Given observed data $\mathbf{x} = (x_1,...,x_n)$ the log-likelihood is:
$$\ell_\mathbf{x}(\theta) = n ( \bar{x} \ln (\theta) - \theta ).$$
This has derivatives:
$$\begin{equation} \begin{aligned}
\frac{d \ell_\mathbf{x}}{d \theta}(\theta)
&= n \bigg( \frac{\bar{x}}{\theta} - 1 \bigg), \\[6pt]
\frac{d^2 \ell_\mathbf{x}}{d \theta^2}(\theta)
&= - \frac{n \bar{x}}{\theta^2} <0. \\[6pt]
\end{aligned} \end{equation}$$
This shows that the log-likelihood function is concave, so the MLE occurs at the unique critical point given by:
$$0 = \frac{d \ell_\mathbf{x}}{d \theta}(\hat{\theta})
\quad \quad \quad \implies \quad \quad \quad
\hat{\theta} = \bar{x}.$$
Since $\mathbb{E}(X) = \theta$ and $\mathbb{V}(X) = \theta$, it follows from the central limit theorem that when $n$ is large, we have the approximate distribution $\hat{\theta} \sim \text{N} ( \theta, \theta/n )$. The estimator is unbiased, and so it has MSE given by:
$$\begin{equation} \begin{aligned}
\text{MSE}(\hat{\theta},\theta)
= \mathbb{V} (\hat{\theta})
= \mathbb{V} (\bar{X} )
= \frac{\theta}{n}.
\end{aligned} \end{equation}$$
Your reasoning is mostly correct.
The joint density of the sample $(X_1,X_2,\ldots,X_n)$ is
\begin{align}
f_{\theta}(x_1,x_2,\ldots,x_n)&=\frac{\theta^n}{\left(\prod_{i=1}^n (1+x_i)\right)^{1+\theta}}\mathbf1_{x_1,x_2,\ldots,x_n>0}\qquad,\,\theta>0
\\\\\implies \ln f_{\theta}(x_1,x_2,\ldots,x_n)&=n\ln(\theta)-(1+\theta)\sum_{i=1}^n\ln(1+x_i)+\ln(\mathbf1_{\min_{1\le i\le n} x_i>0})
\\\\\implies\frac{\partial}{\partial \theta}\ln f_{\theta}(x_1,x_2,\ldots,x_n)&=\frac{n}{\theta}-\sum_{i=1}^n\ln(1+x_i)
\\\\&=-n\left(\frac{\sum_{i=1}^n\ln(1+x_i)}{n}-\frac{1}{\theta}\right)
\end{align}
Thus we have expressed the score function in the form
$$\frac{\partial}{\partial \theta}\ln f_{\theta}(x_1,x_2,\ldots,x_n)=k(\theta)\left(T(x_1,x_2,\ldots,x_n)-\frac{1}{\theta}\right)\tag{1}$$
, which is the equality condition in the Cramér-Rao inequality.
It is not difficult to verify that $$E(T)=\frac{1}{n}\sum_{i=1}^n\underbrace{E(\ln(1+X_i))}_{=1/\theta}=\frac{1}{\theta}\tag{2}$$
From $(1)$ and $(2)$ we can conclude that
- The statistic $T(X_1,X_2,\ldots,X_n)$ is an unbiased estimator of $1/\theta$.
- $T$ satisfies the equality condition of the Cramér-Rao inequality.
These two facts together imply that $T$ is the UMVUE of $1/\theta$.
The second bullet actually tells us that variance of $T$ attains the Cramér-Rao lower bound for $1/\theta$.
Indeed, as you have shown,
$$E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(X_1)\right]=-\frac{1}{\theta^2}$$
This implies that the information function for the whole sample is $$I(\theta)=-nE_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(X_1)\right]=\frac{n}{\theta^2}$$
So the Cramér-Rao lower bound for $1/\theta$ and hence the variance of the UMVUE is
$$\operatorname{Var}(T)=\frac{\left[\frac{d}{d\theta}\left(\frac{1}{\theta}\right)\right]^2}{I(\theta)}=\frac{1}{n\theta^2}$$
Here we have exploited a corollary of the Cramér-Rao inequality, which says that for a family of distributions $f$ parametrised by $\theta$ (assuming regularity conditions of CR inequality to hold), if a statistic $T$ is unbiased for $g(\theta)$ for some function $g$ and if it satisfies the condition of equality in the CR inequality, namely $$\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=k(\theta)\left(T(x)-g(\theta)\right)$$, then $T$ must be the UMVUE of $g(\theta)$. So this argument does not work in every problem.
Alternatively, using the Lehmann-Scheffe theorem you could say that $T=\frac{1}{n}\sum_{i=1}^{n} \ln(1+X_i)$ is the UMVUE of $1/\theta$ as it is unbiased for $1/\theta$ and is a complete sufficient statistic for the family of distributions. That $T$ is compete sufficient is clear from the structure of the joint density of the sample in terms of a one-parameter exponential family. But variance of $T$ might be a little tricky to find directly.
Best Answer
You can use a computation with a few random examples as a sanity check.
Also, you can find the answer by looking up some of the more well known distributions. Hints: You are looking for something with a logarithm and you don't need to worry if the distribution has two parameters (you can set one of them to zero).