Find UMVUE of $\theta$ where $f_X(x\mid\theta) =\theta(1 +x)^{−(1+\theta)}I_{(0,\infty)}(x)$ As a slight modification of my previous problem:

Let $X_1, X_2, . . . , X_n$ be iid random variables having pdf 
$$f_X(x\mid\theta) =\theta(1 +x)^{−(1+\theta)}I_{(0,\infty)}(x)$$
where $\theta >0$. Give the UMVUE of $\theta$, the Cramer-Rao Lower Bound (CRLB) for unbiased estimators of $\theta$ and compute the variance of the UMVUE of $\theta$.

I have that $f_X(x\mid\theta)$ is a full one-parameter exponential family with 
$h(x)=I_{(0,\infty)}$, $c(\theta)=\theta$, $w(\theta)=-(1+\theta)$, $t(x)=\text{log}(1+x)$ 
Since $w'(\theta)=1$ is nonzero on $\Theta$, the CRLB result applies. We have
$$\text{log }f_X(x\mid\theta)=\text{log}(\theta)-(1+\theta)\cdot\text{log}(1+x)$$
$$\frac{\partial}{\partial \theta}\text{log }f_X(x\mid\theta)=\frac{1}{\theta}-\text{log}(1+x)$$
$$\frac{\partial^2}{\partial \theta^2}\text{log }f_X(x\mid\theta)=-\frac{1}{\theta^2}$$
so $$I_1(\theta)=-\mathsf E\left(-\frac{1}{\theta^2}\right)=\frac{1}{\theta^2}$$
and  the CRLB for unbiased estimators of $\tau(\theta)$ is
$$\frac{[\tau'(\theta)]^2}{n\cdot I _1(\theta)} = \frac{\theta^2}{n}[\tau'(\theta)]^2=\boxed{\frac{\theta^2}{n}}$$
As for finding the UMVUE of $\theta$, since $\frac{1}{n}\sum \text{log}(1+X_i)$ is unbiased for $\frac{1}{\theta}$ then perhaps something similar to $\frac{n}{\sum\text{log}(1+X_i)}$ will be unbiased for $\theta$. After finding the expected value, I can hopefully make a slight adjustment to get an unbiased estimator. Let $T=\sum \text{log}(1+X_i)$
$$\mathsf E\left(\frac{n}{T}\right)=n\cdot\mathsf E\left(\frac{1}{T}\right)=n\int_0^{\infty}\frac{1}{t}f_T(t)dx$$
We must next find the distribution of $T$, but first let's find the distribution of $t=\text{log}(1+X)$. Let $Y=\text{log}(1+X)$. Then 
$$\begin{align*}
F_Y(y)
&=\mathsf P(Y\leq y)\\\\
&=\mathsf P(\text{log}(1+X)\leq y)\\\\
&=\mathsf P(1+X\leq e^y)\\\\
&=\mathsf P(X\leq e^y -1)\\\\
&=F_X\left(-(1+e^y-1)^{-\theta}+1\right)\\\\
&=1-e^{-\theta y}
\end{align*}$$
So $Y\sim \text{exp}\left(\frac{1}{\theta}\right)$ and hence $T\sim\text{Gamma}\left(\alpha=n,\beta=\frac{1}{\theta}\right)$
Hence
$$\begin{align*}
\mathsf E\left(\frac{n}{T}\right)
&=n\int_0^{\infty} \frac{1}{t} \frac{\theta^n}{\Gamma(n)}t^{n-1}e^{-\theta t}dt\\\\
&=\frac{n\theta}{n-1}\underbrace{\int_0^{\infty}\frac{\theta^{n-1}}{\Gamma(n-1)}t^{n-2}e^{-\theta t}dt}_{=1}\\\\
&=\frac{n}{n-1}\theta
\end{align*}$$
It follows that $$\frac{n-1}{n}\cdot\frac{n}{\sum\text{log}(1+X_i)}=\boxed{\frac{n-1}{\sum\text{log}(1+X_i)}}$$ is an unbiased estimator of $\theta$ which is a function of the complete sufficient statistic $T$, and so by the Lehmann-Scheffe Theorem, it's the unique UMVUE of $\theta$.
As $\hat{\theta}\sim(n-1)\cdot\text{Inv-Gamma}(n,\theta)$
then 
$$\mathsf{Var}\left(\frac{n-1}{T}\right)=(n-1)^2\cdot\mathsf{Var}\left(\frac{1}{T}\right)=(n-1)^2 \cdot \frac{\theta^2}{(n-1)^2\cdot(n-2)}=\boxed{\frac{\theta^2}{n-2}}$$
Are these valid solutions?
 A: From your previous question, you already have the complete sufficient statistic:
$$T(\mathbf{X}) = \sum_{i=1}^n \ln(1+X_i).$$
The simplest way to find the UMVUE estimator for $\theta$ is to appeal to the Lehmann-Scheffé theorem, which says that any unbiased estimator of $\theta$ which is a function of $T$ is the unique UMVUE.
To find an estimator with these properties, let $T_i = \ln(1+X_i)$ and observe that $T_i \sim \text{Exp}(\theta)$ so that $T \sim \text{Gamma}(n,\theta)$.  Hence, we can use the complete sufficient statistic to form the estimator:
$$\hat{\theta}(\mathbf{X}) = \frac{n-1}{T(\mathbf{X})} = \frac{n-1}{\sum_{i=1}^n \ln(1+X_i)} \sim (n-1) \cdot \text{Inv-Gamma}(n,\theta).$$
From the known moments of the inverse gamma distribution, we have $\mathbb{E}(\hat{\theta}) = \theta$ so we have found an unbiased estimator that is a function of the complete sufficient statistic.  The Lehmann-Scheffé theorem ensures that our estimator is UMVUE for $\theta$.  The variance of the UMVUE can easily be found by appeal to the moments of the inverse gamma distribution.
