UMVUE of $ \frac{1}{\theta}$ coming from $f(x) = \theta x^{\theta - 1}$. Let $X_1, \ldots, X_n$ be i.i.d. sampled from the distribution
$$
f(x; \theta) = \theta x^{\theta - 1},
$$
where $x \in (0, 1)$ and $\theta > 0$.
Show that
$$
T(x_{1}, \ldots, x_{n}) = - \frac{1}{n} \sum_{i = 1}^n \ln(x_i)
$$
is a Unique Minimum Variance Unbiased Estimator (UMVUE) of $\dfrac{1}{\theta} $.
I know that $g(T) = E[h(x) \mid T(x)]$ is UMVUE if $h(x)$ is an unbiased estimator and $T(x)$ is a complete sufficient statistic.
I tried guessing that $\dfrac{1}{x}$ would be a UE of $\dfrac{1}{\theta}$ using $E\left( \dfrac{k}{x} \right) = \int_0^1 \left( \dfrac{k}{x} \right) \cdot \theta \cdot x^{\theta-1} \ dx$, which gives $\dfrac{k \theta}{\theta - 1}$, so $\dfrac{\theta-1}{x}$ is UE of $\theta$.
Here, we can apply invariance property to see that $\dfrac{x}{\theta - 1}$ is UE of $\dfrac{1}{\theta}$.
I also know that the MLE of $\dfrac{1}{\theta}$ is $-\dfrac{1}{n} \sum_{i = 1}^n \ln(x_i)$.
However, I'm having trouble with actually calculating $g(T)$.
 A: $\newcommand{\E}{\operatorname{E}}$
\begin{align}
\E(-\log X) & = \int_0^1 (-\log x) \Big(\theta x^{\theta-1}\,dx\Big) = \int u\,dv = uv - \int v \, du \\[8pt]
& = \left.(-\log x)x^\theta\vphantom{\frac11}\,\right|_0^1 - \int_0^1 x^\theta\Big( \frac{-dx} x \Big) \\[8pt]
& = \int_0^1 x^{\theta-1} \,dx \quad(\text{L'Hopital's rule showed that the first term is 0.}) \\[8pt]
& = \left.\frac {x^\theta}\theta\right|_0^1 = \frac 1 \theta.
\end{align}
That takes care of unbiasedness.
The joint density is
$$
f(x_1,\ldots,x_n) = \theta^n (x_1\cdots x_n)^{\theta-1} \cdot 1
$$
where the "$1$" is a factor that does not depend on $\theta$ and in this case does not depend on $x_1,\ldots,x_n$, but dependence on those would not upset the following conclusion: the product, and therefore the sum of the logarithms, is sufficient.
You need to show that this sufficient statistic admits no nontrivial unbiased estimators of zero, i.e. there is no nonzero function $g(x_1,\ldots,x_n)$, not depending on $\theta$, for which
$$
\int_0^1\cdots\int_0^1 g(x_1,\ldots,x_n)\theta^n(x_1\cdots x_n)^{\theta-1}\,dx_1\cdots dx_n = 0\text{ for all values of $\theta>0$}.
$$
(You can divide both sides of that by $\theta^n$ and it's a little bit simpler.)
Maybe I'll be back later to deal with this integral${}\,\ldots\ldots$
