) = \frac{1}{\prod_{i=0}^n X_i}$. Since the random variables are independent: \begin{equation} E\left[\frac{1}{\prod_{i=0}^n X_i}\right] = \left(E\left[\frac{1}{X}\right]\right)^n = \left(\int_1^\theta \frac{1}{x^2\log (\theta)} dx\right)^n = \frac{1}{\log ^n(\theta)}\left(1-\frac{1}{\theta}\right)^n \neq \log (\theta) \end{equation}
3) The minimum, $T(X_1, \dots, X_n) = X_{(1)}$, whose pdf is given by: \begin{equation} P_T(X_{(1)} \leq Y) = 1 - P_T(X_{(1)} > Y) = 1 - P_X(X_i > Y\quad\forall i\in\{1,\dots,n\}) = 1 - (P_X(X> Y))^n = 1 - (1 - P_X(X \leq Y))^n \implies\\ f_T(x_1,\dots,x_n) = n(1 - P_X(X\leq Y))^{n-1}f_X(x_1,\dots,x_n) = n\left(1-\frac{\log (x)}{\log (\theta)}\right)^{n-1}\frac{1}{x\log (\theta)} \end{equation} The expected value of $T$ is thus: \begin{equation} E[X_{(1)}] =\int_1^\theta xn\left(1-\frac{\log (x)}{\log (\theta)}\right)^{n-1}\frac{1}{x\log (\theta)} dx = \frac{n}{\log ^n(\theta)}\int_1^\theta \log ^{n-1}(x) = \frac{n}{\log ^n(\theta)}\int_1^\theta (\log (\theta)-\log (x))^{n-1} \end{equation} which, again, is not $\log (\theta)$.
Can anyone shine a light on how to calculate de MVUE?
Thank you.
$\int_1^{\theta}g(y) \frac{n \log^{n-1} y}{y\log^{n} \theta} dy =\log \theta$
$\int_1^{\theta}g(y) \frac{n \log^{n-1} y}{y} dy =\log^{n+1} \theta$
$g(\theta) \frac{n \log^{n-1} \theta}{\theta} dy =(n+1)\frac{1}{\theta}\log^{n} \theta$
$g(\theta) =\frac{(n+1)}{n}\log \theta$
$g(Y)=\frac{(n+1)}{n}\log Y$ is an unbiased estimator where $Y=\max(X_1,\cdots ,X_n)$