I apply a nonlinear transformation to a linear estimator $\hat{\alpha}=f(\hat{\theta})$. Then I find Cramer Rao's Lower Bound, and it asymptothically goes to zero. Does it means that my new estimator $\hat{\alpha}$ is efficient? Why?
I guess I should find $E(\hat{\alpha})$ and then see that $E(\hat{\alpha}) \rightarrow E(\hat{\theta})$, but I think it is not the point of the excercise, because it is a quite complicated nonlinearity.
It is natural that the Cramer-Rao lower bound goes to zero as the number of samples increases and does not mean that the estimator is efficient.
According to Cramer-Rao lower bound we may write: $$ \text{Var}(\hat{\alpha})\geq \frac{\left(1+\frac{db}{d\alpha}\right)^2}{nI(\theta)}$$ where $b$ is the bias of estimator. It can be easily seen that the lower bound may go to zero due to the factor $n$ in the denumerator.
In order to show that an estimator is efficient, you should show that it is unbiased and also attains the lower bound in the Cramér–Rao inequality, for all $\theta$. This means that you should show that:
$$ \text{Var}(\hat{\alpha})= \frac{1}{nI(\theta)}$$
The concept of asymptotic efficiency is much more complicated. You may refer to p.21 of this article for more information.
Cramér, Harald, Mathematical methods of statistics, Princeton Mathematical series. 9. Princeton N. J.: Princeton University Press xvi, 575 p. (1946). ZBL0063.01014.