Efficiency of $\hat{\theta}_{MLE}$ from $\operatorname{Beta}(\theta,1)$ I am working on a problem which asks me to discuss the efficiency of the MLE $\hat{\theta}$ given that $X_1,\ldots,X_n \sim_{iid} \operatorname{Beta}(\theta,1) $.
I was able to deduce that 
$$\hat{\theta} = \frac{n}{-\sum_{i=1}^n \ln X_i}$$
and that the Rao-Cramer Lower Bound is
$$RCLB=\frac{\theta^2}{n}$$.
Since $E[\hat{\theta}]=\frac{n}{n-1}\theta$ the MLE is asymptotically  unbiased, and I found that
$$Var[\hat{\theta}]= \frac{n^2\theta^2}{(n-1)^2(n-2)}$$.
What bothers me a little is that I was able to proove that the coefficient of $\theta^2$ is a value that is larger than $1 \over n$ when $1.57 < n$ so I see that this variance is indeed larger than the RCLB.
However, the fact that when $n=1$ and $n=2$ is undefined makes me a bit uneasy.
Is there something that needs to be considered in these cases or is there an assumption that I am missing?
 A: The exact distribution of the MLE is $$\hat \theta \sim \operatorname{InverseGamma}(n,n\theta)$$  with PDF $$f_{\hat \theta}(t) = \frac{(n\theta)^n e^{-n\theta/t}}{t^{n+1} \Gamma(n)}, \quad t > 0.$$  The variance and expectation you computed are correct.  Note that when $n = 1$, this distribution is heavy-tailed, in the sense that for large $t$, $f_{\hat \theta}(t) \sim 1/t^2$, which is like a Cauchy distribution (or equivalently Student's $t$ with $1$ degree of freedom).  Similarly, when $n = 2$, the distribution of $\hat \theta$ has a finite expectation but infinite variance, because for large $t$, $f_{\hat \theta}(t) \sim 1/t^3$.
I do not see why there should be any concern that $$\operatorname{Var}[\hat \theta] > \theta^2/n.$$  The Cramér-Rao lower bound is not always attainable, even for "best" unbiased estimators.  What would be concerning is if an unbiased estimator were to have a variance that is smaller than such a lower bound as this would violate the theorem.
A: If your definition of efficiency is that the estimator must attain the Rao-Cramer lower bound, then yes the MLE is not efficient. Even the minimum variance unbiased estimator of $\theta$, namely $(1-n)/\sum\limits_{i=1}^n \ln X_i$ does not attain the lower bound, but it remains the most efficient estimator of $\theta$ within the unbiased class.
When you are talking about efficiency, you are more often than not concerned with increasing sample size, so cases like $n=1$ or $n=2$ can be safely ignored.
Now efficiency $(\mathrm{e})$ of your MLE is the ratio of the Rao-Cramer bound and variance of the MLE, i.e. $\mathrm{e}=\frac{(n-1)^2(n-2)}{n^3}$, which converges to $1$. Or if you consider the asymptotic variance of the MLE, then $\mathrm{e}\approx 1$ for large $n$. So either way the MLE is asymptotically efficient.
